Próximo SlideShare
Determinant & Matrices.pdf
Cargando en ... 3
1 de 19

### QE Determinant & Matrices(13th).pdf

1.  a bc k a d b a d c bc d k 2 2       L N M O Q P ( ) ( ) = O  a2 + bc + k = 0 = bc + d2 + k = 0 and (a + d)b = (a + d) c = 0 As bc  0, b  0, c  0  a + d = 0  a = –d Also, k = –(a2 + bc) = –(d2 + bc) = – ( (–ad) + bc ) = |A| ] Q.152515/qe The graph of a quadratic polynomial y = ax2 + bx + c (a, b, c  R, a  0) is as shown. Then the incorrect statement(s) is/are (A*) c > 0 (B) b < 0 (C*) product of the roots is negative (D*) sum oftheroots is positive Q.153510/det The value of  lying between  = 0 & = /2 & satisfying the equation :             4 sin 4 1 cos sin 4 sin 4 cos 1 sin 4 sin 4 cos sin 1 2 2 2 2 2 2 = 0 are : (A*) 7 24  (B) 5 24  (C*) 11 24  (D)  24 [ IIT ’88 , 2 ] Q.154513/det If p, q, r, s are in A.P. and f (x)= p x q x p r x q x r x x r x s x s q x             sin sin sin sin sin sin sin sin sin 1 such that 0 2  f (x)dx = – 4 then the common differenceof theA.P. can be : (A*)  1 (B) 1 2 (C*) 1 (D) none [ Start : p = a ; q = a + d ; r = a + 2 d ; s = a + 3 d  f (x) =  2 d2 Also use R1  R1 – R2 and R2  R2 – R3 ] Q.155517/qe If one root of the quadratic equation px2 + qx + r = 0 ) 0 p (  is a surd b a a a   where p, q, r ; a, b are all rationals then the other root is (A*) b a a a   (B) b ) b a ( a a   (C*) b ) b a ( a a   (D) b b a a   [Hint:  = b a a a   =   ) b a ( a b a a a     = b ) b a ( a a   Conjugate of  is b ) b a ( a a    (C) ]
2.  = n 4  ( 1)n  8 Þ independent of A Þ A, B, C, D ] Q.143505/qe If the quadratic equations, x2 + abx + c = 0 and x2 + acx + b = 0 have a common root then the equationcontainingtheirotherroots is/are: (A) x2 + a (b + c) x  a2bc = 0 (B*) x2  a (b + c) x + a2bc = 0 (C) a (b + c) x2  (b + c) x + abc = 0 (D*) a (b + c) x2 + (b + c) x  abc = 0 Q.144503/mat If AB = Aand BA= B, then (A*)A2B =A2 (B*) B2A = B2 (C*)ABA=A (D*) BAB = B [Sol. We have A2B =A(AB) = AA=A2, B2A= B(BA) = BB = B2, ABA=A(BA) =AB =A, and BAB = B(AB) = BA= B] Q.145505/det The solution(s) of the equation x a b a x a b b x = 0 is/are : (A*) x =  (a + b) (B*) x = a (C*) x = b (D)  b [Hint: Use c1  c1 – c2 & then R1  R1 + R2 to get 0 0 a x b a x a x a b x     ( ) = 0. Now open byc1 & factorize] Q.146511/qe The value(s) of 'p' for which the equation ax2  px + ab = 0 and x2  ax  bx + ab = 0 may have a common root, given a, b are non zero real numbers, is (A) a + b2 (B*) a2 + b (C*) a(1 + b) (D) b(1 + a) [Sol. x2  (a + b) x + a b = 0 or (x  a) (x  b) = 0  x = a or b if x = a is the root of other equation , a3  a p + a b = 0  p = a (a + b) if x = b is the root of the other equation , then a b2  p b + a b = 0 p = a (1 + b) ] Q.147505/mat If D1 and D2 are two 3 x 3 diagonal matrices when none of the diagonal element is zero, then (A*) D1D2 is adiagonal matrix (B*) D1D2 = D2D1 (C*) D1 2 + D2 2 is a diagonal matrix (D) none of these [Sol. Let D1 = x y z 1 1 1 0 0 0 0 0 0 L N M M M O Q P P P and D2 = x y z 2 2 2 0 0 0 0 0 0 L N M M M O Q P P P , when x1 , y1 , z1 , x2 , y2 , z2  0 then D1D2 = D2D1 ]
