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JEE Main · 2010 · 11 Apr 2010 · Offline · Differentiation

Let

f : (- 1, 1) \to R

be a differentiable function with

f (0) = - 1

and

f^{'} (0) = 1

. Let

g (x) = [f (2 f (x) + 2)]^{2}

. Then

g^{'} (0) =

(A)-4✓ Correct
(B)0
(C)-2
(D)4

Answer: (A)

Solution

Differentiate

g (x) = [f (2 f (x) + 2)]^{2}

by the chain rule. With

u (x) = 2 f (x) + 2

we have

g^{'} (x) = 2 f (u) \cdot f^{'} (u) \cdot u^{'} (x)

and

u^{'} (x) = 2 f^{'} (x)

, so

g^{'} (x) = 4 f (u) f^{'} (u) f^{'} (x)

. At

x = 0

,

u (0) = 2 f (0) + 2 = 2 (- 1) + 2 = 0

, so

f (u) = f (0) = - 1

and

f^{'} (u) = f^{'} (0) = 1

. Hence

g^{'} (0) = 4 \times (- 1) \times 1 \times 1 = - 4

.

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option

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