We provide determinants practice exercises, instructions, and a learning material that allows learners to study outside of the classroom. We focus on determinants skills mastery so, below you will get all questions that are also asking in the competition exam beside that classroom.
List of determinants Questions
| Question No | Questions | Class |
|---|---|---|
| 1 | Solve: ( left|begin{array}{ccc}mathbf{0} & boldsymbol{a} & -boldsymbol{b} \ -boldsymbol{a} & boldsymbol{0} & -boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{0}end{array}right|=0 ) | 12 |
| 2 | The value of ( (operatorname{adj} A) ) is equal to A . ( 2 A ) в. ( 4 A ) ( c .8 A ) D. ( 16 A ) | 12 |
| 3 | Find the values of ( x, ) if ( left|begin{array}{ll}mathbf{3} & boldsymbol{x} \ boldsymbol{x} & mathbf{1}end{array}right|=left|begin{array}{ll}mathbf{3} & mathbf{2} \ mathbf{4} & mathbf{1}end{array}right| ) | 12 |
| 4 | ( fleft(begin{array}{ccc}2 & -1 & 3 \ -5 & 3 & 1 \ -3 & 2 & 3end{array}right], ) then ( A .(A d j A)= ) A ( cdotleft(A d j cdot A^{T}right) ) B. ( (text {Adj.A}) . A ) c. ( |A| . A ) D. None of these | 12 |
| 5 | set of all values of 2 for which the system of linear 31. The set of all value equations : 2×1 – 2×2 + x3 = 2×1 2x – 3×2 + 2×3 = 2×2 -x + 2×2=hxz has a non-trivial solution [JEEM 2015] (a) contains two elements (b) contains more than two elements (C) is an empty set (d) is a singleton | 12 |
| 6 | 1 2 3 and (adj M) = -5 where 3 -1 To 1 al -1 1 -17 15. Let M= 1 3 and (adi M = 8 -6 2 | 3 6 1 a and b are real numbers. Which of the following options is/are correct ? (JEE Adv. 2019) (a) a+b=3 (6) det (adj M2)=81 © (adjM)-1 + adjM =-M N (d) IfM , then a-B+y = 3 | 12 |
| 7 | ( f(x)=left|begin{array}{ccc}cos x & x & 1 \ 2 sin x & x^{2} & 2 x \ tan x & x & 1end{array}right| . ) The value of ( lim _{x rightarrow 0} frac{f(x)}{x} ) is equal to A . 1 B. – ( c cdot 0 ) D. None of these | 12 |
| 8 | Using properties of determinants, prove that: [ left|begin{array}{ccc} 1+a & 1 & 1 \ 1 & 1+b & 1 \ 1 & 1 & 1+c end{array}right|=a b c+b c+ ] ( c a+a b ) | 12 |
| 9 | Find the value of the determinant: [ left|begin{array}{cc} mathbf{5} & -mathbf{2} \ -mathbf{3} & mathbf{1} end{array}right| ] | 12 |
| 10 | A. square matrix ( boldsymbol{A} ) of order ( mathbf{3}, ) has ( |boldsymbol{A}|= ) ( 5, ) find ( mid A ) adjal | 12 |
| 11 | valuate: ( left|begin{array}{cccc}1 & a & a^{2} & a^{3}+b c d \ 1 & b & b^{2} & b^{3}+c d a \ 1 & c & c^{2} & c^{3}+a b d \ 1 & d & d^{2} & d^{3}+a b cend{array}right| ) | 12 |
| 12 | ( left|begin{array}{cc}cos 15^{circ} & sin 15^{circ} \ sin 15^{circ} & cos 15^{circ}end{array}right|=? ) A . 1 B. c. ( frac{sqrt{3}}{2} ) D. none of these | 12 |
| 13 | Show that the points ( boldsymbol{A}(mathbf{1}, mathbf{2}, mathbf{7}), boldsymbol{B}(mathbf{2}, mathbf{6}, mathbf{3}) ) and ( boldsymbol{C}(mathbf{3}, mathbf{1 0},-mathbf{1}) ) are collinear. | 12 |
| 14 | ( boldsymbol{a} neq boldsymbol{p}, boldsymbol{b} neq boldsymbol{q}, boldsymbol{c} neq boldsymbol{r} ) and ( left|begin{array}{lll}boldsymbol{p} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a} & boldsymbol{q} & boldsymbol{c} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{r}end{array}right|=0 ) then the value of ( frac{boldsymbol{p}}{boldsymbol{p}-boldsymbol{a}}+frac{boldsymbol{q}}{boldsymbol{q}-boldsymbol{b}}+ ) ( frac{r}{r-c} ) is equal to ( A ) B. 2 ( c cdot 3 ) D. | 12 |
| 15 | Using properties of determinant solve 🙁 left|begin{array}{ccc}1 & a & a^{2}-b c \ 1 & b & b^{2}-a c \ 1 & c & c^{2}-a bend{array}right|= ) | 12 |
| 16 | Evaluate. ( mid begin{array}{ccc}frac{1}{z} & frac{1}{z} & -frac{(x+y)}{z^{2}} \ -frac{(y+z)}{x^{2}} & frac{1}{x} & frac{1}{x} \ -frac{y(y+z)}{x^{2} z} & frac{x+2 y+z}{x z} & -frac{y(x+y)}{x z^{2}}end{array} ) | 12 |
| 17 | Expand: ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a}end{array}right| ) | 12 |
| 18 | ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{omega}^{2} & mathbf{1} \ boldsymbol{omega}^{2} & mathbf{1} & boldsymbol{omega}end{array}right] ) Find Det ( mathbf{A} ) | 12 |
| 19 | If ( boldsymbol{A}=left[begin{array}{ccc}2 & 52 & 152 \ 4 & 106 & 358 \ 6 & 162 & 620end{array}right], ) then the determinant of the matrix ( a d j(2 A) ) is equal to: | 12 |
| 20 | ( fleft|begin{array}{ccc}mathbf{6} i & -mathbf{3} i & mathbf{1} \ mathbf{4} & mathbf{3} i & -mathbf{1} \ mathbf{2 0} & mathbf{3} & boldsymbol{i}end{array}right|=boldsymbol{x}+boldsymbol{i} boldsymbol{y}, ) then A ( . x=3, y=1 ) B. ( x=1, y=3 ) c. ( x=0, y=3 ) D. ( x=0, y=0 ) | 12 |
| 21 | Let ( boldsymbol{alpha}, boldsymbol{beta}, boldsymbol{gamma}, boldsymbol{a}, boldsymbol{b}, boldsymbol{c}, boldsymbol{x} boldsymbol{epsilon} boldsymbol{R} ) and let ( Delta= ) ( left|begin{array}{ccc}sin (x+alpha) & cos (x+alpha) & a+x sin alpha \ sin (x+beta) & cos (x+beta) & b+x sin beta \ sin (x+gamma) & cos (x+gamma) & c+x sin gammaend{array}right| ) then ( Delta ) is This question has multiple correct options A. independent of ( x ) B. dependent of ( alpha, beta, ), c. dependent on ( a, b, c, c ) D. ( a ) constant | 12 |
| 22 | If [] denotes the greatest integer less than or equal to the real number under consideration, and ( -1 leq x<0, quad 0 leq ) ( y<1,1 leq z<2, ) the value of the determinant ( left|begin{array}{ccc}{[boldsymbol{x}]+mathbf{1}} & {[boldsymbol{y}]} & {[boldsymbol{z}]} \ {[boldsymbol{x}]} & {[boldsymbol{y}]+mathbf{1}} & {[boldsymbol{z}]} \ {[boldsymbol{x}]} & {[boldsymbol{y}]} & {[boldsymbol{z}]+mathbf{1}}end{array}right| ) is ( A cdot[x] ) в. ( [y] ) ( c cdot[z ) D. none of these | 12 |
| 23 | ( left|begin{array}{ccc}1+a & 1 & 1 \ 1 & 1+b & 1 \ 1 & 1 & 1+cend{array}right|= ) ( boldsymbol{X} boldsymbol{a b c}left(1+frac{1}{boldsymbol{a}}+frac{1}{boldsymbol{b}}+frac{1}{c}right) . ) Find the value of ( boldsymbol{X} ) | 12 |
| 24 | if ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & mathbf{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{0} & boldsymbol{a}end{array}right|=mathbf{0} ) then A. ( a ) is a cube root of 1 B. ( b ) is a cube root of 1 c. ( frac{a}{b} ) is a cube root of 1 D. ( frac{a}{b} ) is a cube root of -1 | 12 |
| 25 | Find the values of ( K ) if Area of the triangle is 4 sq. units and vertices are ( (k 0)(40)(02) ) using determinants. | 12 |
| 26 | If ( boldsymbol{A}=left[begin{array}{cc}-mathbf{4} & -mathbf{1} \ mathbf{3} & mathbf{1}end{array}right], ) then the determinant of the matrix ( left(A^{2016}-2 A^{2015}-A^{2014}right) ) is: A. -175 в. 2014 ( c .2016 ) D. -25 | 12 |
| 27 | If the value of ( left|begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{4} \ -mathbf{1} & mathbf{3} & mathbf{0} \ mathbf{4} & mathbf{1} & mathbf{0}end{array}right| ) is ( k, ) then find ( frac{-boldsymbol{k}}{mathbf{1 3}} ) | 12 |
| 28 | f ( l, m, n ) are ( p^{t h}, q^{t h}, r^{t h} ) terms of ( G . P . ) al positive, then ( left|begin{array}{lll}log l & p & 1 \ log m & q & 1 \ log n & r & 1end{array}right| ) equals A . – B. 2 ( c ) ( D ) | 12 |
| 29 | If ( boldsymbol{x}=boldsymbol{c} boldsymbol{y}+boldsymbol{b} boldsymbol{z}, boldsymbol{y}=boldsymbol{a} boldsymbol{z}+boldsymbol{c} boldsymbol{x}, boldsymbol{z}=boldsymbol{b} boldsymbol{x}+ ) ( a y, ) where ( x, y, z ) are not all zero, then the value of ( a^{2}+b^{2}+c^{2}+2 a b c ) ( mathbf{A} cdot mathbf{0} ) B. c. -1 D. None of these | 12 |
| 30 | Find the integral value of ( x, ) if ( left|begin{array}{ccc}boldsymbol{x}^{2} & boldsymbol{x} & mathbf{1} \ mathbf{0} & boldsymbol{2} & mathbf{1} \ boldsymbol{3} & boldsymbol{1} & boldsymbol{4}end{array}right|=mathbf{2} mathbf{8} ) | 12 |
| 31 | Let ( a, b, c in R ) be such that ( a+b+ ) ( boldsymbol{c}>mathbf{0} ) and ( boldsymbol{a b c}=mathbf{2 .} ) Let ( boldsymbol{A}=left[begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right] ) If ( A^{2}=I, ) then value of ( a^{3}+b^{3}+c^{3} ) is ( A cdot 7 ) B. ( c cdot 0 ) ( D ) | 12 |
| 32 | If the lines ( boldsymbol{a} boldsymbol{x}+boldsymbol{y}+mathbf{1}=mathbf{0}, boldsymbol{x}+boldsymbol{b} boldsymbol{y}+ ) ( mathbf{1}=mathbf{0} & boldsymbol{x}+boldsymbol{y}+boldsymbol{c}=mathbf{0} ) where ( mathbf{a}, mathbf{b} & mathbf{c} ) are distinct real numbers different from are concurrent, then the value of ( frac{1}{1-a}+frac{1}{1-b}+frac{1}{1-c}= ) ( A cdot 4 ) B. 3 ( c cdot 2 ) ( D ) | 12 |
| 33 | If ( a+b+c neq 0 ) and ( left|begin{array}{lll}a & b & c \ b & c & a \ c & a & bend{array}right|=0 ) then using properties of determinants, prove that ( boldsymbol{a}=boldsymbol{b}=boldsymbol{c} ) | 12 |
| 34 | The system of linear equations: ( boldsymbol{lambda} boldsymbol{x}+boldsymbol{2} boldsymbol{y}+boldsymbol{2} boldsymbol{z}=mathbf{5} ) ( 2 lambda x+3 y+5 z= ) ( 4 x+lambda y+6 z=10 ) has A. no solution when ( lambda=2 ) B. a unique solution when ( lambda=-8 ) c. infinitely many solutions when ( lambda=2 ) D. no solution when ( lambda=8 ) | 12 |
| 35 | Find the value of determinant. (i) ( left|begin{array}{cc}cos theta & -sin theta \ sin theta & cos thetaend{array}right| ) (ii) ( left|begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1end{array}right| ) | 12 |
| 36 | If ( alpha=beta+frac{2 pi}{3} . ) then ( A_{theta} ) is maximum when ( gamma ) equals A. ( alpha+pi / 3 ) в. ( alpha-pi / 3 ) c. ( alpha+2 pi / 3 ) D. none of these | 12 |
| 37 | begin{tabular}{c|ccc} ( mathbf{1} ) & ( mathbf{s i n Q} ) & ( mathbf{1} ) \ ( -mathbf{s i n Q} ) & ( mathbf{1} ) & ( mathbf{s i n Q} ) \ ( mathbf{- 1} ) & ( -mathbf{s i n Q} ) & ( mathbf{1} ) end{tabular} | 12 |
| 38 | ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{y} boldsymbol{z} & boldsymbol{z} boldsymbol{x} & boldsymbol{x} boldsymbol{y}end{array}right|=boldsymbol{b}(boldsymbol{x}-boldsymbol{y})(boldsymbol{y}- ) ( z)(z-x) . ) Find | 12 |
| 39 | The points which are not collinear are: A. (0,1),(8,3) and (6,7) B. (4,3),(5,1) and (1,9) C. (2,5),(-1,2) and (4,7) D. (-3,2)(1,-2) and (9,-10) | 12 |
| 40 | ( fleft(a_{1}, a_{2}, a_{3} dots, a_{n} cdot ) are in G.P, then the right. determinant ( Delta= ) ( left|begin{array}{ccc}log a_{n} & log a_{n+1} & log a_{n+2} \ log a_{n+3} & log a_{n+4} & log a_{n+5} \ log a_{n+6} & log a_{n+7} & log a_{n+8}end{array}right| ) is equal to A . 0 B. 1 ( c cdot 2 ) D. | 12 |
| 41 | Consider the system of linear equations in x, y, z: (sin 30) x-y+z=0 (cos 20) x + 4y + 3z=0 2x+ 7y+ 7z=0 Find the values of for which this system has nontrivial solutions. (1986 – 5 Marks) | 12 |
| 42 | ( fleft(alpha, beta text { are the roots of } x^{2}+x+1=0right. ) ( operatorname{then}left|begin{array}{ccc}boldsymbol{y}+mathbf{1} & boldsymbol{beta} & boldsymbol{alpha} \ boldsymbol{beta} & boldsymbol{y}+boldsymbol{alpha} & mathbf{1} \ boldsymbol{alpha} & mathbf{1} & boldsymbol{y}+boldsymbol{beta}end{array}right|=? ) A ( cdot y^{2}-1 ) В . ( yleft(y^{2}-1right) ) c. ( y^{2}-y ) D. ( y^{3} ) | 12 |
| 43 | What are the values of ( x ) that satisfy the equation ( left|begin{array}{ccc}boldsymbol{x} & mathbf{0} & mathbf{2} \ mathbf{2} boldsymbol{x} & mathbf{2} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right|+left|begin{array}{ccc}mathbf{3} boldsymbol{x} & mathbf{0} & mathbf{2} \ boldsymbol{x}^{mathbf{2}} & boldsymbol{2} & mathbf{1} \ mathbf{0} & boldsymbol{1} & mathbf{1}end{array}right|=mathbf{0} ? ) B. ( -1 pm sqrt{3} ) c. ( -1 pm sqrt{6} ) ( mathbf{D} cdot-2 pm sqrt{6} ) | 12 |
| 44 | f ( x+y+z=pi ) and ( Delta=left|begin{array}{ccc}sin 3 x & sin 3 y & sin 3 z \ sin x & sin y & sin z \ cos x & cos y & cos zend{array}right| ) then ( Delta ) equals ( A ) в. ( c . ) ( D ) | 12 |
| 45 | f ( omega ) is a cube root of unity, then ( left|begin{array}{ccc}mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{omega}^{2} & boldsymbol{1} \ boldsymbol{omega}^{2} & boldsymbol{1} & boldsymbol{omega}end{array}right| ) is equal to A . B. ( omega ) ( c cdot omega^{2} ) D. | 12 |
| 46 | An equilateral triangle has each of its sides of length ( 6 mathrm{cm} . ) If ( left(x_{1}, y_{1}right) ;left(x_{2}, y_{2}right) ) ( &left(x_{3}, y_{3}right) ) are its vertices then the value of determinant (in nearest integer value) ( ,left|begin{array}{lll}boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & mathbf{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & mathbf{1}end{array}right| ) is equal to a then find ( frac{boldsymbol{a}}{mathbf{3 1}} ) | 12 |
| 47 | $ If A2 – A+I=0, then the inverse of A is (2) A+1 (b) A (c) A-I [2005] (2) I-A | 12 |
| 48 | If ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{alpha} & boldsymbol{2} \ boldsymbol{2} & boldsymbol{alpha}end{array}right] ) and ( |boldsymbol{A}|^{3}=125 ) then the value of ( boldsymbol{alpha} ) is A. ±1 B. ±2 ( c .pm 3 ) D. ±5 | 12 |
| 49 | 39. If the system of linear equations x+ky+3z=0 3x + ky-2z=0 2x + 4y-3z=0 (JEE M 2018] has a non-zero solution (x, y, z), then XZ 2 is equal to : (2) 10 (6) – 30 (c) 30 (2) – 10 | 12 |
| 50 | Show that the following set of points are collinear. (2,5),(4,6) and (8,8) | 12 |
| 51 | The value of the determinant ( left|begin{array}{ccc}1-alpha & alpha-alpha^{2} & alpha^{2} \ 1-beta & beta-beta^{2} & beta^{2} \ 1-gamma & gamma-gamma^{2} & gamma^{2}end{array}right| ) is equal to A ( cdot(alpha-beta)(beta-gamma)(alpha-gamma) ) В . ( (alpha-beta)(beta-gamma)(gamma-alpha) ) c. ( (alpha-beta)(beta-gamma)(alpha-gamma)(alpha+beta+gamma) ) D. | 12 |
| 52 | √3 1 15. If P= and A= and Q = PAPT and √3 (2005) x=PTQ2005 P then x is equal to fi 2005 (a) 0 1 [4+200573 6015 1 (b) 2005 4-2005/3] 12+√3 1 (c) A 1-1 2-13] w 1 2005 2-√3] (d) 4 2+13 2005 | 12 |
| 53 | ( operatorname{Det}left{begin{array}{lll}2 & 45 & 55 \ 1 & 29 & 32 \ 3 & 68 & 87end{array}right}=dots dots ) A . 45 B. 64 ( c cdot 54 ) D. 32 | 12 |
| 54 | The least value of the product xyz for which the determinant ( left|begin{array}{lll}boldsymbol{x} & mathbf{1} & mathbf{1} \ mathbf{1} & boldsymbol{y} & mathbf{1} \ mathbf{1} & mathbf{1} & boldsymbol{z}end{array}right| ) is non-negative, is : begin{tabular}{l} A ( -16 sqrt{2} ) \ hline end{tabular} В. ( -2 sqrt{2} ) ( c cdot-1 ) D. – – | 12 |
| 55 | Solve for ( boldsymbol{lambda} ) if ( left|begin{array}{ccc}boldsymbol{a}^{2}+boldsymbol{lambda} & boldsymbol{a} boldsymbol{b} & boldsymbol{a c} \ boldsymbol{a b} & boldsymbol{b}^{2}+boldsymbol{lambda} & boldsymbol{b c} \ boldsymbol{a c} & boldsymbol{b c} & boldsymbol{c}^{2}+boldsymbol{lambda}end{array}right|=mathbf{0} ) | 12 |
| 56 | ( left|begin{array}{ccc}boldsymbol{a}^{2}+mathbf{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b} & boldsymbol{b}^{2}+mathbf{1} & boldsymbol{b} boldsymbol{c} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c} boldsymbol{b} & boldsymbol{c}^{2}+mathbf{1}end{array}right|=mathbf{1}+boldsymbol{a}^{2}+ ) | 12 |
| 57 | The number of distinct real values of ( alpha ) for which the vectors ( boldsymbol{alpha}^{2} hat{boldsymbol{i}}-hat{boldsymbol{j}}-hat{boldsymbol{k}},-hat{boldsymbol{i}}- ) ( boldsymbol{alpha}^{2} hat{boldsymbol{j}}-hat{boldsymbol{k}},-hat{boldsymbol{i}}-hat{boldsymbol{j}}-boldsymbol{alpha}^{2} hat{boldsymbol{k}} ) will lie in the same place is ( A cdot 1 ) B . 2 ( c .3 ) D. | 12 |
| 58 | ( fleft|begin{array}{ccc}mathbf{6} i & -mathbf{3} i & mathbf{1} \ mathbf{4} & mathbf{3} i & -mathbf{1} \ mathbf{2 0} & mathbf{3} & boldsymbol{i}end{array}right|=boldsymbol{x}+boldsymbol{i} boldsymbol{y}, ) then A ( . x=3, y=1 ) B. ( x=1, y=3 ) c. ( x=0, y=3 ) D. ( x=0, y=0 ) | 12 |
| 59 | Prove that ( mid begin{array}{ccc}b c-a^{2} & c a-b^{2} & a b \ -b c+c a+a b & b c-c a+a b & b c+c \ (a+b)(a+c) & (b+c)(b+a) & (c+aend{array} ) ( mathbf{3} cdot(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}-boldsymbol{a})(boldsymbol{a}-boldsymbol{b})(boldsymbol{a}+boldsymbol{b}+ ) ( c(a b+b c+c a) ) | 12 |
| 60 | The line ( A x+B y+C=0 ) cuts the circle ( boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{a} boldsymbol{x}+boldsymbol{b} boldsymbol{y}+boldsymbol{c}=boldsymbol{0} ) in ( boldsymbol{P} ) and ( Q ) The line ( boldsymbol{A}^{prime} boldsymbol{x}+boldsymbol{B}^{prime} boldsymbol{y}+boldsymbol{C}^{prime}=mathbf{0} ) cuts the circle ( boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{a}^{prime} boldsymbol{x}+boldsymbol{b}^{prime} boldsymbol{y}+boldsymbol{c}^{prime}=mathbf{0} ) in ( boldsymbol{R} ) and ( S ). If ( P, Q, R, S ) are concyclic, then show that ( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{a}^{prime} & boldsymbol{b}-boldsymbol{b}^{prime} & boldsymbol{c}-boldsymbol{c}^{prime} \ boldsymbol{A} & boldsymbol{B} & boldsymbol{C} \ boldsymbol{A}^{prime} & boldsymbol{B}^{prime} & boldsymbol{C}^{prime}end{array}right|=mathbf{0} ) | 12 |
| 61 | ( operatorname{Let} boldsymbol{A}(mathbf{1}, mathbf{3}), boldsymbol{B}(mathbf{0}, mathbf{0}) ) and ( boldsymbol{C}(boldsymbol{k}, boldsymbol{0}) ) be vertices of a triangle ( A B C ) such that area of ( triangle A B C ) is ( 3 . ) Find the value of ( k ) ( A cdot pm 2 ) B. ±3 ( c .pm 4 ) D. ±1 | 12 |
| 62 | ( f(a, b, c)=left[begin{array}{ccc}a & 0 & 0 \ {[0.3 e m] 0} & b & 0 \ {[0.3 e m] 0} & 0 & cend{array}right] ) such that ( a b c neq 0 ) ( operatorname{then} A^{-1}=operatorname{diag}left(frac{1}{a}, frac{1}{b}, frac{1}{c}right)= ) ( left[begin{array}{ccc}frac{1}{a} & 0 & 0 \ {[0.3 e m] 0} & frac{1}{b} & 0 \ {[0.3 e m] 0} & 0 & frac{1}{c}end{array}right] ) | 12 |
| 63 | If ( D_{x}=25, D=5 ) are the values of the determinants for certain simultaneous equations in ( x ) and ( y, ) find ( x ) | 12 |
| 64 | The value of ( mid begin{array}{cc}mathbf{2} & boldsymbol{a}+boldsymbol{b}+boldsymbol{c}+boldsymbol{d} \ boldsymbol{a}+boldsymbol{b}+boldsymbol{c}+boldsymbol{d} & boldsymbol{2}(boldsymbol{a}+boldsymbol{b})(boldsymbol{c}+boldsymbol{d}) \ boldsymbol{a} boldsymbol{b}+boldsymbol{c} boldsymbol{d} & boldsymbol{a} boldsymbol{b}(boldsymbol{c}+boldsymbol{d})+boldsymbol{c} boldsymbol{d}(boldsymbol{a}+boldsymbol{b})end{array} ) ( mathbf{A} cdot mathbf{0} ) B. ( c .-1 ) D. None of these | 12 |
| 65 | ( left|begin{array}{ccc}2 & -3 & 3 \ 2 & 2 & 3 \ 3 & -2 & 2end{array}right| ) | 12 |
| 66 | Using the properties of determinant and without expanding, prove that: ( left|begin{array}{lll}2 & 7 & 65 \ 3 & 8 & 75 \ 5 & 9 & 86end{array}right|=0 ) | 12 |
| 67 | ( (1,6),(3 .-2) ) and ( (-2, K) ) are collinear points. What is ( boldsymbol{K} ) ? A . -6 B. 2 c. 8 D. 10 E . 18 | 12 |
| 68 | If 07,a2, az ……., an….. are in G.P., then the value of the determinant [2004] logan log an+1 log an+2| log An+3 log an+4 log an+5 . is log an+6 log an+7 log an+8| (a) -2 (b) 1 (c) 2 (d) 0 . | 12 |
| 69 | Find determinant ( boldsymbol{D} boldsymbol{c}=left|begin{array}{ccc}mathbf{1} & mathbf{1} & -mathbf{2} \ mathbf{1} & -mathbf{2} & mathbf{3} \ mathbf{2} & -mathbf{1} & -mathbf{1}end{array}right| ) | 12 |
| 70 | f ( (k, 2-2 k),(-k+1,2 k),(-4- ) ( k, 6-2 k) ) are collinear, then ( k= ) ( A cdot+1 ) B. – ( c cdot-2 ) ( D .2 ) | 12 |
| 71 | Prove that: ( left|begin{array}{ccc}mathbf{0} & boldsymbol{a} & -boldsymbol{b} \ -boldsymbol{a} & boldsymbol{0} & -boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{0}end{array}right|=mathbf{0} ) | 12 |
| 72 | 13. If and|A = 125 then the value of a is (2004S) (a) #1 (b) +2 (c) +3 (d) 5 | 12 |
| 73 | If the points ( boldsymbol{A}(boldsymbol{x}, mathbf{2}), boldsymbol{B}(-mathbf{3},-mathbf{4}) ) and ( C(7,-5) ) are collinear, then the value of ( x ) is : A . -63 B. 63 ( c .60 ) D. – 60 | 12 |
| 74 | ( left|begin{array}{ccc}a^{2}+1 & a b & a c \ a b & b^{2}+1 & b c \ a c & b c & c^{2}+1end{array}right|= ) A . abc B. atb+c c. ( 1+a^{2}+b^{2}+c^{2} ) ( D cdot a b c(1+a+b+c) ) | 12 |
| 75 | f ( x, y, z ) are all different and if ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{1}+boldsymbol{x}^{3} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{1}+boldsymbol{y}^{3} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{1}+boldsymbol{z}^{3}end{array}right|=boldsymbol{0} operatorname{then} 1+boldsymbol{x} boldsymbol{y} boldsymbol{z}= ) A . -1 B. ( c cdot 1 ) D. 2 | 12 |
| 76 | Find the adjoint of matrix ( A= ) ( left[begin{array}{lll}1 & 1 & 2 \ 2 & 3 & 5 \ 2 & 0 & 1end{array}right] ) | 12 |
| 77 | In a triangle ( A B C, ) with usual notations, if ( left|begin{array}{ccc}1 & a & b \ 1 & c & a \ 1 & b & cend{array}right|=0, ) then ( 4 sin ^{2} A+ ) ( 24 sin ^{2} B+36 sin ^{2} C ) is equal to A . 48 B. 50 c. 44 D. 34 | 12 |
| 78 | Let ( triangle boldsymbol{a}=left|begin{array}{ccc}boldsymbol{a}-mathbf{1} & boldsymbol{n} & boldsymbol{6} \ (boldsymbol{a}-mathbf{1})^{2} & boldsymbol{2} boldsymbol{n}^{2} & boldsymbol{4} boldsymbol{n}-boldsymbol{2} \ (boldsymbol{a}-mathbf{1})^{3} & boldsymbol{3} boldsymbol{n}^{3} & boldsymbol{3} boldsymbol{n}^{2}-boldsymbol{3} boldsymbol{n}end{array}right| ) Then ( sum_{a-1}^{n} triangle a ) is equal to ( mathbf{A} cdot mathbf{0} ) В ( cdot(a-1) sum n^{2} ) c. ( (a-1) n sum n ) D. None of these | 12 |
| 79 | If ( boldsymbol{A}_{mathbf{3} times mathbf{3}} ) and ( |boldsymbol{A}| neq mathbf{0} Rightarrow boldsymbol{A} boldsymbol{d} boldsymbol{j}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A})= ) ( mathbf{A} cdot|A|^{2} A ) B ( cdot|A| A ) c. ( frac{A}{|A|} ) D. ( frac{A}{|A|^{2}} ) | 12 |
| 80 | Find the value of the following determinant: ( left|begin{array}{cc}1.2 & 0.03 \ 0.57 & -0.23end{array}right| ) A. -0.266 B. -0.2471 c. -0.2381 D. -0.2931 | 12 |
| 81 | ( left|begin{array}{ccc}1+sin ^{2} theta & sin ^{2} theta & sin ^{2} theta \ cos ^{2} theta & 1+cos ^{2} theta & cos ^{2} theta \ 4 sin 4 theta & 4 sin 4 theta & 1+4 sin 4 thetaend{array}right|= ) ( 0, ) then ( sin 4 theta ) equals to A. ( 1 / 2 ) B. ( c cdot-1 / 2 ) D. -1 | 12 |
| 82 | The roots of the equation ( left|begin{array}{ccc}boldsymbol{x}-mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & boldsymbol{x}-mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1} & boldsymbol{x}-mathbf{1}end{array}right|=mathbf{0} operatorname{are} ) A. 1,2 в. -1,2 c. -1,-2 D. 1,-2 | 12 |
| 83 | Find the value of ( left|begin{array}{ccc}53 & 106 & 159 \ 52 & 65 & 91 \ 102 & 153 & 221end{array}right| ) | 12 |
| 84 | The value of the determinant ( left|begin{array}{lll}k a & k^{2}+a^{2} & 1 \ k b & k^{2}+b^{2} & 1 \ k c & k^{2}+c^{2} & 1end{array}right| ) is A. ( k(a+b)(b+c)(c+a) ) B. ( k a b cleft(a^{2}+b^{2}+c^{2}right) ) c. ( k(a-b)(b-c)(c-a) ) D. ( k(a+b-c)(b+c-a)(c+a-b) ) | 12 |
| 85 | If ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{y}end{array}right] ) for ( boldsymbol{x} neq ) ( mathbf{0}, boldsymbol{y} neq mathbf{0}, ) then ( boldsymbol{D} ) is: A. divisible by neither ( x ) nor ( y ) B. divisible by both ( x ) nor ( y ) c. divisible by ( x ) but not ( y ) D. divisible by ( y ) but not ( x ) | 12 |
| 86 | ( mathbf{f} mathbf{Delta}=left|begin{array}{lll}boldsymbol{b}^{2}-boldsymbol{a} boldsymbol{b} & boldsymbol{b}-boldsymbol{c} & boldsymbol{b} boldsymbol{c}-boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b}-boldsymbol{a}^{2} & boldsymbol{a}-boldsymbol{b} & boldsymbol{b}^{2}-boldsymbol{a} boldsymbol{b} \ boldsymbol{b} boldsymbol{c}-boldsymbol{a} boldsymbol{c} & boldsymbol{c}-boldsymbol{a} & boldsymbol{a} boldsymbol{b}-boldsymbol{a}^{2}end{array}right| ) then ( Delta ) equals A ( cdot(b-c)(c-a)(a-b) ) B. ( a b c(b-c)(c-a)(a-b) ) c. ( (a+b+c)(b-c)(c-a)(a-b) ) D. | 12 |
| 87 | ( mathbf{f}_{mathbf{0}}=left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r}end{array}right| ) and ( boldsymbol{Delta}_{2}=left|begin{array}{lll}boldsymbol{y} & boldsymbol{b} & boldsymbol{q} \ boldsymbol{x} & boldsymbol{a} & boldsymbol{p} \ boldsymbol{z} & boldsymbol{c} & boldsymbol{r}end{array}right| ) then ( Delta_{1} ) is equal to ( A cdot 2 Delta_{2} ) в. ( Delta_{2} ) ( c cdot-Delta_{2} ) D. none of these | 12 |
| 88 | The value of the determinant ( left|begin{array}{ccc}cos alpha & -sin alpha & 1 \ sin alpha & cos alpha & 1 \ cos (alpha+beta) & -sin (alpha+beta) & 1end{array}right| ) A. Independent of ( alpha ) B. Independent of ( beta ) c. Independent of ( alpha ) and ( beta ) D. None of the above | 12 |
| 89 | f ( a neq b neq c, ) prove that the points ( left(a, a^{2}right),left(b, b^{2}right),left(c, c^{2}right) ) can never be collinear. | 12 |
| 90 | ( mathbf{f}left(mathbf{1}+boldsymbol{a} boldsymbol{x}+boldsymbol{b} boldsymbol{x}^{2}right)^{4}=boldsymbol{a}_{mathbf{0}}+boldsymbol{a}_{1} boldsymbol{x}+ ) ( a_{2} x^{2}+ldots ldots . .+a_{8} x^{8}, ) where ( a, b, a_{0}, a_{1}, dots dots a_{8} in R ) such that ( a_{0}+ ) ( a_{1}+a_{2} neq 0 ) and ( left|begin{array}{lll}a_{0} & a_{1} & a_{2} \ a_{1} & a_{2} & a_{0} \ a_{2} & a_{0} & a_{1}end{array}right|=0, ) then the value of ( frac{mathbf{5} a}{b} ) is | 12 |
| 91 | If ( k ) is a scalar and ( A ) is an ( n times n ) square matrix, then ( |boldsymbol{k} boldsymbol{A}|= ) A ( cdot k|A|^{n} ) в. ( k|A| ) c ( cdot k^{n}left|A^{n}right| ) D ( cdot k^{n}|A| ) | 12 |
| 92 | The value of the determinant ( left|begin{array}{c}b^{2}-a b b-c b c-a c \ a b-a^{2} a-b b^{2}-a b \ b c-a c c-a a b-a^{2}end{array}right|= ) A ( . a b c ) B. ( a+b+c ) c. 0 ( mathbf{D} cdot a b+b c+c a ) | 12 |
| 93 | If ( A ) is a ( 3- ) rowed square matrix and ( |boldsymbol{A}|=mathbf{4} ) then ( boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A})=? ) A ( .4 A ) в. ( 16 A ) ( c cdot 64 A ) D. None of these | 12 |
| 94 | Prove that ( left|begin{array}{lll}boldsymbol{a}+boldsymbol{b} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a} & boldsymbol{b}end{array}right|=mathbf{3} boldsymbol{a} boldsymbol{b} boldsymbol{c}-boldsymbol{1} ) ( boldsymbol{a}^{3}-boldsymbol{b}^{3}-boldsymbol{c}^{3} ) | 12 |
| 95 | If ( A=left|begin{array}{ll}2 & 3 \ 6 & 9end{array}right| ) then ( |A|= ) A . B. 1 ( c cdot 2 ) D. | 12 |
| 96 | ( fleft|begin{array}{lll}boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & mathbf{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & 1end{array}right|=left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{b}_{1} & mathbf{1} \ boldsymbol{a}_{2} & boldsymbol{b}_{2} & mathbf{1} \ boldsymbol{a}_{3} & boldsymbol{b}_{3} & 1end{array}right| ), then the two triangles with vertices ( left(x_{1}, y_{1}right),left(x_{2}, y_{2}right),left(x_{3}, y_{3}right) ) and ( left(a_{1}, b_{1}right) ) ( left(a_{2}, b_{2}right),left(a_{3}, b_{3}right) ) must be congruent A. True B. False | 12 |
| 97 | The value of ( left|begin{array}{ccc}(boldsymbol{a}+boldsymbol{d})(boldsymbol{a}+mathbf{2} boldsymbol{d}) & boldsymbol{a}+mathbf{2} boldsymbol{d} & boldsymbol{a} \ mathbf{2} boldsymbol{d}(boldsymbol{a}+mathbf{2} boldsymbol{d}) & boldsymbol{d} & boldsymbol{d} \ mathbf{2} boldsymbol{d}(boldsymbol{a}+mathbf{3} boldsymbol{d}) & boldsymbol{d} & boldsymbol{d}end{array}right| ) A . ( 4 d ) B ( .4 d^{2} ) ( c cdot 4 d^{3} ) D. ( 4 d^{4} ) | 12 |
| 98 | f ( a, b, c ) are non-zero and different from 1, then the value of ( left|begin{array}{ccc}log _{a} 1 & log _{a} b & log _{a} c \ log _{a}left(frac{1}{b}right) & log _{b} 1 & log _{a}left(frac{1}{c}right) \ log _{a}left(frac{1}{c}right) & log _{a} c & log _{c} 1end{array}right| ) A . 0 B. ( 1+log _{a}(a+b+c) ) ( mathbf{c} cdot log _{a}(a b+b c+c a) ) D. 1 ( E cdot log _{a}(a+b+c) ) | 12 |
| 99 | 9. Which of the following values of a satisfy the equation | (1+a)2 (1+2a)2 (1+3a)2 (2+ a)2 (2+2a)2 (2+3a)2 = -648a ? |(3+a)? (3+2a)2 (3+3a)2 | (a) 4 (6) 9 (c) 9 (JEE Adv. 2015) (d) 4 | 12 |
| 100 | Solve: ( left|begin{array}{ccc}mathbf{0} & -mathbf{3} & boldsymbol{x} \ boldsymbol{x}+mathbf{1} & mathbf{3} & mathbf{1} \ mathbf{4} & mathbf{1} & mathbf{5}end{array}right|=mathbf{0} ) | 12 |
| 101 | solve: ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{2} & -mathbf{1} \ boldsymbol{3} & -boldsymbol{1} & boldsymbol{4} \ boldsymbol{x} & boldsymbol{2} & -boldsymbol{5}end{array}right|=mathbf{0} ) | 12 |
| 102 | Assertion Let ( A ) be a ( 2 times 2 ) matrix with real entries. Let ( I ) be the ( 2 times 2 ) identity matrix. Denote by ( t r(A), ) the sum of diagonal entries of ( A ). Assume that ( boldsymbol{A}^{2}=boldsymbol{I} ) If ( boldsymbol{A} neq boldsymbol{I} ) and ( boldsymbol{A} neq-boldsymbol{I}, ) then ( operatorname{det}(boldsymbol{A})=-mathbf{1} ) Reason If ( boldsymbol{A} neq boldsymbol{I} ) and ( boldsymbol{A} neq-boldsymbol{I}, ) then ( boldsymbol{t r}(boldsymbol{A}) neq mathbf{0} ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 103 | Evaluate: ( left|begin{array}{ccc}mathbf{1} & boldsymbol{b} boldsymbol{c} & boldsymbol{a}(boldsymbol{b}+boldsymbol{c}) \ mathbf{1} & boldsymbol{c} boldsymbol{a} & boldsymbol{b}(boldsymbol{c}+boldsymbol{a}) \ boldsymbol{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{c}(boldsymbol{a}+boldsymbol{b})end{array}right| ) ( mathbf{A} cdot mathbf{0} ) B. ( c . a b c ) D. ( a^{2}+b^{2}+c^{2} ) | 12 |
| 104 | x + 4y + z = 0; If the system of linear equations x + 2ay + az = 0 ; x + 3y + bz = 0 ; x + 4y + CZ has a non – zero solution, then a, b, c. (a) satisfy a + 2b + 3c = 0 (b) are in A.P (c) are in G.P (d) are in H.P. | 12 |
| 105 | Find the value of following determinant. ( left|begin{array}{ll}frac{7}{3} & frac{5}{3} \ frac{3}{2} & frac{1}{2}end{array}right| ) | 12 |
| 106 | If the points ( boldsymbol{A}(-2,1), B(a, b) ) and ( C(4,-1) ) are collinear and ( a-b=1 ) find the values of ( a ) and ( b ) A ( . a=1, b=5 ) В. ( a=1, b=0 ) c. ( a=2, b=0 ) D. None of these | 12 |
| 107 | Using properties of determinants, prove the following [ mid begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \ 2 a b & 1-a^{2}+b^{2} & 2 a \ 2 b & -2 a & 1-a^{2}-b^{2} end{array} ] ( left(1+a^{2}+b^{2}right)^{3} ) | 12 |
| 108 | ( left|begin{array}{ccc}mathbf{1}+boldsymbol{x} & mathbf{2} & mathbf{3} \ mathbf{1} & mathbf{2}+boldsymbol{x} & mathbf{3} \ mathbf{1} & mathbf{2} & mathbf{3}+boldsymbol{x}end{array}right|=mathbf{0} operatorname{then} boldsymbol{x}= ) ( A ) B. -1 ( c .-6 ) ( D ) | 12 |
| 109 | Let ( n ) be a positive integer and ( Delta_{r}= ) ( left|begin{array}{ccc}2 r-1 & n & C_{r} & 1 \ n^{2}-1 & 2^{n} & n+1 \ cos ^{2}left(n^{2}right) & cos ^{2} n & cos ^{2}(n+1)end{array}right| ) then ( sum_{r=0}^{n} Delta_{r}=dots ) ( A cdot 0 ) B. ( c cdot 2 ) ( D ) | 12 |
| 110 | (i) Solve the equation ( left|begin{array}{ccc}boldsymbol{x}-mathbf{1} & boldsymbol{2} & boldsymbol{3} \ mathbf{0} & boldsymbol{x}-boldsymbol{2} & boldsymbol{4} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{x}-boldsymbol{3}end{array}right| ) (ii) Show that ( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{b} boldsymbol{c} \ mathbf{1} & boldsymbol{b} & boldsymbol{c a} \ mathbf{1} & boldsymbol{c} & boldsymbol{a b}end{array}right|=left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2} \ mathbf{1} & boldsymbol{b} & boldsymbol{b}^{2} \ mathbf{1} & boldsymbol{c} & boldsymbol{c}^{2}end{array}right| ) | 12 |
| 111 | Prove the following: ( left|begin{array}{lll}boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c}end{array}right|=mathbf{2}left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a}end{array}right| ) | 12 |
| 112 | ( mathbf{a}=left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a}end{array}right|, ) then the value of ( left|begin{array}{ccc}a^{2}-b c & b^{2}-c a & c^{2}-a b \ c^{2}-a b & a^{2}-b c & b^{2}-c a \ b^{2}-c a & c^{2}-a b & a^{2}-b cend{array}right| ) ( A cdot Delta^{2} ) B ( .2 Delta^{2} ) ( c cdot Delta^{3} ) D. none of these | 12 |
| 113 | Solve: Value of ( boldsymbol{D}=left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2}end{array}right| ) is | 12 |
| 114 | If ( a neq 6, b, c ) satisfy ( left|begin{array}{ccc}a & 2 b & 2 c \ 3 & b & c \ 4 & a & bend{array}right|=0 ) then ( a b c= ) ( mathbf{A} cdot a+b+c ) B. ( c cdot b^{3} ) ( mathbf{D} cdot a b+b-c ) | 12 |
| 115 | Using properties of determinants, show that triangle ( A B C ) is isosceles, if : ( mid begin{array}{ccc}mathbf{1} & mathbf{1} \ mathbf{1}+cos boldsymbol{A} & mathbf{1}+cos boldsymbol{B} & mathbf{1} \ cos ^{2} boldsymbol{A}+cos boldsymbol{A} & cos ^{2} boldsymbol{B}+cos boldsymbol{B} & mathbf{c o s}^{2}end{array} ) 0 | 12 |