3. infinitelymanysolutions. Statement-3 : The system x + y+ z = 1, x = y, y = 1 + z is inconsistent. Statement-4 : If two of the equations in a system of three linear equations are inconsistent, then the wholesystemisinconsistent. (A) FFTT (B*)TTFT (C) TTFF (D)TTTF Select the correct alternatives : (More than one are correct) Q.135512/mat D is a 3 x 3diagonal matrix. Which ofthe followingstatements is not true? (A) D = D (B*)AD = DAfor everymatrixAof order 3 x 3 (C*) D–1 if exists is a scalar matrix (D) none of these [Sol. Let D = d d d 1 2 3 0 0 0 0 0 0 L N M M M O Q P P P . Clearly D = D  A is correct Also, AD = a a a a a a a a a 11 12 13 21 22 23 31 32 33 L N M M M O Q P P P d d d 1 2 3 0 0 0 0 0 0 L N M M M O Q P P P = d a d a d a d a d a d a d a d a d a 1 11 2 12 3 13 1 21 2 22 3 23 1 31 2 32 3 33 L N M M M O Q P P P and, DA = d d d 1 2 3 0 0 0 0 0 0 L N M M M O Q P P P a a a a a a a a a 11 12 13 21 22 23 31 32 33 L N M M M O Q P P P = d a d a d a d a d a d a d a d a d a 1 11 1 12 1 13 2 21 2 22 2 23 3 31 3 32 3 33 L N M M M O Q P P P This shows that in generalAD  DA If d1d2d3  0, then D–1 = d d d 1 1 2 1 3 1 0 0 0 0 0 0    L N M M M O Q P P P  (C) is correct ] Q.136501/det The set of equations x – y + 3z = 2 , 2x – y + z = 4 , x – 2y + z = 3 has (A) unique soluton onlyfor = 0 (B*) unique solution for   8 (C)infinitenumber ofsolutions for =8 (D*) no solution for = 8 [Hint: D =  – 8  (B) ; If  = 8 , D = D1 = D2 = D3 ] Q.137501/qe cos is a root of the equation 25x2 + 5x  12 = 0,  1 < x < 0, then the value of sin 2 is : (A*) 24/25 (B)  12/25 (C*)  24/25 (D) 20/25 Q.138501/mat Suppose a1, a2, ....... real numbers, with a1  0. If a1, a2, a3, ..........are inA.P. then (A*)A= a a a a a a a a a 1 2 3 4 5 6 5 6 7 L N M M M O Q P P P is singular (B*) the system of equations a1x + a2y + a3z = 0, a4x + a5y + a6z = 0, a7x + a8y + a9z = 0 has infinite numberofsolutions (C*) B = a ia ia a 1 2 2 1 L N M O Q Pis nonsingular; where i = 1 
4. Q.125138/qe The equations x3 + 5x2 + px + q = 0 and x3 + 7x2 + px + r = 0 have two roots in common. If the third root of each equation is represented by x1 and x2 respectively, then the ordered pair (x1, x2) is : (A*) ( 5,  7) (B) (1,  1) (C) ( 1, 1) (D) (5, 7) [Hint: The common roots must be roots of the equation 2x2 + (r  q) = 0  sum is zero . Hence third root of first is  5 and third root of 2nd is  7 ] Q.12663/det If x = a + 2b satisfies the cubic (a, bR) f (x)= x a b b b x a b b b x a    =0, then its other two roots are (A)realanddifferent (B*)realand coincident (C)imaginary (D)such that oneis real and other imaginary [Hint: Other roots are each equal to (a – b)  (B) ] Q.12768/mat A is a 2 × 2 matrix such that A       1 1 =       2 1 andA A2       1 1 =       0 1 . The sum of the elements ofA, A, is (A) –1 (B) 0 (C) 2 (D*) 5 [Sol. A       1 1 =       2 1 ....(1) and A2       1 1 =       0 1 ....(2) LetAbe given by A=       d c b a . Thefirstequationgives a – b = – 1 ....(3) and c – d = 2 ....(4) For second equation, A2       1 1 =             1 1 A A =             2 1 A =       0 1 . Thisgives – a + 2b = 1 ....(5) and – c + 2d = 0 ....(6) (3) + (5)  b = 0 and a = – 1 (4) + (6)  d = 2 and c = 4 so the sum a + b + c + d = 5 Ans. ] Q.12864/det Three digit numbers x17, 3y6 and 12z where x, y, z are integers from 0 to 9, are divisible bya fixed constant k.Thenthedeterminant 2 y 1 z 6 7 1 3 x must bedivisibleby (A*) k (B) k2 (C) k3 (D) None Q.12969/mat In a square matrixAof order 3, ai i's are the sum of the roots of the equation x2 – (a + b)x + ab= 0; ai, i + 1's are the product of the roots, ai, i – 1's are all unity and the rest of the elements are all zero. The value of the det. (A) is equal to (A) 0 (B) (a + b)3 (C) a3 – b3 (D*) (a2 + b2)(a + b) [Sol. Given a11 = a22 = a33 = a + b a12 = a23 = ab
