Question.1
The determinant D = 
is
(A) 0
(B) independent of q
(C) independent of f
(D) independent of both q
and f
Solution
Apply R1 → R1 + R2
sinf -R3
cosf
Then taking 2cosf
common from R1 & then apply R1 → R1
+ R3
Then D =
2 cosf
Question.2
If f(q) =
then
(A) 
(B) f(q) is purely real
(C) f(p/2) = 2
(D) None of these
Solution
On operating [R1 → R1 - R2 and R3 → R3 - R2]
f(q) = 
= (-1) [(1-eiq) (-1 - e-iq) – 2(-1 -
eiq] = 2(1+ cosq) when q = 
Question.3
If x, y, z are the integers
in A.P, lying between 1 and 9 and x51, y41 and z31 are three digits numbers,
then the value of 
is
(A) x + y + z
(B) x –y + z
(C) 0
(D) None of these
Solution
D =
R2 → R2
–100R3 –10R1
= x –2y + z = 0 (
x, y, z in A.P)
Question.4
If 
= k (xyz), then k is equal to
(A) 4
(B) -4
(C) zero
(D) None of these
Solution
Putting x = 1 y = 1, and z
= 1 on both sides, we get k = 4
Question.5
Let ax7 +
bx6 + cx5 + dx4 + ex3 + fx2
+ gx + h
= 
(A) g = 3 and h = –
5
(B) g = –3 and h = –
5
(C) g = –3 and h =
–9
(D) None of these
Solution
By putting x = 0 on both sides of the
equation we have
= 9
Differentiating both
sides and then putting x = 0, we get g = -5
Question.6
D = 
is
equal to
(A) logxy
logyz logzx
(B) 1
(C) 0
(D) None of these
Solution
= 1(1 – 1) - 
= 0
Question.7
If a, b and c are pth, qth
and rth terms of an HP. then 
=
(A) Term containing a, b,
c, p, q, r
(B) a constant
(C) Zero
(D) None of these
Solution
If A is the first term and
D is the common difference of the corresponding A.P. then.
= A + (p – 1) D
= A + (q – 1) D
= A + (r –1)D
Now D =
abc 
Operating R1→ R1
– D(R2) – (A – D) R3 D = abc
= 0
Question.8
The system of
equations x + 2y + 3z = 4, 2x + 3y + 4z = 5, 3x + 4y + 5z = 6 has
(A) Infinitely many solution
(B) No solution
(C) Unique solutions
(D) None of these
Solution
Since 
, (row are in AP
with common difference 1)
Similarly Dx = Dy = Dz = 0.
Question.9
Let f (n) = 
where the symbols have their
usual meanings. The f (n) is divisible by
(A) n2 +n +1
(B) (n + 1)!
(C) n !
(D) None of these
Solution
f (n) = 
Using C3→C3-C2
and C2→C2-C1
= (n + 1) (n +1) ! – nn! =
n ! [(n+1)2 –n]
Question.10
D
= 
is equal to
(A) 100
(B) 500
(C) 1000
(D) 0
Solution
The determinant is
Skew symmetric of odd order hence is equal to 0