| 116 | Let ( S ) be the sample space of all ( 3 times 3 ) matrices with entries from the set ( {0,1} . ) Let the events ( E_{1}= ) ( {A in S: operatorname{det} A=0} ) and ( E_{2}= ) ( {A in S: S u m text { of entries of } A text { is } 7} ) If a matrix is chosen at random from ( boldsymbol{S} ) then the conditional probability ( boldsymbol{P}left(boldsymbol{E}_{1} mid boldsymbol{E}_{2}right) ) equals. | 12 |
| 117 | ( left|begin{array}{lll}mathbf{a}+mathbf{b} & mathbf{a} & mathbf{b} \ mathbf{a} & mathbf{a}+mathbf{c} & mathbf{c} \ mathbf{b} & mathbf{c} & mathbf{b}+mathbf{c}end{array}right|= ) ( A cdot 4 ) abc B. abç ( c cdot 2 a^{2} b^{2} c^{2} ) D. ( 4 a^{2} b^{2} c^{2} ) | 12 |
| 118 | ( boldsymbol{D}=left|begin{array}{ccc}mathbf{1 8} & mathbf{4 0} & mathbf{8 9} \ mathbf{4 0} & mathbf{8 9} & mathbf{1 9 8} \ mathbf{8 9} & mathbf{1 9 8} & mathbf{4 4 0}end{array}right|= ) ( A ) B. – c. zero ( D ) | 12 |
| 119 | Let ( boldsymbol{f}(boldsymbol{x})=left|begin{array}{ccc}2 cot boldsymbol{x} & -mathbf{1} & mathbf{0} \ mathbf{1} & cot boldsymbol{x} & -mathbf{1} \ mathbf{0} & mathbf{1} & mathbf{2} cot boldsymbol{x}end{array}right| ) then | 12 |
| 120 | ( left|begin{array}{lll}a^{2}+lambda^{2} & a b+c lambda & c a-b lambda \ a b-c lambda & b^{2}+lambda^{2} & b c+a lambda \ c a+b lambda & b c-a lambda & c^{2}+lambda^{2}end{array}right| mid begin{array}{cc}lambda & c \ -c & lambda \ b & -aend{array} ) ( left(1+a^{2}+b^{2}+c^{2}right)^{3}, ) then the value of ( lambda ) is ( A cdot 8 ) в. 2 ( c ) ( D ) | 12 |
| 121 | Using properties of deteminants, prove [ operatorname{that}left|begin{array}{ccc} frac{(boldsymbol{a}+boldsymbol{b})^{2}}{boldsymbol{c}} & boldsymbol{c} & boldsymbol{c} \ boldsymbol{a} & frac{(boldsymbol{b}+boldsymbol{c})^{2}}{boldsymbol{a}} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{b} & frac{(boldsymbol{c}+boldsymbol{a})^{2}}{boldsymbol{b}} end{array}right| ] ( 2(a+b+c)^{3} ) | 12 |
| 122 | Show that ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c}end{array}right|= ) ( boldsymbol{a}^{3}+boldsymbol{b}^{3}+boldsymbol{c}^{3}-boldsymbol{3} boldsymbol{a} boldsymbol{b} boldsymbol{c} ) | 12 |
| 123 | Assertion If ( A=left(begin{array}{ll}cos alpha & sin alpha \ cos alpha & sin alphaend{array}right) ) and ( B= ) ( left(begin{array}{cc}cos alpha & cos alpha \ sin alpha & sin alphaend{array}right) ) then ( A B neq I ) Reason The product of two matrices can never be equal to an identity matrix A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 12 |
| 124 | Solve: ( left|begin{array}{ccc}-1 & x & 2 \ 3 & 4 & -2 \ 4 & x & -3end{array}right|=0 ) | 12 |
| 125 | ( fleft|begin{array}{ccc}6 i & -3 i & 1 \ 4 & 3 i & -1 \ 20 & 3 & iend{array}right|=x+i y ) then A ( . x=3, y=1 ) B. ( x=1, y=3 ) c. ( x=0, y=3 ) D. ( x=0, y=0 ) | 12 |
| 126 | The value of ( left|begin{array}{ll}cos 15^{circ} & sin 15^{circ} \ sin 75^{circ} & cos 75^{circ}end{array}right| ) | 12 |
| 127 | If ( A=left[begin{array}{cc}-8 & 5 \ 2 & 4end{array}right] ) satisfies the equation ( x^{2}+4 x-p=0, ) then ( p= ) A . 64 B. 42 ( c . ) 36 D. 24 | 12 |
| 128 | ( mathrm{ft} mathrm{A}=left[begin{array}{ccc}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4end{array}right] ) then show that ( |3 A|=27|A| ) | 12 |
| 129 | If the points ( (-3,6),(-9, a) ) and (0,15) are collinear, then find a | 12 |
| 130 | f ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{4} & boldsymbol{7} \ boldsymbol{6} & boldsymbol{5}end{array}right], ) find ( |boldsymbol{3} boldsymbol{A}| ) | 12 |
| 131 | If ( A A^{T}=I ) and ( operatorname{det}(A)=1, ) then A. Every element of ( A ) is equal to it’s co-factor B. Every element of A and it’s co-factor are additive inverse of each other C. Every element of A and it’s co-factor are multiplicative inverse of each other D. None of these | 12 |
| 132 | Find the non-zero roots of the equation ( boldsymbol{Delta}=left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{a} boldsymbol{x}+boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b} boldsymbol{x}+boldsymbol{c} \ boldsymbol{a} boldsymbol{x}+boldsymbol{b} & boldsymbol{b} boldsymbol{x}+boldsymbol{c} & boldsymbol{c}end{array}right|=0 ) | 12 |
| 133 | Find the values of ( x, ) if ( left|begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5}end{array}right|=left|begin{array}{ll}boldsymbol{x} & mathbf{3} \ mathbf{2} boldsymbol{x} & mathbf{5}end{array}right| ) | 12 |
| 134 | 3. If 1,0,02 are the cube roots of unity, then 1 on w2n| A=0″ @21 1 is equal to [2003] (a) 0² (b) 0 (0) 1 (d) w | 12 |
| 135 | ( boldsymbol{A}=left[begin{array}{ccc}mathbf{5} & mathbf{5} boldsymbol{a} & boldsymbol{a} \ mathbf{0} & boldsymbol{a} & mathbf{5} boldsymbol{a} \ mathbf{0} & mathbf{0} & mathbf{5}end{array}right] ) If ( left|boldsymbol{A}^{2}right|=mathbf{2 5} ) then ( |boldsymbol{a}|= ) ( A cdot 5 ) В. ( 5^{2} ) ( c ) ( D ) | 12 |
| 136 | f ( boldsymbol{y}=sin boldsymbol{p} boldsymbol{x} ) prove that ( boldsymbol{Delta}=left|begin{array}{lll}boldsymbol{y} & boldsymbol{y}_{1} & boldsymbol{y}_{2} \ boldsymbol{y}_{3} & boldsymbol{y}_{4} & boldsymbol{y}_{5} \ boldsymbol{y}_{6} & boldsymbol{y}_{7} & boldsymbol{y}_{8}end{array}right|=0 ) where ( y_{r}, ) means ( r-t h ) differential coefficient of ( boldsymbol{y} ) | 12 |
| 137 | ( f(x)=left|begin{array}{ccc}cos x & 1 & 0 \ 1 & cos x & 1 \ 0 & 1 & cos xend{array}right| ) the ( f^{prime}left(frac{pi}{3}right) ) equals A ( frac{11 sqrt{3}}{8} ) ( B cdot frac{5 sqrt{3}}{8} ) ( c cdot-frac{5 sqrt{3}}{8} ) D. none of these | 12 |
| 138 | Find the relation between ( a ) and ( b ). If the points ( boldsymbol{P}(1,2), Q(0,0) ) and ( P(a, b) ) are collinear | 12 |
| 139 | Show that ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{y} boldsymbol{z} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{z} boldsymbol{x} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{x} boldsymbol{y}end{array}right|=(boldsymbol{x}-boldsymbol{y})(boldsymbol{y}- ) ( z(z-x)(x y+y z+z x) ) | 12 |
| 140 | If the points (0,4),(4,0) and ( (5, p) ) are collinear, then value of ( p ) is A . -1 B. 7 ( c cdot 6 ) D. | 12 |
| 141 | If ( boldsymbol{u}_{n}= ) then ( boldsymbol{u}_{n}=boldsymbol{a}_{n} boldsymbol{u}_{n-1}+boldsymbol{u}_{n-2} ) f true enter 1 else enter | 12 |
| 142 | Find the largest value of a third-order determinant whose element are 0 or 1 | 12 |
| 143 | Using properties of determinant, prove [ text { that }left|begin{array}{lll} boldsymbol{b}+boldsymbol{c} & boldsymbol{a}-boldsymbol{b} & boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{b}-boldsymbol{c} & boldsymbol{b} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{c}-boldsymbol{a} & boldsymbol{c} end{array}right|=mathbf{3} boldsymbol{a} boldsymbol{b} boldsymbol{c}-boldsymbol{a}^{3}- ] ( b^{3}-c^{3} ) | 12 |
| 144 | If ( boldsymbol{A}=[boldsymbol{a} boldsymbol{i} boldsymbol{j}] ) is a matrix of order ( 2 x boldsymbol{2} ) such that ( |A|=15 ) and cij represents the co factor of aij then find ( a_{21} c_{21}+ ) ( boldsymbol{a}_{22} boldsymbol{c}_{22} ) | 12 |
| 145 | The value of ( left|begin{array}{ccc}boldsymbol{a}+boldsymbol{p} boldsymbol{d} & boldsymbol{a}+boldsymbol{q} boldsymbol{d} boldsymbol{a}+boldsymbol{r} boldsymbol{d} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r} \ boldsymbol{d} & boldsymbol{f} & boldsymbol{d}end{array}right| ) ( mathbf{A} cdot mathbf{0} ) B. – ( c cdot 1 ) D. ( p+q+r ) | 12 |
| 146 | The value of determinant ( left|begin{array}{ccc}mathbf{1 9} & mathbf{6} & mathbf{7} \ mathbf{2 1} & mathbf{3} & mathbf{1 5} \ mathbf{2 8} & mathbf{1 1} & mathbf{6}end{array}right| ) is : A. 150 B. -110 ( c cdot 0 ) D. None of these | 12 |
| 147 | Find the Adjoint matrix of the matrix ( left|begin{array}{lll}1 & 2 & 3 \ 2 & 3 & 2 \ 3 & 3 & 4end{array}right| ) | 12 |
| 148 | Find the value of the determinant with out expanding: ( left|begin{array}{lll}5 & 2 & 3 \ 7 & 3 & 4 \ 9 & 4 & 5end{array}right| ) | 12 |
| 149 | Let ( a, b, c ) be such that ( b(a+c) neq 0 ) ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{a}+mathbf{1} & boldsymbol{a}-mathbf{1} \ -boldsymbol{b} & boldsymbol{b}+mathbf{1} & boldsymbol{b}-mathbf{1} \ boldsymbol{c} & boldsymbol{c}-mathbf{1} & boldsymbol{c}+mathbf{1}end{array}right|+ ) ( left|begin{array}{ccc}boldsymbol{a}+mathbf{1} & boldsymbol{b}+mathbf{1} & boldsymbol{c}-mathbf{1} \ boldsymbol{a}-mathbf{1} & boldsymbol{b}-mathbf{1} & boldsymbol{c}+mathbf{1} \ (-mathbf{1})^{n+mathbf{2}} boldsymbol{a} & (-mathbf{1})^{n+mathbf{1}} boldsymbol{b} & (-mathbf{1})^{n} boldsymbol{c}end{array}right|=mathbf{0} ) then the value of ( n ) is A. Any integer B. zero c. Any even integer D. Any odd integer | 12 |
| 150 | 5 15. Let A = 10 Το 5α α α 5α O 5 then la equals (a) 1/5 (6) 5 (0) 52 [2007] (d) 1 | 12 |
| 151 | f ( x, y z ) are all different and if ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{boldsymbol{2}} & boldsymbol{1}+boldsymbol{x}^{boldsymbol{3}} \ boldsymbol{y} & boldsymbol{y}^{boldsymbol{2}} & boldsymbol{1}+boldsymbol{y}^{boldsymbol{3}} \ boldsymbol{z} & boldsymbol{z}^{boldsymbol{2}} & boldsymbol{1}+boldsymbol{z}^{boldsymbol{3}}end{array}right|=0 ) then ( mathbf{x y z}= ) A . -1 B. ( c . ) D. ±1 | 12 |
| 152 | Evaluate the following determinant : ( = ) ( begin{array}{|lll|}mathbf{6 7} & mathbf{1 9} & mathbf{2 1} \ mathbf{3 9} & mathbf{1 3} & mathbf{1 4} \ mathbf{8 1} & mathbf{2 4} & mathbf{2 6}end{array} ) | 12 |
| 153 | ( begin{array}{|ccc|}text { If } boldsymbol{x}+boldsymbol{y}+boldsymbol{z}= & mathbf{0} text { ,find } \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{x}^{mathbf{2}} & boldsymbol{y}^{mathbf{2}} & boldsymbol{z}^{mathbf{2}} \ boldsymbol{y}+boldsymbol{z} & boldsymbol{z}+boldsymbol{x} & boldsymbol{x}+boldsymbol{y}end{array} ) | 12 |
| 154 | Using the properties of determinant and without expanding, prove that: ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{a} & boldsymbol{x}+boldsymbol{a} \ boldsymbol{y} & boldsymbol{b} & boldsymbol{y}+boldsymbol{b} \ boldsymbol{z} & boldsymbol{c} & boldsymbol{z}+boldsymbol{c}end{array}right|=mathbf{0} ) | 12 |
| 155 | Find ( ^{prime} x^{prime} ) if ( left|begin{array}{ccc}mathbf{4} & boldsymbol{x} & mathbf{6} \ mathbf{2} & mathbf{3} & mathbf{4} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right|=mathbf{1 0} ) | 12 |
| 156 | If ( a>0, b>0, c>0 ) are respectively the ( p^{t h}, q^{t h}, r^{t h} ) terms of a G.P., then the value of the deteminant ( left|begin{array}{lll}log boldsymbol{a} & boldsymbol{p} & 1 \ log boldsymbol{b} & boldsymbol{q} & 1 \ log boldsymbol{c} & boldsymbol{r} & 1end{array}right| ) is A . 1 B. 0 ( c cdot-1 ) D. None of these | 12 |
| 157 | Let ( 0<theta<pi / 2 ) and ( boldsymbol{Delta}(boldsymbol{x}, boldsymbol{theta})=left|begin{array}{ccc}boldsymbol{x} & tan boldsymbol{theta} & cot boldsymbol{theta} \ -tan boldsymbol{theta} & -boldsymbol{x} & mathbf{1} \ cot boldsymbol{theta} & boldsymbol{1} & boldsymbol{x}end{array}right| ) then This question has multiple correct options A ( cdot Delta(0, theta)=0 ) B. ( Deltaleft(x, frac{pi}{4}right)=x-x^{3} ) c. ( operatorname{Min}_{0<theta<pi / 2} Delta(1, theta)=0 ) D. ( Delta(x, theta) ) is independent of ( x ) | 12 |
| 158 | The determinant ( left|begin{array}{ccc}sin alpha & cos alpha & 1 \ sin beta & cos beta & 1 \ sin gamma & cos gamma & 1end{array}right| ) is equal to This question has multiple correct options ( ^{mathrm{A}} cdot_{-4 sin } frac{alpha-beta}{2} sin frac{alpha-gamma}{2} sin frac{gamma-alpha}{2} ) ( mathbf{B} cdot sin alpha+sin beta+sin gamma ) c. ( sin (alpha-beta)+sin (beta-gamma)+sin (gamma-alpha) ) D. none of these | 12 |
| 159 | The vertices of the triangle ( A B C ) are ( (2,1,1),(3,1,2),(-4,0,1) . ) The area of triangle is A ( cdot frac{3 sqrt{38}}{2} ) B. ( sqrt{38} ) c. ( frac{sqrt{38}}{2} ) D. 4 | 12 |
| 160 | ( fleft(a_{1}, a_{2}, a_{3}, dots ) from a geometric right. progression and ( a_{i}>0 ) for all ( i geq 1 ) [ operatorname{then}left|begin{array}{ccc} log boldsymbol{a}_{m} & log boldsymbol{a}_{m+1} & log boldsymbol{a}_{m+2} \ log boldsymbol{a}_{m+3} & log boldsymbol{a}_{m+4} & log boldsymbol{a}_{m+5} \ log boldsymbol{a}_{m+boldsymbol{6}} & log boldsymbol{a}_{m+boldsymbol{7}} & log boldsymbol{a}_{m+8} end{array}right| text { is } ] equal to ( mathbf{A} cdot log a_{m+8}-log a_{m} ) B. ( log a_{m} ) ( mathbf{c} cdot 2 log a_{m+1} ) D. | 12 |
| 161 | ( A_{3 * 3} ) is a non – singular matrix ( Rightarrow ) ( boldsymbol{A}^{2}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A})= ) ( mathbf{A} cdot|A| A ) B. ( I ) c. ( |A| ) D・ ( |A|^{2} I ) | 12 |
| 162 | ( left|begin{array}{ccc}mathbf{0} & boldsymbol{p}-boldsymbol{q} & boldsymbol{p}-boldsymbol{r} \ boldsymbol{q}-boldsymbol{p} & boldsymbol{0} & boldsymbol{q}-boldsymbol{r} \ boldsymbol{r}-boldsymbol{p} & boldsymbol{r}-boldsymbol{q} & boldsymbol{0}end{array}right| ) is equal to A. ( p+q+r ) B. c. ( p-q-r ) ( mathbf{D} cdot-p+q+r ) | 12 |
| 163 | If ( t_{1}, t_{2} ) and ( t_{3} ) distinct. and the points ( left(t_{1} cdot 2 a t_{1}+a t_{1}^{3}right) cdotleft(t_{2} cdot 2 a t_{2}+a t_{2}^{3}right),left(t_{3} cdot 2 aright. ) are collinear, then ( t_{1}+t_{2}+t_{3}= ) A ( cdot t_{1} t_{2} t_{3}=-1 ) B ( cdot t_{1}+t_{2}+t_{3}=t_{1} t_{2} t ) ( mathbf{c} cdot t_{1}+t_{2}+t_{3}=0 ) D. ( t_{1}+t_{2}+t_{3}=-1 ) | 12 |
| 164 | Prove that ( :=(a-b)(b-c)(c-a)(a+ ) ( b+c) ) [ begin{array}{ll}1 & a \ 1 & b \ 1 & c & a^{3}end{array} ] | 12 |
| 165 | Iet a b c be the real numbers. Then following system of equations in x, y and z (1995) IN NI r2 y2 + 3 = 1 has 22 (a) no solution (b) unique solution (c) infinitely many solutions(d) finitely many solutions Q2 x 22 62. x | 12 |
| 166 | f ( x, y, z ) are different and ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{1}+boldsymbol{x}^{2} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{1}+boldsymbol{y}^{2} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{1}+boldsymbol{z}^{2}end{array}right|=mathbf{0} ) then prove that ( mathbf{1}+boldsymbol{x} boldsymbol{y} boldsymbol{z}=mathbf{0} ) | 12 |
| 167 | Prove that [ left|begin{array}{ccc} mathbf{1} & mathbf{1} & mathbf{1}+mathbf{3} boldsymbol{x} \ mathbf{1}+mathbf{3} boldsymbol{y} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+mathbf{3} boldsymbol{z} & mathbf{1} end{array}right|=mathbf{9}(mathbf{3} boldsymbol{x} boldsymbol{y} boldsymbol{z}+ ] ( boldsymbol{x} boldsymbol{y}+boldsymbol{y} boldsymbol{z}+boldsymbol{z} boldsymbol{x} ) | 12 |
| 168 | For what value of ( x ) the matrix ( A ) is singular? ( boldsymbol{A}=left[begin{array}{ll}1+x & 7 \ 3-x & 8end{array}right] ) | 12 |
| 169 | f ( boldsymbol{A}=left|begin{array}{cc}2 & mathbf{3} \ -mathbf{1} & mathbf{2}end{array}right| ) Find ( boldsymbol{A} ) | 12 |
| 170 | Consider the points ( boldsymbol{P}=(-sin (boldsymbol{beta}- ) ( boldsymbol{alpha}),-cos beta), boldsymbol{Q}=(cos (beta-boldsymbol{alpha}), sin beta) ) and ( boldsymbol{R}=(cos (boldsymbol{beta}-boldsymbol{alpha}+boldsymbol{theta}), sin (boldsymbol{beta}-boldsymbol{theta})) ) where ( 0<alpha, beta<frac{pi}{4} ) then A. ( P ) lies on the line segment ( R Q ) B. ( Q ) lies on the line segment ( P R ) c. ( R ) lies on the line segment ( Q P ) D. ( P, Q, R ) are non-collinear | 12 |
| 171 | ( mathbf{A}=left[begin{array}{lll}b^{2} c^{2} & b c & b+c \ c^{2} a^{2} & c a & c+a \ a^{2} b^{2} & a b & a+bend{array}right] ) then ( |A|=? ) A ( cdot a b c ) B . ( a b c-1 ) ( c cdot a b c+1 ) D. | 12 |
| 172 | Evaluate: ( left|begin{array}{cc}x^{2}-x+1 & x-1 \ x+1 & x+1end{array}right| ) | 12 |
| 173 | If ( A ) is a square matrix of order ( 3 times 3 ) such that ( |boldsymbol{A}|=mathbf{5}, ) then find ( |boldsymbol{4} boldsymbol{A}| ) | 12 |
| 174 | ( A_{3 times 3} ) is a matrix such that ( |A|= ) ( boldsymbol{a}, boldsymbol{B}=(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}) ) such that ( |boldsymbol{B}|=boldsymbol{b} . ) Find the value of ( frac{left(a b^{2}+a^{2} b+1right) S}{25} ) where ( frac{1}{2} S=frac{a}{b}+frac{a^{2}}{b^{3}}+frac{a^{3}}{b^{5}}+dots dots dots u p t o infty ) and ( boldsymbol{a}=mathbf{3} ) | 12 |
| 175 | The number of ( boldsymbol{A} ) in ( boldsymbol{T}_{boldsymbol{p}} ) such that ( boldsymbol{A} ) is either symmetric or skew-symmetric or both, and det ( (A) ) divisible by ( p, ) is A ( cdot(p-1)^{2} ) в. ( 2(p-1) ) c. ( (p-1)^{2}+1 ) D. ( 2 p-1 ) | 12 |
| 176 | Consider the following statements in respect of the determinant ( left|begin{array}{cc}cos ^{2} frac{alpha}{2} & sin ^{2} frac{alpha}{2} \ sin ^{2} frac{beta}{2} & cos ^{2} frac{beta}{2}end{array}right| ) where ( alpha, beta ) are complementary angles 1. The value of the determinant is ( frac{1}{sqrt{2}} cos left(frac{alpha-beta}{2}right) ) 2. The maximum value of the determinant is ( frac{mathbf{1}}{sqrt{mathbf{2}}} ) Which of the above statements is/are correct? A. 1 only B. 2 only c. Both 1 and 2 D. Neither 1 nor 2 | 12 |
| 177 | ( mathbf{a}=left|begin{array}{ccc}cos theta / 2 & 1 & 1 \ 1 & cos theta / 2 & -cos theta / 2 \ -cos theta / 2 & 1 & -1end{array}right| ) If the minimun of ( Delta ) is ( m_{1} ) and maximum of ( Delta ) is ( m_{2}, ) then ( left[m_{1}, m_{2}right] ) are related A. [-4,-2] B. [2, 4] ( c cdot[-4,0] ) D. [0, 2] | 12 |
| 178 | ( left|begin{array}{lll}1 & 2 & 3 \ 0 & 2 & 4 \ 0 & 0 & 5end{array}right| ) | 12 |
| 179 | Prove that ( left|begin{array}{lll}boldsymbol{b} boldsymbol{c} & boldsymbol{b} boldsymbol{c}^{prime}+boldsymbol{b}^{prime} boldsymbol{c} & boldsymbol{b}^{prime} boldsymbol{c}^{prime} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c} boldsymbol{a}^{prime}+boldsymbol{c}^{prime} boldsymbol{a} & boldsymbol{c}^{prime} boldsymbol{a}^{prime} \ boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{b}^{prime}+boldsymbol{a}^{prime} boldsymbol{b} & boldsymbol{a}^{prime} boldsymbol{b}^{prime}end{array}right|= ) ( left(boldsymbol{a} boldsymbol{b}^{prime}-boldsymbol{a}^{prime} boldsymbol{b}right)left(boldsymbol{b} boldsymbol{c}-boldsymbol{b}^{prime} boldsymbol{c}right)left(boldsymbol{c} boldsymbol{a}^{prime}-boldsymbol{c}^{prime} boldsymbol{a}right) ) | 12 |
| 180 | Evaluate the determinant to the closest nteger: ( boldsymbol{A}=left[begin{array}{cc}log _{3} mathbf{5 1 2} & log _{4} mathbf{3} \ log _{3} mathbf{8} & log _{4} mathbf{9}end{array}right] ) | 12 |
| 181 | ( mathbf{f} mathbf{Delta}=left|begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{3} \ mathbf{2} & mathbf{0} & mathbf{1} \ mathbf{5} & mathbf{3} & mathbf{8}end{array}right|, ) write the minor of the elements ( a_{22} ) | 12 |
| 182 | ( fleft|begin{array}{lll}boldsymbol{x}+mathbf{1} & boldsymbol{x}+boldsymbol{2} & boldsymbol{x}+boldsymbol{a} \ boldsymbol{x}+mathbf{2} & boldsymbol{x}+boldsymbol{3} & boldsymbol{x}+boldsymbol{b} \ boldsymbol{x}+mathbf{3} & boldsymbol{x}+boldsymbol{4} & boldsymbol{x}+boldsymbol{c}end{array}right|=mathbf{0} ) then show that ( a, b, c ) are in ( A . P ) | 12 |
| 183 | ( mathrm{f}left|begin{array}{lll}boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b}end{array}right|=boldsymbol{t} times operatorname{det} ) of circulant matrix whose elements of first column are ( a, b, c ) then ( t ) equals A. 5 B. 6 ( c cdot-2 ) ( D ) | 12 |
| 184 | (a) JL How many 3 x 3 matrices M with entries from {0, 1, 2} are there, for which the sum of the diagonal entries of M Mis 5? (JEE Adv. 2017) (a) 126 (b) 198 (c) 162 (d) 135 | 12 |
| 185 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) find ( |mathbf{2} boldsymbol{A}| ) | 12 |
| 186 | Evaluate the following deteminants: i) ( left|begin{array}{cc}boldsymbol{x} & -mathbf{7} \ boldsymbol{x} & mathbf{5} boldsymbol{x}+mathbf{1}end{array}right| ) ii) ( mid begin{array}{cc}cos 15^{circ} & sin 15^{circ} \ sin 75^{circ} & cos 75^{circ}end{array} ) | 12 |
| 187 | 20. Let a, b, c be such that b(a + c)+ 0 if [2009 a a+1 a 1 a+1 6+1 c-1 1-6 6+1 6-1 – a-1 6-1 ct1=0, c c-1 c+1 |(–1)n+2a (+1) +1b (-1)” cl then the value of n is : (a) any even integer (b) any odd integer (c) any integer (d) zero | 12 |
| 188 | 15. (2003 – 2 Marks) If Mis a 3 x 3 matrix, where det M=1 and MM=I, where I is an identity matrix, prove that det (M-1)=0. (2004 2 Maula | 12 |
| 189 | Let ( boldsymbol{d} in boldsymbol{R}, ) and ( boldsymbol{A}= ) ( left[begin{array}{ccc}-2 & 4+d & (sin theta)-2 \ 1 & (sin theta)+2 & d \ 5 & (2 sin theta)-d & (-sin theta)+2+2 dend{array}right. ) ( boldsymbol{theta} in[mathbf{0}, mathbf{2} boldsymbol{pi}] . ) If the minimum value of ( operatorname{det}(A) ) is ( 8, ) then a value of d is? A . -7 B. ( 2(sqrt{2}+2) ) ( c .-5 ) D. ( 2(sqrt{2}+1) ) | 12 |
| 190 | ( mathbf{a}_{r}=left|begin{array}{ccc}mathbf{2}^{r}-mathbf{1} & mathbf{2} mathbf{.} mathbf{3}^{r}-mathbf{1} & mathbf{4} . mathbf{5}^{r}-mathbf{1} \ boldsymbol{alpha} & boldsymbol{beta} & boldsymbol{gamma} \ mathbf{2}^{n}-mathbf{1} & mathbf{3}^{n}-mathbf{1} & mathbf{5}^{n}-mathbf{1}end{array}right| ) then find the value of ( sum_{r=1}^{n} Delta_{r} ) ( A ) в. ( alpha beta gamma ) ( mathbf{c} cdot-alpha beta gamma ) ( D ) | 12 |
| 191 | If ( alpha, beta ) and ( gamma ) are the roots of the equation ( x^{3}+p x+q=0 ) then the value of the determinant ( left|begin{array}{lll}boldsymbol{alpha} & boldsymbol{beta} & gamma \ boldsymbol{beta} & gamma & boldsymbol{alpha} \ boldsymbol{gamma} & boldsymbol{alpha} & boldsymbol{beta}end{array}right| ) is 2 ( mathbf{A} cdot underline{p} ) B. ( q ) c. ( p^{2}-2 q ) ( D ) | 12 |
| 192 | One factor of ( Delta= ) ( left|begin{array}{ccc}a^{2}+lambda & a b & a c \ a b & b^{2}+lambda & c b \ c a & c b & c^{2}+lambdaend{array}right| ) ( A cdot lambda^{2} ) B. ( 1 / lambda ) c. ( left(a^{2}+lambdaright)left(b^{2}+lambdaright)left(c^{2}+lambdaright) ) D. none | 12 |
| 193 | Evaluate the following determinant: ( begin{array}{|ccc|}15 & 11 & 7 \ 11 & 17 & 14 \ 10 & 16 & 13end{array} ) | 12 |
| 194 | Find the equation of line passing through the points (3,2) and (-1,3) by using determinants. | 12 |
| 195 | If ( A ) is a square matrix of order ( n times n ) and ( k ) is a scalar, then ( a d j(k A) ) is equal to A ( cdot k^{n-1} a d j A ) в. ( k^{n} ) adj ( A ) c. ( k^{n+1} a d j A ) D. kadj A | 12 |
| 196 | If ( f(x)=left|begin{array}{ccc}a & -1 & 0 \ a x & a & -1 \ a x^{2} & a x & aend{array}right|, ) using properties of determinets find the value of ( f(2 x)-f(x) ? ) | 12 |
| 197 | The repeated factor of the determinant ( left|begin{array}{lll}boldsymbol{y}+boldsymbol{z} & boldsymbol{x} & boldsymbol{y} \ boldsymbol{z}+boldsymbol{x} & boldsymbol{z} & boldsymbol{x} \ boldsymbol{x}+boldsymbol{y} & boldsymbol{y} & boldsymbol{z}end{array}right| ) A. ( z-x ) в. ( x-y ) ( mathbf{c} cdot y-z ) D. none of these | 12 |
| 198 | Evaluate ( left|begin{array}{ll}cos 65^{circ} & sin 65^{circ} \ sin 25^{circ} & cos 25^{circ}end{array}right| ) | 12 |
| 199 | ( operatorname{lf} Delta=left|begin{array}{lll}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}end{array}right| ) and ( A_{2}, B_{2}, C_{2} ) are respectively cofactors of ( a_{2}, b_{2}, c_{2} ) then ( a_{1} A_{2}+b_{1} B_{2}+c_{1} C_{2} ) is equal to ( A cdot-Delta ) B. ( c cdot Delta ) D. none of these | 12 |
| 200 | In the diagram on a lunar eilpse, if the positions od sun,Earth and moon are shown by ( (-4,6),(k,-2) ) and (5,-6) respectively, then find the value of ( mathrm{k} ) | 12 |
| 201 | If ( A=left[begin{array}{ll}alpha & 2 \ 2 & alphaend{array}right] ) and ( left|A^{3}right|=125 ) then ( alpha ) is ( mathbf{A} cdot pm 1 ) B. =2 ( c .pm 3 ) D. ±5 | 12 |
| 202 | Find the values of the following determinants ( left|begin{array}{cc}mathbf{1}+mathbf{3} i & boldsymbol{i}-mathbf{2} \ -boldsymbol{i}-mathbf{2} & mathbf{1}-mathbf{3} boldsymbol{i}end{array}right| ) where ( i=sqrt{-1} ) | 12 |
| 203 | [2005] 10. If a? +62 +62 =-2 and | 1+a’x (1+62)x (1+c%)x f(x) = (1 + a²)x 1 +6²x (1+ c²)x. (1 + a²)x (1+6²)x 1+ c²x then f(x) is a polynomial of degree (a) 1 (b) 0 (c) 3 (d) 2 | 12 |
| 204 | ( left|begin{array}{ccc}left(a^{x}+a^{-x}right)^{2} & left(a^{x}-a^{-x}right)^{2} & 1 \ left(b^{x}+b^{-x}right)^{2} & left(b^{x}-b^{-x}right)^{2} & 1 \ left(c^{x}+c^{-x}right)^{2} & left(c^{x}-c^{-x}right)^{2} & 1end{array}right| ) is equa to ( A ) B. ( 2 a b c ) ( mathbf{c} cdot a^{2} b^{2} c^{2} ) D. None of these | 12 |
| 205 | ( mathbf{A}=left|begin{array}{lll}mathbf{5} & mathbf{3} & mathbf{8} \ mathbf{2} & mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3}end{array}right|, ) write the cofactor of the element ( a_{32} ) | 12 |
| 206 | ( |f(x)|=left[begin{array}{ccc}sin x & operatorname{cosec} x & tan x \ sec x & x sin x & x tan x \ x^{2}-1 & cos x & x^{2}+1end{array}right] ) then ( ldots int_{-a}^{a}|boldsymbol{f}(boldsymbol{x})| boldsymbol{d} ) equals ( A ) в. ( c cdot 2 a ) ( D ) | 12 |
| 207 | Let ( Delta= ) ( left|begin{array}{ccc}sin theta cos phi & sin theta sin phi & cos theta \ cos theta cos phi & cos theta sin phi & -sin theta \ -sin theta sin phi & sin theta cos phi & 0end{array}right|, ) ther ( A cdot Delta ) is independent of ( theta ) B. ( Delta ) is independent of ( phi ) ( c cdot Delta ) is a constant D. ( Delta ) is dependent of ( phi ) | 12 |
| 208 | Find the values of ( x, ) if ( left|begin{array}{cc}mathbf{2 x} & mathbf{5} \ mathbf{8} & boldsymbol{x}end{array}right|=left|begin{array}{ll}mathbf{6} & mathbf{5} \ mathbf{8} & mathbf{3}end{array}right| ) | 12 |
| 209 | ( left|begin{array}{ccc}mathbf{0} & boldsymbol{a} boldsymbol{b}^{2} & boldsymbol{a} boldsymbol{c}^{2} \ boldsymbol{a}^{2} boldsymbol{b} & boldsymbol{0} & boldsymbol{b} boldsymbol{c}^{2} \ boldsymbol{a} boldsymbol{c}^{2} & boldsymbol{b}^{2} boldsymbol{c} & boldsymbol{0}end{array}right|= ) ( mathbf{A} cdot a^{3} b^{3} c^{3}+a^{2} b^{3} c^{4} ) B ( cdot a^{3} b^{3} c^{3} ) ( mathbf{c} cdot 2 a^{3} b^{3} c^{3} ) D・ ( (2 a b c)^{3} ) | 12 |
| 210 | ff ( boldsymbol{D}_{boldsymbol{r}}=left|begin{array}{ccc}boldsymbol{r}-mathbf{1} & boldsymbol{n} & boldsymbol{6} \ (boldsymbol{r}-mathbf{1})^{2} & boldsymbol{2} boldsymbol{n}^{2} & boldsymbol{4} boldsymbol{n}-boldsymbol{2} \ (boldsymbol{r}-mathbf{1})^{3} & boldsymbol{3} boldsymbol{n}^{3} & boldsymbol{3} boldsymbol{n}^{2}-boldsymbol{3} boldsymbol{n}end{array}right| ) then ( sum_{r=1}^{n} D_{r}= ) ( A ) B. c. ( frac{n(n-1)}{2}-r^{2} ) D. ( 2 n-n^{2} ) | 12 |
| 211 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{1} & mathbf{2} \ mathbf{2} & mathbf{3}end{array}right] ) and ( boldsymbol{B}=left[begin{array}{ll}mathbf{1} & mathbf{1} \ mathbf{0} & mathbf{0}end{array}right] ) then what is determinant of AB? ( mathbf{A} cdot mathbf{0} ) B. c. 10 D. 20 | 12 |
| 212 | 12. Given 2x – y + 2z=2, x – 2y+z=-4, x+y+nza en the value of 2 such that the given system of equation has NO solution, is (a) 3 (6) 1 (c) 0 (d) -3 (2004S) | 12 |
| 213 | Consider the matrix ( boldsymbol{A}=left(begin{array}{ll}mathbf{3} & -mathbf{2} \ mathbf{4} & -mathbf{1}end{array}right) ) Then all possible values of ( lambda ) such that the determinant of ( B=A-lambda I ) is 0 where ( boldsymbol{I}=left(begin{array}{ll}mathbf{1} & mathbf{0} \ mathbf{0} & mathbf{1}end{array}right) ) and ( i=sqrt{-mathbf{1}} ) ( mathbf{A} cdot 1 pm 2 i ) B . ( 2 pm 3 i ) c. ( 3 pm 4 i ) ( mathbf{D} cdot 5 pm 6 i ) | 12 |
| 214 | Find the area of ( triangle P Q R ) whose vertices ( operatorname{are} boldsymbol{P}(mathbf{2}, mathbf{1}), boldsymbol{Q}(mathbf{3}, mathbf{4}) ) and ( boldsymbol{R}(mathbf{5}, mathbf{2}) ) | 12 |
| 215 | Evaluate ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ -mathbf{1} & mathbf{1} & -mathbf{1} \ mathbf{1} & mathbf{- 1} & mathbf{1}end{array}right| ) | 12 |
| 216 | The values of lying between O=0 and O=TJ2 and satisfying the equation (1988 – 2 Marks) 1+ sine cose sin 1+ cos2 e sin cos 4sin 40 4sin 40 1+4 sin 40 = 0 are | 12 |
| 217 | ( begin{array}{ccc}(boldsymbol{b}+boldsymbol{c})^{2} & boldsymbol{a}^{2} & boldsymbol{a}^{2} \ boldsymbol{b}^{2} & (boldsymbol{c}+boldsymbol{a})^{2} & boldsymbol{b}^{2} \ boldsymbol{c}^{2} & boldsymbol{c}^{2} & (boldsymbol{a}+boldsymbol{b})^{2}end{array} mid= ) | 12 |
| 218 | Find a value of ( boldsymbol{x} ) if ( left|begin{array}{cc}boldsymbol{x} & boldsymbol{2} \ boldsymbol{1} boldsymbol{8} & boldsymbol{x}end{array}right|=left|begin{array}{cc}boldsymbol{6} & boldsymbol{2} \ boldsymbol{1} boldsymbol{8} & boldsymbol{6}end{array}right| ) | 12 |
| 219 | Find the values of ( k, ) if the points ( boldsymbol{A}(boldsymbol{k}+ ) 1, ( 2 k ) ), ( B(3 k, 2 k+3) ) and ( C(5 k+1,5 k) ) are collinear. This question has multiple correct options A . B. ( frac{1}{2} ) ( c cdot 2 ) D. 2. | 12 |
| 220 | Prove that the points ( (a, 0),(0, b) ) and (1,1) are collinear if ( left(frac{1}{a}+frac{1}{b}=1right) ) | 12 |
| 221 | Let ( A ) be the matrix of order ( 3 times 3 ) such that ( |boldsymbol{A}|=mathbf{1}, boldsymbol{B}=mathbf{2} boldsymbol{A}^{-1} ) and ( boldsymbol{C}=frac{(boldsymbol{a} d boldsymbol{j} boldsymbol{A})}{sqrt[3]{2}} ) then the value of ( left|A B^{2} . C^{3}right| ) is [Note : ( |A| ) represent determinant value of matrix A.] | 12 |
| 222 | The cofactors of elements in second row of the determinant ( left|begin{array}{ccc}mathbf{2} & mathbf{- 1} & mathbf{4} \ mathbf{4} & mathbf{2} & mathbf{- 3} \ mathbf{1} & mathbf{1} & mathbf{2}end{array}right| ) are ( mathbf{A} cdot 5,6,4 ) В. 6,0,-3 c. 5,1,8 ( mathbf{D} cdot 6,0,3 ) | 12 |
| 223 | The number of positive integral solutions of the equation ( left|begin{array}{ccc}boldsymbol{x}^{boldsymbol{3}}+mathbf{1} & boldsymbol{x}^{boldsymbol{2}} boldsymbol{y} & boldsymbol{x}^{boldsymbol{2}} boldsymbol{z} \ boldsymbol{x} boldsymbol{y}^{boldsymbol{2}} & boldsymbol{y}^{boldsymbol{3}}+boldsymbol{1} & boldsymbol{y}^{boldsymbol{2}} boldsymbol{z} \ boldsymbol{x} boldsymbol{z}^{boldsymbol{2}} & boldsymbol{y} boldsymbol{z}^{boldsymbol{2}} & boldsymbol{z}^{boldsymbol{3}}+mathbf{1}end{array}right|=mathbf{1 1} ) is A. B. 3 ( c cdot 6 ) D. 12 | 12 |
| 224 | Evaluate the following determinant: ( left|begin{array}{ccc}1 & 3 & 5 \ 2 & 6 & 10 \ 31 & 11 & 38end{array}right| ) | 12 |
| 225 | ff ( Delta=left|begin{array}{ccc}mathbf{3} & mathbf{5} & mathbf{7} \ mathbf{2} & -mathbf{3} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{2}end{array}right|, ) find it’s value | 12 |
| 226 | Let ( boldsymbol{f}(boldsymbol{x}, boldsymbol{y})=left|begin{array}{lll}boldsymbol{y}-boldsymbol{x}^{2} & boldsymbol{x}-boldsymbol{y}^{2} & boldsymbol{x} boldsymbol{y}-mathbf{1} \ boldsymbol{x}-boldsymbol{y}^{2} & boldsymbol{x} boldsymbol{y}-mathbf{1} & boldsymbol{y}-boldsymbol{x}^{2} \ boldsymbol{x} boldsymbol{y}-mathbf{1} & boldsymbol{y}-boldsymbol{x}^{2} & boldsymbol{x}-boldsymbol{y}^{2}end{array}right| ) find ( boldsymbol{f}(boldsymbol{2}, boldsymbol{2}) ) | 12 |
| 227 | ( left|begin{array}{cc}boldsymbol{x}+mathbf{2} & boldsymbol{x} \ mathbf{4} & mathbf{3}end{array}right|=mathbf{0} ) find the value of ( mathbf{x} ) | 12 |
| 228 | If the points ( (3,-2),(x, 2) ) and (8,8) are collinear find ( 10 x ) using determinant | 12 |
| 229 | If ( (3,2),left(x, frac{22}{5}right),(8,8) ) lie on a line, then ( x ) is equal to A . -5 B. 2 ( c cdot 4 ) D. 5 | 12 |
| 230 | Prove that: ( left|begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \ 2 a+1 & a+2 & 1 \ 3 & 3 & 1end{array}right|=(a-1)^{2} ) | 12 |
| 231 | Prove that ( :left|begin{array}{ccc}boldsymbol{a} & boldsymbol{c} & boldsymbol{a}+boldsymbol{c} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}end{array}right|= ) ( 4 a b c ) | 12 |
| 232 | Without expanding, show that the value of each of the following determinants is zero: ( left|begin{array}{ccc}a & b & c \ a+2 x & b+2 y & c+2 z \ x & y & zend{array}right| ) | 12 |
| 233 | Let ( omega neq 1 ) be a cube root of unity and ( S ) be the set of all non-singular matrices of the form ( left[begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{omega} & boldsymbol{1} & boldsymbol{c} \ boldsymbol{omega}^{2} & boldsymbol{omega} & boldsymbol{1}end{array}right] ) Where each of ( a, b ) and ( c ) is either ( omega ) or ( omega^{2} ). Then the number of distinct matrices in the set ( boldsymbol{S} ) is ( A cdot 2 ) B. 6 ( c cdot 4 ) D. 8 | 12 |
| 234 | How do I find ( boldsymbol{A}=left[begin{array}{ccc}1 & 2 & -2 \ -1 & 3 & 0 \ 0 & -2 & 1end{array}right]=|A|=? ) | 12 |