5. (A) 0 (B*) 3 (C) 6 (D) 12 [Hint: MultiplyR1 byx; R2 byyand R3 byz and divide thedeterminant byxyz xyz 1 z z yz xz z y y y xy z x y x x x 4 3 3 3 4 3 3 3 4    = 11 1 = 1 z z z y 1 y y x x 1 x xyz xyz 3 3 3 3 3 3 3 3 3    = 11 1 use R1 R1 + R2 + R3 D = (x3 + y3 + z3 + 1) 1 z z z y 1 y y 1 1 1 3 3 3 3 3 3   = 11 1 hence x3 + y3 + z3 = 10 (as the det. has the value 1) (2, 1, 1) , (1, 2, 1) , (1, 1, 2)  (B)] Q.117126/qe The number of non-zero solutions of the equation , x2  5 x  (sgn x) 6 = 0 is : (A*) 1 (B) 2 (C) 3 (D) 4 [Hint: If x = 0 we have x2 – 5x = 0  x = 0 or 5  no solution if x > 0 we have x2 – 5x – 6 = 0  (x – 6) (x + 1)  x = 6 is the solution if x < 0 we have x2 – 5x + 6 = 0  (x – 3) (x – 2) = 0  no solution ] Q.11865/mat IfA, B and C are n × n matrices and det(A) = 2, det(B) = 3 and det(C) = 5, then the value of the det(A2BC–1) is equal to (A) 5 6 (B*) 5 12 (C) 5 18 (D) 5 24 [Hint: | A | = 2 ; | B | = 3 ; | C | = 5 det(A2BC–1) = | A2BC–1| = | C | | B | | A | 2 = 5 3 · 4 = 5 12 Ans. ] [12 & 13th test (29-10-2005)] Q.11958/det The equation x 3 2 x 2 1 x x 5 1 x 3 1 x 2 ) x 2 ( ) x 1 ( ) x 1 ( 2 2 2         + 3 x 2 2 x 3 x 2 1 x 2 x 3 ) x 1 ( 1 x 1 x 2 ) x 1 ( 2 2        = 0 (A) has noreal solution (B)has 4 real solutions (C) has two real and twonon-real solutions (D*)has infinitenumberofsolutions ,real ornon-real [Hint: 1st two columns of 1st determinant are same as 1st two rows of 2nd. Hence transpose the 2nd.Add the two determinants and use C1  C1 + C3  D = 0 ] Q.120130/qe The quadraticequation with real co-efficients one of whose complex roots has the real part 12 and modulus13is: (A) x2  12 x + 13 = 0 (B) x2  24 x + 13 = 0 (C*) x2  24 x + 169 = 0 (D) x2  24 x  169 = 0 [Hint: Let the roots be 12 + i ; other roots is 12 – i now z z = | z |2
6. = 2(2A – I) – A (A2 = 2A – I) A3 = 3A – 2I A4 = 3A2 – 2A = 3(2A – I) – 2A (A2 = 2A – I) A4 = 4A – 3I A5 = 5A – 4I  An = nA – (n – 1)I ] Q.110116/qe The value of x satisfying the equation 6 2 3 6 2 3 x a b c x a b c       = 2 6 3 2 6 3 x a b c x a b c       is : (A*) x = ab/c (B) 2ab/c (C) ab/3c (D) ab/2c [Hint: put 6x + 2a = A ; 3b + c = B ; 2x + 6a = C ; b + 3c = D  A B A B   = C D C D   . Now add unityon both sides ; subtract unity from both sides and divide to get A/B = C/D  x = ab/c ] Q.11154/det If a, b, c are real then the value of determinant 1 c bc ac bc 1 b ab ac ab 1 a 2 2 2    = 1 if (A) a + b + c = 0 (B) a + b + c = 1 (C) a + b + c = –1 (D*) a = b = c = 0 [Hint: MultiplyR1 bya, R2 byb & R3 byc & divide the determinant byabc. Now take a, b & c common from c1, c2 & c3. Now use C1  C1 + C2 + C3 to get ] (a2 + b2 + c2 + 1) 1 1 1 1 1 2 2 2 2 2 2 b b b c c c   = 1. Now use c1  c1 – c2 & c2  c2 – c3 we get 1 + a2 + b2 + c2 = 1  a = b = c = 0  (D) ] Q.11263/mat Readthefollowingmathematicalstatements carefully: I. There canexist two triangles such that the sides of one triangle areall less than 1 cm while the sides of theother triangle areall bigger than10metres, but the areaof the first triangleis larger than the areaof second triangle. II. If x,y, z are all different real numbers, then 2 2 2 ) x z ( 1 ) z y ( 1 ) y x ( 1      = 2 x z 1 z y 1 y x 1              . III. log3x·log4x·log5x=(log3x ·log4x)+(log4x ·log5x)+(log5x·log3x)istrueforexactlyforone real valueofx. IV. A matrix has 12 elements. Number ofpossible orders it canhave is six. Nowindicatethecorrect alternatively. (A*) exactlyonestatement is INCORRECT.
