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JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Differentiation

Let

f (x) = x^{5} + 2 e^{x /4}

for all

x \in R

. Consider a function

g (x)

such that

(g \circ f) (x) = x

for all

x \in R

. Then the value of

8 g^{'} (2)

is

(A)16✓ Correct
(B)4
(C)8
(D)2

Answer: (A)

Solution

g

is the inverse of

f (x) = x^{5} + 2 e^{x /4}

, so

g^{'} (2) = \frac{1}{f ^{'} ( g ( 2 ))}

.

Since

f (0) = 0 + 2 = 2

, we have

g (2) = 0

. Also

f^{'} (x) = 5 x^{4} + \frac{1}{2} e^{x /4}

, so

f^{'} (0) = \frac{1}{2}

.

g^{'} (2) = \frac{1}{f ^{'} ( 0 )} = 2 \Rightarrow 8 g^{'} (2) = 16.

Option A.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option

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