| 235 | The number of solutions of equations ( left|begin{array}{ccc}sin 3 theta & -1 & 1 \ cos 2 theta & 4 & 3 \ 2 & 7 & 7end{array}right|=0 ) in ( [0,2 pi] ) is A .2 B. 3 ( c cdot 4 ) D. 5 | 12 |
| 236 | If ( boldsymbol{u}=boldsymbol{a} boldsymbol{x}^{2}+boldsymbol{2 b} boldsymbol{x} boldsymbol{y}+boldsymbol{c} boldsymbol{y}^{2}, boldsymbol{u}^{prime}=boldsymbol{a}^{prime} boldsymbol{x}^{2}+ ) ( 2 b^{prime} x y+c^{prime} y^{2}, ) then prove that ( left|begin{array}{ccc}boldsymbol{y}^{2} & -boldsymbol{x} boldsymbol{y} & boldsymbol{x}^{2} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{prime} & boldsymbol{b}^{prime} & boldsymbol{c}^{prime}end{array}right|= ) ( left|begin{array}{cc}boldsymbol{a x}+boldsymbol{b y} & boldsymbol{b x}+boldsymbol{c y} \ boldsymbol{a}^{prime} boldsymbol{x}+boldsymbol{b}^{prime} boldsymbol{y} & boldsymbol{b}^{prime} boldsymbol{x}+boldsymbol{c}^{prime} boldsymbol{y}end{array}right|= ) ( -frac{1}{y}left|begin{array}{cc}u & u^{prime} \ a x+b y & a^{prime} x+b^{prime} yend{array}right| ) | 12 |
| 237 | zoaluate ( left|begin{array}{ccc}log _{x} x y z & log _{x} y & log _{x} z \ log _{y} x y z & 1 & log _{y} z \ log _{z} x y z & log _{z} y & 1end{array}right| ) | 12 |
| 238 | If ( 1, omega, omega^{2} ) are two cube roots of unity then ( Delta=left|begin{array}{ccc}1 & omega^{n} & omega^{2 n} \ omega^{2 n} & 1 & omega^{n} \ omega^{n} omega^{2 n} & 1end{array}right| ) has the value ( mathbf{A} cdot mathbf{0} ) B. ( c cdot omega^{2} ) D. | 12 |
| 239 | If ( left.left.right|_{2} ^{4} 1right|^{2}=left|begin{array}{cc}3 & 2 \ 1 & xend{array}right|-left|begin{array}{cc}x & 3 \ -2 & 1end{array}right|, ) then ( x= ) ( A cdot 6 ) B. 7 ( c cdot 8 ) D. 16 | 12 |
| 240 | ( left|begin{array}{cc}boldsymbol{x} & mathbf{2} \ mathbf{1 8} & boldsymbol{x}end{array}right|=left|begin{array}{cc}mathbf{6} & mathbf{2} \ mathbf{3} boldsymbol{x} & mathbf{6}end{array}right|, ) then ( boldsymbol{x} ) is equal to ( A cdot 6 ) B. ±6 ( c .-6 ) ( D ) | 12 |
| 241 | ff ( x_{1}, x_{2}, x_{3} ) as well as ( y_{1}, y_{2}, y_{3} ) are in G.P. with same common ratio, then the points ( boldsymbol{P}left(boldsymbol{x}_{1}, boldsymbol{y}_{1}right), boldsymbol{Q}left(boldsymbol{x}_{2}, boldsymbol{y}_{2}right) ) and ( boldsymbol{R}left(boldsymbol{x}_{3}, boldsymbol{y}_{3}right) ) A. lies on a straight line B. lie on an ellipse c. lie on a circle D. are vertices of a triangle | 12 |
| 242 | Show that: ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right|^{2}= ) [ left|begin{array}{ccc} 2 b c-a^{2} & c^{2} & b^{2} \ c^{2} & 2 a c-b^{2} & a^{2} \ b^{2} & a^{2} & 2 a b-c^{2} end{array}right|= ] ( left(a^{3}+b^{3}+c^{3}-3 a b cright)^{2} ) | 12 |
| 243 | Find the value of: ( left|begin{array}{ll}-3 & -5 \ -2 & -1end{array}right| ) | 12 |
| 244 | <Tt, 1. The number of all possible values of 0, where 0 < for which the system of equations (y + 2) cos 30 = (xyz) sin 30 2 cos 30 2 sin 30 x sin 30= = Y Z (xyz) sin 30= (y + 2z) cos 30+ y sin30 has a solution (xo, Yo, zo) with yożo 70 is – +- | 12 |
| 245 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{lll}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{2} & mathbf{1} & mathbf{0} \ mathbf{3} & mathbf{2} & mathbf{1}end{array}right] ) and ( boldsymbol{U}_{1}, boldsymbol{U}_{2}, boldsymbol{U}_{3} ) be column matrices satisfying ( boldsymbol{A} boldsymbol{U}_{1}= ) ( left[begin{array}{l}1 \ 0 \ 0end{array}right], A U_{2}=left[begin{array}{l}2 \ 3 \ 0end{array}right], A U_{3}=left[begin{array}{l}2 \ 3 \ 1end{array}right] . ) If ( U ) is ( 3 times 3 ) matrix whose columns are ( boldsymbol{U}_{1}, boldsymbol{U}_{2}, boldsymbol{U}_{3}, ) then ( |boldsymbol{U}|= ) A . 3 B. -3 ( c cdot frac{3}{2} ) ( D ) | 12 |
| 246 | Without expanding at any stage, evaluate the value of determinant | 12 |
| 247 | ( left|begin{array}{ccc}boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & mathbf{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & 1end{array}right|=left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{b}_{1} & boldsymbol{b}_{2} & boldsymbol{b}_{3} \ boldsymbol{a}_{1} & boldsymbol{a}_{2} & boldsymbol{a}_{3}end{array}right| ) then the two triangles whose vertices are ( left(x_{1}, y_{1}right),left(x_{2}, y_{2}right),left(left(x_{3}, y_{3}right) ) and right. ( left(a_{1}, b_{1}right),left(a_{2}, b_{2}right),left(a_{13}, b_{3}right), ) are A . congruent B. similar c. equal in area D. none of these | 12 |
| 248 | If each row of a determinant of third order of value ( Delta ) is multipled by ( 3, ) then the value of new determinant is A. ( Delta ) B. ( 27 Delta ) c. ( 21 Delta ) D. ( 54 Delta ) | 12 |
| 249 | Adj ( left[begin{array}{ccc}1 & 0 & 2 \ -1 & 5 & -2 \ 0 & 2 & 1end{array}right]= ) ( left[begin{array}{ccc}mathbf{9} & boldsymbol{a} & mathbf{- 2} \ mathbf{- 1} & mathbf{1} & mathbf{0} \ mathbf{- 2} & mathbf{2} & boldsymbol{b}end{array}right] Rightarrowleft[begin{array}{ll}boldsymbol{a} & boldsymbol{b}end{array}right]= ) ( left.begin{array}{ll}text { A. }[-4 & 5end{array}right] ) B ( cdotleft[begin{array}{ll}-4 & -1end{array}right] ) ( mathbf{c} cdotleft[begin{array}{ll}4 & 1end{array}right] ) D・[4 -1 | 12 |
| 250 | Find the value of ( x ) for which the points ( (x,-1),(2,1) ) and (4,5) are collinear. | 12 |
| 251 | If ( mathbf{A} ) is an unitary matrix then ( |boldsymbol{A}| ) is equal to: ( mathbf{A} cdot mathbf{1} ) B. – ( c .pm 1 ) D. 2 | 12 |
| 252 | begin{tabular}{|ccc} If ( boldsymbol{f}(boldsymbol{x})= ) & \ ( mathbf{1} ) & ( boldsymbol{x} ) & ( boldsymbol{x} ) \ ( boldsymbol{2} boldsymbol{x} ) & ( boldsymbol{x}(boldsymbol{x}-mathbf{1}) ) & ( (boldsymbol{x}-1) ) \ ( boldsymbol{3} boldsymbol{x}(boldsymbol{x}-mathbf{1}) ) & ( boldsymbol{x}(boldsymbol{x}-mathbf{1})(boldsymbol{x}-mathbf{2}) ) & ( (boldsymbol{x}+mathbf{1}) ) end{tabular} then ( f(100) ) is equal to A. B. ( c cdot 100 ) D. -100 | 12 |
| 253 | Let ( A ) be a ( 3 times 3 ) matrix and ( B ) be its adjoint matrix. If ( |boldsymbol{B}|=mathbf{6 4}, ) then ( |boldsymbol{A}|= ) ( A cdot pm 2 ) B. ±4 ( c .pm 8 ) D. ±12 | 12 |
| 254 | The value of determinant ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{2} boldsymbol{y} \ boldsymbol{x}+boldsymbol{2} boldsymbol{y} & boldsymbol{x} & boldsymbol{x}+boldsymbol{y} \ boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{2} boldsymbol{y} & boldsymbol{x}end{array}right| ) is: A ( cdot 9 x^{2}(x+y) ) В ( cdot 9 y^{2}(x+y) ) c. ( 3 y^{2}(x+y) ) D. ( 7 x^{2}(x+y) ) | 12 |
| 255 | If ( Delta_{1}=left|begin{array}{ll}mathbf{1} & mathbf{0} \ boldsymbol{a} & boldsymbol{b}end{array}right| ) and ( boldsymbol{Delta}_{2}=left|begin{array}{ll}mathbf{1} & mathbf{0} \ boldsymbol{c} & boldsymbol{d}end{array}right| ) then ( Delta_{2} Delta_{1} ) is equal to ( mathbf{A} cdot a c ) B. ( b d ) c. ( (b-a)(d-c) ) D. none of these | 12 |
| 256 | Evaluate ( sum_{n=1}^{N} U_{n} ) if [ boldsymbol{U}_{boldsymbol{n}}=mid begin{array}{ccc}boldsymbol{n} & mathbf{1} & mathbf{5} \ boldsymbol{n}^{2} & boldsymbol{2} boldsymbol{N}+mathbf{1} & boldsymbol{2} boldsymbol{N}+mathbf{1} \ boldsymbol{n}^{boldsymbol{3}} & boldsymbol{3} boldsymbol{N}^{2} & boldsymbol{3} boldsymbol{N}end{array} ] | 12 |
| 257 | f ( x, y, z ) are complex numbers, then ( Delta= ) ( left|begin{array}{ccc}mathbf{0} & -boldsymbol{y} & -boldsymbol{z} \ overline{boldsymbol{y}} & mathbf{0} & -boldsymbol{x} \ overline{boldsymbol{z}} & overline{boldsymbol{x}} & boldsymbol{0}end{array}right| ) is equal A. Purely real B. Purely imaginary c. complex D. | 12 |
| 258 | ( operatorname{Let} A=left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z}end{array}right| ) and suppose that det. ( (A)=2 ) then the det.(B) equals, where ( boldsymbol{B}=left|begin{array}{ccc}mathbf{4} boldsymbol{x} & mathbf{2} boldsymbol{a} & -boldsymbol{p} \ mathbf{4} boldsymbol{y} & mathbf{2} boldsymbol{b} & -boldsymbol{q} \ boldsymbol{4} boldsymbol{z} & boldsymbol{2} boldsymbol{c} & -boldsymbol{t}end{array}right| ) A. ( operatorname{det}(B)=-2 ) B. ( operatorname{det}(B)=-8 ) c. ( operatorname{det}(B)=-16 ) D. ( operatorname{det}(B)=8 ) | 12 |
| 259 | If ( a, b, c ) are real, then ( f(x)= ) ( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{a}^{2} & boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b} & boldsymbol{x}+boldsymbol{b}^{2} & boldsymbol{b} boldsymbol{c} \ boldsymbol{a} boldsymbol{c} & boldsymbol{b} boldsymbol{c} & boldsymbol{x}+boldsymbol{c}^{2}end{array}right| ) is decreasing in A ( cdotleft(-frac{2}{3}left(a^{2}+b^{2}+c^{2}right), 0right) ) B. ( left(0, frac{2}{3}left(a^{2}+b^{2}+c^{2}right)right) ) ( left(frac{a^{2}+b^{2}+c^{2}}{3}, 0right) ) D. None of these | 12 |
| 260 | Using determinants show that points ( boldsymbol{A}(boldsymbol{a}, boldsymbol{b}+boldsymbol{c}), boldsymbol{B}(boldsymbol{b}, boldsymbol{c}+boldsymbol{a}) ) and ( boldsymbol{C}(boldsymbol{c}, boldsymbol{a}+boldsymbol{b}) ) are col-linear. | 12 |
| 261 | Evaluate the following determinant: ( left|begin{array}{lll}a & h & g \ h & b & f \ g & f & cend{array}right| ) | 12 |
| 262 | Using properties of determinant, prove [ text { that }left|begin{array}{lll} boldsymbol{b}+boldsymbol{c} & boldsymbol{a}-boldsymbol{b} & boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{b}-boldsymbol{c} & boldsymbol{b} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{c}-boldsymbol{a} & boldsymbol{c} end{array}right|=mathbf{3} boldsymbol{a} boldsymbol{b} boldsymbol{c}-boldsymbol{a}^{3}- ] ( b^{3}-c^{3} ) | 12 |
| 263 | ( mathrm{ff}=left[begin{array}{ccc}boldsymbol{a} & mathbf{0} & mathbf{0} \ {[mathbf{0 . 3 e m}] mathbf{0}} & boldsymbol{a} & mathbf{0} \ {[mathbf{0 . 3 e m}] mathbf{0}} & mathbf{0} & boldsymbol{a}end{array}right], ) then the value of |A| |Adj. A| A ( cdot a^{3} ) в. ( a^{6} ) ( c cdot a^{9} ) ( mathbf{D} cdot a^{2} ) | 12 |
| 264 | Solve ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{y}end{array}right| ) ( A ) B. ( c ) D. ( x y ) | 12 |
| 265 | Using the method of slope, show that the following points are collinear: ( (i) A(4,8), B(5,12), C(9,28) ) ( (i i) A(16,-18), B(3,-6), C(-10,6) ) | 12 |
| 266 | If ( A ) is a singular matrix, then ( A(operatorname{adj} A) ) is a A. scalar matrix B. zero matrix c. identity matrix D. orthogonal matrix | 12 |
| 267 | Find the value of ( mathbf{x} ) for which ( left|begin{array}{ll}3 & x \ x & 1end{array}right|= ) ( left|begin{array}{ll}3 & 2 \ 8 & 1end{array}right| ) | 12 |
| 268 | The line ( A x+B y+C=0 ) cuts the circle ( boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{a} boldsymbol{x}+boldsymbol{b} boldsymbol{y}+boldsymbol{c}=boldsymbol{0} ) in ( boldsymbol{P} ) and ( Q ) The line ( boldsymbol{A}^{prime} boldsymbol{x}+boldsymbol{B}^{prime} boldsymbol{y}+boldsymbol{C}^{prime}=mathbf{0} ) cuts the circle ( boldsymbol{x}^{2}+boldsymbol{y}^{2}+boldsymbol{a}^{prime} boldsymbol{x}+boldsymbol{b}^{prime} boldsymbol{y}+boldsymbol{c}^{prime}=mathbf{0} ) in ( boldsymbol{R} ) and ( S ). If ( P, Q, R, S ) are concyclic, then show that ( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{a}^{prime} & boldsymbol{b}-boldsymbol{b}^{prime} & boldsymbol{c}-boldsymbol{c}^{prime} \ boldsymbol{A} & boldsymbol{B} & boldsymbol{C} \ boldsymbol{A}^{prime} & boldsymbol{B}^{prime} & boldsymbol{C}^{prime}end{array}right|=mathbf{0} ) | 12 |
| 269 | Find the value of the following determinant: ( left|begin{array}{cc}mathbf{3} sqrt{mathbf{6}} & -mathbf{4} sqrt{mathbf{2}} \ mathbf{5} sqrt{mathbf{3}} & mathbf{2}end{array}right| ) A. ( 20 sqrt{6} ) B. ( 16 sqrt{6} ) c. ( 26 sqrt{6} ) D. ( 10 sqrt{6} ) | 12 |
| 270 | Evaluate ( Delta=mid begin{array}{cc}cos 2 alpha & sin 2 alpha \ -sin 3 alpha & cos 3 alphaend{array} ) | 12 |
| 271 | ff ( f(x), g(x), h(x) ) are polynomials in ( x ) of ( operatorname{degree} 2 ) and ( F(x)=left|begin{array}{ccc}boldsymbol{f} & boldsymbol{g} & boldsymbol{h} \ boldsymbol{f}^{prime} & boldsymbol{g}^{prime} & boldsymbol{h}^{prime} \ boldsymbol{f}^{prime prime} & boldsymbol{g}^{prime prime} & boldsymbol{h}^{prime prime}end{array}right|, ) then ( F(x) ) is equal to ( A cdot 1 ) B. ( c cdot-1 ) D. ( f(x) . g(x) . h(x) ) | 12 |
| 272 | Let ( t ) be a positive integer and ( Delta_{t}= ) ( left|begin{array}{lll}mathbf{2} t-1 & boldsymbol{m}^{2}-mathbf{1} & cos ^{2}left(boldsymbol{m}^{2}right) \ boldsymbol{m}_{C_{t}} & mathbf{2}^{m} & cos ^{2}(boldsymbol{m}) \ mathbf{1} & boldsymbol{m}+mathbf{1} & cos left(boldsymbol{m}^{2}right)end{array}right| ) then the value of ( sum_{t=0}^{m} Delta_{t} ) is equal to: ( A cdot 2^{m} ) B. ( mathbf{c} cdot 2^{m} cos ^{2}left(2^{m}right) ) D. ( m^{2} ) | 12 |
| 273 | The point ( (-a,-b),(0,0),(a, b) ) and ( left(a^{2}, a bright) ) are A. collinear B. concyclic c. vertices of a rectangle D. vertices of a parallelogram | 12 |
| 274 | ff ( f(x)= ) ( f(x-a)(x-b)(x-c)(x-d) ) then prove that ( Delta=left|begin{array}{cccc}boldsymbol{a} & boldsymbol{x} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{b} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{c} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x} & boldsymbol{d}end{array}right|=boldsymbol{f}(boldsymbol{x})-boldsymbol{x} boldsymbol{f}^{prime}(boldsymbol{x}) ) | 12 |
| 275 | Let ( m ) be a positive integer and ( Delta_{r}= ) ( left|begin{array}{ccc}mathbf{2 r}-mathbf{1} & boldsymbol{m} boldsymbol{C}_{boldsymbol{r}} & mathbf{1} \ boldsymbol{m}^{mathbf{2}}-mathbf{1} & mathbf{2}^{boldsymbol{m}} & boldsymbol{m}+mathbf{1} \ sin ^{2}left(boldsymbol{m}^{2}right) & sin ^{2}(boldsymbol{m}) & sin ^{2}(boldsymbol{m}+mathbf{1})end{array}right| ) ( boldsymbol{r} leq boldsymbol{m}) . ) Then the value of ( sum_{r=0}^{m} boldsymbol{Delta}_{boldsymbol{r}} ) ( A . ) B. ( m^{2}-1 ) ( c cdot 2^{m} ) D ( cdot 2^{m} sin ^{2}left(2^{m}right) ) | 12 |
| 276 | Using the property of determinants and with out expanding prove that ( left|begin{array}{lll}boldsymbol{a}-boldsymbol{b} & boldsymbol{b}-boldsymbol{c} & boldsymbol{c}-boldsymbol{a} \ boldsymbol{b}-boldsymbol{c} & boldsymbol{c}-boldsymbol{a} & boldsymbol{a}-boldsymbol{b} \ boldsymbol{c}-boldsymbol{a} & boldsymbol{a}-boldsymbol{b} & boldsymbol{b}-boldsymbol{c}end{array}right|=mathbf{0} ) | 12 |
| 277 | a b c] 14. If matrix A= b c a where a, b, c are real positive Lc a b numbers, abc = 1 and ATA = 1, then find the value of a3 + b3 + c3 (2003 – 2 Marks) | 12 |
| 278 | ( f(x)=left|begin{array}{ccc}2 x & x^{2} & x^{3} \ x^{2}+2 x & 1 & 3 x+1 \ 2 x & 1-3 x^{2} & 5 xend{array}right| ) then find ( boldsymbol{f}^{prime}(mathbf{1}) ) | 12 |
| 279 | ( left|begin{array}{ccc}mathbf{1} & boldsymbol{w} & boldsymbol{w}^{2} \ boldsymbol{w} & boldsymbol{w}^{2} & mathbf{1} \ boldsymbol{w}^{2} & boldsymbol{w} & mathbf{1}end{array}right| ) Where ( mathbf{w} ) is a complex cube root of unity | 12 |
| 280 | Prove [ left|begin{array}{ccc} x^{2} & y^{2} & z^{2} \ (x+1)^{2} & (y+1)^{2} & (z+1)^{2} \ (x-1)^{2} & (y-1)^{2} & (z-1)^{2} end{array}right|= ] ( -4(x-y)(y-z)(x-z) ) | 12 |
| 281 | If ( f(x)= ) ( left|begin{array}{ccc}(mathbf{1}+mathbf{3} boldsymbol{x})^{2 boldsymbol{a}} & (mathbf{1}+mathbf{2} boldsymbol{x})^{boldsymbol{3} boldsymbol{b}} & mathbf{1} \ mathbf{1} & (mathbf{1}+mathbf{3} boldsymbol{x})^{2 boldsymbol{a}} & (mathbf{1}+mathbf{2} boldsymbol{x})^{boldsymbol{3} boldsymbol{b}} \ (mathbf{1}+mathbf{2} boldsymbol{x})^{boldsymbol{3} boldsymbol{b}} & mathbf{1} & (mathbf{1}+mathbf{3} boldsymbol{x})^{mathbf{2} boldsymbol{a}}end{array}right| ) then A. ( f(x) ) has constant term 1 B. constant term is ( 2 a-3 b ) c. coefficient of ( x ) in ( f(x) ) is zero D. constant term is ( 2 a+3 b ) | 12 |
| 282 | f ( a, b, c ) are real numbers such that ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a}end{array}right|=0, ) then show that either ( boldsymbol{a}+boldsymbol{b}+boldsymbol{c}=boldsymbol{0} ) or, ( boldsymbol{a}=boldsymbol{b}=boldsymbol{c} ) | 12 |
| 283 | 1. If a > 0 and discriminant of ax2+2bx+c is-ve, [2001 a b ax+b b c bx+c ax + bl. bx+c is equal to o (b) (ac-b2)(ax2 +2bx+c) (d) 0 (a) +ve (C) -ve | 12 |
| 284 | Assertion If ( A ) is skew symmetric matrix of order 3 then its determinant should be zero Reason If ( A ) is square matrix, then ( d e t A= ) | 12 |
| 285 | ( begin{array}{lllll}mathbf{0} & & mathbf{c o s} boldsymbol{alpha} & mathbf{c o s} & boldsymbol{beta} \ mathbf{c o s} & boldsymbol{alpha} & mathbf{0} & mathbf{c o s} & gamma \ mathbf{c o s} & boldsymbol{beta} & mathbf{c o s} boldsymbol{gamma} & mathbf{0} & end{array} mid= ) A ( cdot cos alpha+cos beta+cos gamma ) ( mathbf{B} cdot cos alpha cos beta cos gamma ) c. ( 2 cos alpha cos beta cos gamma ) D. ( 2 sum cos alpha cos beta ) | 12 |
| 286 | Solve for ( boldsymbol{x} ) ( left|begin{array}{ccc}1-x & 2 & 3 \ 0 & x & 0 \ 0 & 0 & xend{array}right|=0 ) ( mathbf{A} cdot 1,0,0 ) B. 1,1,0 c. 1,1,1 D. 0,0,0 | 12 |
| 287 | Points (1,5),(2,3) and (-2,-11) are A. Non-collinear B. Collinear c. vertices of equilateral triangle D. Vertices of right angle triangle | 12 |
| 288 | Verify whether the points (1,5),(2,3) ,and (-2,-1) are collinear or not. | 12 |
| 289 | If ( left.A=left[begin{array}{ll}a & b \ c & dend{array}right] text { (where } b neq cright) ) and satisfies the equation ( A^{2}+k I=0, ) then ( mathbf{A} cdot a+d=0 ) B . ( k=-|A| ) c. ( k=|A| ) D. None of the above | 12 |
| 290 | Evaluate the following: ( left|begin{array}{ccc}1 & a & b c \ 1 & b & c a \ 1 & c & a bend{array}right| ) | 12 |
| 291 | Find the equation of the line joining ( A(1,3) ) and ( B(0,0) ) using determinants. | 12 |
| 292 | ( left|begin{array}{ccc}boldsymbol{x}+mathbf{2} & mathbf{2} boldsymbol{x}+mathbf{3} & mathbf{3} boldsymbol{x}+mathbf{4} \ mathbf{2} boldsymbol{x}+mathbf{3} & mathbf{3} boldsymbol{x}+mathbf{4} & mathbf{4} boldsymbol{x}+mathbf{5} \ mathbf{3} boldsymbol{x}+mathbf{5} & mathbf{5} boldsymbol{x}+mathbf{8} & mathbf{1 0} boldsymbol{x}+mathbf{1 7}end{array}right|=mathbf{0} ) then ( x ) is equal to A ( .-1,-2 ) в. 1,2 ( c cdot 1,-2 ) D. -1,2 | 12 |
| 293 | If ( A ) is a matrix of order ( 3 times 3 ) then find ( |boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}| ) where ( |boldsymbol{A}|=boldsymbol{2} ) | 12 |
| 294 | Show that ( begin{array}{l}sin 10^{circ} quad-cos 10^{circ} \ sin 80^{circ} quad cos 80^{circ}end{array} mid=1 ) | 12 |
| 295 | Solve ( left|begin{array}{ccc}1 & 1 & -1 \ 6 & 4 & -5 \ -4 & -2 & 3end{array}right| ) | 12 |
| 296 | ( begin{array}{ccc}(boldsymbol{b}+boldsymbol{c})^{2} & boldsymbol{a}^{2} & boldsymbol{a}^{2} \ boldsymbol{b}^{2} & (boldsymbol{c}+boldsymbol{a})^{2} & boldsymbol{b}^{2} \ boldsymbol{c}^{2} & boldsymbol{c}^{2} & (boldsymbol{a}+boldsymbol{b})^{2}end{array} mid= ) | 12 |
| 297 | Using properties of determinants, find the following: ( left|begin{array}{ccc}boldsymbol{alpha} & boldsymbol{beta} & boldsymbol{gamma} \ boldsymbol{alpha}^{2} & boldsymbol{beta}^{2} & boldsymbol{gamma}^{2} \ boldsymbol{beta}+boldsymbol{gamma} & boldsymbol{gamma}+boldsymbol{alpha} & boldsymbol{alpha}+boldsymbol{beta}end{array}right| ) A ( cdot(alpha+beta)(beta+gamma)(gamma-alpha)(alpha+beta+gamma) ) B . ( (alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma) ) c. D. None of these | 12 |
| 298 | ( left|begin{array}{ccc}1 & sin theta & 1 \ -sin theta & 1 & sin theta \ -1 & -sin theta & 1end{array}right| ) then ( mathbf{A} cdot Delta=0 ) B. ( Delta in(0, infty) ) c. ( Delta in[-1,2] ) D. ( Delta in[2,4] ) | 12 |
| 299 | The number of ordered triplets of positive integral solutions of ( left|begin{array}{ccc}boldsymbol{y}^{3}+mathbf{1} & boldsymbol{y}^{2} boldsymbol{z} & boldsymbol{y}^{2} boldsymbol{x} \ boldsymbol{y} boldsymbol{z}^{2} & boldsymbol{z}^{3}+mathbf{1} & boldsymbol{z}^{2} boldsymbol{x} \ boldsymbol{y} boldsymbol{x}^{2} & boldsymbol{x}^{2} boldsymbol{z} & boldsymbol{x}^{3}+mathbf{1}end{array}right|=mathbf{1 1} ) | 12 |
| 300 | If ( omega ) is a non-real cube root of unity and n is not a multiple of ( 3, ) then ( Delta= ) ( left|begin{array}{ccc}mathbf{1} & boldsymbol{omega}^{n} & boldsymbol{omega}^{2 n} \ boldsymbol{omega}^{2 boldsymbol{n}} & boldsymbol{1} & boldsymbol{omega}^{n} \ boldsymbol{omega}^{boldsymbol{n}} & boldsymbol{omega}^{2 boldsymbol{n}} & boldsymbol{1}end{array}right| ) is equal to ( mathbf{A} cdot mathbf{0} ) B. ( omega ) ( c cdot omega^{2} ) D. | 12 |
| 301 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{2} & mathbf{- 1} \ -mathbf{1} & mathbf{1} & mathbf{2} \ mathbf{2} & mathbf{- 1} & mathbf{1}end{array}right], ) then ( operatorname{det}(operatorname{adj}(operatorname{adj} A)) ) is equal to A ( cdot 14^{4} ) B. ( 14^{text {? }} ) ( c cdot 14^{2} ) D. 14 | 12 |
| 302 | Show that ( (x-a) ) is a factor of ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{x} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a} & boldsymbol{x}end{array}right| ) | 12 |
| 303 | Write minors and cofactors of the elements of following determinants (i) ( left|begin{array}{cc}2 & -4 \ 0 & 3end{array}right| ) (ii) ( left|begin{array}{ll}boldsymbol{a} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{d}end{array}right| ) | 12 |
| 304 | ( fleft(begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4end{array}right], ) then show that [ |mathbf{3} boldsymbol{A}|=mathbf{2 7}|boldsymbol{A}| ] | 12 |
| 305 | If ( A=left[begin{array}{cc}2 & -1 \ -1 & 2end{array}right], ) then the general solution of ( sin theta=left|A^{2}-4 A+3 Iright| ) is A ( . n pi ) B ( cdot 2 n+1 frac{pi}{2} ) c. ( n pi+(-1)^{n} frac{pi}{2} ) D. ( 2 n pi, n epsilon z ) | 12 |
| 306 | ( operatorname{Det}left[begin{array}{ccc}mathbf{4 3} & mathbf{1} & mathbf{6} \ mathbf{3 5} & mathbf{7} & mathbf{4} \ mathbf{1 7} & mathbf{3} & mathbf{2}end{array}right]=dots ) ( A ) B. – – ( c cdot 0 ) ( D ) | 12 |
| 307 | Prove that ( left|begin{array}{ccc}boldsymbol{y} boldsymbol{z}-boldsymbol{x}^{2} & boldsymbol{z} boldsymbol{x}-boldsymbol{y}^{2} & boldsymbol{x} boldsymbol{y}-boldsymbol{z}^{2} \ boldsymbol{z} boldsymbol{x}-boldsymbol{y}^{2} & boldsymbol{x} boldsymbol{y}-boldsymbol{z}^{2} & boldsymbol{y} boldsymbol{z}-boldsymbol{x}^{2} \ boldsymbol{x} boldsymbol{y}-boldsymbol{z}^{2} & boldsymbol{y} boldsymbol{z}-boldsymbol{x}^{2} & boldsymbol{z} boldsymbol{x}-boldsymbol{y}^{2}end{array}right| ) divisible by ( (x+y+z) ) and hence find the quotient | 12 |
| 308 | The remainder when the determinant ( left|begin{array}{lll}2014^{2014} & 2015^{2015} & 2016^{2016} \ 2017^{2017} & 2018^{2018} & 2019^{2019} \ 2020^{2020} & 2021^{2021} & 2022^{2022}end{array}right| ) is divided by 5 is. A . 1 B . 2 ( c cdot 4 ) ( D ) | 12 |
| 309 | Prove that: ( left|begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{y}+boldsymbol{z} \ mathbf{1} & boldsymbol{y} & boldsymbol{z}+boldsymbol{x} \ mathbf{1} & boldsymbol{z} & boldsymbol{x}+boldsymbol{y}end{array}right|=mathbf{0} ) | 12 |
| 310 | Let ( boldsymbol{A}=left[boldsymbol{a}_{i j}right]_{n times n} ) be a square matirx and let ( c_{i j} ) be cofactor of ( a_{i j} ) in A. If ( C=left[c_{i j}right] ) then B . ( |C|=|A|^{n-1} ) c. ( |C|=|A|^{n-2} ) D. none of these | 12 |
| 311 | ( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{b} boldsymbol{c} \ mathbf{1} & boldsymbol{b} & boldsymbol{c a} \ mathbf{1} & boldsymbol{c} & boldsymbol{a b}end{array}right|=boldsymbol{lambda}left|begin{array}{ccc}boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{1} & boldsymbol{1} & boldsymbol{1}end{array}right| ) then ( lambda ) is equal to ( A ) B . – ( c ) ( D .- ) | 12 |
| 312 | Calculate the values of the determinants: ( left|begin{array}{cccc}mathbf{0} & boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ -boldsymbol{x} & boldsymbol{0} & boldsymbol{c} & boldsymbol{b} \ -boldsymbol{y} & -boldsymbol{c} & boldsymbol{0} & boldsymbol{a} \ -boldsymbol{z} & -boldsymbol{b} & -boldsymbol{a} & boldsymbol{0}end{array}right| ) | 12 |
| 313 | If ( boldsymbol{A}=left[begin{array}{ccc}boldsymbol{x} & mathbf{1} & -boldsymbol{x} \ mathbf{0} & mathbf{1} & -mathbf{1} \ boldsymbol{x} & mathbf{0} & mathbf{7}end{array}right] ) and ( operatorname{det}(boldsymbol{A})= ) ( left|begin{array}{ccc}mathbf{3} & mathbf{0} & mathbf{1} \ mathbf{2} & mathbf{- 1} & mathbf{0} \ mathbf{0} & mathbf{6} & mathbf{7}end{array}right| ) then the value of ( boldsymbol{x} ) is A . -3 B. 3 ( c cdot 2 ) D. -8 E. -2 | 12 |
| 314 | Evaluate ( left|begin{array}{ccc}265 & 240 & 219 \ 240 & 225 & 198 \ 219 & 198 & 181end{array}right| ) | 12 |
| 315 | Find the values of ( x, ) if ( left|begin{array}{ll}boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{1} \ boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{2}end{array}right|=left|begin{array}{ll}boldsymbol{4} & -mathbf{1} \ mathbf{1} & mathbf{3}end{array}right| ) | 12 |
| 316 | f ( a, b, c>1, Delta= ) ( left|begin{array}{ccc}log _{a}(a b c) & log _{a} b & log _{a} c \ log _{b}(a b c) & 1 & log _{b} c \ log _{c}(a b c) & log _{c} b & 1end{array}right| ) is ( mathbf{A} cdot mathbf{0} ) B ( cdot log _{a} b+log _{b} c+log _{c} a ) ( mathbf{c} cdot log _{a b c}(a+b+c) ) D. none of these | 12 |
| 317 | The value of the determinant ( left|begin{array}{ccc}left(a^{x}+a^{-x}right)^{2} & left(a^{x}-a^{-x}right)^{2} & 1 \ left(b^{x}+b^{-x}right)^{2} & left(b^{x}-b^{-x}right)^{2} & 1 \ left(c^{x}+c^{-x}right)^{2} & left(c^{x}-c^{-x}right)^{2} & 1end{array}right| ) is ( mathbf{A} cdot mathbf{0} ) B. 2abc ( mathbf{c} cdot a^{2} b^{2} c^{2} ) D. None of these | 12 |
| 318 | what is the value of ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{a}^{2} & boldsymbol{a}^{boldsymbol{3}}-mathbf{1} \ boldsymbol{b} & boldsymbol{b}^{2} & boldsymbol{b}^{3}-mathbf{1} \ boldsymbol{c} & boldsymbol{c}^{2} & boldsymbol{c}^{boldsymbol{3}}-mathbf{1}end{array}right|=? ) | 12 |
| 319 | If ( A ) is a square matrix of order 3 such that ( |boldsymbol{a} boldsymbol{d} boldsymbol{j} cdot boldsymbol{A}|=boldsymbol{3} boldsymbol{6}, ) find ( |boldsymbol{A}| ) | 12 |
| 320 | Without expanding, show that the value of the following determinant is zero: ( left|begin{array}{ccc}a & b & c \ a+2 x & b+2 y & c+2 z \ x & y & zend{array}right| ) | 12 |
| 321 | Prove that [ left|begin{array}{ccc} boldsymbol{x}+boldsymbol{y}+mathbf{2} boldsymbol{z} & boldsymbol{x} & boldsymbol{y} \ boldsymbol{z} & boldsymbol{y}+boldsymbol{z}+mathbf{2} boldsymbol{x} & boldsymbol{y} \ boldsymbol{z} & boldsymbol{x} & boldsymbol{z}+boldsymbol{x}+boldsymbol{2} boldsymbol{y} end{array}right| ] ( 2(x+y+z)^{3} ) | 12 |
| 322 | The adjoint of the matrix ( boldsymbol{A}= ) [ left[begin{array}{ccc} 1 & 1 & 1 \ 2 & 1 & -3 \ -1 & 2 & 3 end{array}right] text { is } ] ( A ) [ left.begin{array}{l|lcc} & 9 & -1 & -4 \ frac{1}{11} & -3 & 4 & 5 \ & 5 & -3 & -1 end{array}right] ] B. [ left[begin{array}{ccc} 9 & 1 & -4 \ 3 & 4 & -5 \ 5 & 3 & -1 end{array}right] ] ( c ) [ left[begin{array}{ccc} 9 & -3 & 5 \ -1 & 4 & -3 \ -4 & 5 & -1 end{array}right] ] D. [ left[begin{array}{ccc} 9 & -1 & -4 \ -3 & 4 & 5 \ 5 & -3 & -1 end{array}right] ] | 12 |
| 323 | If ( boldsymbol{A}=int_{1}^{sin theta} frac{boldsymbol{t}}{1+boldsymbol{t}^{2}} boldsymbol{d} boldsymbol{t} ) and ( B=int_{1}^{operatorname{cosec} theta} frac{1}{tleft(1+t^{2}right)} d t, ) then the value of determinant ( left|begin{array}{ccc}boldsymbol{A} & boldsymbol{A}^{2} & boldsymbol{B} \ boldsymbol{e}^{boldsymbol{A}+boldsymbol{B}} & boldsymbol{B}^{2} & -mathbf{1} \ mathbf{1} & boldsymbol{A}^{2}+boldsymbol{B}^{2} & -mathbf{1}end{array}right| ) is ( A cdot sin theta ) B. ( operatorname{cosec} theta ) ( c ) D. | 12 |
| 324 | If ( left(begin{array}{lll}1 & 2 & 4 \ 1 & 3 & 5 \ 1 & 4 & aend{array}right) ) is singular, the value of ( a ) is A ( . a=-6 ) B. ( a=5 ) c. ( a=-5 ) ( mathbf{D} cdot a=6 ) | 12 |
| 325 | The value of the determinant ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & mathbf{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{b} & boldsymbol{0} & boldsymbol{a}end{array}right| ) is equal to A ( cdot a^{3}-b^{3} ) В . ( a^{3}+b^{3} ) ( c cdot 0 ) D. none of these | 12 |
| 326 | For a positive numbers ( x, y ) and ( z ) the numerical value of the determinant ( left[begin{array}{ccc}1 & log _{x} y & log _{x} z \ log _{y} x & 1 & log _{y} z \ log _{z} x & log _{z} y & 1end{array}right] ) is: A. 0 B. 1 ( mathbf{c} cdot log _{e} x y z ) ( mathbf{D} cdot-log _{e} x y z ) | 12 |
| 327 | ( mathbf{A}=left|begin{array}{ccc}cos frac{theta}{2} & mathbf{1} & mathbf{1} \ mathbf{1} & cos frac{theta}{2} & -cos frac{theta}{2} \ -cos frac{theta}{2} & mathbf{1} & -mathbf{1}end{array}right| ) the minimum of ( Delta ) is ( m_{1} ) and maximum of ( Delta ) is ( m_{2} ) then ( left[m_{1}, m_{2}right] ) are related to A. [4, 2] B. [2,4] ( c cdot[4,0] ) D. [0,2] | 12 |
| 328 | ( left|begin{array}{ccc}boldsymbol{a}+boldsymbol{x} & boldsymbol{a}-boldsymbol{x} & boldsymbol{a}-boldsymbol{x} \ boldsymbol{a}-boldsymbol{x} & boldsymbol{a}+boldsymbol{x} & boldsymbol{a}-boldsymbol{x} \ boldsymbol{a}-boldsymbol{x} & boldsymbol{a}-boldsymbol{x} & boldsymbol{a}+boldsymbol{x}end{array}right|=mathbf{0} ) then the non-zero value of ( x=dots ) ( A ) B. ( 3 a ) ( c .2 a ) D. ( 4 a ) | 12 |
| 329 | Calculate the values of the determinants: ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{h} & boldsymbol{g} \ boldsymbol{h} & boldsymbol{b} & boldsymbol{f} \ boldsymbol{g} & boldsymbol{f} & boldsymbol{c}end{array}right| ) | 12 |
| 330 | Let ( triangle_{1}, triangle_{2}, triangle_{3}, dots, triangle_{k} ) be the set of third-order determinants that can be made with the distinct nonzero real numbers ( a_{1}, a_{2}, a_{3}, dots a_{9}, ) and ( Sigma a_{i}=0 ) Then This question has multiple correct options ( mathbf{A} cdot k=9 ! ) B . ( sum_{i=1}^{k} Delta_{i}=0 ) C . at least one ( Delta_{i}=0 ) D. none of these | 12 |
| 331 | ( left|begin{array}{cc}mathbf{0} & boldsymbol{a}-boldsymbol{b} \ -boldsymbol{a} mathbf{0}-boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} mathbf{0}end{array}right|=mathbf{0} ) | 12 |
| 332 | The value of the determinant [ left|begin{array}{ccc} mathbf{1} & cos (boldsymbol{alpha}-boldsymbol{beta}) & mathbf{c o s} boldsymbol{alpha} \ cos (boldsymbol{alpha}-boldsymbol{beta}) & mathbf{1} & cos beta \ cos boldsymbol{alpha} & cos beta & mathbf{1} end{array}right| ] A ( cdot alpha^{2}+beta^{2} ) B ( cdot alpha^{2}-beta^{2} ) ( c .1 ) ( D ) | 12 |
| 333 | frue Enter ( ^{prime} 1^{prime} ) else ( ^{prime} 0^{prime} ) ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{y} & boldsymbol{x}+boldsymbol{y} \ boldsymbol{y} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x} \ boldsymbol{x}+boldsymbol{y} & boldsymbol{x} & boldsymbol{y}end{array}right|=-boldsymbol{2}(boldsymbol{x}+ ) ( boldsymbol{y})left(boldsymbol{x}^{2}+boldsymbol{y}^{2}-boldsymbol{x} boldsymbol{y}right) ) | 12 |
| 334 | Let ( Delta(x)= ) ( left|begin{array}{ccc}(boldsymbol{x}-mathbf{2}) & (boldsymbol{x}-mathbf{1})^{2} & boldsymbol{x}^{mathbf{3}} \ (boldsymbol{x}-mathbf{1}) & boldsymbol{x}^{2} & (boldsymbol{x}+mathbf{1})^{3} \ boldsymbol{x} & (boldsymbol{x}+mathbf{1})^{2} & (boldsymbol{x}+mathbf{2})^{3}end{array}right| ) Then the coefficient of ( x ) in ( Delta(x) ) is ( -k . ) Find ( k ) ( A cdot-1 ) B. 2 ( c cdot-2 ) ( D ) | 12 |
| 335 | ( mathbf{f} boldsymbol{u}_{boldsymbol{r}}=boldsymbol{a}_{boldsymbol{r}} boldsymbol{x}+boldsymbol{b}_{boldsymbol{r}} boldsymbol{y}+, boldsymbol{c}_{boldsymbol{r}}=mathbf{0}(r=1,2,3 ) be the three sides of a triangle them the equations of the circumcircle of this triangle is ( mid begin{array}{ccc}frac{mathbf{1}}{boldsymbol{u}_{1}} & frac{mathbf{1}}{boldsymbol{u}_{2}} & frac{mathbf{1}}{boldsymbol{u}_{3}} \ boldsymbol{a}_{2} boldsymbol{a}_{3}-boldsymbol{b}_{2} boldsymbol{b}_{3} & boldsymbol{a}_{3} boldsymbol{a}_{1}-boldsymbol{b}_{3} boldsymbol{b}_{1} & boldsymbol{a}_{1} boldsymbol{a}_{2}-boldsymbol{b}_{4} \ boldsymbol{a}_{2} boldsymbol{b}_{3}+boldsymbol{a}_{3} boldsymbol{b}_{2} & boldsymbol{a}_{3} boldsymbol{b}_{1}+boldsymbol{a}_{1} boldsymbol{b}_{3} & boldsymbol{a}_{1} boldsymbol{b}_{2}+boldsymbol{a}_{2}end{array} ) ( =0 ) | 12 |
| 336 | If ( A ) is a skew-symmetric matrix of order 3 then find ( |A| ) | 12 |
| 337 | 14. 11 If D= 1 1 1 1+ x 1 1 1 1+y for x = 0, y 70, then D is (a) divisible by x but not y (b) divisible by y but not x (c) divisible by neither x nor y (d) divisible by both x and y | 12 |