7. Q.10159/mat If everyelement of a squarenon singular matrixAis multiplied byk and the new matrix is denoted by B then |A–1| and | B–1| are related as (A) | A–1| = k | B–1| (B) | A–1| = k 1 | B–1| (C*) | A–1| = kn | B–1| (D) | A–1| = k–n | B–1| where n is order of matrices. [Hint: verifybytaking a square matrixA=       d c b a or       1 0 0 1 ] Q.102109/qe If the quadratic polynomial, y = (cot ) x2 + 2   sin  x + 1 2 tan  ,   [0, 2 ], can take negative values for all x R, then the value of  must in the interval : (A*)  5 6   ,       (B)  5 6   ,        11 6 2   ,       (C)  11 6 2   ,       (D)   0 6 ,   5 6   ,       [Sol. b2 – 4ac < 0 and a < 0 hence cot  < 0 i.e.   2nd and 4th quadrant and 4sin  – 2 tan  cot  < 0 2 sin  < 1  sin  < 2 1    2nd and 4th quadrant hence a          , 6 5 ] Q.10350/det If f  (x) = p n 2 mx p n 2 mx n 2 mx p n p n n p mx p mx mx          then y = f(x) represents (A*) astraight lineparallel tox- axis (B) astraight line parallel toy-axis (C) parabola (D)astraightlinewithnegativeslope [Hint: R3  R3 – 2R2 hence 2 identical rows  f  (x) = 0  f(x) = constant ] Q.10460/mat Let A=           1 1 1 3 1 2 1 1 1 and 10B =            3 2 1 0 5 2 2 4 . If B is the inverse of matrixA, then  is (A) – 2 (B) – 1 (C) 2 (D*) 5 [Sol. B = A–1  AB = I B =                    3 2 1 10 0 2 1 5 1 10 2 5 2 ; now                   10 3 0 10 1 10 0 2 1 5 1 10 2 5 2           1 1 1 3 1 2 1 1 1 =         1 0 0 0 1 0 0 0 1
8. sec2 · sec2 · sec2 is (A) p2 + r2 + 2rp + 1 (B*) p2 + r2 – 2rp + 1 (C) p2 – r2 – 2rp + 1 (D) None [Sol.   tan = p ;    tan · tan = 0 ;   tan = r now sec2 · sec2 · sec2 = (1 + tan2) (1 + tan2) (1 + tan2) = 1 +   2 tan +    2 2 tan · tan + tan2 · tan2· tan2 now   2 tan =  2 tan   – 2    tan · tan = p2    2 2 tan · tan =  2 tan · tan    – 2 tan  · tan  · tan      tan = 0 – 2rp   2 tan = r2    2 sec = 1 + p2 – 2rp + r2 = 1 + (p – r)2 ] Q.9269/det Let A=                          2 2 2 2 2 2 2 2 2 y x z 1 ) x yz ( 2 ) y zx ( 2 ) x yz ( 2 x z y 1 ) z xy ( 2 ) y zx ( 2 ) z xy ( 2 z y x 1 then det.Ais equal to (A) (1 + xy+ yz + zx)3 (B*) (1 + x2 + y2 + z2)3 (C) (xy+ yz + zx)3 (D) (1 + x3 + y3 + z3)2 [Hint: multiply R2 by z and R3 by y and use R1  R1 – R2 + R3 Objective approach : put z = y = 0 then choices are A= 1 ; B = (1 + x2)3 ; C = 0 ; D = (1 + x3)2 and determinant comes out to be (1 + x2)3  (B)] Q.93102/qe If the equation a (x – 1)2 + b(x2 – 3x + 2) + x – a2 = 0 is satisfied for all x  R then the number of ordered pairs of (a, b) can be (A) 0 (B*) 1 (C) 2 (D)infinite [Sol. equation is an identity  coefficient of x2 = 0= coefficient of x = constant term  a + b = 0 ....(1) – 2a – 3b + 1 = 0 ....(2) and a + 2b – a2 = 0 ....(3) from (1) and (2) a = – 1 and b = 1 whichalsosatisfies(3)  (a, b) = (–1, 1)  (B) ] Q.9444/det The following system of equations 3x – 7y + 5z = 3; 3x + y + 5z = 7 and 2x + 3y + 5z = 5 are (A)consistentwithtrivialsolution (B*)consistentwithuniquenontrivialsolution (C)consistentwithinfinitesolution (D)inconsistentwithnosolution [Hint: D  0  consistency] Q.9556/mat IfA1,A3, .....A2n – 1 are n skew symmetric matrices of same order then B =      n 1 r 1 r 2 1 r 2 ) A )( 1 r 2 ( willbe (A)symmetric (B*)skewsymmetric (C)neithersymmetricnorskewsymmetric (D) data is adequate [Sol. B = A1 + 3 3 A 3 + ..... (2n – 1) 1 n 2 1 n 2 ) A (  
