
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) ]

 =
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 ]

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


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

(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

= 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.

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

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
( 


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

[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 ]

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

(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

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]

(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

(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.

(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]

 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

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