| 338 | ( f(x)= ) [ begin{array}{ccc} cos ^{x} & cos x sin x & -sin x \ cos x sin x & sin ^{x} & cos x \ sin x & -cos x & 0 end{array} ] Prove that ( f(x)=1 ) is an identity | 12 |
| 339 | Prove that ( triangle= ) ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{a} boldsymbol{alpha}+boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b} boldsymbol{alpha}+boldsymbol{c} \ boldsymbol{a} boldsymbol{alpha}+boldsymbol{b} & boldsymbol{b} boldsymbol{alpha}+boldsymbol{c} & boldsymbol{0}end{array}right|=boldsymbol{0} ) if ( mathbf{a}, mathbf{b}, mathbf{c} ) ( operatorname{arcin} mathrm{G} . mathrm{P} ) | 12 |
| 340 | ( mathbf{f}left|begin{array}{ccc}boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b}end{array}right|=boldsymbol{t} times operatorname{det} ) of circulant matrix whose elements of first column are ( a, b, c ) then ‘t’ equals A . 5 B. 6 ( c cdot-2 ) ( D ) | 12 |
| 341 | Three lines ( boldsymbol{p} boldsymbol{x}+boldsymbol{q} boldsymbol{y}+boldsymbol{r}=mathbf{0}, boldsymbol{q} boldsymbol{x}+boldsymbol{r} boldsymbol{y}+ ) ( boldsymbol{p}=mathbf{0} ) and ( boldsymbol{r} boldsymbol{x}+boldsymbol{p} boldsymbol{y}+boldsymbol{q}=mathbf{0} ) are concurrent if This question has multiple correct options A ( . p+q+r=0 ) B . ( p^{2}+q^{2}+r^{2}=p r+q r+p q ) c. ( p^{3}+q^{3}+r^{3}=3 p q r ) D. none of these | 12 |
| 342 | ( operatorname{Solve}left|begin{array}{ccc}mathbf{0} & boldsymbol{p}-boldsymbol{q} & boldsymbol{p}-boldsymbol{r} \ boldsymbol{q}-boldsymbol{p} & boldsymbol{0} & boldsymbol{q}-boldsymbol{r} \ boldsymbol{r}-boldsymbol{p} & boldsymbol{r}-boldsymbol{q} & boldsymbol{0}end{array}right|= ) ( mathbf{A} cdot p+q+r ) в. ( p q+q r+r p ) ( c cdot c ) D. ( p^{2}+q^{2}+r^{2} ) | 12 |
| 343 | Without expanding prove that the determinant ( left|begin{array}{ccc}sin A & operatorname{Cos} A & sin (A+theta) \ sin B & cos B & sin (B+theta) \ sin C & cos C & sin (c+theta)end{array}right|=0 ) | 12 |
| 344 | If planes ( boldsymbol{x}-boldsymbol{c} boldsymbol{y}-boldsymbol{b} boldsymbol{z}=boldsymbol{0}, boldsymbol{c} boldsymbol{x}-boldsymbol{y}+ ) ( boldsymbol{a} boldsymbol{z}=mathbf{0} ) and ( boldsymbol{b} boldsymbol{x}+boldsymbol{a} boldsymbol{y}-boldsymbol{z}=mathbf{0} ) pass through a straight line then ( a^{2}+b^{2}+ ) ( boldsymbol{c}^{2}= ) A. ( 1-a b c ) B . ( a b c-1 ) c. ( 1-2 a b c ) D. ( 2 a b c-1 ) | 12 |
| 345 | 1. Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset of A consisting of all determinants with value 1. Let C be the subset of A consisting of all determinants with value-1. Then (a C is empty (1981 – 2 Marks) (b) B has as many elements as C (c) A= BUC (d) B has twice as many elements as elements as C | 12 |
| 346 | ( A=left[begin{array}{lll}3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3end{array}right], ) then ( operatorname{Adj}(A) ) ( A cdot 3 A ) B. ( 6 A ) ( mathrm{c} cdot 9 A^{T} ) D. ( 2 A^{text {? }} ) | 12 |
| 347 | 9. (1992 – 4 For a fixed positive integer n, if n! (n+1)! (n + 2)!| D= (n+1)! (n+ 2)! (n+3)! |(n+2)! (n+3)! (n+4)! then show that – -4 is divisible by n. n13 | 12 |
| 348 | Find the values of ( a ) and ( b ) so that the points ( (boldsymbol{a}, boldsymbol{b}, mathbf{3}),(mathbf{2}, mathbf{0},-mathbf{1}) ) and (1,-1,-3) are collinear. This question has multiple correct options A ( . a=4, b=2 ) В. ( a=0, b=2 ) c. ( a=4, b=-2 ) begin{tabular}{l} D. ( a=-4, b=-2 ) \ hline end{tabular} | 12 |
| 349 | 1. For what value of k do the following system of equations possess a non trivial (i.e., not all zero) solution over the set of rationals Q? x + ky + 3z=0 3x + ky-2z=0 2x+3y – 4z=0 For that value of k, find all the solutions for the system. | 12 |
| 350 | Prove that ( left|begin{array}{lll}mathbf{1} & boldsymbol{x} & boldsymbol{x}^{3} \ mathbf{1} & boldsymbol{y} & boldsymbol{y}^{3} \ mathbf{1} & boldsymbol{z} & boldsymbol{z}^{3}end{array}right|=(boldsymbol{x}+boldsymbol{y}+ ) ( z(x-y)(y-z)(z-x) ) | 12 |
| 351 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & mathbf{2} \ mathbf{2} & mathbf{1}end{array}right] ) then ( boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{A})=? ) ( mathbf{A} cdotleft[begin{array}{cc}1 & -2 \ -2 & 1end{array}right] ) в. ( left[begin{array}{cc}2 & 1 \ 1 & 1end{array}right] ) c. ( left[begin{array}{cc}1 & -2 \ -2 & -1end{array}right] ) D. ( left[begin{array}{cc}-1 & 2 \ 2 & -1end{array}right] ) | 12 |
| 352 | Solve:det ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{y} boldsymbol{z} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{z} boldsymbol{x} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{x} boldsymbol{y}end{array}right| ) | 12 |
| 353 | If ( |boldsymbol{A}|=mathbf{3} ) and ( boldsymbol{A}^{-1}=left[begin{array}{cc}mathbf{3} & -mathbf{1} \ -mathbf{5} & mathbf{2} \ hline mathbf{3} & mathbf{3}end{array}right] ) then ( boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}=? ) ( mathbf{A} cdotleft[begin{array}{cc}9 & 3 \ -5 & -2end{array}right] ) В. ( left[begin{array}{cc}9 & -3 \ -5 & 2end{array}right] ) begin{tabular}{lll} C. ( left[begin{array}{cc}-9 & 3 \ 5 & -2end{array}right] ) \ hline end{tabular} D. ( left[begin{array}{cc}9 & -3 \ 5 & -2end{array}right] ) | 12 |
| 354 | The value of ( Delta=left|begin{array}{llll}1^{2} & 2^{2} & 3^{2} & 4^{2} \ 2^{2} & 3^{2} & 4^{2} & 5^{2} \ 3^{2} & 4^{2} & 5^{2} & 6^{2} \ 4^{2} & 5^{2} & 6^{2} & 7^{2}end{array}right| ) equals to ( mathbf{A} cdot mathbf{0} ) B ( cdot 1^{2}+2^{2}+3^{2}+4^{2}+ldots+7^{2} ) ( c cdot 1 ) D. – | 12 |
| 355 | The area of the triangle whose vertices ( operatorname{are} A(1,2,3), B(2,-1,1) ) and ( boldsymbol{C}(mathbf{1}, mathbf{2},-mathbf{4}) ) is A . ( 7 sqrt{10} ) sq units B. ( frac{1}{2} sqrt{10} ) sq units c. ( frac{7}{2} sqrt{10} ) sq units D. None of these | 12 |
| 356 | If ( left(x_{1}-x_{2}right)^{2}+left(y_{1}-y_{2}right)^{2}=a^{2} ) ( left(x_{2}-x_{3}right)^{2}+left(y_{2}-y_{3}right)^{2}=b^{2} ) ( left(x_{3}-x_{1}right)^{2}+left(y_{3}-y_{1}right)^{2}=c^{2} ) and ( kleft|begin{array}{lll}boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & mathbf{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & mathbf{1}end{array}right|=(boldsymbol{a}+boldsymbol{b}+boldsymbol{c})(boldsymbol{b}+boldsymbol{c}- ) ( a)(c+a-b) times(a+b-c), ) then the value of ( k ) is ( A ) в. 2 ( c ) D. none of these | 12 |
| 357 | ( left|begin{array}{ccc}x^{2}+x & x+1 & x-2 \ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \ x^{2}+2 x+3 & 2 x-1 & 2 x-1end{array}right|= ) ( A x+B ) then A. A and B are independent of ( x ) B. A and B are dependent of ( x ) C. A dependent on x but B does not depend on x D. B depends on x but A does not depend on x | 12 |
| 358 | Let ( 0(0,0), P(3,4), Q(6,0) ) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR,PQR, OQR are of equal area. The coordinates of ( mathrm{R} ) are ( ^{mathrm{A}} cdotleft(frac{4}{3}, 3right) ) в. ( left(3, frac{2}{3}right) ) c. ( left(3, frac{4}{3}right) ) D ( cdotleft(frac{4}{3}, frac{2}{3}right) ) | 12 |
| 359 | If ( omega ) is a complex cube root of unity, the ( left|begin{array}{ccc}mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{omega}^{2} & boldsymbol{1} \ boldsymbol{omega}^{2} & boldsymbol{1} & boldsymbol{omega}end{array}right| ) is equal to A . -1 B. ( c cdot 0 ) D. | 12 |
| 360 | ( f(x)=left|begin{array}{ccc}cos x & x & 1 \ 2 sin x & x^{2} & 2 x \ tan x & x & 1end{array}right| ) then ( lim _{x rightarrow 0} f(x)= ) ( A cdot O ) B. ( c cdot-2 ) ( D ) | 12 |
| 361 | If ( f(x)=left|begin{array}{cc}x & lambda \ 2 lambda & xend{array}right|, ) then ( f(lambda x)-f(x) ) is equal to: A ( cdot xleft(lambda^{2}-1right) ) B ( cdot 2 lambdaleft(x^{2}-1right) ) C ( cdot lambda^{2}left(x^{2}-1right) ) D. ( lambdaleft(x^{2}-1right) ) E ( cdot x^{2}left(lambda^{2}-1right) ) | 12 |
| 362 | Value of ( left|begin{array}{lll}sin alpha & cos alpha & sin alpha+cos beta \ sin beta & cos alpha & sin beta+cos beta \ sin gamma & cos alpha & sin gamma+cos betaend{array}right| ) is ( A cdot sin alpha+sin beta+sin gamma ) B. ( cos alpha+cos beta+cos gamma ) ( mathbf{c} cdot sin alpha-sin (alpha+beta)-cos alpha+cos (gamma+beta) ) ( D ) | 12 |
| 363 | Find the of ( lambda, ) so that the matrix ( left[begin{array}{cc}mathbf{5}-boldsymbol{lambda} & boldsymbol{lambda}+mathbf{1} \ mathbf{2} & mathbf{4}end{array}right] ) may be singular | 12 |
| 364 | 28. If P= 1 a 37 1 3 3 is the adjoint of a 3 x 3 matrix A and 2 4 4 JA) = 4, then a is equal to : (a) 4 (6) 11 [JEEM 2013 (d) 0 (C) 5 | 12 |
| 365 | U TIUTUU ay = 0, az + y = 0 and If the system of equations x + ay = 0, az + y = ax + z = 0 has infinite solutions, then the value of a 15 (2003) (a) -1 (b) 1 (c) o (d) no real values | 12 |
| 366 | ff ( alpha, beta ) are the roots of ( left|begin{array}{lll}x & 1 & 2 \ 0 & 1 & 1 \ 1 & x & 2end{array}right|=0 ) ( operatorname{then} boldsymbol{alpha}^{boldsymbol{n}}+boldsymbol{beta}^{boldsymbol{n}}=? ) A. в. ( c cdot 2 ) D. 2n | 12 |
| 367 | If ( D_{p}=left|begin{array}{ccc}boldsymbol{p} & mathbf{1 5} & mathbf{8} \ boldsymbol{p}^{2} & mathbf{2 5} & mathbf{9} \ boldsymbol{p}^{mathbf{3}} & mathbf{4 5} & mathbf{1 0}end{array}right|, ) then ( operatorname{det}left(boldsymbol{D}_{1}+right. ) ( D_{2} ) ) is equal to ( mathbf{A} cdot mathbf{0} ) B . 25 ( c .625 ) D. None of these | 12 |
| 368 | Find the values of ( x ) for which ( left|begin{array}{ll}3 & x \ x & 1end{array}right|= ) ( left|begin{array}{ll}mathbf{3} & mathbf{2} \ mathbf{4} & mathbf{1}end{array}right| ) | 12 |
| 369 | 6. x +1 If fx) = 2x x (x-1) (x+1)x then 3x(x – 1) x(x – 1) (x – 2) (x+1) x(x – 1)| f(100) is equal to (1999 – 2 Marks) (a) o (6) 1 (c) 100 (d) -100 | 12 |
| 370 | The number of distinct real roots of the quation ( left|begin{array}{ccc}cos x & sin x & sin x \ sin x & cos x & sin x \ sin x & sin x & cos xend{array}right|=0 ) in the interval ( left[-frac{pi}{4}, frac{pi}{4}right] ) is ( A ) B. 4 ( c cdot 2 ) ( D ) | 12 |
| 371 | ( fleft|begin{array}{ccc}6 i & -3 i & 1 \ 4 & 3 i & -1 \ 20 & 3 & iend{array}right|=x+i y ) then A ( . x=3, y=1 ) B. ( x=1, y=3 ) c. ( x=0, y=3 ) D. ( x=0, y=0 ) | 12 |
| 372 | Let ( A, B, C, D, E ) be the interior angles of convex pentagon and ( boldsymbol{Delta}=left|begin{array}{ccc}cos boldsymbol{A} & sin boldsymbol{A} & sin (boldsymbol{A}+boldsymbol{D}+boldsymbol{E}) \ cos boldsymbol{B} & sin boldsymbol{B} & sin (boldsymbol{B}+boldsymbol{D}+boldsymbol{E}) \ cos boldsymbol{C} & sin boldsymbol{C} & sin (boldsymbol{C}+boldsymbol{D}+boldsymbol{E})end{array}right| ) find ( Delta(pi / 3)+Delta^{prime}(pi / 6) ) | 12 |
| 373 | Find the area of the triangle formed by the lines ( x=3, y=2 ) and ( 3 x+4 y=29 ) | 12 |
| 374 | ( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2} \ mathbf{1} & boldsymbol{b} & boldsymbol{b}^{2} \ mathbf{1} & boldsymbol{c} & boldsymbol{c}^{2}end{array}right|=(boldsymbol{a}-boldsymbol{b})(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}-boldsymbol{a}) ) | 12 |
| 375 | ( fleft(begin{array}{ccc}-1 & -3 & -3 \ 3 & 1 & -3 \ 3 & -3 & 1end{array}right] ) then adj ( (A) ) is A. [ =4left[begin{array}{ccc}-2 & 3 & 3 \ -3 & 2 & -3 \ -3 & 3 & 2end{array}right] ] B. ( quadleft[begin{array}{ccc}-2 & 3 & 3 \ 3 & 2 & -3 \ -3 & -3 & 2end{array}right] ) C. [ =4left[begin{array}{ccc}-2 & -3 & 3 \ -3 & 2 & -3 \ -3 & -3 & 2end{array}right] ] D. [ =4left[begin{array}{ccc}-2 & 3 & 3 \ -3 & 2 & -3 \ -3 & -3 & 2end{array}right] ] | 12 |
| 376 | ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{0}end{array}right|=mathbf{0}, ) then ( boldsymbol{a}, boldsymbol{b}, boldsymbol{c} ) ( operatorname{are} ) in A. A.P B. G.P. ( c . ) н. D. None of these | 12 |
| 377 | Prove the following: ( left|begin{array}{ccc}boldsymbol{a}^{2}+boldsymbol{b}^{2} & boldsymbol{c} & boldsymbol{c} \ boldsymbol{c} & frac{boldsymbol{b}^{2}+boldsymbol{c}^{2}}{boldsymbol{a}} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{b} & frac{boldsymbol{c}^{2}+boldsymbol{a}^{2}}{boldsymbol{b}}end{array}right|=boldsymbol{4 a b c} ) | 12 |
| 378 | ( operatorname{Matrix} boldsymbol{A}=left[begin{array}{ccc}boldsymbol{x} & boldsymbol{3} & boldsymbol{2} \ boldsymbol{1} & boldsymbol{y} & boldsymbol{4} \ boldsymbol{2} & boldsymbol{2} & boldsymbol{z}end{array}right], ) if ( boldsymbol{x} boldsymbol{y} boldsymbol{z}=boldsymbol{6} mathbf{0} ) and ( 8 x+4 y+3 z=20, ) then ( A(a d j A) ) is equal to A. ( left|begin{array}{lll}64 & 0 & 0 \ 0 & 64 & 0 \ 0 & 0 & 64end{array}right| ) B. ( mid begin{array}{ccc}68 & 0 & 0 \ 0 & 68 & 0 \ 0 & 0 & 68end{array} ) begin{tabular}{l|lll} 38 & 0 & 0 \ 0 & 38 & 0 \ 0 & 0 & 38 end{tabular} D. ( mid begin{array}{ccc}32 & 0 & 0 \ 0 & 32 & 0 \ 0 & 0 & 32end{array} ) | 12 |
| 379 | Find the values of ( x, ) if ( left|begin{array}{cc}mathbf{3 x} & mathbf{7} \ mathbf{2} & mathbf{4}end{array}right|=mathbf{1 0} ) | 12 |
| 380 | If ( boldsymbol{D}_{boldsymbol{r}}=left|begin{array}{ccc}boldsymbol{r} & boldsymbol{x} & frac{boldsymbol{n}(boldsymbol{n}+mathbf{1})}{mathbf{2}} mid \ boldsymbol{2 r}-boldsymbol{1} & boldsymbol{y} & boldsymbol{n}^{2} \ boldsymbol{3 r}-boldsymbol{2} & boldsymbol{z} & frac{boldsymbol{n}(boldsymbol{3} boldsymbol{n}-mathbf{1})}{mathbf{2}}end{array}right|, ) then ( sum_{r=1}^{n} D_{r} ) is equal to A ( cdot frac{1}{6} n(n+1)(2 n+1) ) B. ( frac{1}{4} n^{2}(n+1)^{2} ) ( c cdot 0 ) D. none of these | 12 |
| 381 | The co-ordinates of the vertices ( A, B, C ) of a triangle are ( (mathbf{6}, mathbf{3}),(-mathbf{3}, mathbf{5}),(mathbf{4},-mathbf{2}) ) respectively and ( P ) is any point ( (x, y) ) then the ratio of areas of triangles PBC and ABC is A ( cdot|x-y-2|: 7 ) B cdot | ( x+y+2 mid: 7 ) c. ( |x+y-2|: 7 ) D. None of these | 12 |
| 382 | For what value of ( x ) the matrix ( A ) is singular? ( boldsymbol{A}=left[begin{array}{ll}mathbf{1}+boldsymbol{x} & mathbf{7} \ mathbf{3}-boldsymbol{x} & mathbf{8}end{array}right] ) A ( cdot frac{12}{15} ) в. ( frac{13}{15} ) c. ( frac{14}{15} ) D. none of these | 12 |
| 383 | The value of the determinant ( Delta=left|begin{array}{lll}log x & log y & log z \ log 2 x & log 2 y & log 2 z \ log 3 x & log 3 y & log 3 zend{array}right| ) ( A cdot 0 ) ( mathbf{B} cdot log (x y z) ) ( mathbf{C} cdot log (6 x y z) ) ( mathbf{D} cdot 6 log (x y z) ) | 12 |
| 384 | If ( f(x)= ) ( left|begin{array}{ccc}cos ^{2} x & cos x cdot sin x & -sin x \ cos x cdot sin x & sin ^{2} x & cos x \ sin x & -cos x & 0end{array}right|, ) then for all ( x epsilon R, ) the value of ( f(x)= ) ( mathbf{A} cdot mathbf{0} ) B. ( c ) D. None of the above | 12 |
| 385 | For what values of ( m ) will the expression ( y^{2}+2 x y+2 x+m y-3 ) be capable of resolution into two rational factors? | 12 |
| 386 | The maximum value of ( left|begin{array}{ccc}1+sin ^{2} x & cos ^{2} x & 4 cos 2 x \ sin ^{2} x & 1+cos ^{2} x & 4 sin 2 x \ sin ^{2} x & cos ^{2} x & 1+4 sin 2 xend{array}right| ) ( A ) B. ( c .5 ) ( D ) | 12 |
| 387 | If ( D=left|begin{array}{cc}3 sqrt{5} & 6 \ 5 & mend{array}right|=0 ) Find the value of ( boldsymbol{m} ) A . ( sqrt{5} ) B. ( 4 sqrt{5} ) ( c cdot sqrt{3} ) D. ( 2 sqrt{5} ) | 12 |
| 388 | 13. Prove that for all values of o, sine cose sin20 | sin ( 1 + 2 ) cos( 1+ 2+ sin(20 + 4) = 0 sin(0-20) cos(0 – 21) sin( 20 – 49) COS | 12 |
| 389 | Find the value of ( x, ) if ( left|begin{array}{cc}boldsymbol{x}+mathbf{2} & boldsymbol{x} \ boldsymbol{x}-boldsymbol{4} & boldsymbol{x}+mathbf{3}end{array}right|=left|begin{array}{cc}boldsymbol{4} & boldsymbol{2} \ -boldsymbol{2} & boldsymbol{5}end{array}right| ) | 12 |
| 390 | The value of ( left|begin{array}{lllll}mathbf{1} & mathbf{0} & mathbf{0} & mathbf{0} & mathbf{0} \ mathbf{2} & mathbf{2} & mathbf{0} & mathbf{0} & mathbf{0} \ mathbf{4} & mathbf{4} & mathbf{3} & mathbf{0} & mathbf{0} \ mathbf{5} & mathbf{5} & mathbf{5} & mathbf{4} & mathbf{0} \ mathbf{6} & mathbf{6} & mathbf{6} & mathbf{6} & mathbf{5}end{array}right| ) ( A cdot 6 ) B. 5 c. ( 1.2^{2} .3 .4^{3} .5^{4} .6^{4} ) D. None of these | 12 |
| 391 | Find the value of following determinant. ( left|begin{array}{cc}-1 & 7 \ 2 & 4end{array}right| ) | 12 |
| 392 | Find the equation of line joining (1,2) and (3,6) using determinants. | 12 |
| 393 | If ( operatorname{in} ) a ( Delta A B C ; frac{cos A}{7}=frac{cos B}{19}= ) ( frac{cos C}{25}=k, ) then ( left|begin{array}{ccc}-1 / k & 25 & 19 \ 25 & -1 / k & 7 \ 19 & 7 & -1 / kend{array}right|= ) ( A ) B. ( c cdot 2 ) ( D ) | 12 |
| 394 | The number of distinct real roots of ( left|begin{array}{lll}sin x & cos x & cos x \ cos x & sin x & cos x \ cos x & cos x & sin xend{array}right|=0 ) in the nterval ( -frac{pi}{4}2 ) | 12 |
| 395 | Find values of ( k ) if area of triangle is 4 sq. units and vertices are (i) ( (k, 0),(4,0),(0,2) ) (ii) ( (2,0),(0,4),(0, k) ) | 12 |
| 396 | A determinant of second order is made with the elements 0 and ( 1 . ) Find the number of determinants with non- negative values. | 12 |
| 397 | ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{2}-boldsymbol{b} boldsymbol{c} & boldsymbol{b}^{2}-boldsymbol{c} boldsymbol{a} & boldsymbol{c}^{2}-boldsymbol{a} boldsymbol{b}end{array}right|= ) ( A . ) ( B ) ( c . a b c ) D. ( (a-b),(b-c),(c-a) ) | 12 |
| 398 | Consider three points ( boldsymbol{P}=(-sin (boldsymbol{beta}- ) ( boldsymbol{alpha}),-cos beta), boldsymbol{Q}=(cos (beta-boldsymbol{alpha}), sin beta) ) and ( boldsymbol{R}=(cos (boldsymbol{beta}-boldsymbol{alpha}+boldsymbol{theta}), sin (boldsymbol{beta}-boldsymbol{theta})) ) where ( 0<alpha, beta, theta<frac{pi}{4} . ) Then A. ( P ) lies on the line segment ( R Q ) B. ( Q ) lies on the line segment ( P R ) c. ( R ) lies on the line segment ( Q P ) D. ( P, Q, R ) are non-collinear | 12 |
| 399 | The value of ( left|begin{array}{ccc}boldsymbol{y} boldsymbol{z} & boldsymbol{z} boldsymbol{x} & boldsymbol{x} boldsymbol{y} \ boldsymbol{p} & boldsymbol{2} boldsymbol{q} & boldsymbol{3} boldsymbol{r} \ boldsymbol{1} & boldsymbol{1} & boldsymbol{1}end{array}right| ) where ( boldsymbol{x}, boldsymbol{y}, boldsymbol{z} ) are respectively, ( p t h,(2 q) t h, a n d(3 r) t h ) terms of an H.P., is A . -1 B. c. 1 D. none of these | 12 |
| 400 | Calculate the value of the following determinant: ( left|begin{array}{cccc}mathbf{1}+boldsymbol{a} & mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{b} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{c} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{d}end{array}right| ) | 12 |
| 401 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & log _{boldsymbol{b}} boldsymbol{a} \ log _{boldsymbol{a}} boldsymbol{b} & mathbf{1}end{array}right] ) then ( |boldsymbol{A}| ) is equal to A. 0 B ( cdot log _{a} b ) ( c cdot-1 ) ( mathbf{D} cdot log _{b} a ) | 12 |
| 402 | Prove that ( left|begin{array}{lll}1 ! & 2 ! & 3 ! \ 2 ! & 3 ! & 4 ! \ 3 ! & 4 ! & 5 !end{array}right|=4 ! ) | 12 |
| 403 | Let ( A ) be a square matrix of order 3 write the value of ( |2 A|, ) where ( |A|=4 ) | 12 |
| 404 | Find determinant of ( left|begin{array}{lll}mathbf{1} & mathbf{0} & mathbf{2} \ mathbf{2} & mathbf{1} & mathbf{0} \ mathbf{3} & mathbf{2} & mathbf{1}end{array}right| ) | 12 |
| 405 | ( mathbf{f} mathbf{Delta}=left|begin{array}{ccc}-boldsymbol{a} & mathbf{2} boldsymbol{b} & mathbf{0} \ mathbf{0} & -boldsymbol{a} & mathbf{2} boldsymbol{b} \ mathbf{2} boldsymbol{b} & mathbf{0} & -boldsymbol{a}end{array}right|=mathbf{0}, ) then A ( cdot frac{1}{b} ) is a cube root of unity B. ( a ) is one of the cube roots of unity ( mathrm{c} . b ) is one of the cube roots of 8 D. ( frac{a}{b} ) is a cube root of 8 | 12 |
| 406 | Prove the following: [ left|begin{array}{ccc} boldsymbol{a}^{2} & boldsymbol{b} boldsymbol{c} & boldsymbol{a} boldsymbol{c}+boldsymbol{c}^{2} \ boldsymbol{a}^{2}+boldsymbol{a} boldsymbol{b} & boldsymbol{b}^{2} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a b} & boldsymbol{b}^{2}+boldsymbol{b} boldsymbol{c} & boldsymbol{c}^{2} end{array}right|=boldsymbol{4} boldsymbol{a}^{2} boldsymbol{b}^{2} boldsymbol{c}^{2} ] | 12 |
| 407 | ( left|begin{array}{lll}boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c}end{array}right|=boldsymbol{K}left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right| ) ( operatorname{then} K= ) ( A ) B. 2 ( c ) ( D ) | 12 |
| 408 | f ( alpha, beta, gamma ) are the roots of the equation ( x^{3}+p x+q=0 ) then the value of the determinant ( left|begin{array}{lll}boldsymbol{alpha} & boldsymbol{beta} & gamma \ boldsymbol{beta} & gamma & boldsymbol{alpha} \ boldsymbol{gamma} & boldsymbol{alpha} & boldsymbol{beta}end{array}right| ) is ( mathbf{A} cdot mathbf{q} ) B. c. ( p ) D. ( p^{2}-2 q ) | 12 |
| 409 | ( fleft(a_{1}, a_{2}, a_{3}, dots . . ) are in G.P. then the value right. of determinant ( left|begin{array}{ccc}log a_{n} & log a_{n+1} & log a_{n+2} \ log a_{n+3} & log a_{n+4} & log a_{n+5} \ log a_{n+6} & log a_{n+7} & log a_{n+8}end{array}right| ) equal ( mathbf{A} cdot mathbf{0} ) B. 1 ( c cdot 2 ) D. 4 | 12 |
| 410 | What is the area of the triangle formed by the points ( (a, c+a),(a, c) ) and ( (-a, c-a) ? ) A. ( -a^{2} ) B. ( frac{1}{a^{2}} ) c. ( a^{2}+a ) D. zero | 12 |
| 411 | f the determinant ( D= ) [ left|begin{array}{ccc} mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{alpha}+boldsymbol{beta} & boldsymbol{alpha}^{2}+boldsymbol{beta}^{2} & mathbf{2} boldsymbol{alpha} boldsymbol{beta} \ boldsymbol{alpha}+boldsymbol{beta} & mathbf{2} boldsymbol{alpha} boldsymbol{beta} & boldsymbol{alpha}^{2}+boldsymbol{beta}^{2} end{array}right| text { and } ] ( boldsymbol{D}_{1}=left|begin{array}{ccc}mathbf{1} & mathbf{0} & mathbf{0} \ mathbf{0} & boldsymbol{alpha} & boldsymbol{beta} \ mathbf{0} & boldsymbol{beta} & boldsymbol{alpha}end{array}right|, ) then find the determinant of ( D_{2} ) such that ( D_{2}=frac{D}{D_{1}} ) | 12 |
| 412 | f ( boldsymbol{A}+boldsymbol{B}+boldsymbol{C}=boldsymbol{pi}, ) then ( left|begin{array}{ccc}tan (boldsymbol{A}+boldsymbol{B}+boldsymbol{C}) & tan boldsymbol{B} & tan C \ tan (boldsymbol{A}+boldsymbol{C}) & boldsymbol{0} & tan boldsymbol{A} \ tan (boldsymbol{A}+boldsymbol{B}) & -tan boldsymbol{A} & boldsymbol{0}end{array}right| ) equal to ( mathbf{A} cdot mathbf{0} ) B. ( c . ) tan AtanBtan D. -2 | 12 |
| 413 | With out expanding show that ( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2} \ mathbf{1} & boldsymbol{b} & boldsymbol{b}^{2} \ mathbf{1} & boldsymbol{c} & boldsymbol{c}^{2}end{array}right|=(boldsymbol{a}-boldsymbol{b})(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}-boldsymbol{a}) ) | 12 |
| 414 | If ( a, b, c ) are positive and are the ( p ) th ( , q t h ) and ( r t h ) terms, respectively, of a G.P. ( operatorname{then} Delta=left|begin{array}{lll}log a & p & 1 \ log b & q & 1 \ log c & r & 1end{array}right| ) is ( A ) B. ( log (a b c) ) c. ( -(p+q+r) ) D. none of these | 12 |
| 415 | Find area of triangle with vertices at the point given in each of the following ( (mathrm{i})(1,0),(6,0),(4,3) ) ( (i i)(2,7),(1,1),(10,8) ) ( (text { iii) }(-2,-3),(3,2),(-1,-8) ) | 12 |
| 416 | 23. Consider the system of linear equations ; x₂ + 2x₂ + x₃ = 3 2x + 3x₂+xz=3 3x, + 5×2 + 2×3 = 1 The system has (a) exactly 3 solutions (b) a unique solution (C) no solution (d) infinite number of solutions | 12 |
| 417 | begin{tabular}{|ccc} if ( boldsymbol{f}(boldsymbol{x})= ) & \ ( mathbf{1} ) & ( boldsymbol{x} ) & ( boldsymbol{x}+ ) \ ( boldsymbol{2} boldsymbol{x} ) & ( boldsymbol{x}(boldsymbol{x}-mathbf{1}) ) & ( (boldsymbol{x}+ ) \ ( boldsymbol{3} boldsymbol{x}(boldsymbol{x}-mathbf{1}) ) & ( boldsymbol{x}(boldsymbol{x}-mathbf{1})(boldsymbol{x}-mathbf{2}) ) & ( (boldsymbol{x}+mathbf{1}) boldsymbol{x} ) end{tabular} then ( f(100) ) is equal to ( mathbf{A} cdot mathbf{0} ) B. 1 ( c .100 ) D. – | 12 |
| 418 | Using the properties of determinants & without expanding ( left[begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{q}+boldsymbol{r} & boldsymbol{y}+boldsymbol{z} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{r}+boldsymbol{p} & boldsymbol{z}+boldsymbol{x} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{p}+boldsymbol{q} & boldsymbol{x}+boldsymbol{y}end{array}right]=boldsymbol{2}left[begin{array}{ccc}boldsymbol{a} & boldsymbol{p} & boldsymbol{x} \ boldsymbol{b} & boldsymbol{q} & boldsymbol{y} \ boldsymbol{c} & boldsymbol{r} & boldsymbol{z}end{array}right] ) | 12 |
| 419 | If each element in a row of a determinant is multiplied by the same factor ( r, ) then the value of the determinant: A . Is multiplied by ( r^{3} ) B. Is increased by ( 3 r ) c. Remains unchanged D. Is multiplied by ( r ) | 12 |
| 420 | 16. Let A be a 2 x 2 matrix with real entries. Let I be the 2 x 2 identity matrix. Denote by tr(A), the sum of diagonal entries of a. Assume that A2 = 1. [2008] Statement-1: IfA #I and A#-I, then det(A)=-1 Statement-2: If A #I and A#-I, then tr (A) +0. (a) Statement-1 is false, Statement-2 is true (b) Statement-1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false | 12 |
| 421 | STATEMENT 1: In a ( Delta A B C, a, b, c ) denotes lengths of the sides and ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right|=mathbf{0} ) then the triangle is equilateral triangle. STATEMENT 2: Sum of three non- negative numbers ( =0 Rightarrow ) each number is zero. A. Statement-1 is true, Statement-2 is true, Statement-2 is correct explanation of Statement- B. Statement-1 is true, Statement-2 is true, Statement-2 is not correct explanation for Statement- c. Statement-1 is true, Statement-2 is false D. Statement-1 is false, Statement-2 is true | 12 |
| 422 | ff ( x_{i}=a_{i} b_{i} c_{i}, i=1,2,3 ) are three-digit positive integers such that each ( x_{i} ) is a multiple of ( 19, ) then for some integer ( n ) prove that ( left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{a}_{2} & boldsymbol{a}_{3} \ boldsymbol{b}_{1} & boldsymbol{b}_{2} & boldsymbol{b}_{3} \ boldsymbol{c}_{1} & boldsymbol{c}_{2} & boldsymbol{c}_{3}end{array}right| ) is divisible by 19 | 12 |
| 423 | Evaluate the determinant: ( left|begin{array}{ccc}mathbf{8} & mathbf{2} & mathbf{7} \ mathbf{1 2} & mathbf{3} & mathbf{5} \ mathbf{1 6} & mathbf{4} & mathbf{3}end{array}right| ) | 12 |
| 424 | 2. Let a>0, d>0. Find the value of the determinant (1996 – 5 Marks) a(a +d) (a+d)(a +2d) (a + d) (a +d)(a +2d) (a +2d)(a +3d) (a + 2d) (a +2d)(a +3d) (a +3d)(a +40) | 12 |
| 425 | Evaluate : [ boldsymbol{Delta}=left|begin{array}{ccc} cos alpha cos beta & cos alpha sin beta & -sin alpha \ -sin beta & cos beta & 0 \ sin alpha cos beta & sin alpha sin beta & cos alpha end{array}right| ] | 12 |
| 426 | ( left|begin{array}{ccc}-mathbf{2} boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{a}+boldsymbol{c} \ boldsymbol{b}+boldsymbol{a} & -boldsymbol{2 b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{c}+boldsymbol{b} & -boldsymbol{2 c}end{array}right|=boldsymbol{4}(boldsymbol{b}+boldsymbol{c})(boldsymbol{c}+ ) ( boldsymbol{a})(boldsymbol{a}+boldsymbol{b}) ) | 12 |
| 427 | ( left|begin{array}{ccc}mathbf{8} & mathbf{- 5} & mathbf{1} \ mathbf{5} & boldsymbol{x} & mathbf{3} \ mathbf{6} & mathbf{3} & mathbf{1}end{array}right|=mathbf{2} ) then what is the value of ( x ) ? A . 44 B. 55 c. 61 D. 84 | 12 |
| 428 | ( boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{1} \ mathbf{2} & mathbf{1} & -mathbf{3} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right] ) and ( boldsymbol{B}= ) ( left[begin{array}{ccc}4 & 2 & 2 \ -5 & 0 & alpha \ 1 & -2 & 3end{array}right] ) If ( B ) is the adjoint of ( A ) then ( alpha ) equals A .2 в. ( c cdot-2 ) D. | 12 |
| 429 | If ( A ) is ( 3 x 3 ) matrix and ( operatorname{det} operatorname{adj}(A)=k ) then ( operatorname{det}(operatorname{adj} 2 A)= ) ( A cdot 2 k ) в. 8k ( c cdot 16 k ) D. ( 64 k^{2} ) | 12 |
| 430 | First row of the matrix ( boldsymbol{A} ) is ( left[begin{array}{lll}mathbf{1} & mathbf{3} & mathbf{2}end{array}right] . ) [ begin{array}{l} boldsymbol{a d j}(boldsymbol{A})= \ {left[begin{array}{ccc} -mathbf{2} & mathbf{4} & boldsymbol{a} \ -mathbf{1} & mathbf{2} & mathbf{1} \ mathbf{3} boldsymbol{a} & mathbf{- 5} & mathbf{- 2} end{array}right]} end{array} ] then a possible value of ( operatorname{det}(boldsymbol{A}) ) is A . 1 B . 2 ( c cdot-1 ) D. -2 | 12 |
| 431 | 33. The system of linear equations x+2y-z=0 ax-y-z=0 xty-dz=0 has a non-trivial solution for: (a) exactly two values of 2. (b) exactly three values of . (c) infinitely many values of 2. (d) exactly one value of 2. | 12 |
| 432 | f ( 2 s=a+b+c, ) prove that ( left|begin{array}{ccc}a^{2} & (s-a)^{2} & (s-a)^{2} \ (s-b)^{2} & b^{2} & (s-b)^{2} \ (s-c)^{2} & (s-c)^{2} & c^{2}end{array}right| ) ( =2 s^{3}(s-a)(s-b)(s-c) ) | 12 |
| 433 | If ( boldsymbol{A}=left[begin{array}{ccc}5 & 5 x & x \ 0 & x & 5 x \ 0 & 0 & 5end{array}right] ) and ( left|A^{2}right|=25 ) then ( |x| ) is equal to? A ( cdot frac{1}{5} ) B. 5 ( c cdot 5^{2} ) D. | 12 |
| 434 | Find the ratio in which the line segment joining the points ( P(x, 2) ) divides the line segment joining the points ( boldsymbol{A}(mathbf{1} mathbf{2}, mathbf{5}) ) and ( B(4,-3) . ) Also find the value of ( x ) | 12 |
| 435 | The value of the determinant ( left|begin{array}{ccc}5 & 5 & 14 \ ^{5} C_{1} & ^{5} C_{4} & 1 \ ^{5} C_{2} & ^{5} C_{5} & 1end{array}right| ) is ( mathbf{A} cdot mathbf{0} ) B. -576 c. 80 D. none of these | 12 |
| 436 | Find the values of the following determinants where ( i=sqrt{-1} ) ( (i)left|begin{array}{cc}2 i & -3 i \ i^{3} & -2 i^{5}end{array}right| ) ( (i i)left|begin{array}{cc}mathbf{1}+mathbf{3} i & boldsymbol{i}-mathbf{2} \ boldsymbol{i}+mathbf{2} & mathbf{1}-mathbf{3} iend{array}right| ) | 12 |
| 437 | f ( a, b, c epsilon R, ) find the number of real roots of the equation given by ( Delta=0 ) where ( boldsymbol{Delta}=left|begin{array}{ccc}boldsymbol{x} & boldsymbol{c} & boldsymbol{- b} \ -boldsymbol{c} & boldsymbol{x} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{-} boldsymbol{a} & boldsymbol{x}end{array}right| ) A . 0 B. ( c cdot 2 ) D. 3 | 12 |
| 438 | ( left|begin{array}{cc}boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{1} \ boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{2}end{array}right|=left|begin{array}{cc}mathbf{4} & -mathbf{1} \ mathbf{1} & mathbf{3}end{array}right|, ) then write the value of ( x ) | 12 |
| 439 | If the points ( (k, 2 k),(3 k, 3 k) ) and (3,1) are collinear then the value of ( k ) is A ( cdot frac{7}{9} ) B. ( frac{2}{3} ) c. ( frac{-2}{3} ) D. ( frac{-1}{3} ) | 12 |
| 440 | Let ( boldsymbol{Delta}_{1}=left|begin{array}{ccc}mathbf{1} & cos boldsymbol{alpha} & cos beta \ cos boldsymbol{alpha} & boldsymbol{1} & cos gamma \ cos beta & cos gamma & 1end{array}right| ) and ( Delta_{2}=left|begin{array}{ccc}0 & cos alpha & cos beta \ cos alpha & 0 & cos gamma \ cos beta & cos gamma & 0end{array}right| ) If ( Delta_{1}=Delta_{2}, ) find ( sin ^{2} alpha+sin ^{2} beta+sin ^{2} gamma ) | 12 |
| 441 | Solve this ( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{b}-boldsymbol{c} & boldsymbol{2} boldsymbol{a} & boldsymbol{2} boldsymbol{a} \ boldsymbol{2} boldsymbol{b} & boldsymbol{b}-boldsymbol{c}-boldsymbol{a} & boldsymbol{2} boldsymbol{b} \ boldsymbol{2} boldsymbol{c} & boldsymbol{2} boldsymbol{c} & boldsymbol{c}-boldsymbol{a}-boldsymbol{b}end{array}right| ) | 12 |
| 442 | Assertion The area of the triangle formed by the points ( boldsymbol{A}(mathbf{2 0 0 7}, mathbf{2 0 0 9}), boldsymbol{B}(mathbf{2 0 0 8}, mathbf{2 0 1 1}) ) ( C(2009,2010) ) will be same as the area formed by the points ( boldsymbol{P}(mathbf{0}, mathbf{0}) ) ( Q(1,2), R(2,1) ) Reason The area of the triangle remains same w.r.t to transition of co-ordinate axes. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 12 |
| 443 | ( operatorname{det}left{begin{array}{lll}18 & 40 & 89 \ 40 & 89 & 198 \ 89 & 198 & 440end{array}right}= ) A . -8 B. – ( c .-1 ) ( D ) | 12 |
| 444 | Solve: ( mid begin{array}{ccc}x^{2}-1 & x^{2}+2 x+1 & 2 x^{2}+3 x \ 2 x^{2}+x-1 & 2 x^{2}+5 x-3 & 4 x^{2}+4 x \ 6 x^{2}-x-2 & 6 x^{2}-7 x+2 & 12 x^{2}-5end{array} ) 0. Let sum of all values of ( x ) be ( k ). Find ( -2 k ) | 12 |
| 445 | Find the minor of ( left[begin{array}{ccc}2 & 7 & 3 \ -4 & 3 & -1 \ 0 & -3 & 7end{array}right] ) | 12 |