9. abc 1 [ 2b2 (a2 c2) – 2a2 (– b2 c2) ] = abc 4 abc c b a 4 2 2 2  ] Q.8246/mat GivenA=       2 2 3 1 ; I =       1 0 0 1 . IfA– I is a singular matrix then (A)    (B*) 2 – 3 – 4 = 0 (C) 2 + 3 + 4 = 0 (D) 2 – 3 – 6 = 0 [Hint: A – I =       2 2 3 1 –         0 0 =           2 2 3 1 sinceA– I is singular  det. (A– I) = 0; now     2 2 3 1 = (1 – ) (2 – ) – 6 = 2 – 3 – 4 hence 2 – 3 – 4 = 0 ] Q.8395/qe The values of 'a' for which the quadratic equation (a2 – a – 2)x2 + 2ax + a3 –27 =0 has roots of opposite signs are (A) (–1, 2) U (3, ) (B*) (–, –1) U (2, 3) (C) R/(–1, 2) (D) R/(2, 3) Q.8438/det If the system of equations, a2 x  ay = 1  a & bx + (3  2b)y = 3 + a possess a unique solution x = 1, y = 1 then : (A*) a = 1 ; b =  1 (B) a =  1 , b = 1 (C) a = 0 , b = 0 (D) none [Hint: put x = 1 & y = 1 and solve for a & b . In B & C system has infinite solutions] Q.8547/mat Let A =                1 sin 1 sin 1 sin 1 sin 1 , where 0   < 2, then (A) Det (A) = 0 (B) Det A  (0, ) (C*) Det (A)  [2, 4] (D) Det A  [2, ) [Sol. |A| = 1 sin 1 sin 1 sin 1 sin 1        = 1(1 + sin2) – sin(– sin + sin) + (1 + sin2) = 2 (1 + sin2) | sin |  1  –1  sin  1  0  sin2  1  1  1 + sin2  2  2  2(1 + sin2)  4  | A |  [2, 4] ] Q.8696/qe Three roots of the equation, x4  px3 + qx2  rx + s = 0 are tanA, tanB & tanC whereA, B, C are the angles ofatriangle . Thefourth root ofthebiquadratic is : (A*) p r q s    1 (B) p r q s    1 (C) p r q s    1 (D) p r q s    1 [Hint: Let the fourth root be tan D Now tan ( A) =     tan tan tan tan tan tan tan A A B C A B A    1 tan D = p r q s    1 ] Q.8739/det Number of value of 'a' for which the system of equations, a2 x + (2  a) y = 4 + a2
10. [Sol. Let X =         d c b a X2 =             2 2 d bc cd ac bd ab bc a a2 + bc = 1 ab + bd = 1  b(a + d) = 1 ac + cd = 2  c (a + d) = 2  2 1 c b   c = 2b bc + d2 = 3  (d2 – a2) = 2  (d – a) (a + d) = 2 d – a = 2b (using bc = 1 – a2) a + d = 1/b __________________ 2d = 2b + 1/b 2a = 1/b – 2b d = b + 1/2b a = 1/2b – b c = 2b 3 b 2 1 b 4 1 b 2 2 2            2 b 4 1 b 3 2 2   3x + x 4 1 = 2  b = 6 1  or b = 2 1  Matrices are         2 2 2 / 1 0 ;            2 2 2 / 1 0 ;          6 / 4 6 / 2 6 / 1 6 / 2 Alternative: nlinear equationin n variable haveexactlyone solution but n equation not linearin n variables willhavemorethanonesolutionherewewillhave4equationsin4variableofdegreehigherthan1hence morethan2solutions.] Q.7792/qe If both the roots of the equation, (3a + 1) x2  (2a + 3b)x + 3 = 0 are infinite then : (A) a =  ; b = 0 (B) a = 0 ; b =  (C*) a =  1/3 ; b = 2/9 (D) a = ; b =  Q.7836/det If x,y, z are not all simultaneouslyequal tozero, satisfying the system of equations (sin 3)x  y + z = 0 (cos 2 ) x + 4 y + 3 z = 0 2 x + 7 y + 7 z = 0 then the numberofprincipal values of  is (A) 2 (B) 4 (C*) 5 (D) 6 [Hint:  = n  + ( 1)n .  6 or n  ] [ IIT ’86 , 5 ]
11. f (x) = x c 1 x ) b 1 ( ) c b a ( x x 2 1 x ) c 1 ( x b 1 ) c b a ( x x 2 1 x ) c 1 ( x ) b 1 ( ) c b a ( x x 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2                   (as a2 + b2 + c2 = – 2) x c 1 x ) b 1 ( 1 x ) c 1 ( x b 1 1 x ) c 1 ( x ) b 1 ( 1 2 2 2 2 2 2       R2  R2 – R1 & R3  R3 – R1 x 1 x 1 0 0 x 1 0 x ) c 1 ( x ) b 1 ( 1 2 2      f (x) = (1 – x)2 = 1 – 2x + x2  (C) ] Q.7041/mat Matrix A =         z 2 2 4 y 1 2 3 x , if x y z = 60 and 8x + 4y + 3z = 20 , then A (adj A) is equal to (A)         64 0 0 0 64 0 0 0 64 (B)         88 0 0 0 88 0 0 0 88 (C*)         68 0 0 0 68 0 0 0 68 (D)         34 0 0 0 34 0 0 0 34 [Sol. A. adj A = |A | I |A | = xyz – 8x – 3 (z – 8) + 2 (2 – 2y) |A | = xyz – (8x + 3z + 4y) + 28  60 – 20 + 28 = 68  (C) ] Q.7184/qe If the roots ofthe cubic, x3 + ax2 +bx + c =0 are threeconsecutive positive integers. Thenthe value of 1 b a2  is equal to (A*) 3 (B) 2 (C) 1 (D) 1/3 [Sol. n, n + 1, n + 2 sum = 3(n + 1) = – a  a2 = 9(n + 1)2 sum of the roots taken 2 at atime = + b  n(n + 1) + (n + 1)(n + 2) + (n + 2)n + 1 = b + 1 (adding 1 both sides) n2 + n + n2 + 3n + 2 + n2 + 2n + 1 = b + 1  b + 1 = 3n2 + 6n + 3 = 3(n + 1)2 b + 1 = 3(n + 1)2 = 3 a2 ;  1 b a2  = 3  (A) ] Q.7234/det The values of  for whichthe followingequations sinx – cosy + (+1)z = 0; cosx + siny– z = 0; x +( + 1)y + cos z = 0 havenontrivialsolution,is (A)  = n,  R – {0} (B)  = 2n, is anyrational number
