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Solved Objective Question on Determinant Set 1

Posted on - 04-01-2017

Math

IIT JEE

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

 
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