| 446 | If ( (8,1),(k,-4),(2,-5) ) are collinaer, then ( k= ) ( mathbf{A} cdot mathbf{1} ) B . 2 ( c .3 ) D. 4 | 12 |
| 447 | The value of ( left|begin{array}{ccc}a_{1} x+b_{1} y & a_{2} x+b_{2} y & a_{3} x+b_{3} y \ b_{1} x+a_{1} y & b_{2} x+a_{2} y & b_{3} x+a_{3} y \ b_{1} x+a_{1} & b_{2} x+a_{2} & b_{3} x+a_{3}end{array}right| ) is equal to A ( cdot x^{2}+y^{2} ) B. ( mathbf{c} cdot a_{1} a_{2} a_{3} x^{2}+b_{1} b_{2} b_{3} y^{2} ) D. none of these | 12 |
| 448 | Solve ( left|begin{array}{ccc}boldsymbol{a}^{2}+mathbf{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b} & boldsymbol{b}^{2}+mathbf{1} & boldsymbol{b} boldsymbol{c} \ boldsymbol{a} boldsymbol{c} & boldsymbol{b} boldsymbol{c} & boldsymbol{c}^{2}+1end{array}right|=boldsymbol{a}^{2}+boldsymbol{b}^{2}+ ) ( c^{2}+1 ) | 12 |
| 449 | If ( boldsymbol{A}=left(begin{array}{ccc}1 & 2 & 1 \ -1 & 0 & 3 \ 2 & -1 & 1end{array}right) ) then characteristic equation is given by A. ( -lambda^{3}+2 lambda^{2}-4 lambda+18=0 ) 0 B . ( lambda^{3}+2 lambda^{2}+4 lambda+18=0 ) c. ( 2 lambda^{3}-lambda^{2}+6 lambda-2=0 ) D. None of these | 12 |
| 450 | Write the value of the determinant ( left|begin{array}{cc}mathbf{3} & mathbf{- 1} \ mathbf{2} & mathbf{1}end{array}right| ) | 12 |
| 451 | If ( f(x)=left|begin{array}{ccc}a & -1 & 0 \ a x & a & -1 \ a x^{2} & a x & aend{array}right|, ) by using properties of determinants, find the value ( boldsymbol{f}(mathbf{2 x})-boldsymbol{f}(boldsymbol{x}) ) | 12 |
| 452 | If the determinant ( left|begin{array}{ccc}boldsymbol{b}-boldsymbol{c} & boldsymbol{c}-boldsymbol{a} & boldsymbol{a}-boldsymbol{b} \ boldsymbol{b}^{prime}-boldsymbol{c}^{prime} & boldsymbol{c}^{prime}-boldsymbol{a}^{prime} & boldsymbol{a}^{prime}-boldsymbol{b}^{prime} \ boldsymbol{b}^{prime prime}-boldsymbol{c}^{prime prime} & boldsymbol{c}^{prime prime}-boldsymbol{a}^{prime prime} & boldsymbol{a}^{prime prime}-boldsymbol{b}^{prime prime}end{array}right| ) is expressible as ( mleft|begin{array}{lll}a & b & c \ a^{prime} & b^{prime} & c^{prime} \ a^{prime prime} & b^{prime prime} & c^{prime prime}end{array}right|, ) then the value of ( mathrm{m} ) is A . – B. ( c ) ( D ) | 12 |
| 453 | Find ( left|begin{array}{lll}log e & log e^{2} & log e^{3} \ log e^{2} & log e^{3} & log e^{4} \ log e^{3} & log e^{4} & log e^{5}end{array}right| ) ( A ) B. ( c .4 log e ) ( mathbf{D} cdot 5 log e ) | 12 |
| 454 | If ( boldsymbol{A}=left[begin{array}{lll}boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{a}end{array}right], ) then the value of ( |boldsymbol{A}||boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}| ) is ( boldsymbol{a}^{boldsymbol{k}} ) What is k? | 12 |
| 455 | ( 19, 1U MOHON 17. The number of 3 x 3 matrices A whose entries are either 0 or 1 and for which the system A y = 0 has exactly two (2010) distinct solutions, is (a) 0 (b) 22–1 (c) 168 (d) 2 10 Lot 1 1 | 12 |
| 456 | ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{x}^{2} & boldsymbol{y}^{2} & boldsymbol{z}^{2} \ boldsymbol{x}^{boldsymbol{3}} & boldsymbol{y}^{boldsymbol{3}} & boldsymbol{z}^{boldsymbol{3}}end{array}right|=boldsymbol{x} boldsymbol{y} boldsymbol{z}(boldsymbol{x}-boldsymbol{y})(boldsymbol{y}-boldsymbol{z})(boldsymbol{z} ) ( boldsymbol{x}) ) A. True B. Falss | 12 |
| 457 | ff ( a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} in R ) and are such ( operatorname{that} a_{i} b_{j} neq 1 ) for ( 1 leq i, j leq 3, ) then ( begin{array}{|lll|}frac{1-a_{1}^{3} b_{1}^{3}}{1-a_{1} b_{1}} & frac{1-a_{1}^{3} b_{2}^{3}}{1-a_{1} b_{2}} & frac{1-a_{1}^{3} b_{3}^{3}}{1-a_{1} b_{3}} \ frac{1-a_{2}^{3} b_{1}^{3}}{1-a_{2} b_{1}} & frac{1-a_{2}^{3} b_{2}^{3}}{1-a_{2} b_{2}} & frac{1-a_{2}^{3} b_{3}^{3}}{1-a_{2} b_{3}} \ frac{1-a_{3}^{3} b_{1}^{3}}{1-a_{3} b_{1}} & frac{1-a_{3}^{3} b_{2}^{3}}{1-a_{3} b_{2}} & frac{1-a_{3}^{3} b_{3}^{3}}{1-a_{3} b_{3}}end{array} ) either ( a_{1}<a_{2}<a_{3} ) and ( b_{1}<b_{2}a_{2} a_{3} ) and ( boldsymbol{b}_{1}>boldsymbol{b}_{2}>boldsymbol{b}_{3} ) then show ( left(a_{1}-a_{2}right)left(a_{2}-a_{3}right)left(a_{3}-right. ) ( left.boldsymbol{a}_{1}right)left(boldsymbol{b}_{1}-boldsymbol{b}_{2}right)left(boldsymbol{b}_{2}-boldsymbol{b}_{3}right)left(boldsymbol{b}_{3}-boldsymbol{b}_{1}right)<mathbf{0} ) | 12 |
| 458 | Find the solution set of ( left|begin{array}{ccc}2+x & 2-x & 2-x \ 2-x & 2+x & 2-x \ 2-x & 2-x & 2+xend{array}right|=0 ) | 12 |
| 459 | ( mathrm{ff}=left[begin{array}{ccc}boldsymbol{a} & mathbf{0} & mathbf{0} \ {[mathbf{0 . 3 e m}] mathbf{0}} & boldsymbol{a} & mathbf{0} \ {[mathbf{0 . 3 e m}] mathbf{0}} & mathbf{0} & boldsymbol{a}end{array}right], ) then the value of IAdj. Al is equal to A ( cdot a^{3} ) в. ( a^{6} ) ( c cdot a^{9} ) ( mathbf{D} cdot a^{2} ) | 12 |
| 460 | ( A ) and ( B ) are two points and ( C ) is any point collinear with ( A ) and ( B . ) IF ( A B= ) ( mathbf{1 0}, boldsymbol{B C}=mathbf{5}, ) then ( boldsymbol{A} boldsymbol{C} ) is equal to: A. either 15 or 5 B. necessarily 5 c. necessarily 16 D. none of these | 12 |
| 461 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{1} & mathbf{2} \ mathbf{3} & mathbf{4}end{array}right], ) then ( boldsymbol{A}^{-mathbf{1}}= ) A ( cdot frac{-1}{2}left[begin{array}{cc}4 & -2 \ -3 & 1end{array}right] ) в. ( frac{1}{2}left[begin{array}{cc}4 & -2 \ -3 & 1end{array}right] ) c. ( left[begin{array}{cc}-2 & 4 \ 1 & 3end{array}right] ) D. ( left[begin{array}{ll}2 & 4 \ 1 & 3end{array}right] ) | 12 |
| 462 | If ( f(x) ) and ( g(x) ) are functions such that ( boldsymbol{f}(boldsymbol{x}+)=boldsymbol{f}(boldsymbol{x}) boldsymbol{g}(boldsymbol{y})+boldsymbol{g}(boldsymbol{x}) boldsymbol{f}(boldsymbol{y}), ) then the value of ( left|begin{array}{lll}boldsymbol{f}(boldsymbol{alpha}) & boldsymbol{g}(boldsymbol{alpha}) & boldsymbol{f}(boldsymbol{alpha}+boldsymbol{theta}) \ boldsymbol{f}(boldsymbol{beta}) & boldsymbol{g}(boldsymbol{beta}) & boldsymbol{f}(boldsymbol{beta}+boldsymbol{theta}) \ boldsymbol{f}(boldsymbol{gamma}) & boldsymbol{g}(boldsymbol{gamma}) & boldsymbol{f}(boldsymbol{gamma}+boldsymbol{theta})end{array}right| ) A ( cdot f(alpha) cdot f(beta) cdot f(gamma) ) B. ( mathbf{c} cdot g(alpha) cdot g(beta) cdot g(gamma) ) D. None of these | 12 |
| 463 | ( operatorname{Let} A=left|begin{array}{ll}mathbf{5} & mathbf{0} \ mathbf{1} & mathbf{0}end{array}right| ) and ( boldsymbol{B}=left|begin{array}{ll}mathbf{2 0} & mathbf{5} mid \ -mathbf{1} & mathbf{0}end{array}right| ) ( mathbf{4} boldsymbol{A}+mathbf{5} boldsymbol{B}-boldsymbol{C}=mathbf{0}, ) then ( boldsymbol{C} ) is ( mathbf{A} cdotleft|begin{array}{cc}5 & 25 \ -1 & 0end{array}right| ) ( mathbf{B} cdotleft|begin{array}{cc}120 & 25 \ -1 & 0end{array}right| ) ( mathbf{c} cdotleft|begin{array}{cc}5 & -1 \ 0 & 25end{array}right| ) ( mathbf{D} cdotleft|begin{array}{cc}5 & 25 \ -1 & 5end{array}right| ) | 12 |
| 464 | ( operatorname{Let} boldsymbol{A}=left(begin{array}{cc}boldsymbol{x}+mathbf{2} & mathbf{3} boldsymbol{x} \ boldsymbol{3} & boldsymbol{x}+mathbf{2}end{array}right), boldsymbol{B}= ) ( left(begin{array}{cc}boldsymbol{x} & mathbf{0} \ mathbf{5} & boldsymbol{x}+mathbf{2}end{array}right) . ) Then all solutions of the equation ( operatorname{det}(A B)=0 ) is A. 1,-1,0,2 B. 1,4,0,-2 c. 1,-1,4,3 D. -1,4,0,3 | 12 |
| 465 | The value of the determinant ( mid begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \ 2 a b & 1-a^{2}+b^{2} & 2 a \ 2 b & -2 a & 1+a^{2}-b^{2}end{array} ) is equal to ( A ) B ( cdotleft(1+a^{2}+b^{2}right) ) c. ( left(1+a^{2}+b^{2}right)^{2} ) D. ( left(1+a^{2}+b^{2}right)^{3} ) | 12 |
| 466 | ( mathbf{f} A=left[begin{array}{cc}frac{-mathbf{1}+i sqrt{mathbf{3}}}{mathbf{2} i} & frac{-mathbf{1}-i sqrt{mathbf{3}}}{mathbf{2} i} \ frac{mathbf{1}+i sqrt{mathbf{3}}}{mathbf{2} i} & frac{mathbf{1}-i sqrt{mathbf{3}}}{mathbf{2} i}end{array}right], i= ) ( sqrt{-i} ) and ( f(x)=x^{2}+2, ) then ( f(A) ) is equal to ( ^{A} cdotleft(frac{5-i sqrt{3}}{2}right)left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) ( ^{text {В }} cdotleft(frac{3-i sqrt{3}}{2}right)left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) c. ( left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) D. ( (2+i sqrt{3})left[begin{array}{ll}1 & 0 \ 0 & 1end{array}right] ) | 12 |
| 467 | ( left|begin{array}{lll}a^{2}+2 a & 2 a+1 & 1 \ 2 a+1 & a+2 & 1 \ 3 & 3 & 1end{array}right|= ) ( A cdot(1-a)^{3} ) В – ( (a-1)^{2} ) c. ( (a-1)^{3} ) D・ ( (a+1)^{2} ) | 12 |
| 468 | Let ( A ) be a square matrix of order ( 3 times 3 ) then ( |k A| ) is equal to ( mathbf{A} cdot k mid A ) в. ( k^{2}|A| ) c. ( k^{3}|A| ) D. ( 3 k|A| ) | 12 |
| 469 | Evaluate the following ( begin{array}{|ccc|}15 & 11 & 7 \ 11 & 17 & 14 \ 10 & 16 & 13end{array} ) | 12 |
| 470 | Evaluate the following determinant: ( left|begin{array}{ccc}1 & 4 & 9 \ 4 & 9 & 16 \ 9 & 16 & 25end{array}right| ) | 12 |
| 471 | If ( omega ) is an imaginary cube root of unity,then the value of ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} boldsymbol{omega}^{2} & boldsymbol{a} boldsymbol{omega} \ boldsymbol{b} boldsymbol{c} & boldsymbol{c} & boldsymbol{b} boldsymbol{omega}^{2} \ boldsymbol{c} boldsymbol{omega}^{2} & boldsymbol{a} boldsymbol{omega} & boldsymbol{c}end{array}right|, ) is ( mathbf{A} cdot a^{3}+b^{3}+c^{3} ) B . ( a^{2} b-b^{2} c ) ( c cdot 0 ) D. ( a^{3} b+b^{3}+3 a b c ) | 12 |
| 472 | ( left|begin{array}{ccc}text { If } & \ cos (boldsymbol{A}+boldsymbol{B}) & -sin (boldsymbol{A}+boldsymbol{B}) & cos 2 boldsymbol{B} \ sin boldsymbol{operatorname { s o s }} boldsymbol{operatorname { s i n }} boldsymbol{A} & boldsymbol{operatorname { s i n } boldsymbol { operatorname { m o s } }}end{array}right| ) ( =0 ) then ( B= ) ( A cdot(2 n+1) frac{pi}{2} ) B. ( n pi ) ( c cdot(2 n+1) pi ) D. 2 nn | 12 |
| 473 | If the lines ( boldsymbol{p}_{1} boldsymbol{x}+boldsymbol{q}_{1} boldsymbol{y}=mathbf{1}, boldsymbol{p}_{2} boldsymbol{x}+boldsymbol{q}_{2} boldsymbol{y}= ) 1 and ( p_{3} x+q_{3} y=1 ) be concurrent show that the points ( left(p_{1}, q_{1}right),left(p_{2}, q_{2}right) ) and ( left(p_{3}, q_{3}right) ) are collinear A. vertices of right angle triangle B. vertices of an equilateral triangle c. vertices of an isosceles triangle D. Collinear | 12 |
| 474 | If points ( (a, 0),(0, b) ) and ( (x, y) ) are collinear, prove that ( frac{x}{a}+frac{y}{b}=1 ) | 12 |
| 475 | If ( left|begin{array}{ccc}2 a & x_{1} & y_{1} \ 2 b & x_{2} & y_{2} \ 2 c & x_{3} & y_{3}end{array}right|=frac{a b c}{2} neq 0, ) then the area of the triangle whose vertices are ( left(frac{x_{1}}{a}, frac{y_{1}}{a}right),left(frac{x_{2}}{b}, frac{y_{2}}{b}right) ) and ( left(frac{x_{3}}{c}, frac{y_{3}}{c}right) ) is A ( cdot frac{1}{4} a b c ) B. ( frac{1}{8} a b c ) ( c cdot frac{1}{4} ) D. ( frac{1}{8} ) E ( cdot frac{1}{12} ) | 12 |
| 476 | If ( omega ) is cube root of unity, then ( boldsymbol{Delta}=left|begin{array}{cccc}boldsymbol{x}+mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{x}+boldsymbol{omega}^{2} & mathbf{1} \ boldsymbol{omega}^{2} & boldsymbol{1} & boldsymbol{x}+boldsymbol{omega}end{array}right|= ) ( mathbf{A} cdot x^{3}+1 ) B. ( x^{3}+omega ) c. ( x^{3}+omega^{2} ) D. ( x^{3} ) | 12 |
| 477 | If ( boldsymbol{A}=left|begin{array}{ll}mathbf{0} & mathbf{0} \ mathbf{1} & mathbf{1}end{array}right| ) then the value of ( boldsymbol{A}+ ) ( boldsymbol{A}^{2}+boldsymbol{A}^{3}+ldots+boldsymbol{A}^{n}=? ) ( A cdot A ) B. nA c. ( (n+1) A ) ( D ) | 12 |
| 478 | Find the values of ( x, ) if ( left|begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5}end{array}right|=left|begin{array}{ll}boldsymbol{x} & mathbf{3} \ mathbf{2} boldsymbol{x} & mathbf{5}end{array}right| ) | 12 |
| 479 | 76. Consider the system of linear equations in x, y, and z: (sin 30) x – y + z = 0 (cos 20) x + 4y + 3z = 0 2 x + 7y + z = 0 Which of the following can be the values of O for which the system has a non-trivial solution? a. nit+ (-1)” īc/6, Vnez b. nn + (-1)” 7/3, ne z c. nit+ (-1)” Tt/9, ne z d. none of these | 12 |
| 480 | Consider the system of linear equations in x, y, z: (sin 30) x-y+z=0 (cos 20) x+4y + 3z=0 2x+7y+z=0 | 12 |
| 481 | f maximum and minimum values of ( boldsymbol{D}=left|begin{array}{ccc}mathbf{1} & -cos boldsymbol{theta} & mathbf{1} \ cos boldsymbol{theta} & mathbf{1} & -cos boldsymbol{theta} \ mathbf{1} & cos boldsymbol{theta} & mathbf{1}end{array}right| ) are ( p ) and respectively, then the value of ( 2 p+3 q ) is A . 16 B. 6 ( c .14 ) ( D ) | 12 |
| 482 | ( A ) is a ( 3 times 3 ) matrix and ( B ) is its adjoint matrix. If the determinant of ( B ) is 64 then the det ( A ) is ( A cdot 4 ) B. ±4 ( c .pm 8 ) D. 8 | 12 |
| 483 | ( mathrm{f}left|begin{array}{ccc}boldsymbol{x}^{2}+boldsymbol{x} & boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{2} \ mathbf{2} boldsymbol{x}^{2}+mathbf{3} boldsymbol{x}-mathbf{1} & mathbf{3} boldsymbol{x} & mathbf{3} boldsymbol{x}-mathbf{3} \ boldsymbol{x}^{mathbf{2}}+mathbf{2} boldsymbol{x}+mathbf{3} & mathbf{2} boldsymbol{x}-mathbf{1} & mathbf{2} boldsymbol{x}-mathbf{1}end{array}right|= ) ( A x-12, ) then the value of ( A ) is ( A cdot 12 ) в. 24 ( c .-12 ) ( D .-24 ) | 12 |
| 484 | Let ( boldsymbol{P}=left[boldsymbol{a}_{i j}right] ) be a ( boldsymbol{3} times boldsymbol{3} ) matrix and let ( Q=left(b_{i j}right) ) where ( b_{i j}=2^{i+j} a_{i j} ) for ( 1 leq ) ( i, j leq 3 . ) If the determinant of ( P ) is 2 then the determinant of the matrix ( Q ) is? A ( cdot 2^{10} ) B . ( 2^{11} ) ( c cdot 2^{12} ) D. ( 2^{13} ) | 12 |
| 485 | A point ( boldsymbol{P}(mathbf{2},-mathbf{1}) ) is equidistant from points ( (a, 7) ) and ( (-3, a) ). Find ( a ) | 12 |
| 486 | The value of ( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{b}-boldsymbol{c} & boldsymbol{2} boldsymbol{a} & boldsymbol{2} boldsymbol{a} \ boldsymbol{2} boldsymbol{b} & boldsymbol{b}-boldsymbol{c}-boldsymbol{a} & boldsymbol{2} boldsymbol{b} \ boldsymbol{2} boldsymbol{c} & boldsymbol{2} boldsymbol{c} & boldsymbol{c}-boldsymbol{a}-boldsymbol{b}end{array}right| ) will n ( mathbf{2 C} ) A ( cdot(a+b+c)^{2} ) B ( cdot(a+b+c)^{3} ) c. ( (a-b-c)^{2} ) D ( cdot(a+b-c)^{2} ) | 12 |
| 487 | ( mathbf{1 f} mathbf{A}=left[begin{array}{ccc}mathbf{1} & mathbf{5} & mathbf{- 6} \ mathbf{- 8} & mathbf{0} & mathbf{4} \ mathbf{3} & mathbf{- 7} & mathbf{2}end{array}right] ) then the cofactors of the elements 3,-7,2 are p,q,r respectively their ascending order is ( mathbf{A} cdot mathbf{p}, mathbf{r}, mathbf{q} ) B. ( mathrm{q}, mathrm{r}, mathrm{p} ) ( c cdot p, q, r ) ( D cdot r, p, q ) | 12 |
| 488 | Find the value of ( k ) if ( boldsymbol{A}(mathbf{4}, mathbf{1 1}), boldsymbol{B}(mathbf{2}, mathbf{5}), boldsymbol{C}(boldsymbol{6}, boldsymbol{k}) ) are collinear points. | 12 |
| 489 | f ( x ) is a positive integer, then ( left|begin{array}{ccc}boldsymbol{x} ! & (boldsymbol{x}+mathbf{1}) ! & (boldsymbol{x}+mathbf{2}) ! \ (boldsymbol{x}+mathbf{1}) ! & (boldsymbol{x}+mathbf{2}) ! & (boldsymbol{x}+mathbf{3}) ! \ (boldsymbol{x}+mathbf{2}) ! & (boldsymbol{x}+mathbf{3}) ! & (boldsymbol{x}+mathbf{4}) !end{array}right| ) is equa to A ( cdot 2 x !(x+1) ! ) в. ( 2 x !(x+1) !(x+2) ) c. ( 2 x !(x+3) ! ) D. ( 2(x+1) !(x+2) !(x+3) ! ) | 12 |
| 490 | ( mathbf{f} mathbf{Delta}=left|begin{array}{lll}boldsymbol{a} & mathbf{0} & mathbf{0} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right| ) then ( left|begin{array}{lll}boldsymbol{p}^{2} boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{p} boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{p c} & boldsymbol{a} & boldsymbol{b}end{array}right| ) is equal to A ( . p Delta ) B . ( p^{2} Delta ) c. ( p^{3} Delta ) D. ( 2 p Delta ) | 12 |
| 491 | Prove the following: ( left|begin{array}{ccc}boldsymbol{b} boldsymbol{c} & boldsymbol{b} boldsymbol{c}^{prime}+boldsymbol{b}^{prime} boldsymbol{c} & boldsymbol{b}^{prime} boldsymbol{c}^{prime} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c} boldsymbol{a}^{prime}+boldsymbol{c}^{prime} boldsymbol{a} & boldsymbol{c}^{prime} boldsymbol{a}^{prime} \ boldsymbol{a b} & boldsymbol{a b}^{prime}+boldsymbol{d}^{prime} boldsymbol{b} & boldsymbol{d}^{prime} boldsymbol{b}^{prime}end{array}right|= ) [ left(b c^{prime}-b^{prime} cright)left(c a^{prime}-c^{prime} aright)left(a b^{prime}-d bright) ] | 12 |
| 492 | ( boldsymbol{B}_{1}+boldsymbol{B}_{2}+ldots ldots+boldsymbol{B}_{49} ) is equal to A. ( B_{0} ) В. ( 7 B_{0} ) ( mathbf{c} cdot 49 B_{0} ) D. 49 | 12 |
| 493 | ( left|begin{array}{ccc}mathbf{2}^{3} & mathbf{3}^{3} & mathbf{3 . 2}^{mathbf{2}}+mathbf{3 . 2}+mathbf{1} \ mathbf{3}^{mathbf{3}} & mathbf{4}^{mathbf{3}} & mathbf{3 . 3}^{mathbf{2}}+mathbf{3 . 3}+mathbf{1} \ mathbf{4}^{mathbf{3}} & mathbf{5}^{mathbf{3}} & mathbf{3 . 4}^{mathbf{2}}+mathbf{3 . 4}+mathbf{1}end{array}right| ) is equal to ( A ) B. ( c cdot 2 ) D. 3 | 12 |
| 494 | Suppose ( x, y, z ) are positive integers ( (neq ) 1) if ( Delta=operatorname{det} ) of ( left[begin{array}{ccc}1 & log _{x} y & log _{x} z \ log _{y} x & 1 & log _{y} z \ sin (x+y) & -cos (x+y) & sin ^{2} zend{array}right] ) then ( Delta ) is 1 ) Independent of ( x ) II) Independent of ( y ) IIII Independent of ( z ) The which of the above statement is are correct A. only land II D. all the three I,II,III | 12 |
| 495 | ( operatorname{et} Delta_{1}=left|begin{array}{ccc}x & y & x+y \ y & x+y & x \ x+y & x & yend{array}right| ) ( boldsymbol{Delta}_{2}=left|begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{y} \ mathbf{1} & boldsymbol{x}+boldsymbol{y} & boldsymbol{y} \ mathbf{1} & boldsymbol{x} & boldsymbol{x}+boldsymbol{y}end{array}right| ) ( boldsymbol{Delta}_{3}=left|begin{array}{ccc}boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{2} boldsymbol{y} \ -boldsymbol{x} & boldsymbol{x} & boldsymbol{0} \ boldsymbol{0} & -boldsymbol{x} & boldsymbol{x}end{array}right| ) ( boldsymbol{Delta}_{4}=left|begin{array}{lll}1 & boldsymbol{x} & boldsymbol{x}^{2} \ mathbf{1} & boldsymbol{y} & boldsymbol{y}^{2} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right| ) then | 12 |
| 496 | Find the maximum value of ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+sin boldsymbol{theta} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+mathbf{c} mathbf{o} mathbf{s} boldsymbol{theta}end{array}right| ) | 12 |
| 497 | The minors and cofactors of -4 and 9 in determinant ( left|begin{array}{ccc}-1 & -2 & 3 \ -4 & -5 & -6 \ -7 & 8 & 9end{array}right| ) are respectively A. 42,( 42 ; 3,3 ) В. -42,( 42 ;-3,-3 ) c. 42,( -42 ; 3,-3 ) D. 42,3: 42,3 | 12 |
| 498 | The straight lines ( mathbf{x}+mathbf{2 y}-mathbf{9}=mathbf{0}, mathbf{3 x}+ ) ( 5 y-5=0 ) and ( a x+b y-1=0 ) are concurrent if the straight line ( 22 x- ) ( 35 y-1=0 ) passes through the point ( A cdot(a, b) ) в. ( (b, a) ) c. ( (-a, b) ) D. (-a,-b) | 12 |
| 499 | The value of the determinant ( Delta= ) A ( a_{1} a_{2}+b_{1} b_{2} ) a В ( cdotleft(a_{1} a_{2} a_{3}right)+left(b_{1} b_{2} b_{3}right) ) c. ( a_{1} a_{2} b_{1} b_{2}+a_{2} a_{3} b_{2} b_{3}+a_{3} a_{1} b_{3} b_{1} ) D. none of these | 12 |
| 500 | If ( boldsymbol{A}=left[begin{array}{cc}boldsymbol{a} & boldsymbol{p} \ boldsymbol{b} & boldsymbol{q} \ boldsymbol{c} & boldsymbol{r}end{array}right]_{mathbf{3} times mathbf{2}} ) then determinant ( left(A A^{T}right) ) is equal to ( mathbf{A} cdot mathbf{0} ) B. ( a^{2}+b^{2}+c^{2} ) c. ( p^{2}+q^{2}+r^{2} ) D. ( p^{2}+q^{2} ) | 12 |
| 501 | Evaluate ( left|begin{array}{ccc}cos alpha cos beta & cos alpha sin beta & -sin alpha \ -sin beta & cos beta & 0 \ sin alpha cos beta & sin alpha sin beta & cos alphaend{array}right| ) | 12 |
| 502 | Prove the following identities ( left|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{a}^{2} \ boldsymbol{a}^{2} & mathbf{1} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a}^{2} & mathbf{1}end{array}right|=left(boldsymbol{a}^{3}-mathbf{1}right)^{2} ) | 12 |
| 503 | If ( boldsymbol{a}=sin boldsymbol{theta}, boldsymbol{b}=sin (boldsymbol{theta}+boldsymbol{2} boldsymbol{pi} / mathbf{3}), boldsymbol{c}= ) ( sin (theta+4 pi / 3), x=cos theta, y= ) ( cos (theta+2 pi / 3), z=cos (theta+4 pi / 3) ) then value of ( boldsymbol{Delta}=left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{b} boldsymbol{c} & boldsymbol{c} boldsymbol{a} & boldsymbol{a} boldsymbol{b}end{array}right| ) is A . 1 ( overline{8} ) B. ( frac{3 sqrt{3}}{4} ) ( c cdot frac{3 sqrt{3}}{8} ) D. ( frac{sqrt{3}}{4} ) | 12 |
| 504 | The value of the determinant ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ m_{mathbf{C}} & m+mathbf{C}_{1} & m+mathbf{2} \ m & mathbf{C}_{1} \ m & boldsymbol{m}+mathbf{1} & boldsymbol{C}_{2}end{array}right| mathbf{~ i s ~ e q u a l ~ t o ~} ) ( mathbf{A} cdot mathbf{1} ) B. – ( c cdot 0 ) D. None of these | 12 |
| 505 | ( left|begin{array}{ccc}x^{2}+x & x+1 & x+2 \ x^{2}+3 x-1 & 3 x & 3 x-3 \ x^{2}+2 x+3 & 2 x-1 & 2 x-1end{array}right|= ) ( A x+B ) where ( A ) and ( B ) are determinants of order ( 3 . ) Then ( A+2 B ) is equal to | 12 |
| 506 | Find the values of ( x, ) if ( left|begin{array}{ll}mathbf{2} & mathbf{3} \ mathbf{4} & mathbf{5}end{array}right|=left|begin{array}{ll}boldsymbol{x} & mathbf{3} \ mathbf{2} boldsymbol{x} & mathbf{5}end{array}right| ) | 12 |
| 507 | If ( A ) is a ( 3 times 3 ) matrix and ( operatorname{det}(3 A)= ) ( boldsymbol{k}{boldsymbol{d e t}(boldsymbol{A})}, boldsymbol{k}= ) ( mathbf{A} cdot mathbf{9} ) B. 6 c. 1 D. 27 | 12 |
| 508 | Which of the following are correct in respect of the system of equations ( boldsymbol{x}+ ) ( boldsymbol{y}+boldsymbol{z}=mathbf{8}, boldsymbol{x}-boldsymbol{y}+mathbf{2} boldsymbol{z}=boldsymbol{6} ) and ( boldsymbol{3} boldsymbol{x}- ) ( boldsymbol{y}+mathbf{5} boldsymbol{z}=boldsymbol{k} ? ) 1. They have no solution, if ( k=15 ) 2. They have infinitely many solutions, if ( k=20 ) 3. They have unique solution, if ( k=25 ) Select the correct answer using the code given below ( A cdot 1 ) and 2 only B. 2 and 3 only C. 1 and 3 only D. 1,2 and 3 | 12 |
| 509 | The graph of ( f(x) ) is shown above in the ( x y ) -plane. The points ( (0,3),(5 b, b) ) and ( (10 b,-b) ) are on the line described by ( f(x) . ) If ( b ) is a positive constant, find the coordinates of point ( C ) ( mathbf{A} cdot(5,1) ) B. (10,-1) c. (15,-0.5) (15, ( -0.5) ) D. (20,-2) | 12 |
| 510 | Applying ( boldsymbol{C}_{1} rightarrow boldsymbol{C}_{1}-^{8} boldsymbol{C}_{3}, ) we getWithout expanding the determinant, prove that ( left|begin{array}{ccc}mathbf{4 1} & mathbf{1} & mathbf{5} \ mathbf{7 9} & mathbf{7} & mathbf{9} \ mathbf{2 9} & mathbf{5} & mathbf{3}end{array}right|=mathbf{0} ) | 12 |
| 511 | If ( A ) is a singular matrix, then adj ( A ) is A. non- singular B. singular c. symmetric D. not defined | 12 |
| 512 | ( left|begin{array}{cc}mathbf{2} & -mathbf{4} \ mathbf{9} & boldsymbol{d}-mathbf{3}end{array}right|=mathbf{4} ) then ( boldsymbol{d}= ) A . 1 B. -11 c. 12 D. -13 | 12 |
| 513 | s I a-1 n 6 Let Aa=/(a 1)2 2n² 4n-2 (a – 1)3 3n² 3n² – 3n Show that Aa = c , a constant. a = 1 | 12 |
| 514 | . If ( A, B, C ) are angles of angle and ( mid begin{array}{ccc}mathbf{1} & mathbf{1} \ mathbf{1}+boldsymbol{s} mathbf{i} boldsymbol{n} boldsymbol{A} & mathbf{1}+boldsymbol{operatorname { s i n }} boldsymbol{B} & mathbf{1}+ \ boldsymbol{s i n} boldsymbol{A}+boldsymbol{s i n}^{2} boldsymbol{A} & boldsymbol{s i n} boldsymbol{B}+boldsymbol{s i n}^{2} boldsymbol{B} & boldsymbol{s i n} boldsymbol{C}end{array} ) ( =0 ) then triangle is isosceles II. If ( boldsymbol{a}=mathbf{1}+mathbf{2}+mathbf{4}+— ) upto ( mathbf{n} ) terms ( boldsymbol{b}=mathbf{1}+mathbf{3}+mathbf{9}+— ) up to ( mathbf{n} ) terms ( c=1+5+25+—- ) up to ( n ) terms then ( Deltaleft|begin{array}{ccc}a & 2 b & 4 c \ 2 & 2 & 2 \ 2^{n} & 3^{n} & 5^{n}end{array}right|=0 ) A. I, II both are true B. only lis true c. only I I is true D. neither of them are true | 12 |
| 515 | f ( P(x, y) ) is such that ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{y} & mathbf{1} \ boldsymbol{x}_{1} & boldsymbol{y}_{1} & mathbf{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & boldsymbol{1}end{array}right|+left|begin{array}{ccc}boldsymbol{x} & boldsymbol{y} & mathbf{1} \ boldsymbol{x}_{1} & boldsymbol{y}_{1} & boldsymbol{1} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & 1end{array}right|=mathbf{0} ) then the line through ( A ) and ( P ) is A. median of ( Delta A B C ) B. bisector of ( angle A ) c. altitude through vertex ( A ) D. perpendicular bisector of the side ( B C ) | 12 |
| 516 | Solve for ( boldsymbol{x} ) ( left|begin{array}{ccc}boldsymbol{x}-mathbf{2} & mathbf{2} boldsymbol{x}-mathbf{3} & mathbf{3} boldsymbol{x}-mathbf{4} \ boldsymbol{x}-mathbf{4} & mathbf{2} boldsymbol{x}-mathbf{9} & mathbf{3} boldsymbol{x}-mathbf{1} mathbf{6} \ boldsymbol{x}-mathbf{8} & mathbf{2} boldsymbol{x}-mathbf{2 7} & mathbf{3} boldsymbol{x}-mathbf{6 4}end{array}right|=mathbf{0} ) ( A cdot frac{7}{4} ) B. ( frac{28}{3} ) ( c cdot frac{28}{13} ) D. ( frac{14}{0} ) | 12 |
| 517 | Prove that: ( left|begin{array}{ccc}-a^{2} & a b & a c \ a b & -b^{2} & b c \ a c & b c & -c^{2}end{array}right|=4 a^{2} b^{2} c^{2} ) | 12 |
| 518 | 11. For all values of A, B, C and P, Q, R show that (1994 – 4 Marks) cos(A-P) cos(A-2) cos(A – R) cos(B-P) cos(B-Q) cos(B – R) = 0 cos(C-P) cos(C-0) cos(C-R) | 12 |
| 519 | If the points ( (k, 2-2 k)(1-k, 2 k) ) and ( (-k ) ( -4,6-2 x) ) be collinear the possible values of k are A ( cdot frac{1}{2} ) B. ( frac{1}{2} ) ( c ) D. – – | 12 |
| 520 | Find the value of ( x ) if ( left|begin{array}{ccc}boldsymbol{x}-mathbf{2} & mathbf{2} boldsymbol{x}-mathbf{3} & mathbf{3} boldsymbol{x}-mathbf{4} \ boldsymbol{x}-mathbf{4} & mathbf{2} boldsymbol{x}-mathbf{9} & mathbf{3} boldsymbol{x}-mathbf{1 6} \ boldsymbol{x}-mathbf{8} & mathbf{2} boldsymbol{x}-mathbf{2 7} & mathbf{3} boldsymbol{x}-mathbf{6 4}end{array}right|=mathbf{0} ? ) | 12 |
| 521 | Show that ( triangle A B C ) is an isosceles triangle, if the determinant ( mid begin{array}{ccc}mathbf{1} & mathbf{1} \ mathbf{1}+cos boldsymbol{A} & mathbf{1}+cos boldsymbol{B} & mathbf{1} \ cos ^{2} boldsymbol{A}+cos boldsymbol{A} & cos ^{2} boldsymbol{B}+cos boldsymbol{B} & mathbf{c o s}^{2}end{array} ) ( mathbf{D} ) | 12 |
| 522 | Let A=(12) – ( a and B=1 0 , (3 4 10 b) 0,6 EN. Then [2006] (a) there cannot exist any B such that AB = BA (b) there exist more then one but finite number of B’s such that AB=BA (c) there exists exactly one B such that AB=BA (d) there exist infinitely many B’s such that AB=BA | 12 |
| 523 | ( mid begin{array}{cccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} & boldsymbol{d} \ boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{a}+boldsymbol{b}+boldsymbol{c} & boldsymbol{a}+boldsymbol{b}+boldsymbol{c} \ boldsymbol{a} & boldsymbol{2} boldsymbol{a}+boldsymbol{b} & boldsymbol{3} boldsymbol{a}+boldsymbol{2} boldsymbol{b}+boldsymbol{c} & boldsymbol{4} boldsymbol{a}+boldsymbol{3} boldsymbol{b}+boldsymbol{2} \ boldsymbol{a} & boldsymbol{3} boldsymbol{a}+boldsymbol{b} & boldsymbol{6} boldsymbol{a}+boldsymbol{3} boldsymbol{b}+boldsymbol{c} & boldsymbol{1} boldsymbol{0} boldsymbol{a}+boldsymbol{6} boldsymbol{b}+boldsymbol{3}end{array} ) | 12 |
| 524 | 17. Let x e R and let ( 11 17 [2 x x P=10 2 2 8 = 0 4 ol and R= PQP-1 10 0 3 X x 6] Then which of the following options is/are correct? (JEE Adv. 2019) [2 x x (a) det R=det 0 4 0 +8, for all x ER [x x 5 (b) For x = 1, there exists a unit vector ai+Bj+ył for To There exists a real number x such that PQ=QP (C) (d) For x = 0, if = a [b] a . then a+b=5 [b] | 12 |
| 525 | Find the values of ( x, ) if ( left|begin{array}{cc}mathbf{2 x} & mathbf{5} \ mathbf{8} & boldsymbol{x}end{array}right|=left|begin{array}{ll}mathbf{6} & mathbf{5} \ mathbf{8} & mathbf{3}end{array}right| ) | 12 |
| 526 | If in the determinant ( Delta= ) ( left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{b}_{1} & boldsymbol{c}_{1} \ boldsymbol{a}_{2} & boldsymbol{b}_{2} & boldsymbol{c}_{2} \ boldsymbol{a}_{3} & boldsymbol{b}_{3} & boldsymbol{c}_{3}end{array}right|, boldsymbol{A}_{i}, boldsymbol{B}_{i}, boldsymbol{C}_{i} ) etc. be the co- factors of ( a_{i}, b_{i}, c_{i} ) etc., then which of the following relations is incorrect? A ( cdot a_{1} A_{1}+b_{1} B_{1}+c_{1} C_{1}=Delta ) B . ( a_{2} A_{2}+b_{2} B_{2}+c_{2} C_{2}=Delta ) c. ( a_{3} A_{3}+b_{3} B_{3}+c_{3} C_{3}=Delta ) D. ( a_{1} A_{2}+b_{1} B_{2}+c_{1} C_{2}=Delta ) | 12 |
| 527 | f ( a neq b neq c ) such that ( left|begin{array}{ccc}boldsymbol{a}^{3}-mathbf{1} & boldsymbol{b}^{mathbf{3}}-mathbf{1} & boldsymbol{c}^{mathbf{3}}-mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{mathbf{2}} & boldsymbol{b}^{mathbf{2}} & boldsymbol{c}^{mathbf{2}}end{array}right|=mathbf{0} ) ther A ( . a b+b c+c a=0 ) B . ( a+b+c=0 ) ( c cdot a b c=1 ) D. ( a+b+c=1 ) | 12 |
| 528 | Find the values of ( x, ) if ( left|begin{array}{ll}boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{1} \ boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{2}end{array}right|=left|begin{array}{cc}mathbf{4} & mathbf{- 1} \ mathbf{1} & mathbf{3}end{array}right| ) | 12 |
| 529 | ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r} \ boldsymbol{p} & boldsymbol{q} & boldsymbol{r}+mathbf{1}end{array}right| ) is equal to A ( cdot q-p ) B. ( q+p ) ( c cdot q ) ( D ) | 12 |
| 530 | Find the values of ( x ), if ( left|begin{array}{cc}boldsymbol{x}+mathbf{1} & boldsymbol{x}-mathbf{1} \ boldsymbol{x}-mathbf{3} & boldsymbol{x}+mathbf{2}end{array}right|=left|begin{array}{cc}mathbf{4} & mathbf{- 1} \ mathbf{1} & mathbf{3}end{array}right| ) | 12 |
| 531 | If ( boldsymbol{A}=left[begin{array}{lll}boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{a}end{array}right], ) then the value of ( |boldsymbol{A}||mathbf{A} mathbf{d} mathbf{j} boldsymbol{A}| ) A ( cdot a^{3} ) в. ( a^{6} ) ( c cdot a^{9} ) ( D cdot a^{2} ) | 12 |
| 532 | Find the values of the following determinant, where ( boldsymbol{i}=sqrt{-1} ) ( left|begin{array}{cc}mathbf{2} boldsymbol{i} & -mathbf{3} i \ boldsymbol{i}^{mathbf{3}} & -mathbf{2} boldsymbol{i}^{5}end{array}right| ) | 12 |
| 533 | 23. Let @ =- 23. Let o – + , then the value of the det . , then the value of the det. 1 -1-0- (2002 – 2 Marks (a) 30 © 36? (b) 30(0-1) (d) 30(1-w) | 12 |
| 534 | The points ( (-a,-b),(0,0),(a, b) ) and ( left(a^{2},right. ) ab) are A. collinear B. vertices of a rectangle c. vertices of a parallelogram D. None of these | 12 |
| 535 | The value of ( begin{array}{|ccc|}1 & cos alpha-sin alpha & cos alpha+sin alpha \ 1 & cos beta-sin beta & cos beta+sin beta \ 1 & cos gamma-sin gamma & cos gamma+sin gammaend{array} mid ) A. ( left|begin{array}{ccc}1 & cos alpha & sin alpha \ 1 & cos beta & sin beta \ 1 & cos gamma & sin gammaend{array}right| ) B. ( left|begin{array}{ccc}1 & cos alpha & sin alpha \ 1 & cos beta & sin beta \ 1 & cos gamma & sin gammaend{array}right| ) ( mathbf{c} cdotleft|begin{array}{lll}cos alpha & cos beta & 1 \ cos beta & cos gamma & 1 \ cos gamma & cos alpha & 1end{array}right| ) D. ( left|begin{array}{ccc}1 & cos alpha & sin alpha \ 1 & cos beta & sin beta \ 1 & cos gamma & sin gammaend{array}right| ) | 12 |
| 536 | Match the entries of List – ( A ) and List ( -B ) | 12 |
| 537 | be a square matrix all of whose entries are integers. [2008] then A- exists but all its entries are not 18. Let A be a square matrix all Then which one of the following is true? (a) If det A=+1, then A-1 exists but all its entri necessarily integers (b) If det A++1, then A-1 exists and all its entries are non integers (©) If det A = + 1, then A-1 exists but all its entries are integers (d) If det A=+1, then A-1 need not exists | 12 |
| 538 | ( f f(x)=left|begin{array}{ccc}sin x & 1 & 0 \ 1 & 2 sin x & 1 \ 0 & 1 & 2 sin xend{array}right| ) then ( int_{-pi / 2}^{pi / 2} f(x) d x ) equals to ( A ) B. ( c ) D. ( frac{3 pi}{2} ) | 12 |
| 539 | Find the value of the determinant: [ left|begin{array}{ccc} cos (theta+phi) & -sin (theta+phi) & cos 2 phi \ sin theta & cos theta & sin phi \ -cos theta & sin theta & cos phi end{array}right| ] | 12 |
| 540 | If ( A ) is a square matrix of order 3 with ( |A|=4, ) then write the value of ( |-2 A| ) | 12 |