12. (A) order of B is 3 × 4 (B*) order of BA is 4 × 4 (C) order of BA is 3 × 3 (D) BA is undefined [Hint: A = 3 × 4 ; A = 4×3 As A B is defined  let order of B = 3 × n now BA = (3 ×n) × (4 × 3)  n = 4  order of B is 3 × 4  order of B = 4 × 3 order of B A = (4×3) × (3 × 4) = 4 × 4 Ans] Q.6276/qe If , &  are the roots of the equation, x3  x  1 = 0 then, 1 1     + 1 1     + 1 1     has the value equal to : (A) zero (B)  1 (C*)  7 (D) 1 [Hint: Let , ,  be the roots of 2nd  = A ;  = B ;  = C Let y = 1 1   x x  x = 1 y 1 y    y y         1 1 3  y y   1 1  1 = 0  y3 + 7y2  y + 1 = 0  1 1      =  7 ] Q.6330/det If the system of equations x+2y+3z = 4, x +py+2z = 3, x +4y+z = 3 has an infinite number of solutions, then : (A) p = 2 ,  = 3 (B) p = 2 ,  = 4 (C) 3 p = 2  (D*) none of these [Hint: For 2nd and 3rd equation , 1 = p/4 = 2/ = 1 p = 4 ;  = 1 ] Q.6439/mat If A =             2 2 sin cos sin cos sin cos ; B =             2 2 sin cos sin cos sin cos are such that,AB is a null matrix, then which of the following should necessarily be an odd integral multiple of 2  . (A)  (B)  (C*)  –  (D)  +  [Hint: AB =               2 2 sin cos sin cos sin cos               2 2 sin cos sin cos sin cos =                                     2 2 2 2 2 2 2 2 sin sin cos sin cos sin cos sin sin cos sin cos sin cos sin cos sin cos cos sin cos sin cos cos =                             ) cos( sin sin ) cos( cos sin ) cos( sin cos ) cos( cos cos   –  must be an odd integral multiple of /2  (C) ] Q.6580/qe For every x  R, the polynomial x8  x5 + x2  x + 1 is : (A*) positive (B) neverpositive (C) positiveas well as negative (D) negative [Hint: for x  1 E = x5 (x3  1) + (x  1) + 1 > 0 for 1 < x < 0 , E = (1  x) + x2 (1  x3) + x8 > 0
13. is       0 a 0 0 ] Q.5370/qe The absolute term in the quadratic expression x k x k k n   F H G I K J  F H G I K J   1 1 1 1 as n is (A*) 1 (B) –1 (C) 0 (D) 1/2 [Sol.      n 1 k n ) 1 k ( k 1 Lim =           n 1 k 1 k 1 k 1 2 1 =         1 n 1 1 2 1  absolute term =           1 n 1 1 2 1 Lim n = 2 1 ] Q.5426/det If a, b, c are all different and a a a b b b c c c 3 4 3 4 3 4 1 1 1    = 0, then : (A*) abc (ab + bc + ca) = a + b + c (B) (a + b + c) (ab + bc + ca) = abc (C) abc (a + b + c) = ab + bc + ca (D) none of these [Hint: Split the determinant into 2 & then evaluate R1  R1 – R2 & R2  R2 – R3] Q.5535/mat Givethe correctorder ofinitials T orF for followingstatements. Use T if statement is trueandFifit isfalse. Statement-1 : IfAis an invertible 3 × 3 matrix and B is a 3 × 4 matrix, thenA–1B is defined Statement-2 : It is never true thatA+ B,A– B, andAB are all defined. Statement-3 :Everymatrix none of whose entries arezero is invertible. Statement-4 : Everyinvertible matrix is square and has no two rows the same. (A) TFFF (B) TTFF (C*) TFFT (D)TTTF Q.5672/qe Number of values of the parameter  [0, 2] for which the quadratic function, (sin )x2 + 2 cos x + 1 2 (cos  + sin ) is the square of a linear function is (A*) 2 (B) 3 (C) 4 (D) 1 [Hint : Let f (x) =  2 b x sin   now compare the coefficient and eliminate b. divide by cos2 to get (tan  – 1) (tan + 2) = 0   = 4  or  – tan–1( 2) ] Q.5728/det If  is one ofthe imaginary cube roots ofunity, thenthe valueof the determinant 1 1 1 3 2 3 2       = (A) 1 (B) 2 (C*) 3 (D) none [Hint: Put 3 = 1 1 1 1 1 1 2 2     and open by R1 to get (1 – 2) + (1 – ) = 3]