| 541 | If the points ( (a, 0),(0, b) ) and (1,1) are collinear, then ( frac{1}{a}+frac{1}{b} ) equal to – A . 1 B. 2 ( c cdot 3 ) D. 4 | 12 |
| 542 | Assertion Points ( boldsymbol{P}(-sin (boldsymbol{beta}- ) ( boldsymbol{alpha}),-cos beta), boldsymbol{Q}(cos (boldsymbol{beta}-boldsymbol{alpha}), sin beta) ) and ( boldsymbol{R}(cos (boldsymbol{beta}-boldsymbol{alpha}+boldsymbol{theta}), sin (boldsymbol{beta}-boldsymbol{theta})), ) where ( beta=frac{pi}{4}+frac{alpha}{2} ) are non-collinear. Reason Three given points are non-collinear if they form a triangle of non-zero area. A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but the Reason is correct | 12 |
| 543 | Let the three din the three digit numbers A 28, 309, and 62 C, where A, B, Care integers between 0 and 9, be divisible by a fixed integer k Show that the determinant 8 12 9 C is divisible B 2 (1990) – 4 Marks) by ki | 12 |
| 544 | If ( D=left|begin{array}{ccc}a^{2}+1 & a b & a c \ b a & b^{2}+1 & b c \ c a & c b & c^{2}+1end{array}right| ) then ( D= ) | 12 |
| 545 | ( boldsymbol{A}=left[begin{array}{lll}mathbf{4} & -mathbf{2} & mathbf{5}end{array}right], boldsymbol{B}=left[begin{array}{l}mathbf{2} \ mathbf{0} \ mathbf{3}end{array}right], ) then ( boldsymbol{A} boldsymbol{d} boldsymbol{j}(boldsymbol{B} boldsymbol{A})= ) ( A cdotleft{begin{array}{lll}0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{array}right} ) ( mathbf{B} cdotleft{begin{array}{lll}8 & -4 & 10 \ 0 & 0 & 0 \ 12 & -6 & 15end{array}right} ) ( mathbf{c} cdotleft{begin{array}{ccc}8 & 0 & 12 \ -4 & 0 & -6 \ 10 & 0 & 5end{array}right} ) D. None of the above | 12 |
| 546 | If ( A ) and ( B ) are square matrices of order 3 such that ( |boldsymbol{A}|=-mathbf{1},|boldsymbol{B}|=mathbf{3}, ) then ( |mathbf{3} boldsymbol{A} boldsymbol{B}|= ) A . -9 B. -81 c. -27 D. 81 | 12 |
| 547 | ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{a}-boldsymbol{c} & boldsymbol{a}-boldsymbol{b} \ boldsymbol{b}-boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{b}-boldsymbol{a} \ boldsymbol{c}-boldsymbol{b} & boldsymbol{c}-boldsymbol{a} & boldsymbol{a}+boldsymbol{b}end{array}right|= ) A. ( 4 a b c ) B. ( 6 a b c ) ( c .8 a b c ) D. ( 2 a b c ) | 12 |
| 548 | If ( Delta=left|begin{array}{lll}mathbf{5} & mathbf{3} & mathbf{8} \ mathbf{2} & mathbf{0} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3}end{array}right|, ) white the cofactor of the element ( a_{32} ) | 12 |
| 549 | If ( boldsymbol{x}^{boldsymbol{a}} boldsymbol{y}^{boldsymbol{b}}=boldsymbol{e}^{boldsymbol{m}}, boldsymbol{x}^{boldsymbol{c}} boldsymbol{y}^{boldsymbol{d}}=boldsymbol{e}^{boldsymbol{n}}, Delta_{mathbf{1}}= ) ( left|begin{array}{ll}boldsymbol{m} & boldsymbol{b} \ boldsymbol{n} & boldsymbol{d}end{array}right|, triangle_{2}=left|begin{array}{ll}boldsymbol{a} & boldsymbol{m} \ boldsymbol{c} & boldsymbol{n}end{array}right| ) and ( Delta_{boldsymbol{3}}=left|begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right| ) the value of ( x ) and ( y ) are respectively A ( cdot frac{Delta_{1}}{Delta_{3}} ) and ( frac{Delta_{2}}{Delta_{3}} ) B. ( frac{Delta_{2}}{Delta_{1}} ) and ( frac{Delta_{3}}{Delta_{1}} ) ( ^{mathbf{c}} cdot log left(frac{Delta_{1}}{Delta_{3}}right) a n d log left(frac{Delta_{2}}{Delta_{3}}right) ) ( mathbf{D} cdot e^{Delta_{1} / Delta_{3}} ) and ( e^{Delta_{2} / Delta_{3}} ) | 12 |
| 550 | If ( A ) is a square matrix of order 3 and ( |boldsymbol{A}|=4 . ) Find the value of ( |mathbf{2 A}| ) | 12 |
| 551 | |6i -3i 1 7. If 4 3i -1 = x+iy, then (1998 – 2 Marks) en 20 3 il (a) x=3, y=2 (b)x=1, y=3 (C) x=0, y=3 (d) x=0, y=0 | 12 |
| 552 | ( fleft|begin{array}{ccc}mathbf{1} & boldsymbol{a} & boldsymbol{b} boldsymbol{c} \ mathbf{1} & boldsymbol{b} & boldsymbol{c a} \ mathbf{1} & boldsymbol{c} & boldsymbol{a b}end{array}right|=boldsymbol{lambda}left|begin{array}{ccc}boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ mathbf{1} & boldsymbol{1} & boldsymbol{1}end{array}right|, ) then ( lambda ) is equal to ( A cdot 1 ) B. -1 ( c cdot 2 ) ( D .-3 ) | 12 |
| 553 | Show that the area of the triangle on the Argand diagram formed by the complex numbers z, iz and z+iz is — Iz. | 12 |
| 554 | ( mathbf{f} mathbf{Delta}=left|begin{array}{lll}boldsymbol{a}_{11} & boldsymbol{a}_{12} & boldsymbol{a}_{13} \ boldsymbol{a}_{21} & boldsymbol{a}_{22} & boldsymbol{a}_{23} \ boldsymbol{a}_{31} & boldsymbol{a}_{32} & boldsymbol{a}_{33}end{array}right| ) and ( boldsymbol{c}_{i j}= ) ( (-1)^{i+j} ) (determinant obtained by deleting ith row and jth column), then ( left|begin{array}{lll}c_{11} & c_{12} & c_{13} \ c_{21} & c_{22} & c_{23} \ c_{31} & c_{32} & c_{33}end{array}right|=Delta^{2} ) ( left|begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{x}^{2} \ boldsymbol{x} & boldsymbol{x}^{2} & mathbf{1} \ boldsymbol{x}^{2} & boldsymbol{1} & boldsymbol{x}end{array}right|=mathbf{7} ) and ( boldsymbol{Delta}= ) ( left|begin{array}{ccc}x^{3}-1 & 0 & x-x^{4} \ 0 & x-x^{4} & x^{3}-1 \ x-x^{4} & x^{3}-1 & 0end{array}right|, ) then A ( . Delta=7 ) B. ( Delta=343 ) c. ( Delta=-49 ) D. ( Delta=49 ) | 12 |
| 555 | Prove that ( left[begin{array}{ccc}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}end{array}right]= ) ( (a-b)(b-c)(c-a) ) | 12 |
| 556 | Let a, b, c be positive and not all equal. Show that the value а ь с of the determinant b ca is negative с а ь | 12 |
| 557 | ( fleft|begin{array}{lll}boldsymbol{x} & boldsymbol{2} & boldsymbol{8} \ boldsymbol{2} & boldsymbol{8} & boldsymbol{x} \ boldsymbol{8} & boldsymbol{x} & boldsymbol{2}end{array}right|=left|begin{array}{lll}boldsymbol{3} & boldsymbol{x} & boldsymbol{7} \ boldsymbol{x} & boldsymbol{7} & boldsymbol{3} \ boldsymbol{7} & boldsymbol{3} & boldsymbol{x}end{array}right|= ) ( left|begin{array}{ccc}mathbf{5} & mathbf{5} & boldsymbol{x} \ mathbf{5} & boldsymbol{x} & mathbf{5} \ boldsymbol{x} & mathbf{5} & mathbf{5}end{array}right|=mathbf{0} ) then ( boldsymbol{x} ) is equal to ( mathbf{A} cdot mathbf{0} ) B. -10 ( c .3 ) D. None of these | 12 |
| 558 | Match the entries in column I with column II | 12 |
| 559 | if ( a, b, c ) are unequal what is the condition that the value of following determinat is zero ( delta=left|begin{array}{lll}a & a^{2} & a^{3}+1 \ b & b^{2} & b^{3}+1 \ c & c^{2} & c^{3}+1end{array}right| ) A ( .1+a b c=0 ) B. ( a+b+c+1=0 ) c. ( (a-b)(b-c)(c-a)=0 ) D. None of these | 12 |
| 560 | Find the value of the determinant without expansion ( left|begin{array}{ccc}b^{2}-a b & b-c & b c-a c \ a b-a^{2} & a-b & b^{2}-a b \ b c-a c & c-a & a b-a^{2}end{array}right| ) | 12 |
| 561 | Show that the points ( boldsymbol{A}(-mathbf{3}, mathbf{3}), boldsymbol{B}(mathbf{0}, mathbf{0}) ) ( C(3,-3) ) are collinear | 12 |
| 562 | ( left|begin{array}{lll}boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} \ boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c}end{array}right|=boldsymbol{K}left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{b}end{array}right| ) ( operatorname{then} K= ) A . 1 B. 2 ( c ) ( D ) | 12 |
| 563 | If ( a^{2}+b^{2}+c^{2}=-2 ) and ( f(x)= ) ( left|begin{array}{ccc}mathbf{1}+boldsymbol{a}^{2} boldsymbol{x} & left(mathbf{1}+boldsymbol{b}^{2}right) boldsymbol{x} & left(mathbf{1}+boldsymbol{c}^{2}right) boldsymbol{x} \ left(mathbf{1}+boldsymbol{a}^{2}right) boldsymbol{x} & mathbf{1}+boldsymbol{b}^{2} boldsymbol{x} & left(mathbf{1}+boldsymbol{c}^{2}right) boldsymbol{x} \ left(mathbf{1}+boldsymbol{a}^{2}right) boldsymbol{x} & left(mathbf{1}+boldsymbol{b}^{2}right) boldsymbol{x} & mathbf{1}+boldsymbol{c}^{mathbf{2}} boldsymbol{x}end{array}right| ) then ( f(x) ) is a polynomial of degree ( A cdot ) B. ( c .3 ) ( D ) | 12 |
| 564 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{0} & mathbf{1} \ -mathbf{1} & mathbf{0}end{array}right] ) then determinant of ( [boldsymbol{A}] ) is ( mathbf{A} cdot mathbf{1} ) B. – c. 0 D. | 12 |
| 565 | If the value of the determinant ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{1} & boldsymbol{1} \ boldsymbol{1} & boldsymbol{b} & boldsymbol{1} \ boldsymbol{1} & boldsymbol{1} & boldsymbol{c}end{array}right| ) is positive, then A ( . a b c>1 ) B . ( a b c>-8 ) c. ( a b c-2 ) | 12 |
| 566 | f ( A+B+C=pi ), then ( left|begin{array}{ccc}sin (boldsymbol{A}+boldsymbol{B}+boldsymbol{C}) & sin boldsymbol{B} & cos boldsymbol{C} \ -sin boldsymbol{B} & boldsymbol{0} & boldsymbol{t a n} boldsymbol{A} \ cos (boldsymbol{A}+boldsymbol{B}) & -boldsymbol{t a n} boldsymbol{A} & boldsymbol{0}end{array}right| ) equals A . B. 2 ( sin B tan A cos C ) ( c ) D. none of these | 12 |
| 567 | Find the value of ( K ) if the point ( A(2,3), B ) ( (4, K) ) and ( C(6,-3) ) are collinear? | 12 |
| 568 | For how many real values of ( ^{prime} m^{prime} ) the points ( boldsymbol{A}(boldsymbol{m}+mathbf{1}, mathbf{1}), boldsymbol{B}(boldsymbol{2 m}+mathbf{1}, boldsymbol{3}) ) and ( C(2 m+2,2 m) ) are collinear. | 12 |
| 569 | Let ( mathbf{P} ) and ( mathbf{Q} ) be ( mathbf{3} times mathbf{3} ) matrices with ( boldsymbol{P} neq boldsymbol{Q} . ) If ( boldsymbol{P}^{3}=boldsymbol{Q}^{3} ) and ( boldsymbol{P}^{2} boldsymbol{Q}=boldsymbol{Q}^{2} boldsymbol{P} ) then determinant of ( left(P^{2}+Q^{2}right) ) is equal to: A . -2 B. ( c cdot 0 ) D. – | 12 |
| 570 | ( left|begin{array}{ccc}boldsymbol{x}^{boldsymbol{n}} & boldsymbol{x}^{boldsymbol{n}+mathbf{2}} & boldsymbol{x}^{boldsymbol{n}+mathbf{3}} \ boldsymbol{y}^{boldsymbol{n}} & boldsymbol{y}^{boldsymbol{n}+mathbf{2}} & boldsymbol{y}^{boldsymbol{n}+boldsymbol{3}} \ boldsymbol{z}^{boldsymbol{n}} & boldsymbol{z}^{boldsymbol{n}+mathbf{2}} & boldsymbol{z}^{boldsymbol{n}+mathbf{3}}end{array}right|=(boldsymbol{x}-boldsymbol{y})(boldsymbol{y}- ) ( (z-x)left(frac{1}{x}+frac{1}{y}+frac{1}{z}right), ) then the value of ( n ) is ( A ) B. – ( c cdot 1 ) ( D ) | 12 |
| 571 | Find the integral value of ( x, ) if ( left|begin{array}{ccc}boldsymbol{x}^{2} & boldsymbol{x} & mathbf{1} \ mathbf{0} & boldsymbol{2} & mathbf{1} \ boldsymbol{3} & boldsymbol{1} & boldsymbol{4}end{array}right|=mathbf{2} mathbf{8} ) | 12 |
| 572 | The value of ( left|begin{array}{llll}boldsymbol{p} & boldsymbol{0} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{a} & boldsymbol{q} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{r} & boldsymbol{0} \ boldsymbol{d} & boldsymbol{e} & boldsymbol{f} & boldsymbol{s}end{array}right| ) is ( A cdot p+q+r+s ) B. ( c cdot a b+c d+e f ) D. pqrs | 12 |
| 573 | Prove that: [ left|begin{array}{ccc} boldsymbol{a}-boldsymbol{b}-boldsymbol{c} & boldsymbol{2} boldsymbol{a} & boldsymbol{2} boldsymbol{a} \ boldsymbol{2} boldsymbol{b} & boldsymbol{b}-boldsymbol{c}-boldsymbol{a} & boldsymbol{2} boldsymbol{b} \ boldsymbol{2} boldsymbol{c} & boldsymbol{2} boldsymbol{c} & boldsymbol{c}-boldsymbol{a}-boldsymbol{b} end{array}right|= ] ( (a+b+c)^{3} ) | 12 |
| 574 | The coefficient of ( x^{2} ) in the expansion of the determinant ( left|begin{array}{ccc}x^{2} & x^{3}+1 & x^{5}+2 \ x^{3}+3 & x^{2}+x & x^{3}+x^{4} \ x+4 & x^{3}+x^{4} & 2^{3}end{array}right| ) is A . -10 B. -8 ( c .-2 ) D. – 6 E. | 12 |
| 575 | The characteristic equation of a matrix ( A ) is ( lambda^{3}-5 lambda^{2}-3 lambda+2 I=0 ) then ( |boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}|= ) ( mathbf{A} cdot mathbf{4} ) B . 25 c. 9 D. 30 | 12 |
| 576 | If ( P=left[begin{array}{lll}1 & c & 3 \ 1 & 3 & 3 \ 2 & 4 & 4end{array}right] ) is the adjoint of a ( 3 times 3 ) matrix ( Q ) and ( operatorname{det} .(Q)=4, ) then ( c ) is equal to. ( mathbf{A} cdot mathbf{0} ) B. 4 c. 5 D. 11 | 12 |
| 577 | ( left|begin{array}{cc}sin ^{2} theta & cos ^{2} theta \ -cos ^{2} theta & sin ^{2} thetaend{array}right|= ) ( mathbf{A} cdot cos 2 theta ) в. ( frac{1}{2}left(1+cos ^{2} 2 thetaright) ) c. ( frac{1}{2}left(1-sin ^{2} 2 thetaright) ) D. ( frac{1}{2} sin ^{2} 2 theta ) | 12 |
| 578 | ( mathbf{f} mathbf{f}_{mathbf{3}}=left[begin{array}{ccc}mathbf{0} & mathbf{1} & mathbf{- 1} \ mathbf{2} & mathbf{1} & mathbf{3} \ mathbf{3} & mathbf{2} & mathbf{1}end{array}right], ) then ( left[boldsymbol{A}(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}) boldsymbol{A}^{-1}right] boldsymbol{A}= ) ( mathbf{A} cdotleft[begin{array}{lll}6 & 0 & 0 \ 0 & 6 & 0 \ 0 & 0 & 6end{array}right] ) ( mathbf{B} cdotleft[begin{array}{lll}4 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 4end{array}right] ) ( mathbf{C} cdotleft[begin{array}{lll}2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2end{array}right] ) D. ( I ) | 12 |
| 579 | Find the value of ( lambda ) if the following equations are consistent ( boldsymbol{x}+boldsymbol{y}-mathbf{3}=mathbf{0} ) ( (1+lambda) x+(2+lambda) y-8=0 ) ( boldsymbol{x}-(mathbf{1}+boldsymbol{lambda}) boldsymbol{y}+(boldsymbol{2}+boldsymbol{lambda})=mathbf{0} ) | 12 |
| 580 | 18. sinx cos x cos x sin x cos xl The number of distinct real roots of COS X cos x cos x sin x VI s is = 0 in the interval (a) 0 (6) 2 (20015) (d) 3 ( 1 | 12 |
| 581 | The value of ( left|begin{array}{ccc}-a^{2} & a b & a c \ a b & -b^{2} & b c \ a c & b c & -c^{2}end{array}right| ) is A . a perfect cube B. zero c. a perfect square D. negative | 12 |
| 582 | Evaluate ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{x}^{2} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{y}^{2} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{z}^{3}end{array}right| ) | 12 |
| 583 | f ( p lambda^{4}+p lambda^{3}+p lambda^{2}+s lambda+t= ) ( left|begin{array}{ccc}lambda^{2}+3 lambda & lambda+1 & lambda+3 \ lambda+1 & 2-lambda & lambda-4 \ lambda-3 & lambda+4 & 3 lambdaend{array}right|, ) then value of t is A . 16 B . 18 ( c .17 ) D. 19 | 12 |
| 584 | The value of determinant ( left|begin{array}{lll}1 / a & b c & a^{3} \ 1 / b & c a & b^{3} \ 1 / c & a b & c^{3}end{array}right| ) A ( cdot a^{2} b^{2} c^{2}(a-b)(b-c)(c-a) ) B. 0 c ( cdot(a-b)(b-c)(c-a) ) D. None of the above | 12 |
| 585 | Solve ( left.boldsymbol{D}=mid begin{array}{ccc}mathbf{1} & -mathbf{2} & mathbf{1} \ mathbf{2} & mathbf{1} & -mathbf{1} \ mathbf{1} & mathbf{3} & mathbf{1}end{array}right] ) | 12 |
| 586 | f ( triangle_{1}= ) ( mid begin{array}{ccc}a_{1}^{2}+b_{1}+c_{1} & a_{1} a_{2}+b_{2}+c_{2} & a_{1} a_{3}+c \ b_{1} b_{2}+c_{1} & b_{2}^{2}+c_{2} & b_{2} b_{3} \ c_{3} c_{1} & c_{3} c_{2} & c_{3}^{2}end{array} ) and ( triangle_{2}=left|begin{array}{lll}a_{1} & b_{1} & c_{1} \ a_{2} & b_{2} & c_{2} \ a_{3} & b_{3} & c_{3}end{array}right|, ) then ( frac{triangle_{1}}{triangle_{2}} ) is equal to A ( cdot a_{1} b_{2} c_{3} ) B. ( a_{1} a_{2} a_{3} ) ( c cdot a_{3} b_{2} c ) D. ( a_{1} b_{1} c_{1}+a_{2} b_{2} c_{2}+a_{3} b_{3} c_{3} ) | 12 |
| 587 | In each of the following find the value of ( k, ) for which the points are collinear. (ii) ( (8,1),(k,-4),(2,-5) ) | 12 |
| 588 | ( (k, k),(2,3) ) and (4,-1) are collimear So find the value of ( k ) | 12 |
| 589 | ( mathbf{f}_{mathbf{1}}=left|begin{array}{ccc}mathbf{7} & boldsymbol{x} & mathbf{2} \ -mathbf{5} & boldsymbol{x}+mathbf{1} & mathbf{3} \ mathbf{4} & boldsymbol{x} & mathbf{7}end{array}right|, boldsymbol{Delta}_{2}= ) ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{2} & boldsymbol{7} \ boldsymbol{x}+boldsymbol{1} & boldsymbol{3} & -boldsymbol{5} \ boldsymbol{x} & boldsymbol{7} & boldsymbol{4}end{array}right| ) then ( boldsymbol{Delta}_{1}-boldsymbol{Delta}_{2}=boldsymbol{0} ) for ( mathbf{A} cdot x=2 ) B. all real ( x ) c. ( x=0 ) D. none of these | 12 |
| 590 | Find ( left|begin{array}{ll}cos alpha & -sin alpha \ sin alpha & cos alphaend{array}right| ) | 12 |
| 591 | Ta 1 o7 [a 1 17 [f] 16. If A= 1 b d ,B= 0 d c ,U = 8 V = 0 ,x = y 1 b c f g h [h lol and AX = U has infinitely many solutions, prove that BX=V has no unique solution. Also show that if afd + 0, then BX= V has no solution. (2004 – 4 Marks) | 12 |
| 592 | ( f(x)=left|begin{array}{ccc}x & cos x & e^{x^{2}} \ sin x & x^{2} & sec x \ tan x & 1 & 2end{array}right| . ) Find ( boldsymbol{f}(mathbf{0}) ? ) ( mathbf{A} cdot mathbf{0} ) ( B ) ( c .-r ) D. ( 2 pi ) | 12 |
| 593 | Using properties of determinant, prove the following: [ mid begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \ 2 a b & 1-a^{2}+b^{2} & 2 a \ 2 b & -2 a & 1-a^{2}-b^{2} end{array} ] ( left(1+a^{2}+b^{2}right)^{3} ) | 12 |
| 594 | Without expanding a determinant at any stage, show that x²+x x+1 x-2 2×2 + 3x -1 3x 3x – 3) = xA+B , where A and B are x2 + 2x + 3 2x-1 2x – 1 | 12 |
| 595 | 6. Let A= (1 2 (1 -1 1 1 1) ( 4 -3. and 10 B = -5 1 (1 2 0 – 2) a . If B is 3) 1 the inverseof matrix A, then a is (a) 5 (6) 1 (c) 2 [2004] (d) 2 | 12 |
| 596 | If ( a_{k}, b_{k}, c_{k} epsilon R ) for ( k=1,2,3 ) and ( left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{b}_{1} & boldsymbol{c}_{1} \ boldsymbol{a}_{2} & boldsymbol{b}_{2} & boldsymbol{c}_{2} \ boldsymbol{a}_{3} & boldsymbol{b}_{3} & boldsymbol{c}_{3}end{array}right| ) then sum of the roots of the equation ( left|begin{array}{lll}a_{1}+i b_{1} x & i a_{1} x+b_{1} & c_{1} \ a_{2}+i b_{2} x & i a_{2} x+b_{2} & c_{2} \ a_{3}+i b_{3} x & i a_{3} x+b_{3} & c_{3}end{array}right| ) is ( mathbf{A} cdot a_{1}+a_{2}+a_{3} ) B. ( b_{1}+b_{2}+b_{3} ) ( mathbf{c} cdot a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3} ) D. none of these | 12 |
| 597 | Solve the determinant ( left|begin{array}{ll}boldsymbol{x} & boldsymbol{y} \ -boldsymbol{y} & boldsymbol{x}end{array}right| ) | 12 |
| 598 | Write the minors and cofactors of each element of the first column of the following matrices ( boldsymbol{A}=left[begin{array}{lll}1 & a & b c \ 1 & b & c a \ 1 & c & a bend{array}right] ) | 12 |
| 599 | ( mid begin{array}{ccc}mathbf{2} boldsymbol{a}_{1} boldsymbol{b}_{1} & boldsymbol{a}_{1} boldsymbol{b}_{1}+boldsymbol{a}_{2} boldsymbol{b}_{1} & boldsymbol{a}_{1} boldsymbol{b}_{3}+boldsymbol{a}_{3} boldsymbol{b}_{1} \ boldsymbol{a}_{1} boldsymbol{b}_{2}+boldsymbol{a}_{2} boldsymbol{b}_{1} & boldsymbol{2} boldsymbol{a}_{2} boldsymbol{b}_{2} & boldsymbol{a}_{2} boldsymbol{b}_{3}+boldsymbol{a}_{3} boldsymbol{b}_{2} \ boldsymbol{a}_{1} boldsymbol{b}_{3}+boldsymbol{a}_{3} boldsymbol{b}_{1} & boldsymbol{a}_{3} boldsymbol{b}_{2}+boldsymbol{b}_{3} boldsymbol{a}_{2} & boldsymbol{2} boldsymbol{a}_{3} boldsymbol{b}_{3}end{array} ) ( A ) B. ( mathbf{c} cdot a_{1} b_{1} a_{2} b ) D. ( a_{1} b_{1}+a_{2} b_{2} ) | 12 |
| 600 | If ( boldsymbol{A}=left|begin{array}{ll}mathbf{1 0} & mathbf{2} \ mathbf{3 0} & mathbf{6}end{array}right| ) then ( |boldsymbol{A}|= ) ( A cdot O ) B. 10 ( c cdot 12 ) D. 60 | 12 |
| 601 | ( fleft(begin{array}{lll}1 & 0 & 1 \ 0 & 1 & 2 \ 0 & 0 & 4end{array}right], ) then show that [ |mathbf{3} boldsymbol{A}|=mathbf{2 7}|boldsymbol{A}| ] | 12 |
| 602 | If every element of third order determinant of ( Delta ) is multiplied by 5 then value of new determinant equals to, A ( . Delta ) B. ( 5 Delta ) c. ( 25 Delta ) D. ( 125 Delta ) | 12 |
| 603 | 27. Let P and Q be 3 x 3 matrices P+Q. If P3= Q and P20 = Q2P then determinant of (P2 +Q) is equal to: [2012] (a) -2 (6) 1 (c) o (d) -1. | 12 |
| 604 | Expand: ( left|begin{array}{ccc}1 & -7 & 3 \ 5 & -6 & 0 \ 1 & 2 & -3end{array}right| ) | 12 |
| 605 | Evaluate : ( Delta=left|begin{array}{ccc}0 & sin alpha & -cos alpha \ -sin alpha & 0 & sin beta \ cos alpha & -sin beta & 0end{array}right| ) | 12 |
| 606 | ( left|begin{array}{ccc}boldsymbol{a}+boldsymbol{b} & boldsymbol{a}+mathbf{2} boldsymbol{b} & boldsymbol{a}+boldsymbol{3} boldsymbol{b} \ boldsymbol{a}+mathbf{2} boldsymbol{b} & boldsymbol{a}+boldsymbol{3} boldsymbol{b} & boldsymbol{a}+boldsymbol{4} boldsymbol{b} \ boldsymbol{a}+boldsymbol{4} boldsymbol{b} & boldsymbol{a}+boldsymbol{5} boldsymbol{b} & boldsymbol{a}+boldsymbol{6} boldsymbol{b}end{array}right|= ) A ( cdot a^{2}+b^{2}+c^{2}-3 a b ) B. ( 3 a b c ) ( c cdot 3 a+5 b ) ( D ) | 12 |
| 607 | ( mathrm{f}left|begin{array}{ccc}boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{y}+boldsymbol{z} \ mathbf{2} boldsymbol{x} & mathbf{3} boldsymbol{x}+mathbf{2} boldsymbol{y} & boldsymbol{4} boldsymbol{x}+boldsymbol{3} boldsymbol{y}+boldsymbol{2} boldsymbol{z} \ boldsymbol{3} boldsymbol{x} & boldsymbol{6} boldsymbol{x}+boldsymbol{3} boldsymbol{y} & boldsymbol{1} mathbf{0} boldsymbol{x}+boldsymbol{6} boldsymbol{y}+boldsymbol{3} boldsymbol{z}end{array}right|=mathbf{6} boldsymbol{4} ) then find ( x ) | 12 |
| 608 | If ( [mathbf{x}] ) stands greatest integer ( leq mathbf{x} ) then the value of ( left|begin{array}{ccc}{[boldsymbol{e}]} & {[boldsymbol{pi}]} & {left[boldsymbol{pi}^{2}-boldsymbol{6}right]} \ {[boldsymbol{pi}]} & boldsymbol{pi}^{2}-boldsymbol{6} & {[boldsymbol{e}]} \ {left[boldsymbol{pi}^{2}-boldsymbol{6}right]} & {[boldsymbol{e}]} & {[boldsymbol{pi}]}end{array}right| ) equals ( A cdot-8 ) B. 8 ( c cdot-1 ) ( D ) | 12 |
| 609 | Calculate the values of the determinants: ( left|begin{array}{cccc}mathbf{0} & mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & boldsymbol{b}+boldsymbol{c} & boldsymbol{a} & boldsymbol{a} \ mathbf{1} & boldsymbol{b} & boldsymbol{c}+boldsymbol{a} & boldsymbol{b} \ mathbf{1} & boldsymbol{c} & boldsymbol{c} & boldsymbol{a}+boldsymbol{b}end{array}right| ) | 12 |
| 610 | Let ( Delta= ) ( left|begin{array}{ccc}sin theta cos phi & sin theta sin phi & cos theta \ cos theta cos phi & cos theta sin phi & -sin theta \ -sin theta sin phi & sin theta cos phi & 0end{array}right|, ) then A. ( Delta ) is independent of ( theta ) B. ( Delta ) is independent of ( phi ) ( mathrm{c} cdot Delta ) is a constant D. none of these | 12 |
| 611 | 37. 14- 37. IfA= ), then adj(x2 + 124) is equal to , then adj (3A2 + 12A) is equal to : (JEE M 2017] [ 72 -847 L-63 51 a) -84 51 a [51 84] () [84 72 | 12 |
| 612 | If ( boldsymbol{A}=left[begin{array}{cc}-mathbf{5} & mathbf{2} \ mathbf{1} & -mathbf{3}end{array}right], ) then adj ( mathbf{A} ) is equal to A. ( left[begin{array}{ll}-3 & -2 \ -1 & -5end{array}right] ) (年) 0 В. ( left[begin{array}{cc}3 & -2 \ -1 & 5end{array}right] ) c. ( left[begin{array}{ll}5 & 1 \ 2 & 3end{array}right] ) D. ( left[begin{array}{ll}3 & 2 \ 1 & 5end{array}right] ) | 12 |
| 613 | Let ( boldsymbol{A}=left[begin{array}{cc}mathbf{5} & mathbf{8} \ mathbf{8} & mathbf{1 3}end{array}right] ) then show that ( boldsymbol{A} ) satisfies the equation ( x^{2}-18 x+1= ) ( mathbf{0} ) | 12 |
| 614 | ( operatorname{Let} boldsymbol{D}_{1}=left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c} & boldsymbol{d} & boldsymbol{c}+boldsymbol{d} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{a}-boldsymbol{b}end{array}right|, quad boldsymbol{D}_{2}= ) ( left|begin{array}{ccc}boldsymbol{a} & boldsymbol{c} & boldsymbol{a}+boldsymbol{c} \ boldsymbol{b} & boldsymbol{d} & boldsymbol{b}+boldsymbol{d} \ boldsymbol{a} & boldsymbol{c} & boldsymbol{a}+boldsymbol{b}+boldsymbol{c}end{array}right|, ) then the value of ( left|frac{boldsymbol{D}_{1}}{boldsymbol{D}_{2}}right|, ) where ( boldsymbol{b} neq mathbf{0} ) and ( boldsymbol{a} boldsymbol{d} neq boldsymbol{b} boldsymbol{c}, ) is | 12 |
| 615 | f ( a, b, c ) are ( p ) th ( , q ) thand ( r ) th terms of a ( mathrm{GP}, ) then ( left|begin{array}{lll}log boldsymbol{a} & boldsymbol{p} & mathbf{1} \ log boldsymbol{b} & boldsymbol{q} & mathbf{1} \ log boldsymbol{c} & boldsymbol{r} & mathbf{1}end{array}right| ) is equal to ( mathbf{A} cdot mathbf{0} ) B. c. ( log a b c ) D. none of these | 12 |
| 616 | ( f(a, b, c text { are in } A P ) then Prove that ( left|begin{array}{ccc}boldsymbol{x}+mathbf{2} & boldsymbol{x}+mathbf{3} & boldsymbol{x}+mathbf{2} boldsymbol{a} \ boldsymbol{x}+mathbf{3} & boldsymbol{x}+mathbf{4} & boldsymbol{x}+mathbf{2} boldsymbol{b} \ boldsymbol{x}+mathbf{4} & boldsymbol{x}+mathbf{5} & boldsymbol{x}+mathbf{2} boldsymbol{c}end{array}right|=mathbf{0} ) | 12 |
| 617 | If ( boldsymbol{A}=left[begin{array}{ll}boldsymbol{alpha} & boldsymbol{2} \ boldsymbol{2} & boldsymbol{alpha}end{array}right] ) and ( left|boldsymbol{A}^{3}right|=mathbf{1 2 5}, ) then the value of ( alpha ) is ( A cdot pm 1 ) ( B cdot pm 2 ) ( c cdot pm 3 ) ( D .pm 5 ) | 12 |
| 618 | Find ( (5 sqrt{3}+3 sqrt{2}) Delta ) where ( Delta=left|begin{array}{ccc}sqrt{13}+sqrt{3} & sqrt{5} & 2 sqrt{5} \ sqrt{15}+sqrt{26} & sqrt{10} & 5 \ 3+sqrt{65} & 5 & sqrt{15}end{array}right| ) | 12 |
| 619 | ( f ) ( a, b, c ) are all different and if ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{a}^{2} & mathbf{1}+boldsymbol{a}^{3} \ boldsymbol{b} & boldsymbol{b}^{2} & mathbf{1}+boldsymbol{b}^{3} \ boldsymbol{c} & boldsymbol{c}^{2} & boldsymbol{1}+boldsymbol{c}^{3}end{array}right|=mathbf{0} ) then ( -boldsymbol{a} boldsymbol{b} boldsymbol{c}= ) | 12 |
| 620 | If the entries in a ( 3 times 3 ) determinant are either 0 or 1 , then the greatest value of their determinats is: A B. 2 ( c cdot 3 ) D. | 12 |
| 621 | Evaluate the following determinants without expansion as far as possible. ( left|begin{array}{ccc}1 & b c & b c(b+c) \ 1 & c a & c a(c+a) \ 1 & a b & a b(a+b)end{array}right| ) | 12 |
| 622 | If ( left|begin{array}{lll}a & a^{3} & a^{4}-1 \ b & b^{3} & b^{4}-1 \ c & c^{3} & c^{4}-1end{array}right|=0 ) and ( a, b, c ) are all distinct then ( a b c(a b+b c+c a) ) is equal to A ( . a+b+c ) в. ( a b c ) c. 0 D. none of these | 12 |
| 623 | If ( A ) is a square matrix of order ( n ), then ( |mathbf{A} mathbf{d} mathbf{j} A| ) is ( mathbf{A} cdot|A|^{2} ) B cdot ( |A|^{n} ) C ( cdot|A|^{n-1} ) D・ |A| | 12 |
| 624 | The value of determinant ( left|begin{array}{lll}mathbf{1} & boldsymbol{x} & boldsymbol{y}+boldsymbol{z} \ mathbf{1} & boldsymbol{y} & boldsymbol{z}+boldsymbol{x} \ mathbf{1} & boldsymbol{z} & boldsymbol{x}+boldsymbol{y}end{array}right| ) ( mathbf{A} cdot mathbf{0} ) B. ( x+y+z ) c. ( 1+x+y+z ) D. ( (x-y)(y-z)(z-x) ) | 12 |
| 625 | Match the statements in Column I with statements in column II | 12 |
| 626 | Find the value of the determinant: [ left|begin{array}{ccc} cos (theta+phi) & -sin (theta+phi) & cos 2 phi \ sin theta & cos theta & sin phi \ -cos theta & sin theta & cos phi end{array}right| ] | 12 |
| 627 | If ( m ) and ( p ) are positive ( (m geq p) ) and ( boldsymbol{Delta}(boldsymbol{m}, boldsymbol{p})= ) ( left|begin{array}{ccc}m & C_{p} & m \ m+1 & C_{p} \ m+2 & m+1 & C_{p+1} & m+1 \ m_{p} & m+2 & m_{p+2} \ & & m+2end{array}right| ) and ( m_{boldsymbol{p}}=mathbf{0} ) if ( mathbf{m}<mathbf{p}, ) then This question has multiple correct options A ( . Delta(2,1) / Delta(1,0)=4 ) B. ( Delta(4,3) / Delta(3,2)=2 ) ( mathbf{c} cdot Delta(4,3) / Delta(2,1)=5 ) D. ( Delta(4,3) / Delta(1,0)=10 ) | 12 |
| 628 | Evaluate the following: ( left|begin{array}{ccc}1 & a & b c \ 1 & b & c a \ 1 & c & a bend{array}right| ) | 12 |
| 629 | Find the values of ( x, ) if (i) ( left|begin{array}{ll}2 & 4 \ 5 & 1end{array}right|=left|begin{array}{ll}2 x & 4 \ 6 & xend{array}right| ) (ii) ( left|begin{array}{ll}2 & 3 \ 4 & 5end{array}right|= ) ( left|begin{array}{ll}boldsymbol{x} & boldsymbol{3} \ boldsymbol{2} boldsymbol{x} & boldsymbol{5}end{array}right| ) | 12 |
| 630 | The value of ( A cdot O ) B. ( 30^{text {th }} ) ( c cdot 30^{-x} ) D. None of these | 12 |
| 631 | Assertion Let ( p<0 ) and ( alpha_{1}, alpha_{2}, dots, alpha_{9} ) be the nine roots of ( boldsymbol{x}^{mathbf{9}}=boldsymbol{p}, ) then ( boldsymbol{Delta}=left|begin{array}{lll}boldsymbol{alpha}_{1} & boldsymbol{alpha}_{2} & boldsymbol{alpha}_{3} \ boldsymbol{alpha}_{4} & boldsymbol{alpha}_{5} & boldsymbol{alpha}_{6} \ boldsymbol{alpha}_{4} & boldsymbol{alpha}_{8} & boldsymbol{alpha}_{9}end{array}right|=0 ) Reason If two rows of a determinant are identical, then determinant equals zero A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 632 | ut 7 Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x=cy + bz, y=az + cx, and z=bx + ay. Then a2 + b2 + c2 + 2abc is equal to [2008] (a) 2 (b) 1 (c) 0 (d) 1 | 12 |
| 633 | x – 4 2x 2x 38. If 2X X-4 2x 28 2X X-4 =(A+Bx)(x- A)2 , then the ordered pair (A, B) is equal to : (a) (-4, 3) (b) (-4,5) [JEE M 2018] (d) (-4,-5) (c) (4,5) | 12 |
| 634 | Let ( Delta_{mathrm{o}}=left[begin{array}{lll}a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33}end{array}right] ) and let ( Delta_{1} ) denote the determinant formed by the cofactors of elements of ( Delta_{0} ) and ( Delta_{2} ) denote the determinant formed by the cofactor of ( Delta_{1}, ) similarly ( Delta_{n} ) denotes the determinant formed by the cofactors of ( Delta_{n-1} ) then the determinant value of ( Delta_{n} ) is ( mathbf{A} cdot Delta_{0}^{2 n} ) B . ( Delta_{0}^{2^{2}} ) ( mathbf{C} cdot Delta_{0}^{n^{2}} ) D. ( Delta^{2}_{0} ) | 12 |
| 635 | Find the value of determinant, ( left|begin{array}{ccc}1 & a & a^{2}-b c \ 1 & b & b^{2}-c a \ 1 & c & c^{2}-a bend{array}right| ) | 12 |
| 636 | 14. [1 o o [10 of A= 0 1 1 and 1 = 0 1 0 and To 2 4. Lo o 1 dI), then the value of c and d are (20055) (a) (6,-11) (b) (6,11) (C) (-6,11) (d) (6,-11) | 12 |
| 637 | ( mathbf{y}=sin mathbf{x}, boldsymbol{y}_{n}=frac{boldsymbol{d}^{n}(sin boldsymbol{x})}{boldsymbol{d} boldsymbol{x}^{n}} ) ( operatorname{then}left|begin{array}{lll}boldsymbol{y} & boldsymbol{y}_{1} & boldsymbol{y}_{2} \ boldsymbol{y}_{3} & boldsymbol{y}_{4} & boldsymbol{y}_{5} \ boldsymbol{y}_{6} & boldsymbol{y}_{7} & boldsymbol{y}_{8}end{array}right|=? ) ( A cdot-sin x ) B. ( c cdot sin x ) D. ( cos ) | 12 |
| 638 | If ( triangle(r)=left|begin{array}{cc}r & r^{3} \ 1 & n(n+1)end{array}right|, ) then ( sum_{r=1}^{n} triangle(r) ) is equal to ( ^{A} cdot sum_{i=1}^{n} r ) B. ( sum_{n=1}^{n} r ) ( c cdot sum^{n} ) D. ( sum_{n=1}^{n} r^{4} ) | 12 |
| 639 | ( begin{array}{c}text { If } boldsymbol{A}+boldsymbol{B}+boldsymbol{C}=boldsymbol{pi}, text { then } \ left|begin{array}{ccc}sin (boldsymbol{A}+boldsymbol{B}+boldsymbol{C}) & sin boldsymbol{B} & cos C \ -sin boldsymbol{0} & boldsymbol{operatorname { t a n }} boldsymbol{A}end{array}right|= \ cos (boldsymbol{A}+boldsymbol{B}) quad-boldsymbol{operatorname { t a n } boldsymbol { A }} & boldsymbol{0}end{array} mid ) | 12 |
| 640 | If ( left|begin{array}{lll}a & a^{3} & a^{4}-1 \ b & b^{3} & b^{4}-1 \ c & c^{3} & c^{4}-1end{array}right|=0 ) and ( a, b, c ) are all distinct, then ( a b c(a b+b c+c a) ) is equal to A. ( a+b+c ) B. abc ( c cdot 0 ) D. none of these | 12 |
| 641 | ( mathbf{f} mathbf{Delta}=left|begin{array}{cc}mathbf{f}(boldsymbol{x}) & boldsymbol{f}left(frac{mathbf{1}}{boldsymbol{x}}right)+boldsymbol{f}(boldsymbol{x}) \ mathbf{1} & boldsymbol{f}left(frac{mathbf{1}}{boldsymbol{x}}right)end{array}right|=mathbf{0} ) where; ( boldsymbol{f}(boldsymbol{x})=boldsymbol{a}+boldsymbol{b} boldsymbol{x}^{n} ) and ( boldsymbol{f}(boldsymbol{2})=mathbf{1 7}, ) then ( boldsymbol{f}(mathbf{5}) ) is: A . 126 в. 326 ( c .428 ) D. 626 | 12 |
| 642 | ( fleft|begin{array}{ccc}boldsymbol{x}+mathbf{1} & mathbf{3} & mathbf{5} \ mathbf{2} & boldsymbol{x}+mathbf{2} & mathbf{5} \ mathbf{2} & mathbf{3} & boldsymbol{x}+mathbf{4}end{array}right|=mathbf{0}, ) then ( boldsymbol{x}=? ) A . 1,9 В. -1,9 c. -1,-9 D. 1,-9 | 12 |
| 643 | If ( A=left[begin{array}{ll}1 & 1 \ 0 & 1end{array}right], ) then ( operatorname{det}left(A+A^{2}+right. ) ( left.boldsymbol{A}^{3}+boldsymbol{A}^{4}+boldsymbol{A}^{5}right) ) is ( mathbf{A} cdot mathbf{1} ) B. 32 c. 25 D. 30 | 12 |
| 644 | If the vectors ( vec{a}, vec{b}, vec{c} ) are coplanar, then the value of ( left|begin{array}{ccc}overrightarrow{boldsymbol{a}} & overrightarrow{boldsymbol{b}} & overrightarrow{boldsymbol{c}} \ overrightarrow{boldsymbol{a}} cdot overrightarrow{boldsymbol{a}} & overrightarrow{boldsymbol{a}} cdot overrightarrow{boldsymbol{b}} & overrightarrow{boldsymbol{a}} cdot overrightarrow{boldsymbol{c}} \ overrightarrow{boldsymbol{b}} cdot overrightarrow{boldsymbol{a}} & overrightarrow{boldsymbol{b}} cdot overrightarrow{boldsymbol{b}} & overrightarrow{boldsymbol{b}} cdot overrightarrow{boldsymbol{c}}end{array}right|= ) ( mathbf{A} cdot mathbf{1} ) B. ( c cdot-1 ) D. ( vec{a}+vec{b}+vec{c} ) | 12 |
| 645 | What is the value of ( y ) if ( (y, 3),(-5,6) ) and (-8,8) are collinear? A . -1 B . 2 c. ( frac{1}{2} ) D. ( -frac{1}{2} ) | 12 |