14. (A*)           7 / 2 7 / 6 7 / 3 7 / 6 7 / 3 7 / 2 7 / 3 7 / 2 7 / 6 (B)           7 / 2 7 / 6 7 / 3 7 / 6 7 / 3 7 / 2 7 / 3 7 / 2 7 / 6 (C)             7 / 2 7 / 6 7 / 3 7 / 6 7 / 3 7 / 2 7 / 3 7 / 2 7 / 6 (D)            7 / 3 7 / 2 7 / 6 7 / 3 7 / 2 7 / 2 7 / 3 7 / 2 7 / 6 [Hint: Matrix           3 2 1 3 2 1 3 2 1 c c c b b b a a a isorthogonalif 1 c b a 2 i 2 i 2 i       ; 0 a c c b b a i i i i i i        (A) ] Q.4764/qe Number ofintegral values of x satisfying the inequality 64 27 4 3 2 x 10 x 6          is (A) 6 (B*) 7 (C) 8 (D)infinite [Hint: 6x + 10 – x2 > 3  x2 – 6x – 7 < 0 (x + 1) (x – 7) < 0  0, 1, 2, 3, 4, 5, 6  B ] Q.4824/det The determinant 1 1 1             a x a y a z b x b y b z c x c y c z = (A*) (1 + a + b + c) (1 + x + y+ z)  3 (ax + by+ cz) (B) a(x+y) + b(y+z) + c(z+x)  (xy+yz+zx) (C) x (a + b) + y(b + c) + z (c + a)  (ab + bc + ca) (D) none of these [Hint: 1 + a + b + c = k and use R1 + R2 + R3 we get D = z c 1 y c x c z b y b 1 x b z 3 k y 3 k x 3 k            = k z c 1 y c x c z b y b 1 x b 1 1 1         + z c 1 y c x c z b y b 1 x b z y x 3         now proceed ] Q.4932/mat Which of the following statements is incorrect for a square matrixA. ( |A|  0) (A) IfAis adiagonal matrix,A–1 willalso beadiagonal matrix (B)IfAisasymmetricmatrix,A–1 willalsobeasymmetricmatrix (C*) IfA–1 =A Ais an idempotent matrix
15. (AB) (adj.B · adj A) = |B| |A| In or |AB| In ....(2) from (1) and (2) adj (AB) = (adj B) · (adjA) ] Q.3751/qe The graph of a quadratic polynomial y = ax2 + bx + c (a, b, c  R) with vertex on y-axis is as shown in the figure. Then which one of the following statement is INCORRECT? (A) Product of the roots of the corresponding quadratic equation is positive. (B)Discriminantofthequadraticequationisnegative. (C*) Nothingdefinite can be saidabout the sum ofthe roots, whether positive,negative or zero. (D)Both roots ofthe quadratic equation arepurelyimaginary. [Sol. Rootsarepurelyimaginary i.e. i  and – i   sum of roots = 0 incorrect (C) product of roots = – i2 2 = 2  product > 0 ; a c > 0  c = + ve ] Q.3818/det If r a 4 a 3 a q a 3 a 2 a p a 2 a 1 a          = 0 , then p, q, r are in : (A*) AP (B) GP (C) HP (D) none [Hint: Use R2  R2 – R1 & R3  R3 – R2 & then c1  c1 – c2 to get q r 1 1 p q 1 1 p a 2 a 1      open by c1 to get p + r = 2q ] Q.3955/qe The number of solution of the equation e2x + ex + e–2x + e–x = 3(e–2x + ex) is (A) 0 (B) 2 (C*) 1 (D) more than 2 [Hint: x = ln 2] Q.4028/mat LetA= x x x x x x x x x    L N M M M O Q P P P    , thenA A–1 exists if (A) x  0 (B)  0 (C*) 3x +  0,  0 (D) x  0,  0 [Sol. We have |A| = x x x x x x x x x       = 3 3 3 x x x x x x x x x           = (3x +)     x x 1 x x 1 x x 1 = (3x + ) 1 0 0 0 0 x x   = 2(3x + ) [Take 3x +  common and use R2  R2 – R1, R3  R3– R1] Thus,A–1 will exist if  0 and 3x +   0 ] Q.4156/qe Let a, b, c be the three roots of the equation x3 + x2 – 333x – 1002 = 0 then the value of a3 + b3 + c3.
16. (A) 33 (B) 0 (C*) 21 (D) none [Hint: Put x = 0 & then evaluate] Q.2436/qe Let P (x) = kx3 + 2k2x2 + k3. Find the sum of all real numbers k for which x – 2 is a factor of P(x). (A) 4 (B) 8 (C) – 4 (D*) – 8 [Hint: put x = 2, P (2) = 0, k3 + 8k2 + 8k = 0  k1 + k2 + k3 = – 8 ] Q.2515/mat IfAand B are invertiblematrices, which one ofthe followingstatements is not correct (A) Adj. A = |A|A–1 (B) det (A–1) = |det (A)|–1 (C*) (A + B)–1 = B–1 + A–1 (D) (AB)–1 = B–1A–1 [Sol. AA–1 = I  | AA–1 | = | I | = 1 hence |A| |A–1| = 1  |A–1| = | A | 1  (B) is correct ] Q.2639/qe The sum of all the value of m for which the roots x1 and x2 of the quadratic equation x2 – 2mx + m = 0 satisfythe condition 2 2 2 1 3 2 3 1 x x x x    , is (A) 4 3 (B) 1 (C) 4 9 (D*) 4 5 [Hint: x1 + x2 = 2m ; x1 x2 = m (x1 + x2)3 – 3x1x2(x1 + x2) = (x1 + x2)2 – 2x1x2 8m3 – 3m(2m) = 4m2 – 2m 8m3 – 10m2 + 2m = 0 2m(4m2 – 5m + 1) = 0  m = 0 (m – 1)(4m – 1) = 0  m = 1 or m = 1/4 ] Q.2713/det If D = 1 c cb ca bc 1 b ba ac ab 1 a 2 2 2    then D = (A*) 1 + a2 + b2 + c2 (B) a2 + b2 + c2 (C) (a + b + c)2 (D) none [Hint: MultiplyR1 bya, R2 byb & R3 byc & divide thedeterminant byabc. Nowtake a, b &c common from c1, c2 & c3. Now use C1  C1 + C2 + C3 to get] (a2 + b2 + c2 + 1) 1 1 1 1 1 2 2 2 2 2 2 b b b c c c   . Now use c1  c1 – c2 & c2  c2 – c3 to get the value as 1.] Q.2819/mat IfA=       d c b a satisfies the equation x2 – (a + d)x + k = 0, then (A) k = bc (B) k = ad (C) k = a2 + b2 + c2 + d2 (D*) ad–bc [Sol. We have A2 =       d c b a       d c b a =           2 2 d bc cd ac bd ab bc a ; (a + d)A = =           ) d a ( d ) d a ( c ) d a ( b ) d a ( a ;       k 0 0 k  A2 –(a + d)A =         da bc 0 0 ad bc = (bc – ad) I As A2 – (a + d)A + kI = 0, we get (bc –ad)I + kI = 0  k = ad – bc]