| 646 | ( fleft(a_{1}, a_{2}, a_{3}, dots, a_{n}, dots a r e text { in } G P ) then the right. determinant ( Delta= ) ( left|begin{array}{ccc}log a_{n} & log a_{n+1} & log a_{n+2} \ log a_{n+3} & log a_{n+4} & log a_{n+5} \ log a_{n+6} & log a_{n+7} & log a_{n+8}end{array}right| ) is equal to A . 0 B. 1 ( c cdot 2 ) D. | 12 |
| 647 | If ( A ) is an invertible matrix of order ( n ) then the determinant of adj ( A ) is equal to: ( mathbf{A} cdot|A|^{n} ) В ( cdot|A|^{n+1} ) c. ( |A|^{n-1} ) D. ( |A|^{n+2} ) | 12 |
| 648 | The number of distinct real roots of the [ text { guation, }left|begin{array}{ccc} cos x & sin x & sin x \ sin x & cos x & sin x \ sin x & sin x & cos x end{array}right|=0 ] in the interval ( left[-frac{pi}{4}, frac{pi}{4}right] ) is/are ( A ) B . 2 ( c cdot 1 ) ( D ) | 12 |
| 649 | Suppose ( f(x) ) is a function satisfying the following conditions. ( (a) f(0)=2, f(1)=1 ) (b) ( f(x) ) has a maximum at ( x=5 / 2, ) and (c) for all ( x in R ) ( boldsymbol{f}^{prime}(boldsymbol{x})= ) ( mid begin{array}{ccc}2 a x & 2 a x-1 & 2 a x+b+ \ b & b+1 & -1 \ 2(a x+b) & 2 a x+2 b+1 & 2 a x+bend{array} ) where ( a, b ) are constants, then | 12 |
| 650 | The value of the determinant ( left|begin{array}{ccc}b^{2}-a b & b-c & b c-a c \ a b-a^{2} & a-b & b^{2}-a b \ b c-a c & c-a & a b-a^{2}end{array}right| ) A ( . a b c ) B. ( a+b+c ) ( c ) ( mathbf{D} cdot a b+b c+c a ) | 12 |
| 651 | ( operatorname{Det}left{begin{array}{ccc}1^{2} & 2^{2} & 3^{2} \ 2^{2} & 3^{2} & 4^{2} \ 3^{2} & 4^{2} & 5^{2}end{array}right}=dots ) A . -8 B. -7 ( c .-6 ) D. ( -21 / 4 ) | 12 |
| 652 | 24. The number of values of k for which the linear equations 4x + ky + 2z=0, kx + 4y+z=0 and 2x +2y+z= 0 possess a non-zero solution is [2011] (a) 2 (b) 1 (C) zero (d) 3 | 12 |
| 653 | Find the value of ( x ) if ( left|begin{array}{ccc}3 & 5 & x \ 2 & 4 & 1 \ -1 & 2 & 3end{array}right|=0 ) | 12 |
| 654 | ( left|begin{array}{ccc}frac{1}{a} & a^{2} & b c \ frac{1}{b} & b^{2} & c a \ frac{1}{c} & c^{2} & a bend{array}right|= ) A ( cdot ) abc B. ( a+b+c ) ( c cdot 0 ) D. ( 4 a b c ) | 12 |
| 655 | When the determinant ( left|begin{array}{lll}cos 2 x & sin ^{2} x & cos 4 x \ sin ^{2} x & cos 2 x & cos ^{2} x \ cos 4 x & cos ^{2} x & cos 2 xend{array}right| ) is expanded in powers of ( sin x, ) then the constant term in that expression is ( A ) B. ( c cdot-1 ) D. | 12 |
| 656 | ( f e^{i theta}=cos theta+i sin theta, ) find the value of ( -left|begin{array}{ccc}mathbf{1} & boldsymbol{e}^{i boldsymbol{pi} / mathbf{3}} & boldsymbol{e}^{i boldsymbol{pi} / mathbf{4}} \ boldsymbol{e}^{-i boldsymbol{pi} / mathbf{3}} & mathbf{1} & boldsymbol{e}^{boldsymbol{i} mathbf{2} boldsymbol{pi} / mathbf{3}} \ boldsymbol{e}^{-boldsymbol{i} boldsymbol{pi} / mathbf{4}} & boldsymbol{e}^{-boldsymbol{i} mathbf{2} boldsymbol{pi} / mathbf{3}} & mathbf{1}end{array}right|-mathbf{2}^{frac{1}{2}} ) | 12 |
| 657 | ( f(a+b+c=0, ) then one root of ( left|begin{array}{ccc}boldsymbol{a}-boldsymbol{x} & boldsymbol{c} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{b}-boldsymbol{x} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{a} & boldsymbol{c}-boldsymbol{x}end{array}right|=0 ) is ( A cdot a+b ) B. ( c cdot b+c ) D. atca | 12 |
| 658 | If ( D_{P}=left|begin{array}{ccc}boldsymbol{P} & mathbf{1 5} & mathbf{8} \ boldsymbol{P}^{2} & mathbf{3 5} & mathbf{9} \ boldsymbol{P}^{3} & mathbf{2 5} & mathbf{1 0}end{array}right|, ) then ( boldsymbol{D}_{1}+ ) ( D_{2}+D_{3}+D_{4}+D_{5} ) is equal to A. -29000 в. -25000 c. 25000 D. none of these | 12 |
| 659 | 19. Let A be a 2 x 2 matrix Statement-1: adj (adj A)=A [2009] Statement-2: (adj A F|A| (a) Statement-1 is true, Statement-2 is true. Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (C) Statement -1 is false, Statement-2 is true. (d) Statement-1 is true, Statement -2 is true. Statement-2 is a correct explanation for Statement-1. | 12 |
| 660 | Find the value of the following determinants. ( left|begin{array}{cc}mathbf{5} & mathbf{3} \ -mathbf{7} & mathbf{0}end{array}right| ) | 12 |
| 661 | The value of ( triangle= ) ( begin{array}{cc}mathbf{2} & boldsymbol{a}+boldsymbol{r}+mathbf{2} \ boldsymbol{a}+boldsymbol{r}+mathbf{2} & mathbf{2}(boldsymbol{a}+mathbf{1})(boldsymbol{r}+mathbf{1}) \ boldsymbol{a}+boldsymbol{r} & boldsymbol{a}(boldsymbol{r}+mathbf{1})+boldsymbol{r}(boldsymbol{a}+mathbf{1})end{array} ) ( A ) в. ( -2 a(r+1) ) c. ( a(a r+r+a) ) D. -1 | 12 |
| 662 | Find equation of line joining (1,2) and (3,6) using determinants. | 12 |
| 663 | ( mathbf{f} boldsymbol{A}=left|begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{1} \ mathbf{0} & mathbf{2} & -mathbf{3} \ mathbf{2} & mathbf{1} & mathbf{0}end{array}right| ) and ( boldsymbol{B}=(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}) ) and ( C=5 A, ) then ( frac{|a d j B|}{|C|} ) is? ( mathbf{A} cdot mathbf{5} ) B . 25 c. -1 D. | 12 |
| 664 | , then prove that Illustration 3.84 If x+y+z= sin x sin y sin z cosx cos y cos z = 0. [cos x cos y cos z | 12 |
| 665 | 4. [1 4 4 If the adjoint of a 3 x 3 matrix P is 2 1 7 , then the [ 113] (2012) possible value(s) of the determinant of Pis (are) (a) 2 (6) -1 (c) 1 (d) 2 | 12 |
| 666 | Find the value of the determinant ( left|begin{array}{cc}mathbf{2} boldsymbol{i} & -mathbf{3} boldsymbol{i} \ boldsymbol{i}^{3} & -mathbf{2} boldsymbol{i}end{array}right| ) where ( boldsymbol{i}=sqrt{-boldsymbol{i}} ) | 12 |
| 667 | Evaluate ( mid begin{array}{ccc}cos left(x+x^{2}right) & sin left(x+x^{2}right) & -cos (x+1) \ sin left(x-x^{2}right) & cos left(x-x^{2}right) & sin (x-x) \ sin 2 x & 0 & sin 2 x^{2}end{array} ) A ( cdot sin left(2 x+2 x^{2}right) ) B. ( -sin left(2 x+2 x^{2}right) ) ( c cdot cos left(2 x+2 x^{2}right) ) D. ( -cos left(2 x+2 x^{2}right) ) | 12 |
| 668 | If the system of linear equations ( 2 x+2 a y+a z=0 ) ( 2 x+3 b y+b z=0 ) ( 2 x+4 c y+c z=0 ) where ( a, b, c epsilon R ) are non-zero and distinct; has a non-zero solution, then: A ( cdot frac{1}{a}, frac{1}{b}, frac{1}{c} ) are in A.P B . ( a+b+c=0 ) c. ( a, b, c ) are in A.P ( mathbf{D} cdot a, b, c ) are in G.P | 12 |
| 669 | If ( A ) is an ( n times n ) non-singular matrix, then ( |boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A}| ) is: ( mathbf{A} cdot|A|^{n} ) B ( cdot|A|^{n+1} ) c. ( |A|^{n-1} ) D. ( |A|^{n-2} ) | 12 |
| 670 | ( mathbf{f} mathbf{a}^{2}+boldsymbol{b}^{2}+boldsymbol{c}^{2}=-mathbf{2} ) and ( boldsymbol{f}(boldsymbol{x})= ) ( left|begin{array}{ccc}mathbf{1}+boldsymbol{a}^{2} boldsymbol{x} & left(mathbf{1}+boldsymbol{b}^{2}right) boldsymbol{x} & left(mathbf{1}+boldsymbol{c}^{2}right) boldsymbol{x} \ left(mathbf{1}+boldsymbol{a}^{2}right) boldsymbol{x} & mathbf{1}+boldsymbol{b}^{2} boldsymbol{x} & left(mathbf{1}+boldsymbol{c}^{2}right) boldsymbol{x} \ left(mathbf{1}+boldsymbol{a}^{2}right) boldsymbol{x} & left(mathbf{1}+boldsymbol{b}^{2}right) boldsymbol{x} & mathbf{1}+boldsymbol{c}^{2} boldsymbol{x}end{array}right| ) Then, ( f(x) ) is a polynomial of degree A .2 B. 3 ( c .4 ) ( D ) | 12 |
| 671 | If 3 points ( A(1, a, b), B(a, 2, b), C(a, b, 3) ) are collinear, then find ( a+b=? ) | 12 |
| 672 | Consider the following statements: 1. Determinant is a square matrix. 2. Determinant is a number associated with a square matrix. Which of the above statements is/are correct? A . 1 only B. 2 only c. Both 1 and 2 D. Neither 1 nor 2 | 12 |
| 673 | Solve: ( left|begin{array}{ccc}mathbf{0} & -mathbf{3} & boldsymbol{x} \ boldsymbol{x}+mathbf{1} & mathbf{3} & mathbf{1} \ mathbf{4} & mathbf{1} & mathbf{5}end{array}right|=mathbf{0} ) | 12 |
| 674 | If the determinant ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{b} & boldsymbol{a} boldsymbol{t}-boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b} boldsymbol{t}-boldsymbol{c} \ boldsymbol{2} & boldsymbol{1} & boldsymbol{0}end{array}right|=0 ) if ( a, b, c ) are in A. ( A . P ) в. G.Р. c. ( H . P ) D. ( k=1 / 2 ) | 12 |
| 675 | ( operatorname{Show}left|begin{array}{ccc}boldsymbol{a x} & boldsymbol{b y} & boldsymbol{c z} \ boldsymbol{x}^{2} & boldsymbol{y}^{2} & boldsymbol{z}^{2} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right|=left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{y} boldsymbol{z} & boldsymbol{z} boldsymbol{x} & boldsymbol{x} boldsymbol{y}end{array}right| ) | 12 |
| 676 | ( left(begin{array}{lll}mathbf{7} & mathbf{1} & mathbf{5} \ mathbf{8} & mathbf{0} & mathbf{0}end{array}right)left(begin{array}{l}mathbf{2} \ mathbf{3} \ mathbf{1}end{array}right)+mathbf{5}left(begin{array}{l}mathbf{1} \ mathbf{0}end{array}right) ) is equal to A ( cdotleft(begin{array}{l}16 \ 27end{array}right. ) B. ( left(begin{array}{l}27 \ 16end{array}right. ) c. ( left(begin{array}{l}15 \ 16end{array}right. ) D. ( left(begin{array}{l}16 \ 15end{array}right. ) | 12 |
| 677 | Calculate the values of the determinants: ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{c}+boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{c} & boldsymbol{a}+boldsymbol{b}end{array}right| ) | 12 |
| 678 | Evaluate the determinant : ( left|begin{array}{ccc}mathbf{1}+boldsymbol{a} & mathbf{0} & mathbf{0} \ mathbf{0} & mathbf{1}+boldsymbol{a} & mathbf{0} \ mathbf{0} & mathbf{0} & mathbf{1}+boldsymbol{a}end{array}right| ) | 12 |
| 679 | ( left|begin{array}{ccc}(boldsymbol{b}+boldsymbol{c})^{2} & boldsymbol{a}^{2} & boldsymbol{a}^{2} \ boldsymbol{b}^{2} & (boldsymbol{c}+boldsymbol{a})^{2} & boldsymbol{b}^{2} \ boldsymbol{c}^{2} & boldsymbol{c}^{2} & (boldsymbol{a}+boldsymbol{b})^{2}end{array}right| ) is equal to ( mathbf{A} cdot a b c(a+b+c)^{2} ) B ( cdot 2 a b c(a+b+c)^{2} ) ( mathbf{c} cdot 2 a b c(a+b+c)^{3} ) ( mathbf{D} cdot 2 a b c(a+b+c) ) | 12 |
| 680 | Prove the following identities : [ begin{array}{ccc} boldsymbol{a}+boldsymbol{b}+mathbf{2 c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{b}+boldsymbol{c}+mathbf{2} boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{c}+boldsymbol{a}+boldsymbol{2} boldsymbol{b} end{array} mid= ] ( mathbf{2}(boldsymbol{a}+boldsymbol{b}+boldsymbol{c})^{3} ) | 12 |
| 681 | Prove that: ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{c} & boldsymbol{a} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b} & boldsymbol{c}end{array}right|=(boldsymbol{a}+boldsymbol{b}+ ) ( c)(a-c)^{2} ) | 12 |
| 682 | Suppose ( a, b, c epsilon R ) and ( a b c=1 . ) If ( A= ) ( left[begin{array}{ccc}2 a & b & c \ b & 2 c & a \ c & a & 2 bend{array}right] ) is such that ( |A|left|A^{prime}right|=64 ) and ( |A|>0, ) then find the value of ( left(a^{3}+b^{3}+c^{3}right)^{4} ) ( left(A^{prime} text { denotes transpose of a matrix } A .right) ) A . 1 B. 2 ( c .3 ) ( D ) | 12 |
| 683 | Find the integral value of ( x, ) if ( left|begin{array}{ccc}boldsymbol{x}^{2} & boldsymbol{x} & mathbf{1} \ mathbf{0} & boldsymbol{2} & mathbf{1} \ boldsymbol{3} & boldsymbol{1} & boldsymbol{4}end{array}right|=mathbf{2} mathbf{8} ) | 12 |
| 684 | ( mathbf{f}left|begin{array}{lll}boldsymbol{x}^{boldsymbol{k}} & boldsymbol{x}^{boldsymbol{k}+boldsymbol{2}} & boldsymbol{x}^{boldsymbol{k}+boldsymbol{3}} \ boldsymbol{y}^{boldsymbol{k}} & boldsymbol{y}^{boldsymbol{k}+boldsymbol{2}} & boldsymbol{y}^{boldsymbol{k}+boldsymbol{3}} \ boldsymbol{z}^{boldsymbol{k}} & boldsymbol{z}^{boldsymbol{k}+boldsymbol{2}} & boldsymbol{z}^{boldsymbol{k}+boldsymbol{3}}end{array}right| ) 三室 ( (x-y)(y-z)(z-x)left(frac{1}{x}+frac{1}{y}+frac{1}{z}right) ) then A. ( k=-3 ) B. ( k=-1 ) ( mathbf{c} cdot k=1 ) D. ( k=3 ) | 12 |
| 685 | For positive numbers ( boldsymbol{x}, boldsymbol{y}, boldsymbol{z} ) the numerical value of the determinant ( left|begin{array}{ccc}mathbf{1} & log _{x} boldsymbol{y} & log _{boldsymbol{x}} boldsymbol{z} \ log _{boldsymbol{y}} boldsymbol{x} & boldsymbol{3} & log _{boldsymbol{y}} boldsymbol{z} \ log _{boldsymbol{z}} boldsymbol{x} & log _{boldsymbol{z}} boldsymbol{y} & boldsymbol{5}end{array}right| ) is ( mathbf{A} cdot mathbf{0} ) B. ( log x log y log z ) ( c .1 ) D. 8 | 12 |
| 686 | If ( I ) is the unit matrix of order ( n, ) where ( k neq 0 ) is a constant then ( operatorname{adj}(k I)= ) A ( cdot k^{n}(operatorname{adj} I) ) B. ( k(operatorname{adj} I) ) c. ( k^{2}(operatorname{adj} I) ) D. ( k^{n-1}(text { adj } I) ) | 12 |
| 687 | Find the value of: ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{x}^{2} & boldsymbol{y}^{2} & boldsymbol{z}^{2} \ boldsymbol{x}^{3} & boldsymbol{y}^{3} & boldsymbol{z}^{3}end{array}right| ) ( mathbf{A} cdot mathbf{0} ) B. ( (x-y)(y-z)(z-x)(x y+y z+z x) ) C. ( (x+y x+z x)(x+y)(y+z)(z+x) ) D. ( (x+y+z) ) | 12 |
| 688 | If the area of the triangle formed by ( (0,0),(a, 0) ) and ( left(frac{1}{2}, aright) ) is equal to ( frac{1}{2} s q ) unit, then the values of ( a ) are : ( A ldots pm 2 ) B. ±3 ( c .pm 1 ) D. ±4 ( mathrm{E} cdot pm 5 ) | 12 |
| 689 | If ( boldsymbol{A}=left[begin{array}{rrr}3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1end{array}right] ) then find ( boldsymbol{A} boldsymbol{d} boldsymbol{j}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A}) ) A. ( left[begin{array}{lll}3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1end{array}right] ) В. ( left[begin{array}{ccc}3 & 3 & 4 \ 2 & -3 & -4 \ 0 & -1 & 1end{array}right] ) ( begin{array}{llll}text { C. } & {left[begin{array}{ccc}3 & 3 & 4 \ 2 & -3 & 4 \ 0 & 1 & 1end{array}right]}end{array} ) D. ( left[begin{array}{ccc}3 & -3 & 4 \ 2 & -3 & -4 \ 0 & 1 & 1end{array}right] ) | 12 |
| 690 | A set of points which do not lie on the same line are called as A. collinear B. non-collinear c. concurrent D. square | 12 |
| 691 | If ( boldsymbol{A}(boldsymbol{x})=left|begin{array}{ccc}boldsymbol{x}+mathbf{1} & mathbf{2} boldsymbol{x}+mathbf{1} & mathbf{3} boldsymbol{x}+mathbf{1} \ mathbf{2} boldsymbol{x}+mathbf{1} & mathbf{3} boldsymbol{x}+mathbf{1} & boldsymbol{x}+mathbf{1} \ boldsymbol{3} boldsymbol{x}+mathbf{1} & boldsymbol{x}+mathbf{1} & boldsymbol{2} boldsymbol{x}+mathbf{1}end{array}right| ) ( operatorname{then} int_{0}^{1} boldsymbol{A}(boldsymbol{x}) boldsymbol{d} boldsymbol{x}= ) A . -15 ( B cdot frac{-1}{2} ) ( c .-30 ) ( D ) | 12 |
| 692 | ( fleft|begin{array}{ll}boldsymbol{x} & boldsymbol{y} \ boldsymbol{4} & boldsymbol{2}end{array}right|=mathbf{7} ) and ( left|begin{array}{ll}boldsymbol{2} & boldsymbol{3} \ boldsymbol{y} & boldsymbol{x}end{array}right|=boldsymbol{4} ) then A ( cdot x=-3, y=-frac{5}{2} ) B. ( x=-frac{5}{2}, y=-3 ) c. ( _{x}=-3, y=frac{5}{2} ) D. ( x=-frac{5}{2}, y=3 ) | 12 |
| 693 | 5. The parameter, on which the value of the determinant 1 a cos(p-d)x cos px sin(p-d)x sin px cos(p+d)x sin(p+.d)x does not depend upon is (a) a (b) p (c) d (1997 – 2 Marks) (d) x | 12 |
| 694 | Evaluate the determinant : ( left|begin{array}{cc}cos theta & -sin theta \ sin theta & cos thetaend{array}right| ) | 12 |
| 695 | Prove the following: [ left|begin{array}{ccc} boldsymbol{a}+boldsymbol{b}+mathbf{2} boldsymbol{c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{b}+boldsymbol{c}+mathbf{2} boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{c}+boldsymbol{a}+mathbf{2} boldsymbol{b} end{array}right|= ] ( mathbf{2}(boldsymbol{a}+boldsymbol{b}+boldsymbol{c})^{3} ) | 12 |
| 696 | f ( a, b, c ) are real numbers such that ( left|begin{array}{ccc}boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{b}+boldsymbol{c} & boldsymbol{c}+boldsymbol{a}end{array}right|=0, ) then show that either ( boldsymbol{a}+boldsymbol{b}+boldsymbol{c}=boldsymbol{0} ) or, ( boldsymbol{a}=boldsymbol{b}=boldsymbol{c} ) | 12 |
| 697 | Let ( omega=-frac{1}{2}+i frac{sqrt{3}}{2}, ) then the value of the determinant ( left|begin{array}{ccc}1 & 1 & 1 \ 1 & -1-omega^{2} & omega^{2} \ 1 & omega^{2} & omega^{4}end{array}right|, ) is This question has multiple correct options ( A .3 omega ) В. ( 3 omega(omega-1) ) ( c cdot 3 omega^{2} ) D. ( 3(-2 omega-1) ) | 12 |
| 698 | Adj ( left(A d jleft[begin{array}{cc}2 & -3 \ 4 & 6end{array}right]right)= ) A ( cdotleft[begin{array}{cc}2 & -3 \ 4 & 6end{array}right] ) в. ( left[begin{array}{cc}6 & 3 \ -4 & 2end{array}right] ) с. ( left[begin{array}{cc}-6 & 3 \ -4 & -2end{array}right] ) D ( cdotleft[begin{array}{cc}-6 & -3 \ 4 & -2end{array}right] ) | 12 |
| 699 | If ( boldsymbol{A}=left|begin{array}{ll}mathbf{0} & mathbf{0} \ mathbf{1} & mathbf{1}end{array}right| ) then the value of ( boldsymbol{A}+ ) ( boldsymbol{A}^{2}+boldsymbol{A}^{3}+ldots+boldsymbol{A}^{n}=? ) ( A cdot A ) B. nA c. ( (n+1) A ) ( D ) | 12 |
| 700 | Prove the following: ( left|begin{array}{llll}boldsymbol{x} & boldsymbol{a} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{x} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a} & boldsymbol{x} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a} & boldsymbol{a} & boldsymbol{x}end{array}right|=(boldsymbol{x}+boldsymbol{3} boldsymbol{a})(boldsymbol{x}-boldsymbol{a})^{3} ) | 12 |
| 701 | 2. The values of a lying between 0 = 0 and 0 = 7/2 and satisfying the equation 1+sine cose 4sin 40 sin 1+cos e 4 sin 40 = 0 are sin’e cosạe 1+4 sin 40 a. 77/24 & b. 57/24 08 c. 117/24 d. Tt/24 (IIT-JEE 1988) | 12 |
| 702 | ( left|begin{array}{ccc}cos C & tan A & 0 \ sin B & 0 & -tan A \ 0 & sin B & cos Cend{array}right| ) has the value ( mathbf{A} cdot mathbf{0} ) B. ( c cdot sin A sin B cos B ) D. none of these | 12 |
| 703 | ( mathbf{f} boldsymbol{x}, boldsymbol{y}, boldsymbol{z} in boldsymbol{R} & boldsymbol{Delta}= ) ( left|begin{array}{ccc}boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x}+boldsymbol{y}+boldsymbol{z} \ mathbf{2} boldsymbol{x} & mathbf{5} boldsymbol{x}+mathbf{2} boldsymbol{y} & mathbf{7} boldsymbol{x}+mathbf{5} boldsymbol{y}+mathbf{2} z \ mathbf{3} boldsymbol{x} & mathbf{7} boldsymbol{x}+mathbf{3} boldsymbol{y} & mathbf{9} boldsymbol{x}+mathbf{7} boldsymbol{y}+mathbf{3} boldsymbol{z}end{array}right|=-mathbf{1} mathbf{6} ) then value of ( x ) is A . -2 B. -3 ( c cdot 2 ) ( D ) | 12 |
| 704 | If ( A ) is a ( 3 * 3 ) singular matrix then ( boldsymbol{A}(boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A})= ) ( A cdot ) Det ( A ) B. ( c cdot c ) D. ±1 | 12 |
| 705 | If ( A, B, C ) are the angles of a triangle, then the value of determinant ( left|begin{array}{ccc}-1+cos B & cos C+cos B & cos B \ cos C+cos A & -1+cos A & cos A \ -1+cos B & -1+cos A & -1end{array}right| ) ( mathbf{A} cdot mathbf{0} ) B. ( c .-1 ) ( D ) | 12 |
| 706 | 20. Tf Pisa 3 x 3 matrix such that pl = 2P+1, where PT is the transpose of P and I is the 3 x 3 identity matrix, then there x] rol exists a column matrix X = y 20 such that z] [0] (2012) (b) PX=X (a) PX=0 0 (c) PX= 2X (d) PX=-X | 12 |
| 707 | Minor ( m_{33} ) of the determinant ( left|begin{array}{ccc}2 & 3 & 5 \ 2 & -1 & 8 \ 1 & 2 & 4end{array}right| ) is | 12 |
| 708 | 2. l, m, n are the pth, qth and th term of a G. P. all positive, log 1 p 1 then log m q 1 equals [2002] log nr 1 (a) -1 (b) 2 (C) 1 (d) 0 | 12 |
| 709 | ( mathbf{A}=left[begin{array}{ccc}-boldsymbol{q} boldsymbol{r} & boldsymbol{p}(boldsymbol{q}+boldsymbol{r}) & boldsymbol{p} boldsymbol{r}+boldsymbol{p} boldsymbol{q} \ boldsymbol{p} boldsymbol{q}+boldsymbol{q} boldsymbol{r} & -boldsymbol{p} boldsymbol{r} & boldsymbol{p} boldsymbol{q}+boldsymbol{q} boldsymbol{r} \ boldsymbol{q} boldsymbol{r}+boldsymbol{p} boldsymbol{r} & boldsymbol{q} boldsymbol{r}+boldsymbol{p} boldsymbol{r} & boldsymbol{-} boldsymbol{p} boldsymbol{q}end{array}right] ) then ( |boldsymbol{A}| ) equals: ( ^{A} cdotleft(sum p qright)^{2} ) ( ^{mathrm{B}}left(sum p^{2} q^{2}right)^{2} ) ( c cdotleft[sum(q r)^{3}right. ) ( ^{mathrm{D}} cdotleft(sum p qright)^{frac{3}{2}} ) | 12 |
| 710 | Let ( a, b, c ) be real numbers with ( a^{2}+ ) ( b^{2}+c^{2}=1 ) show that the equation ( mid begin{array}{ccc}boldsymbol{a} boldsymbol{x}-boldsymbol{b} boldsymbol{y}-boldsymbol{c} & boldsymbol{b} boldsymbol{x}+boldsymbol{a} boldsymbol{y} & boldsymbol{c} boldsymbol{x}+ \ boldsymbol{b} boldsymbol{x}+boldsymbol{a} boldsymbol{y} & boldsymbol{-} boldsymbol{a} boldsymbol{x}+boldsymbol{b} boldsymbol{y}-boldsymbol{c} & boldsymbol{c} boldsymbol{y}+ \ boldsymbol{c} boldsymbol{x}+boldsymbol{a} & boldsymbol{c} boldsymbol{y}+boldsymbol{b} & boldsymbol{-} boldsymbol{a} boldsymbol{x}-boldsymbol{b}end{array} ) represents a straight line. | 12 |
| 711 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{3} & mathbf{2} \ mathbf{1} & mathbf{4}end{array}right], ) then ( boldsymbol{A}(boldsymbol{A} boldsymbol{d} boldsymbol{j} cdot boldsymbol{A}) ) equals- A. ( left[begin{array}{cc}10 & 0 \ 0 & 10end{array}right] ) В. ( left[begin{array}{cc}0 & 10 \ 10 & 0end{array}right] ) c. ( left[begin{array}{ll}10 & 1 \ 1 & 10end{array}right] ) D. none of these | 12 |
| 712 | ( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{alpha} & boldsymbol{beta} & gamma \ boldsymbol{alpha} & boldsymbol{x}+boldsymbol{beta} & gamma \ boldsymbol{alpha} & boldsymbol{beta} & boldsymbol{x}+boldsymbol{gamma}end{array}right|=mathbf{0}, ) then ( boldsymbol{x} ) is equal to ( mathbf{A} cdot 0,-(alpha+beta+gamma) ) B. ( 0, alpha+beta+gamma ) c. ( 1, alpha+beta+gamma ) D. ( 0, alpha^{2}+beta^{2}+gamma^{2} ) | 12 |
| 713 | Find ( left|begin{array}{lll}mathbf{4} & mathbf{6} & mathbf{2} \ mathbf{3} & mathbf{0} & mathbf{9} \ mathbf{7} & mathbf{1} & mathbf{5}end{array}right| ) | 12 |
| 714 | ( mathbf{f} mathbf{A}=left[begin{array}{lll}mathbf{1} & mathbf{5} & mathbf{- 6} \ mathbf{- 8} & mathbf{0} & mathbf{4} \ mathbf{3} & mathbf{- 7} & mathbf{2}end{array}right], ) then the cofactor of ( -7= ) A . 44 B. 43 ( c cdot 40 ) D. 39 | 12 |
| 715 | Let ( f(x)=left|begin{array}{ccc}sec ^{2} x & 1 & 1 \ cos ^{2} x & cos ^{2} x & operatorname{cosec}^{2} x \ 1 & cos ^{2} x & cot ^{2} xend{array}right| ) then | 12 |
| 716 | Solve the following determinant : ( begin{array}{|ccc|}15 & 11 & 7 \ 11 & 17 & 14 \ 10 & 16 & 13end{array} ) | 12 |
| 717 | (2000 – TV US) 25. Let a, b, c be real numbers with a2 + b2 + c2= 1. Show that the equation Jax – by-c bx + ay cxta b6+ay ax + by-c cytb axta cy+b 1=0 – ax – bytc. represents a straight line (20016 Marts) | 12 |
| 718 | A matrix A of order ( 3 times 3 ) has determinant 6. What is the value of ( |mathbf{3} boldsymbol{A}| ) ( ? ) | 12 |
| 719 | If ( a neq b neq c, ) the value of ( x ) which satisfies the question ( left|begin{array}{ccc}mathbf{0} & boldsymbol{x}-boldsymbol{a} & boldsymbol{x}-boldsymbol{b} \ boldsymbol{x}+boldsymbol{a} & mathbf{0} & boldsymbol{x}-boldsymbol{c} \ boldsymbol{x}+boldsymbol{b} & boldsymbol{x}+boldsymbol{c} & mathbf{0}end{array}right|=mathbf{0} ) is ( mathbf{A} cdot x=0 ) B. ( x=a ) ( mathbf{c} cdot x=b ) D. ( x=c ) | 12 |
| 720 | Find the value of the following determinant: ( left|begin{array}{cc}-4 & frac{-6}{35} \ 7 & frac{35}{5} \ 5 & 5end{array}right| ) A ( cdot frac{15}{34} ) в. ( frac{32}{45} ) c. ( frac{25}{33} ) D. ( frac{38}{35} ) | 12 |
| 721 | Eliminating ( a, b, c ) from ( x=frac{a}{b-c}, y= ) ( frac{b}{c-a}, z=frac{c}{a-b} ) we get ( mathbf{A} cdotleft|begin{array}{lll}1 & -x & x \ 1 & -y & y \ 1 & -z & zend{array}right|=0 ) B. ( left|begin{array}{ccc}1 & -x & x \ 1 & 1 & -y \ 1 & z & 1end{array}right|=0 ) c. ( left|begin{array}{ccc}1 & -x & x \ y & 1 & -y \ -z & z & 1end{array}right|=0 ) D. none of these | 12 |
| 722 | Using the factor theorem it is found that ( b+c, c+a ) and ( a+b ) are three factors of the determinant ( left|begin{array}{ccc}-2 a & a+b & a+c \ b+a & -2 b & b+c \ c+a & c+b & -2 cend{array}right| . ) The other factor in the value of the determinant is A . 4 B. 2 ( mathbf{c} cdot a+b+c ) D. none of these | 12 |
| 723 | ( left|begin{array}{lll}mathbf{1} & boldsymbol{omega} & boldsymbol{omega}^{2} \ boldsymbol{omega} & boldsymbol{omega}^{2} & mathbf{1} \ boldsymbol{omega}^{2} & mathbf{1} & boldsymbol{omega}end{array}right|=ldots ) ( A ) B. ( c cdot 2 ) D. -1 | 12 |
| 724 | Numbers of ways in which 75600 can be resolved as product of two divisors which are relatively prime? A .44 B. 8 ( c .9 ) D. 16 | 12 |
| 725 | If ( a, b, c ) all are non-zero and unequal and ( left|begin{array}{ccc}mathbf{1}+boldsymbol{a} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{b} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{c}end{array}right|=mathbf{0}, ) then ( 1+frac{1}{a}+frac{1}{b}+frac{1}{c} ) is equal to? | 12 |
| 726 | Prove that ( left|begin{array}{ccc}mathbf{1}+boldsymbol{a} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{b} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{c}end{array}right|= ) ( boldsymbol{a b c}left(mathbf{1}+frac{mathbf{1}}{boldsymbol{a}}+frac{mathbf{1}}{boldsymbol{b}}+frac{mathbf{1}}{boldsymbol{c}}right) ) | 12 |
| 727 | f ( boldsymbol{A}+boldsymbol{B}+boldsymbol{C}=boldsymbol{pi}, ) then ( left|begin{array}{ccc}sin (boldsymbol{A}+boldsymbol{B}+boldsymbol{C}) & sin boldsymbol{B} & cos boldsymbol{C} \ -sin boldsymbol{B} & boldsymbol{0} & tan boldsymbol{A} \ cos (boldsymbol{A}+boldsymbol{B}) & -tan boldsymbol{A} & 0end{array}right| ) equal to ( mathbf{A} cdot mathbf{0} ) B. ( 2 sin B tan A cos C ) ( c .1 ) D. None of these | 12 |
| 728 | Evaluate the following determinant: [ left|begin{array}{ccc} 102 & 18 & 36 \ 1 & 3 & 4 \ 17 & 3 & 6 end{array}right| ] | 12 |
| 729 | Evaluate the following: ( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda}end{array}right| ) | 12 |
| 730 | Evaluate the following determinant: ( left|begin{array}{ccc}1 & -3 & 2 \ 4 & -1 & 2 \ 3 & 5 & 2end{array}right| ) | 12 |
| 731 | f the matrix ( boldsymbol{A}=left[begin{array}{ccc}mathbf{6} & boldsymbol{x} & mathbf{2} \ mathbf{2} & mathbf{- 1} & mathbf{2} \ mathbf{- 1 0} & mathbf{5} & mathbf{2}end{array}right] ) is singular matrix. Find the value of ( x ) | 12 |
| 732 | How that the points ( boldsymbol{P}(-boldsymbol{2}, boldsymbol{3}, boldsymbol{5}), boldsymbol{Q}(boldsymbol{1}, boldsymbol{2}, boldsymbol{3}) ) and ( boldsymbol{R}(boldsymbol{7}, boldsymbol{0},-boldsymbol{1}) ) are collinear. | 12 |
| 733 | Evaluate ( left[begin{array}{cc}sqrt{mathbf{3}} & sqrt{mathbf{5}} \ -sqrt{mathbf{5}} & mathbf{3} sqrt{mathbf{3}}end{array}right] ) | 12 |
| 734 | ( boldsymbol{I f} mathbf{A}=left[begin{array}{ll}mathbf{1} & mathbf{3} \ mathbf{2} & mathbf{1}end{array}right], ) then the determinant ( mathbf{A}^{2}-2 mathbf{A}: ) A . 5 B . 25 ( c .-5 ) D. -25 | 12 |
| 735 | Using properties of determinants, prove the following: ( left|begin{array}{ccc}a^{2} & b c & a c+c^{2} \ a^{2}+a b & b^{2} & a c \ a b & b^{2}+b c & c^{2}end{array}right|=4 a^{2} b^{2} c^{2} ) | 12 |
| 736 | rove: ( left|begin{array}{ccc}-cos alpha & sin beta & 0 \ 0 & -sin alpha & cos alpha \ sin alpha & 0 & -sin betaend{array}right|=0 ) | 12 |
| 737 | Evaluate the determinants ( left|begin{array}{cc}2 & 4 \ -5 & -1end{array}right| ) | 12 |
| 738 | Evaluate the following: ( left|begin{array}{ccc}boldsymbol{x} & mathbf{1} & mathbf{1} \ mathbf{1} & boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{1} & boldsymbol{x}end{array}right| ) | 12 |
| 739 | Using the properties of determinants, find the value of ( left|begin{array}{ccc}mathbf{0} & boldsymbol{a} & -boldsymbol{b} \ -boldsymbol{a} & boldsymbol{0} & -boldsymbol{c} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{0}end{array}right| ) | 12 |
| 740 | If a + p, b #9, C # r and TP a a b cl q c = 0. Then find the b rl P value of (1991 – 4 Marks) p-aq-br-c | 12 |
| 741 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & -mathbf{3} \ mathbf{- 4} & mathbf{1}end{array}right], ) then adj ( left(mathbf{3} boldsymbol{A}^{2}+right. ) ( 12 A) ) is equal to. ( mathbf{A} cdotleft[begin{array}{cc}72 & -84 \ -63 & 51end{array}right] ) В. ( left[begin{array}{ll}51 & 63 \ 84 & 72end{array}right] ) c. ( left[begin{array}{ll}51 & 84 \ 63 & 72end{array}right] ) D. ( left[begin{array}{cc}72 & -63 \ -84 & 51end{array}right] ) | 12 |
| 742 | Maximum value of a second order determinant whose every element is either 0,1 or 2 only is: A. B. ( c cdot 2 ) D. 4 | 12 |
| 743 | Find the determinant: ( left|begin{array}{ccc}1 & x & x^{2} \ 1 & y & y^{2} \ 1 & z & z^{2}end{array}right| ) | 12 |
| 744 | Write the value of the determinant [ left[begin{array}{cc} boldsymbol{p} & boldsymbol{p}+mathbf{1} \ boldsymbol{p}-mathbf{1} & boldsymbol{p} end{array}right] ] when ( p=1342 ) | 12 |
| 745 | ( mathbf{a}=left|begin{array}{ccc}boldsymbol{a} & mathbf{5}-boldsymbol{i} & mathbf{7}+boldsymbol{i} \ mathbf{5}+boldsymbol{i} & boldsymbol{b} & mathbf{3}+boldsymbol{i} \ mathbf{7}-boldsymbol{i} & boldsymbol{3}-boldsymbol{i} & boldsymbol{c}end{array}right|, ) then ( boldsymbol{Delta} ) is always A . real B. imaginary ( c cdot 0 ) D. None of these | 12 |
| 746 | Evaluate the following determinant: ( left|begin{array}{ccc}1 & 3 & 5 \ 2 & 6 & 10 \ 31 & 11 & 38end{array}right| ) | 12 |
| 747 | If ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) is a ( mathbf{4} times mathbf{4} ) matrix and ( boldsymbol{c}_{i j} ) is the co-factor of the element ( boldsymbol{a}_{boldsymbol{i} j} ) in ( |boldsymbol{A}| ) then the expression ( a_{11} c_{11}+a_{12} c_{12}+ ) ( boldsymbol{a}_{13} boldsymbol{c}_{13}+boldsymbol{a}_{14} boldsymbol{c}_{14} ) equals ( mathbf{A} cdot mathbf{0} ) B. – c. 1 D. ( |A| ) | 12 |
| 748 | toppr Q Type your question List I The value of the determinant A. ( quadleft|begin{array}{lll}x+2 & x+3 & x+5 \ x+4 & x+6 & x+9 \ x+8 & x+11 & x+15end{array}right| ) is If one of the roots of the equation B. ( left|begin{array}{ccc}7 & 6 & x^{2}-13 \ 2 & x^{2}-13 & 2 \ x^{2}-13 & 3 & 7end{array}right|=0 ) is ( x+2 ), then the sum of all other five roots is The value of C. ( quadleft|begin{array}{ccc}sqrt{6} & 2 i & 3+sqrt{6} \ sqrt{12} & sqrt{3}+sqrt{8} i & 3 sqrt{2}+sqrt{6} i \ sqrt{18} & sqrt{2}+sqrt{12} i & sqrt{27}+2 iend{array}right| ) If ( f(theta)= ) D. ( quadleft|begin{array}{ccc}cos ^{2} theta & cos theta sin theta & -sin theta \ cos theta sin theta & sin ^{2} theta & cos theta \ sin theta & -cos theta & 0end{array}right| ) then ( f(pi / 3) ) ( mathbf{A} cdot A-i v, B-i i i, C-i i, D-i ) B . ( A-i, B-i i i, C-i i, D-i v ) c. ( A-i i, B-i i i, C-i v, D-i ) D. ( A-i i i, B-i v, C-i i, D-i ) | 12 |
| 749 | If ( boldsymbol{a} boldsymbol{x}_{1}^{2}+boldsymbol{b} boldsymbol{y}_{1}^{2}+boldsymbol{c} boldsymbol{z}_{1}^{2}=boldsymbol{a} boldsymbol{x}_{2}^{2}+boldsymbol{b} boldsymbol{y}_{2}^{2}+ ) ( boldsymbol{c} boldsymbol{z}_{2}^{2}=boldsymbol{a} boldsymbol{x}_{3}^{2}+boldsymbol{b} boldsymbol{y}_{3}^{2}+boldsymbol{c} boldsymbol{z}_{3}^{2}=boldsymbol{d} ) and ( boldsymbol{a} boldsymbol{x}_{2} boldsymbol{x}_{3}+boldsymbol{b} boldsymbol{y}_{2} boldsymbol{y}_{3}+boldsymbol{c} boldsymbol{z}_{2} boldsymbol{z}_{3}=boldsymbol{a} boldsymbol{x}_{3} boldsymbol{x}_{1}+ ) ( boldsymbol{b} boldsymbol{y}_{3} boldsymbol{y}_{1}+boldsymbol{c} boldsymbol{z}_{3} boldsymbol{z}_{1}=boldsymbol{a} boldsymbol{x}_{1} boldsymbol{x}_{2}+boldsymbol{b} boldsymbol{y}_{1} boldsymbol{y}_{2}+ ) ( boldsymbol{c} boldsymbol{z}_{1} boldsymbol{z}_{2}=boldsymbol{f}, ) then prove that ( left|begin{array}{lll}boldsymbol{x}_{1} & boldsymbol{y}_{1} & boldsymbol{z}_{1} \ boldsymbol{x}_{2} & boldsymbol{y}_{2} & boldsymbol{z}_{2} \ boldsymbol{x}_{3} & boldsymbol{y}_{3} & boldsymbol{z}_{3}end{array}right|=(boldsymbol{d}- ) ( f)left[frac{d+2 f}{a b c}right]^{1 / 2}(boldsymbol{a}, boldsymbol{b}, boldsymbol{c} neq mathbf{0}) ) | 12 |
| 750 | Prove that ( left|begin{array}{ccc}boldsymbol{a}^{2}+mathbf{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{a} boldsymbol{c} \ boldsymbol{a} boldsymbol{b} & boldsymbol{b}^{2}+mathbf{1} & boldsymbol{b} boldsymbol{c} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c b} & boldsymbol{c}^{2}+1end{array}right|= ) ( mathbf{1}+boldsymbol{a}^{2}+boldsymbol{b}^{2}+boldsymbol{c}^{2} ) | 12 |
| 751 | Assertion ( begin{array}{ccc}mathbf{Delta}= & & \ & sin pi & cos (boldsymbol{x}+boldsymbol{pi} / mathbf{4}) & tan (boldsymbol{x}- \ sin (boldsymbol{x}-boldsymbol{pi} / mathbf{4}) & -cos (boldsymbol{pi} / mathbf{2}) & log (boldsymbol{x} \ cot (boldsymbol{pi} / mathbf{4}+boldsymbol{x}) & log (boldsymbol{y} / boldsymbol{x}) & tan end{array} ) 0 Reason A skew symmetric determinant of odd order equals 0 A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Both Assertion and Reason are incorrect | 12 |
| 752 | The system of equations a x + y +z= a – 1 x + ay+z= a -1 x+y+ az= a – 1 has infinite solutions, if a is (a) -2 (c) not-2 2 (b) either – 2 or 1 (d) 1 | 12 |
| 753 | If ( n ) is a positive integer, then the value of the determinant ( left|begin{array}{ccc}boldsymbol{a}^{n}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+1}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+mathbf{2}}-boldsymbol{x} \ boldsymbol{a}^{boldsymbol{n}+mathbf{3}}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+mathbf{4}}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+mathbf{5}}-boldsymbol{x} \ boldsymbol{a}^{boldsymbol{n}+mathbf{6}}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+mathbf{7}}-boldsymbol{x} & boldsymbol{a}^{boldsymbol{n}+boldsymbol{8}}-boldsymbol{x}end{array}right|= ) ( mathbf{A} cdot mathbf{1} ) B. ( c .-1 ) D. None of these | 12 |
| 754 | f ( x, y ) and ( z ) are all distinct and ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{x}^{2} & boldsymbol{1}+boldsymbol{x}^{3} \ boldsymbol{y} & boldsymbol{y}^{2} & boldsymbol{1}+boldsymbol{y}^{3} \ boldsymbol{z} & boldsymbol{z}^{2} & boldsymbol{1}+boldsymbol{z}^{3}end{array}right|=mathbf{0}, ) then the value of ( x y z ) is. A . -4 B. – ( c .-2 ) D. – – | 12 |
| 755 | Assertion ( operatorname{Let} boldsymbol{A}=left[begin{array}{ccc}mathbf{0} & boldsymbol{c} & -boldsymbol{b} \ -boldsymbol{c} & boldsymbol{0} & boldsymbol{a} \ boldsymbol{b} & boldsymbol{-} boldsymbol{a} & boldsymbol{0}end{array}right] ) and ( boldsymbol{t} in boldsymbol{C} ) ( boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{t} boldsymbol{I}-boldsymbol{A})=boldsymbol{t}^{2} boldsymbol{I}+boldsymbol{t} boldsymbol{A}+boldsymbol{A}^{2}+left(boldsymbol{a}^{2}+right. ) ( left.b^{2}+c^{2}right) I ) Reason ( boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A}=boldsymbol{A}^{2}+left(boldsymbol{a}^{2}+boldsymbol{b}^{2}+boldsymbol{c}^{2}right) boldsymbol{I} ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 756 | Evaluate ( left|begin{array}{cc}sin 60^{circ} & cos 60^{circ} \ -sin 30^{circ} & cos 30^{circ}end{array}right| ) | 12 |
| 757 | For what value of ( x, ) will the points ( (-1, x) ) (-3,2) and (-4,4) lie on a line? ( A cdot-3 ) B. 3 ( c cdot-2 ) D. | 12 |
| 758 | If ( boldsymbol{A} ) is matrix of order ( boldsymbol{3}, ) then ( boldsymbol{d} boldsymbol{e} boldsymbol{t}(boldsymbol{k} boldsymbol{A}) ) is: A ( cdot k^{3} d e t(A) ) B . ( k^{2} d e t(A) ) c. ( k d e t(A) ) D. ( operatorname{det}(A) ) | 12 |
| 759 | If ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{a}^{2} & mathbf{1}+boldsymbol{a}^{3} \ boldsymbol{b} & boldsymbol{b}^{2} & mathbf{1}+boldsymbol{b}^{3} \ boldsymbol{c} & boldsymbol{c}^{2} & boldsymbol{1}+boldsymbol{c}^{3}end{array}right|=mathbf{0} ) and vectors ( left(1, a, a^{2}right),left(1, b, b^{2}right) ) and ( left(1, c, c^{2}right) ) are non coplanar then ( a b c ) equals A . -1 B. 1 ( c cdot 0 ) D. | 12 |
| 760 | 8. Let @=- XV . Then the value of the determinant -1-02 (2002) (a) 30 (b) 3ola -1) (c) 302 (d) 3011–@) Ti | 12 |
| 761 | Prove that ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{b} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c}end{array}right|=boldsymbol{a}(boldsymbol{b}-boldsymbol{c})(boldsymbol{a}-boldsymbol{b}) ) Hence find the value of ( left|begin{array}{lll}mathbf{3} & mathbf{3} & mathbf{3} \ mathbf{3} & mathbf{5} & mathbf{5} \ mathbf{3} & mathbf{5} & mathbf{7}end{array}right| ) | 12 |
| 762 | Let ( n ) and ( r ) be two positive integers such that ( n geq r+2 . ) Suppose ( Delta(n, r)= ) ( left|begin{array}{ccc}n & C_{r} & ^{n} C_{r+1} & ^{n} C_{r+2} \ ^{n+1} C_{r} & ^{n+1} C_{r+1} & ^{n+1} C_{r+2} \ ^{n+2} C_{r} & ^{n+2} C_{r+1} & ^{n+2} C_{r+2}end{array}right| ) Show that ( Delta(n, r)=frac{n+2}{n+2} C_{3} Delta(n-2, r-1) ) Hence or otherwise, A ( cdot frac{left(^{n+2} C_{3}right)left(^{n+1} C_{3}right) ldots . .left(^{n-r+3} C_{3}right)}{left.left(r+2 C_{3}right)left(r+1 C_{3}right) ldots . .^{3} C_{3}right)} ) B. ( -frac{left(^{n+2} C_{3}right)left(^{n+1} C_{3}right) ldots . .left(^{n-r+3} C_{3}right)}{left(r+2 C_{3}right)left(^{r+1} C_{3}right) ldotsleft(^{3} C_{3}right)} ) c. ( frac{(n+3}{(r+3)}(sqrt[n+2]{(r+2)}) ldots . .left(^{n-r+3} C_{3}right) ) D. ( -frac{left.left(^{n+3} C_{3}right)(n+2)_{3}right) ldots . .left(^{n-r+3} C_{3}right)}{left(r+3 C_{3}right)left(r+2 C_{3}right) ldotsleft(^{3} C_{3}right)} ) | 12 |
| 763 | Show that ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2}end{array}right|=(boldsymbol{a}-boldsymbol{b})(boldsymbol{b}- ) ( c)(c-a) ) | 12 |
| 764 | Calculate the values of the determinants: ( left|begin{array}{cccc}mathbf{7} & mathbf{1 3} & mathbf{1 0} & mathbf{6} \ mathbf{5} & mathbf{9} & mathbf{7} & mathbf{4} \ mathbf{8} & mathbf{1 2} & mathbf{1 1} & mathbf{7} \ mathbf{4} & mathbf{1 0} & mathbf{6} & mathbf{3}end{array}right| ) | 12 |
| 765 | Consider the matrix ( A, B, C, D ) with order ( 2 times 3,3 times 4,4 times 4,4 times 2 ) respectively. Let ( boldsymbol{x}=left(boldsymbol{alpha} boldsymbol{A} boldsymbol{B} boldsymbol{gamma} boldsymbol{C}^{2} boldsymbol{D}right)^{3} ) where ( alpha ) and ( gamma ) are scalars. Let ( |x|= ) ( kleft|A B C^{2} Dright|^{3} . ) Then the value of ( k ) A ( cdot alpha^{6} gamma^{6} ) в. ( alpha gamma ) c. ( alpha^{3} gamma^{3} ) D. ( alpha^{6} gamma^{12} ) | 12 |
| 766 | Which of these are correct? (a) If any two rows or columns of a determinant are identical, then the value of the determinant is zero. (b) If the corresponding rows and columns of a determinant are interchanged, then the value of the determinant does not change. (c) If any two rows (or column) of a determinant are interchanged, then the value of the determinant changes in ( operatorname{sign} ) A ( cdot(a) ) and ( (b) ) B. ( (b) ) and ( (c) ) c. ( (a) ) and ( (c) ) D. ( (a),(b) ) and ( (c) ) | 12 |
| 767 | ( begin{array}{ccc}sin ^{2} x & cos ^{2} x & 1 \ cos ^{2} x & sin ^{2} x & 1 \ -10 & 12 & 2end{array} mid= ) ( mathbf{A} cdot mathbf{0} ) B. ( 12 cos ^{2} x-10 sin ^{2} x ) c. ( 12 sin ^{2} x-10 cos ^{2} x-2 ) D. ( 10 sin 2 x ) | 12 |
| 768 | Using properties of determinants, prove that [ boldsymbol{omega}left|begin{array}{ccc} boldsymbol{x} & boldsymbol{y} & boldsymbol{z} \ boldsymbol{x}^{2} & boldsymbol{y}^{2} & boldsymbol{z}^{2} \ boldsymbol{y}+boldsymbol{z} & boldsymbol{z}+boldsymbol{x} & boldsymbol{x}+boldsymbol{y} end{array}right|=(boldsymbol{x}- ] ( boldsymbol{y})(boldsymbol{y}-boldsymbol{z})(boldsymbol{z}-boldsymbol{x})(boldsymbol{x}+boldsymbol{y}+boldsymbol{z}) ) | 12 |
| 769 | Using properties of determinants, prove [ operatorname{that}left[begin{array}{ccc} boldsymbol{b}+boldsymbol{c} & boldsymbol{q}+boldsymbol{r} & boldsymbol{y}+boldsymbol{z} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{r}+boldsymbol{p} & boldsymbol{z}+boldsymbol{x} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{p}+boldsymbol{q} & boldsymbol{x}+boldsymbol{y} end{array}right]= ] ( mathbf{2}left[begin{array}{lll}boldsymbol{a} & boldsymbol{p} & boldsymbol{x} \ boldsymbol{b} & boldsymbol{q} & boldsymbol{y} \ boldsymbol{c} & boldsymbol{r} & boldsymbol{z}end{array}right] ) | 12 |
| 770 | If area of a triangle is 35 sq units with vertices (2,-6),(5,4) and ( (k, 4), ) then find the value of ( k ) | 12 |
| 771 | ( left|begin{array}{ccc}a^{2} & b^{2} & c^{2} \ (a+1)^{2} & (b+1)^{2} & (c+1)^{2} \ (a-1)^{2} & (b-1)^{2} & (c-1)^{2}end{array}right|= ) ( k(a-b)(b-c)(c-a), ) then find the value of ( -boldsymbol{k} ) | 12 |
| 772 | Evaluate [ left|begin{array}{ccc} 2 & 7 & 3 \ -4 & 3 & -1 \ 0 & -3 & 7 end{array}right| ] | 12 |
| 773 | Prove that: [ begin{array}{ccc} boldsymbol{a}+boldsymbol{b}+mathbf{2 c} & boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{b}+boldsymbol{c}+mathbf{2} boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{a} & boldsymbol{c}+boldsymbol{a}+boldsymbol{2} boldsymbol{b} end{array} mid= ] ( mathbf{2}(boldsymbol{a}+boldsymbol{b}+boldsymbol{c})^{3} ) | 12 |
| 774 | If ( boldsymbol{A}=left|begin{array}{ll}mathbf{3} & mathbf{4} \ mathbf{1} & mathbf{2}end{array}right|, ) find the value of ( mathbf{3}|boldsymbol{A}| ) | 12 |
| 775 | Let the three digit numbers ( A 28,3 B 9 ) and ( 62 C, ) where ( A, B, C ) are integers between 0 and 9 , be divisible by a fixed integer ( k, ) Show that the determinant ( left|begin{array}{lll}boldsymbol{A} & boldsymbol{3} & boldsymbol{6} \ boldsymbol{8} & boldsymbol{9} & boldsymbol{C} \ boldsymbol{2} & boldsymbol{B} & boldsymbol{2}end{array}right| ) is also divisible by the same integer ( boldsymbol{k} ) | 12 |
| 776 | ( P, Q, R ) are three collinear points. The coordinates of ( mathrm{P} ) and ( mathrm{R} ) are (3,4) and (11 10) respectively and PQ is equal to 2.5 units. Coordinates of ( Q ) are- A ( cdot(5,11 / 2) ) B. (11,5/2) c. ( (5,-11 / 2) ) D. ( (-5,11 / 2) ) | 12 |
| 777 | ( fleft(begin{array}{cc}x+y & x-y \ 2 x+z & x+zend{array}right)=left(begin{array}{cc}0 & 0 \ 1 & 1end{array}right), ) then the values of ( x, y ) and ( z ) are respectively ( mathbf{A} cdot 0,0,1 ) в. 1,1,0 ( mathrm{c} cdot-1,0,0 ) ( mathbf{D} cdot 0,0,0 ) | 12 |
| 778 | If the system of equations x-ky – z= 0, kx – y – z=0, x + y – z = 0 has a non-zero solution, then the possible values of k are (20005) (a) -1,2 (6) 1,2 (c) 0,1 (d) -1, 1 | 12 |
| 779 | ( mathbf{f} boldsymbol{A}=left[begin{array}{ccc}mathbf{1} & mathbf{- 1} & mathbf{2} \ mathbf{3} & mathbf{0} & -mathbf{2} \ mathbf{1} & mathbf{0} & mathbf{3}end{array}right] ) verify that ( boldsymbol{A}(boldsymbol{a} boldsymbol{d} boldsymbol{j} boldsymbol{A})=|boldsymbol{A}| boldsymbol{I} ) | 12 |
| 780 | The value of determinant is [ left|begin{array}{ccc} boldsymbol{a}+boldsymbol{b} & boldsymbol{a}+mathbf{2 b} & boldsymbol{a}+mathbf{3} boldsymbol{b} \ boldsymbol{a}+mathbf{2} boldsymbol{b} & boldsymbol{a}+mathbf{3} boldsymbol{b} & boldsymbol{a}+mathbf{4} boldsymbol{b} \ boldsymbol{a}+mathbf{4} boldsymbol{b} & boldsymbol{a}+mathbf{5} boldsymbol{b} & boldsymbol{a}+boldsymbol{6 b} end{array}right| ] | 12 |
| 781 | Using properties of determinants, prove the following: ( left(begin{array}{ccc}mathbf{1} & boldsymbol{x} & boldsymbol{x} \ mathbf{2} boldsymbol{x} & boldsymbol{x}(boldsymbol{x}-mathbf{1}) & boldsymbol{x}(boldsymbol{x} boldsymbol{x} \ mathbf{3} boldsymbol{x}(mathbf{1}-boldsymbol{x}) & boldsymbol{x}(boldsymbol{x}-mathbf{1})(boldsymbol{x}-mathbf{2}) & boldsymbol{x}(boldsymbol{x}+mathbf{1})end{array}right. ) ( 6 x^{2}left(1-x^{2}right) ) | 12 |
| 782 | A square matrix ( mathbf{B} ) of order ( mathbf{3}, ) has ( |boldsymbol{B}|= ) ( mathbf{7}, ) find ( mid boldsymbol{B} ) adj ( mathbf{B} mid ) | 12 |
| 783 | Find the maximum value of ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+sin boldsymbol{theta} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+cos boldsymbol{pi}end{array}right| ) | 12 |
| 784 | Assertion consider the determinant ( Delta= ) ( left|begin{array}{lll}a_{1}+b_{1} x^{2} & a_{1} x^{2}+b_{1} & c_{1} \ a_{2}+b_{2} x^{2} & a_{2} x^{2}+b_{2} & c_{2} \ a_{3}+b_{3} x^{2} & a_{3} x^{2}+b_{3} & c_{3}end{array}right|=0, ) where ( boldsymbol{x}_{i}, boldsymbol{y}_{i}, boldsymbol{z}_{i} in boldsymbol{R} quad(boldsymbol{i}=mathbf{1}, boldsymbol{2}, boldsymbol{3}), quad boldsymbol{x} in boldsymbol{R} ) The values of ( x ) satisfying ( Delta=0 ) are ( boldsymbol{x}=mathbf{1},-mathbf{1} ) Reason ( left|begin{array}{lll}boldsymbol{a}_{1} & boldsymbol{b}_{1} & boldsymbol{c}_{1} \ boldsymbol{a}_{2} & boldsymbol{b}_{2} & boldsymbol{c}_{2} \ boldsymbol{a}_{3} & boldsymbol{b}_{3} & boldsymbol{c}_{3}end{array}right|=mathbf{0}, ) then ( boldsymbol{Delta}=mathbf{0} ) A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B. Both Assertion and Reason are correct but Reason not the correct explanation for Assertion c. Assertion is correct but Reason is incorrect D. Assertion is incorrect but Reason is correct | 12 |
| 785 | If ( boldsymbol{A}=left[begin{array}{ll}2 & 0 \ 0 & 3end{array}right], ) then show that ( |3 A|= ) ( mathbf{9}|boldsymbol{A}| ) | 12 |
| 786 | f ( f(x)= ) ( left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{2} boldsymbol{x} & boldsymbol{x}-mathbf{1} & boldsymbol{x} \ mathbf{3} boldsymbol{x}(boldsymbol{x}-mathbf{1}) & (boldsymbol{x}-mathbf{1})(boldsymbol{x}-mathbf{2}) & boldsymbol{x}(boldsymbol{x}-mathbf{1})end{array}right| ) then ( boldsymbol{f}(mathbf{5 0})= ) 4.0 B. 2 ( c_{1} ) ( D ) ( E ) | 12 |
| 787 | If ( a neq b neq c, ) the value of ( x ) which satisfies the question ( left|begin{array}{ccc}mathbf{0} & boldsymbol{x}-boldsymbol{a} & boldsymbol{x}-boldsymbol{b} \ boldsymbol{x}+boldsymbol{a} & mathbf{0} & boldsymbol{x}-boldsymbol{c} \ boldsymbol{x}+boldsymbol{b} & boldsymbol{x}+boldsymbol{c} & mathbf{0}end{array}right|=mathbf{0} ) is ( mathbf{A} cdot x=0 ) B. ( x=a ) ( mathbf{c} cdot x=b ) D. ( x=c ) | 12 |
| 788 | uet (Padj ())=2 (u) uute 12. Let a, 2. uR. Consider the system ax +2y=a 3x – 2y = u T the system of linear equations ment(s) is (are) correct? (JEE Adv. 2016) tem has a unique solution for all 1= -3, then the system has infinitely many solutio for all values of 2 and u. (b) Ifa -3, then the system has a unique solution values of 2 andu. © If 2 +u = 0, then the system has infinitely many solutions for a = -3. (d) If a+u = 0, then the system has no solution for a=-3. | 12 |
| 789 | ( mathrm{f}left|begin{array}{ccc}boldsymbol{x} & boldsymbol{2} & boldsymbol{x} \ boldsymbol{x}^{2} & boldsymbol{x} & boldsymbol{6} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{6}end{array}right|=boldsymbol{a} boldsymbol{x}^{4}+boldsymbol{b} boldsymbol{x}^{3}+boldsymbol{c} boldsymbol{x}^{2}+ ) ( boldsymbol{d} boldsymbol{x}+boldsymbol{e}, ) then ( boldsymbol{5} boldsymbol{a}+boldsymbol{4} boldsymbol{b}+boldsymbol{3} boldsymbol{c}+boldsymbol{2} boldsymbol{d}+boldsymbol{e} ) is equal to A. 11 B. -11 c. 12 D. -12 E . 13 | 12 |
| 790 | If ( boldsymbol{A}=left[begin{array}{ll}mathbf{4} & mathbf{2} \ mathbf{3} & mathbf{3}end{array}right], ) then adj ( (operatorname{adj} boldsymbol{A}) ) is equal to ( mathbf{A} cdotleft[begin{array}{cc}3 & -2 \ -3 & 4end{array}right] ) в. ( left[begin{array}{ll}4 & 2 \ 3 & 3end{array}right] ) D. None of these | 12 |
| 791 | If ( boldsymbol{A}=left[begin{array}{cc}mathbf{2} & -mathbf{3} \ mathbf{4} & mathbf{1}end{array}right], ) then adjoint of matrix ( A ) is A. ( left[begin{array}{cc}1 & 3 \ -4 & 2end{array}right] ) В. ( left[begin{array}{cc}1 & -3 \ -4 & 2end{array}right] ) c. ( left[begin{array}{cc}1 & 3 \ 4 & -2end{array}right] ) D. ( left[begin{array}{cc}-1 & -3 \ -4 & 2end{array}right] ) | 12 |
| 792 | Show that ( left|begin{array}{lll}boldsymbol{b} boldsymbol{c} & boldsymbol{b}+boldsymbol{c} & mathbf{1} \ boldsymbol{c} boldsymbol{a} & boldsymbol{c}+boldsymbol{a} & mathbf{1} \ boldsymbol{a} boldsymbol{b} & boldsymbol{a}+boldsymbol{b} & mathbf{1}end{array}right|= ) ( (boldsymbol{a}-boldsymbol{b})(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}-boldsymbol{a}) ) | 12 |
| 793 | If ( |boldsymbol{A}|=2, ) where ( boldsymbol{A} ) is a ( mathbf{3} times mathbf{3} ) matrix. Then find ( |boldsymbol{3} boldsymbol{A}| ) | 12 |
| 794 | Find cofactors of the elements of the ( operatorname{matrix} boldsymbol{A}=left[begin{array}{ll}-1 & 2 \ -3 & 4end{array}right] ) | 12 |
| 795 | Prove that ( left|begin{array}{ccc}a^{2} & b c & a c+c^{2} \ a^{2}+a b & b^{2} & a c \ a b & b^{2}+b c & c^{2}end{array}right| ) ( =4 a^{2} b^{2} c^{2} ) | 12 |
| 796 | If ( d ) is the determinant of a square matrix A of order ( n ), then the determinant of its adjoint is: A . ( d^{n} ) B. ( d^{n-1} ) ( c cdot d^{n-2} ) D. ( d ) | 12 |
| 797 | In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into n determinants, where n has the value ( mathbf{A} cdot mathbf{1} ) B. 9 c. 16 D. 24 | 12 |
| 798 | Using properties of determinants prove that : ( left|begin{array}{lll}boldsymbol{x} & boldsymbol{a} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{x} & boldsymbol{a} \ boldsymbol{a} & boldsymbol{a} & boldsymbol{x}end{array}right|=(boldsymbol{x}+mathbf{2} boldsymbol{a})(boldsymbol{x}-boldsymbol{a})^{2} ) | 12 |
| 799 | Let ( boldsymbol{A}=left[boldsymbol{a}_{i j}right] ) and ( boldsymbol{B}=left[boldsymbol{b}_{boldsymbol{i} j}right] ) be two ( boldsymbol{3} times boldsymbol{3} ) real matrices such that ( b_{i j}= ) ( (3)^{(i+j-2)} a_{i j}, ) where ( i, j=1,2,3 . ) If the determinant of ( B ) is 81 , then the determinant of A is : A. ( 1 / 3 ) B. 3 c. ( 1 / 81 ) D. ( 1 / 9 ) | 12 |
| 800 | Evaluate the following determinant: ( left|begin{array}{lll}boldsymbol{a} & boldsymbol{h} & boldsymbol{g} \ boldsymbol{h} & boldsymbol{b} & boldsymbol{f} \ boldsymbol{g} & boldsymbol{f} & boldsymbol{c}end{array}right| ) | 12 |
| 801 | 30. If A is a 3 x 3 non-singular matrix such that AA’ = A’A and B=A-1 A’, then BB’ equals: [JEE M 2014 (a) B-1 (b) (B-1)’ (©) I+B (d) I | 12 |
| 802 | ( left|begin{array}{ccc}1 & a & a^{2}-b c \ 1 & b & b^{2}-c a \ 1 & c & c^{2}-a bend{array}right| ) is equal to ( A ) B . ( sum a^{2}(b-c) ) c. ( 2 sum a^{2}(b-c) ) D. ( -2 sum a b(a-b) ) | 12 |
| 803 | 43. Let a and ß be the roots of the equation x2 + x + 1 = 0. Then for y#0 in R, y+1 a B1 a y+B 1 is equal to: [JEEM 2019-9 April (M) IB 1 yta (a) y(y2 – 1) (b) y(y – 3) C) y (d) y – 1 | 12 |
| 804 | ( f x neq 0 ) and ( left|begin{array}{ccc}1 & x & 2 x \ 1 & 3 x & 5 x \ 1 & 3 & 4end{array}right|=0, ) then ( x= ) ( A ) B. – ( c cdot 2 ) ( D ldots-2 ) | 12 |
| 805 | Evaluate the following: ( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda}end{array}right| ) | 12 |
| 806 | If ( boldsymbol{x}+boldsymbol{a}+boldsymbol{b}+boldsymbol{c}=boldsymbol{0}, ) then what is the value of ( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a} & boldsymbol{x}+boldsymbol{b} & boldsymbol{c} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{x}+boldsymbol{c}end{array}right| ? ) ( mathbf{A} cdot mathbf{0} ) B ( cdot(A+B+C)^{2} ) c. ( A^{2}+B^{2}+C^{2} ) D. ( A+B+C-2 ) | 12 |
| 807 | Evaluate the following: ( left|begin{array}{ccc}boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{lambda}end{array}right| ) | 12 |
| 808 | Calculate the values of the determinants: ( left|begin{array}{cccc}mathbf{1} & mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{2} & mathbf{3} & mathbf{4} \ mathbf{1} & mathbf{3} & mathbf{6} & mathbf{1 0} \ mathbf{1} & mathbf{4} & mathbf{1 0} & mathbf{2 0}end{array}right| ) | 12 |
| 809 | ( operatorname{Det}left{begin{array}{ccc}-2 a & a+b & c+a \ b+a & -2 b & b+c \ c+a & c+b & -2 cend{array}right}= ) ( A cdot(a+b)(b+c)(c+a) ) В . ( (a-b)(b-c)(c-a) ) ( c cdot 4(a+b)(b+c)(c+a) ) D. ( 4(a-b)(b-c)(c-a) ) | 12 |
| 810 | [ begin{array}{c} operatorname{Let}left|begin{array}{ccc} a^{2}+1 & a b & a c \ a b & b^{2}+1 & b c \ a c & b c & c^{2}+1 end{array}right|=k+ \ a^{2}+b^{2}+c^{2} end{array} ] then ( 4 k ) is | 12 |
| 811 | Evaluate ( left|begin{array}{cc}sqrt{mathbf{6}} & sqrt{mathbf{5}} \ sqrt{mathbf{2 0}} & sqrt{mathbf{2 4}}end{array}right| ) | 12 |
| 812 | Find the value of ( x ) if ( left|begin{array}{ccc}mathbf{3} & mathbf{4} & mathbf{1} \ mathbf{0} & boldsymbol{x} & mathbf{8} \ mathbf{3} & mathbf{- 1} & mathbf{4}end{array}right|=mathbf{0} ) | 12 |
| 813 | ( boldsymbol{A}=left[begin{array}{ccc}mathbf{5} & mathbf{5} boldsymbol{alpha} & boldsymbol{alpha} \ mathbf{0} & boldsymbol{alpha} & mathbf{5} boldsymbol{alpha} \ mathbf{0} & mathbf{0} & mathbf{5}end{array}right] ; ) If ( left|boldsymbol{A}^{2}right|=mathbf{2 5}, ) then ( |boldsymbol{alpha}|= ) A . 5 B ( .5^{2} ) ( c cdot 1 ) ( D ) ( therefore ) | 12 |
| 814 | Solve for ( mathbf{y}:left|begin{array}{ccc}boldsymbol{x}+boldsymbol{y} & boldsymbol{x} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x}+boldsymbol{y} & boldsymbol{x} \ boldsymbol{x} & boldsymbol{x} & boldsymbol{x}+boldsymbol{y}end{array}right|= ) ( mathbf{1 6}(boldsymbol{3} boldsymbol{x}+boldsymbol{4}) ) | 12 |
| 815 | Show that: ( left|begin{array}{ccc}boldsymbol{a}^{2} & boldsymbol{b}^{2} & boldsymbol{c}^{2} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ mathbf{1} & mathbf{1} & mathbf{1}end{array}right|=-(boldsymbol{a}-boldsymbol{b})(boldsymbol{b}-boldsymbol{c})(boldsymbol{c}- ) ( boldsymbol{a}) ) | 12 |
| 816 | ( boldsymbol{A}=left|begin{array}{ccc}mathbf{1} & mathbf{1} & mathbf{1} \ mathbf{1} & mathbf{1}+boldsymbol{x} & mathbf{1} \ mathbf{1} & mathbf{1} & mathbf{1}+boldsymbol{y}end{array}right|= ) A . ( x y ) в. ( x+y ) c ( . x-y ) D. ( x^{2} y^{2} ) | 12 |
| 817 | The value of which of the following determinants can be non-zero? ( mathbf{A} cdotleft|begin{array}{lll}a_{1}+a_{2} & a_{2} & a_{1} \ a_{4}+a_{5} & a_{5} & a_{4} \ a_{7}+a_{8} & a_{8} & a_{7}end{array}right| ) ( mathbf{B} cdotleft|begin{array}{lll}a_{1}+2 a_{2} & a_{2} & a_{3} \ a_{4}+2 a_{5} & a_{5} & a_{6} \ a_{7}+2 a_{8} & a_{8} & a_{9}end{array}right| ) ( mathbf{C} cdotleft|begin{array}{lll}k a_{4} & k a_{5} & k a_{6} \ a_{4} & a_{5} & a_{6} \ a_{7} & a_{8} & a_{9}end{array}right| ) D. None of these | 12 |
| 818 | Solve ( left|begin{array}{ccc}mathbf{1} & boldsymbol{b} boldsymbol{c} & boldsymbol{a}(boldsymbol{b}+boldsymbol{c}) \ mathbf{1} & boldsymbol{c} boldsymbol{a} & boldsymbol{b}(boldsymbol{c}+boldsymbol{a}) \ mathbf{1} & boldsymbol{a} boldsymbol{b} & boldsymbol{c}(boldsymbol{a}+boldsymbol{b})end{array}right| ) ( mathbf{A} cdot mathbf{0} ) B. ( c cdot a b c ) D. ( a b+b c+c a ) | 12 |
| 819 | ( mathbf{f} boldsymbol{z}=left|begin{array}{ccc}mathbf{3}+mathbf{3} boldsymbol{i} & mathbf{5}-boldsymbol{i} & mathbf{7}-mathbf{3} i \ boldsymbol{i} & mathbf{2} boldsymbol{i} & -mathbf{3} i \ boldsymbol{3}-mathbf{2} boldsymbol{i} & mathbf{5}+boldsymbol{i} & mathbf{7}+mathbf{3} boldsymbol{i}end{array}right| ), then A. z is purely real B. z is purely imaginary ( c cdot 0 ) D. none of these | 12 |
| 820 | Using properties of determinants, prove [ text { the following: }left|begin{array}{ccc} mathbf{1} & mathbf{1} & mathbf{1} \ boldsymbol{a} & boldsymbol{b} & boldsymbol{c} \ boldsymbol{a}^{mathbf{3}} & boldsymbol{b}^{mathbf{3}} & boldsymbol{c}^{3} end{array}right|=(boldsymbol{a}- ] ( b)(b-c)(c-a)(a+b+c) ) | 12 |
| 821 | 2. If (71) is a cube root of unity, then 1 1+i+02 02 (1995 -i -i +0-1 -1 (a) 0 (6) 1 (c) i (d) w | 12 |
| 822 | ( operatorname{Let} boldsymbol{A}=left[begin{array}{ll}boldsymbol{a} & boldsymbol{b} \ boldsymbol{c} & boldsymbol{d}end{array}right] ) and ( boldsymbol{f}(boldsymbol{lambda})=operatorname{det}(boldsymbol{A}- ) ( boldsymbol{lambda} boldsymbol{I}), ) then This question has multiple correct options A ( cdot f(lambda)=lambda^{2}-(a+d) lambda+a d-b c ) в. ( f(A)=O ) c. ( A^{2}=O ) implies ( A^{r}=O forall r geq 2 ) D. none of these | 12 |
| 823 | 7. The number of distinct real roots of sin x cos x cos x VI s cos x cos x sin x cos x = 0 in the interval cos x sin x b. 2 d. 3 (IIT-JEE 2001) a. 0 c. 1 | 12 |
| 824 | Consider the determinant ( Delta= ) ( left|begin{array}{lll}a_{1} & a_{2} & a_{3} \ b_{1} & b_{2} & b_{3} \ c_{1} & c_{2} & c_{3}end{array}right| ) ( M_{i j}= ) Minor of the element of ( i^{t h} ) row ( & ) ( j^{t h} ) column ( C_{i j}= ) Cofactor of element of ( i^{t h} ) row ( & ) ( j^{t h} ) column ( a_{2} cdot C_{12}+b_{2} cdot C_{22}+c_{2} cdot C_{32} ) is equal to A. B. ( Delta ) ( c cdot 2 Delta ) D. ( Delta^{2} ) | 12 |
| 825 | ( operatorname{Let} boldsymbol{D}_{boldsymbol{r}}=left|begin{array}{ccc}mathbf{2}^{boldsymbol{r}-mathbf{1}} & mathbf{2} mathbf{.} boldsymbol{3}^{boldsymbol{r}-mathbf{1}} & boldsymbol{4} mathbf{.} mathbf{5}^{boldsymbol{r}-mathbf{1}} \ boldsymbol{alpha} & boldsymbol{beta} & boldsymbol{gamma} \ mathbf{2}^{boldsymbol{n}}-mathbf{1} & boldsymbol{3}^{boldsymbol{n}}-mathbf{1} & mathbf{5}^{boldsymbol{n}}-mathbf{1}end{array}right| ) Then, the value of ( sum_{r=1}^{n} D_{r} ) is ( mathbf{A} cdot alpha beta mathbf{gamma} ) B . ( 2^{n} alpha+3^{n} beta+4^{n} gamma ) ( c cdot 2 alpha+3 beta+4 gamma ) D. None of these | 12 |
| 826 | ( f(x)=left|begin{array}{lll}a^{-x} & e^{x ln a} & x^{2} \ a^{-3 x} & e^{3 x ln a} & x^{4} \ a^{-5 x} & e^{5 x ln a} & 1end{array}right|, ) then A. ( f(x) cdot f(-x)=0 ) B. ( f(x)-f(-x)=0 ) c. ( f(x)+f(-x)=0 ) D. None of these | 12 |
| 827 | Prove that [ left|begin{array}{ccc} boldsymbol{x}+boldsymbol{y}+mathbf{2} boldsymbol{z} & boldsymbol{x} & boldsymbol{y} \ boldsymbol{z} & boldsymbol{y}+boldsymbol{z}+mathbf{2} boldsymbol{x} & boldsymbol{y} \ boldsymbol{z} & boldsymbol{x} & boldsymbol{z}+boldsymbol{x}+boldsymbol{2} boldsymbol{y} end{array}right| ] ( 2(x+y+z)^{3} ) | 12 |
| 828 | ( left|begin{array}{ccc}1 ! & 2 ! & 3 ! \ 2 ! & 3 ! & 4 ! \ 3 ! & 4 ! & 5 !end{array}right|=2016 K ) then value of ( boldsymbol{K} ) is A . 24 B. 84 c. ( frac{1}{24} ) D. ( frac{1}{84} ) | 12 |
| 829 | If the lines ( 3 x+2 y-5=0,2 x-5 y+ ) ( mathbf{3}=mathbf{0}, mathbf{5 x}+mathbf{b y}+mathbf{c}=mathbf{0} ) are concurrent then ( mathbf{b}+mathbf{c}= ) A. 7 B. – – ( c cdot 6 ) D. | 12 |
| 830 | Find the values of the following determinants ( left|begin{array}{cc}mathbf{2} boldsymbol{i} & -mathbf{3} i \ boldsymbol{i}^{3} & -mathbf{2} boldsymbol{i}^{5}end{array}right| ) where ( boldsymbol{i}=sqrt{-mathbf{1}} ) | 12 |
| 831 | Evaluate the following: ( left|begin{array}{ccc}1 & a & b c \ 1 & b & c a \ 1 & c & a bend{array}right| ) | 12 |
| 832 | [ mathbf{f} boldsymbol{a}_{mathbf{1}} boldsymbol{f}_{mathbf{1}}(boldsymbol{x})+boldsymbol{a}_{mathbf{2}} boldsymbol{f}_{mathbf{2}}(boldsymbol{x})+boldsymbol{a}_{mathbf{3}} boldsymbol{f}_{mathbf{3}}(boldsymbol{x})=mathbf{0} ] where ( a_{1}, a_{2}, a_{3} ) are constants (not all zero) and ( f_{1}, f_{2}, f_{3} ) are twice differentiable functions.Then ( D= ) ( left|begin{array}{ccc}boldsymbol{f}_{1}(boldsymbol{x}) & boldsymbol{f}_{2}(boldsymbol{x}) & boldsymbol{f}_{3}(boldsymbol{x}) \ boldsymbol{D} boldsymbol{f}_{1}(boldsymbol{x}) & boldsymbol{D} boldsymbol{f}_{2}(boldsymbol{x}) & boldsymbol{D} boldsymbol{f}_{3}(boldsymbol{x}) \ boldsymbol{D}^{2} boldsymbol{f}_{1}(boldsymbol{x}) & boldsymbol{D}^{2} boldsymbol{f}_{2}(boldsymbol{x}) & boldsymbol{D}^{2} boldsymbol{f}_{3}(boldsymbol{x})end{array}right| ) qual to ( D f_{1}(x)=frac{d}{d x} f_{1} ) | 12 |
| 833 | The value of ( frac{1}{x-y}left|begin{array}{ccc}1 & 0 & 0 \ 3 & x^{3} & 1 \ 5 & y^{3} & 1end{array}right| ) is ( mathbf{A} cdot x+y ) B . ( x^{2}-x y+y^{2} ) c. ( x^{2}+x y+y^{2} ) D. ( x^{3}-y^{3} ) | 12 |
| 834 | ( fleft(a_{1}, a_{2}, a_{3}, dots, a_{n}, dots ) are in GP, then right. the value of the determinant ( begin{array}{|ccc|}log a_{n} & log a_{n+1} & log a_{n+2} \ log a_{n+3} & log a_{n+4} & log a_{n+5} \ log a_{n+6} & log a_{n+7} & log a_{n+8}end{array} mid ), is A . 0 B. 1 ( c cdot 2 ) ( D .-2 ) | 12 |
| 835 | If ( boldsymbol{A}=left[begin{array}{ll}2 & 5 \ 2 & 1end{array}right] ) and ( B=left[begin{array}{cc}4 & -3 \ 2 & 5end{array}right], ) verify that ( |boldsymbol{A B}|=|boldsymbol{A}||boldsymbol{B}| ) | 12 |
| 836 | Find the value of ( lambda ) for which the points ( (6,-1,2),(8,-7, lambda) ) and (5,2,4) are collinear. | 12 |
| 837 | ( left|begin{array}{ccc}mathbf{1}+boldsymbol{i} & mathbf{1}-boldsymbol{i} & mathbf{1} \ mathbf{1}-boldsymbol{i} & boldsymbol{i} & mathbf{1}+boldsymbol{i} \ boldsymbol{i} & mathbf{1}+boldsymbol{i} & mathbf{1}-boldsymbol{i}end{array}right| ) is a A. real number B. irrational number c. complex member D. Purely imaginary | 12 |
| 838 | If ( A B C ) is a triangle, then the vectors ( (-1, cos C, cos B),(cos C,-1, cos A) ) and ( (cos B, cos C,-1) ) are A. linearly independent for all triangles B. linearly dependent for all triangles c. linearly independent for all isosceles triangles D. none of these | 12 |
| 839 | Using properties of determinant, prove [ text { that }left|begin{array}{lll} boldsymbol{b}+boldsymbol{c} & boldsymbol{a}-boldsymbol{b} & boldsymbol{a} \ boldsymbol{c}+boldsymbol{a} & boldsymbol{b}-boldsymbol{c} & boldsymbol{b} \ boldsymbol{a}+boldsymbol{b} & boldsymbol{c}-boldsymbol{a} & boldsymbol{c} end{array}right|=mathbf{3} boldsymbol{a} boldsymbol{b} boldsymbol{c}-boldsymbol{a}^{3}- ] ( b^{3}-c^{3} ) | 12 |
| 840 | ( mathbf{f}left|begin{array}{ccc}boldsymbol{a} & boldsymbol{b} & boldsymbol{a} boldsymbol{alpha}+boldsymbol{b} \ boldsymbol{b} & boldsymbol{c} & boldsymbol{b} boldsymbol{alpha}+boldsymbol{c} \ boldsymbol{a} boldsymbol{alpha}+boldsymbol{b} & boldsymbol{b} boldsymbol{alpha}+boldsymbol{c} & boldsymbol{0}end{array}right|=mathbf{0} . ) Prove that ( a, b, c ) are in G.P. or ( alpha ) is a root of ( boldsymbol{a} boldsymbol{x}^{2}+boldsymbol{2} boldsymbol{b} boldsymbol{x}+boldsymbol{c}=mathbf{0} ) | 12 |
| 841 | Find the value of following determinant. ( left|begin{array}{cc}-1 & 7 \ 2 & 4end{array}right| ) | 12 |
| 842 | Prove that: ( 2left|begin{array}{cc}8 & -5 \ -2 & 6end{array}right|=left|begin{array}{cc}14 & -2 \ -4 & 6end{array}right| ) | 12 |
| 843 | If ( boldsymbol{A}=left|begin{array}{lll}boldsymbol{a} & boldsymbol{0} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{a} & boldsymbol{0} \ boldsymbol{0} & boldsymbol{0} & boldsymbol{a}end{array}right|, ) then the value of ( |boldsymbol{A}||boldsymbol{a} boldsymbol{d} boldsymbol{j}(boldsymbol{A})| ) is A ( cdot a^{3} ) в. ( a^{6} ) ( c cdot a^{9} ) D. ( a^{2} ) | 12 |
| 844 | The points (2,-3),(4,3) and ( (5, k / 2) ) are on the same straight line. The value(s) of k is (are): A . 12 B. -12 ( c .pm 12 ) D. 12 or 6 | 12 |
| 845 | ( f(x)=left|begin{array}{ccc}a^{-x} & e^{x log _{e} a} & x^{2} \ a^{-3 x} & e^{3 x log _{e} a} & x^{4} \ a^{-5 x} & e^{5 x log _{e} a} & 1end{array}right|, ) then A ( cdot g(x)+g(-x)=0 ) B . ( g(x)-g(-x)=0 ) C. ( g(x) times g(-x)=0 ) D. none of these | 12 |
| 846 | If ( A ) is a square matrix such that ( A^{2}= ) ( A, ) then ( |A| ) equals ( mathbf{A} cdot 0 ) or 1 B. – 2 or 2 ( c .-3 ) or 3 D. None of these | 12 |
| 847 | h non-zero entries and let A2=I, (d) less wall Let A be a 2 x 2 matrix with non-zero entries where I is 2 x 2 identity matrix. Define TT(A) = sum of diagonal elements of A and A=determinant of matrix A. Statement-1: Tr(A)=0. Statement-2: A=1. (a) Statement -1 is true, Statement -2 is true, is not a correct explanation for Statement -1. (b) Statement -1 is true, Statement -2 is false. (C) Statement -1 is false, Statement -2 is true. (d) Statement – 1 is true, Statement 2 is true; Statement- is a correct explanation for Statement-1. [2010] ment-2 is true ; Stctement-2 | 12 |
| 848 | If ( A ) is an idempotent matrix satisfying, ( (I-0.4 A)^{-1}=I-alpha A, ) where ( I ) is the unit matrix of the same order as that of ( A, ) then the value of ( |9 alpha| ) is equal to | 12 |
| 849 | ( f x neq 0 ) and ( left|begin{array}{ccc}1 & x & 2 x \ 1 & 3 x & 5 x \ 1 & 3 & 4end{array}right|=0, ) then ( x= ) ( A ) B. – ( c cdot 2 ) ( D ldots-2 ) | 12 |
| 850 | Find the value of the determinant ( left|begin{array}{ccc}1 & 0 & 0 \ 2 & cos x & sin x \ 3 & sin x & cos xend{array}right| ) ( mathbf{A} cdot cos 2 x ) B. ( c cdot 0 ) D. ( sin 2 x ) | 12 |
| 851 | ( mathbf{A}=left[begin{array}{ccc}mathbf{1}^{2} & mathbf{2}^{mathbf{2}} & mathbf{3}^{2} \ mathbf{2}^{mathbf{2}} & mathbf{3}^{2} & mathbf{4}^{2} \ mathbf{3}^{mathbf{2}} & mathbf{4}^{mathbf{2}} & mathbf{5}^{mathbf{2}}end{array}right], ) then ( |boldsymbol{A} boldsymbol{d} boldsymbol{j} boldsymbol{A}|= ) ( A ) B . 16 ( c cdot 64 ) D. 128 | 12 |
| 852 | The adjoint of the matric ( boldsymbol{A}= ) ( left[begin{array}{lll}1 & 0 & 2 \ 2 & 1 & 0 \ 0 & 3 & 1end{array}right] ) is A. ( left[begin{array}{ccc}-1 & 6 & 2 \ -2 & 1 & -4 \ 6 & 3 & 1end{array}right] ) B. [ left[begin{array}{ccc} 1 & 6 & -2 \ -2 & 1 & 4 \ 6 & -3 & 1 end{array}right] ] ( c ) [ left[begin{array}{ccc} 6 & 1 & 2 \ 4 & -1 & 2 \ 6 & 3 & -1 end{array}right] ] D. [ left[begin{array}{ccc} -6 & 2 & 1 \ 4 & -2 & 1 \ 3 & 1 & -6 end{array}right] ] | 12 |
| 853 | The straight lines ( imath_{1}, imath_{2} ) and ( imath_{3} ) are parallel and lie in the same plane. A total of ( mathrm{m} ) points are taken on the line ( imath_{1} ) n points on ( imath_{2}, ) and ( mathrm{k} ) points on ( imath_{3} . ) How many triangles are there whose vertices are at these points? | 12 |
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