17.  Emin occurs when x = 15  Emin = 15 Ans.] Q.126/det If the system of equations ax +y+z = 0 , x +by+ z = 0 & x +y+cz = 0 (a, b, c  1) has a non- trivial solution, then the value of 1 1 1 1 1 1      a b c is : (A)  1 (B) 0 (C*) 1 (D) none of these [Hint: c 1 1 1 b 1 1 1 a Use R1  R1 – R2 & R2  R2 – R1 & open by C1 to get (1 – a) [(1 – b)c + (1 – c)] + (1 – b) (1 – c) = 0 divide by (1 – a) (1 – b) (1 – c) to get the result] Q.1322/qe If a, b, c are real numbers satisfying the condition a + b + c = 0 then the roots of the quadratic equation 3ax2 + 5bx + 7c = 0 are : (A) positive (B) negative (C*) real & distinct (D) imaginary [Hint: D = 25b2 – 84 ac = 25(a + c)2 – 84ac using b = –(a + c) = 21[(a+c)2 – 4ac] + 4(a+c)2 > 0 ] Q.146/mat Consider the matricesA=           5 2 1 2 0 3 1 6 4 , B=          2 1 1 0 4 2 , C =         2 1 3 . Out of the given matrix products (i) (AB)TC (ii) CTC(AB)T (iii)CTAB and (iv)ATABBTC (A) exactlyone is defined (B) exactlytwo are defined (C*) exactlythree are defined (D) allfour are defined [Hint: (i), (iii)and(iv) are correct ] Q.1524/qe If the difference ofthe roots of the equation, x2 +ax + b = 0 is equal to the difference of the roots of the equation x2 + bx + a = 0 then : (A) a + b = 4 (B*) a + b =  4 (C) a  b = 4 (D) a  b =  4 Q.167/det The value of a for which the system of equations ; a3x +(a+1)3 y +(a+2)3 z = 0 , ax +(a+ 1)y + (a +2)z = 0 & x +y+ z = 0 has a non-zero solution is : (A) 1 (B) 0 (C*)  1 (D) none of these [Hint: Use c2  c2 – c1 & c3  c3 – c2 & then open by R3.] Q.1729/qe Suppose a, b, and c are positive numbers such that a + b + c = 1. Then the maximum value of ab + bc + ca is (A*) 3 1 (B) 4 1 (C) 2 1 (D) 3 2 [Sol. a2 + b2 + c2 = 1 – 2 ab ....(1) also (a – b)2  0 etc. hence a2 + b2 + c2  ab + bc + ca 1 – 2ab  ab 1  3ab
18. Question bank on Quadratic Eqaution, Determinant & Matrices Select the correct alternative : (Only one is correct) Q.14/qe If a2 + b2 + c2 = 1 then ab + bc + ca lies in the interval : (A) 1 2 2 ,       (B) [ 1, 2] (C*)        1 2 1 , (D)        1 1 2 , [Hint:  (a–b)2 > 0  2 a2 – 2 ab > 0  ab <  a2  ab + bc + ca < 1 Also note that (a + b + c)2 > 0 ] Q.21/detThe valueofthedeterminant a a nx n x n x nx n x n x 2 1 1 2 1 2 cos( ) cos( ) cos( ) sin( ) sin( ) sin( )     is independentof : (A*) n (B) a (C) x (D) a , n and x [Hint: Directlyopen byR1 to get a form of sin (A – B) etc. ] Q.32/mat Ais an involutarymatrix given by A=            4 3 3 4 3 4 1 1 0 then theinverse of 2 A will be (A*) 2A (B) 2 A 1  (C) 2 A (D)A2 [Hint: Aisinvolutary  A2 = I  A = A–1 also (KA)–1 = k 1 (A)–1 ; hence 1 A 2 1        = 2(A)–1  2A ] Q.45/qe If P(x) = ax2 + bx + c & Q(x) =  ax2 + dx + c, where ac  0, then P(x) . Q(x) = 0 has (A) exactlyone real root (B*) atleast two real roots (C) exactlythree real roots (D) all four are real roots . [Hint: D1 : b2  4 ac & D2 : d2 + 4 a c . Hence atleast one of either D1 or D2 is zero] Q.52/detIf a, b, c are all different from zero & 1 1 1 1 1 1 1 1 1    a b c = 0 , then the value of a1 + b1 + c1 is (A) abc (B) a1 b1 c1 (C) a  b  c (D*)  1 [Hint: C1  C1 – C2 & C2  C2 – C3 & then open by R1 to get ab + abc + ac + bc = 0 ; divided by abc] Q.63/mat If Aand B are symmetric matrices, thenABAis (A*)symmetricmatrix (B)skewsymmetric (C)diagonalmatrix (D)scalarmatrix [Sol. We have (ABA) =ABA =ABAABAis symmetric] Q.711/qe Let a > 0, b > 0 & c > 0 . Then both the roots of the equation ax2 + bx + c = 0