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MHTCET 15 Apr 2026 Shift 1 Paper

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Question : 123 of 150

Marks: +1, -0

Let

f(x)=3 x^{3}+4 e^{\frac{x}{2}}

g(x)=f^{-1}(x)

. Then the value of

g^{\prime}(4)

is equal to...

4
2
1
$\frac{1}{2}$

Solution:

Concept:

For an inverse function,

g'(x)=\frac{1}{f'(g(x))}

.

Explanation:

We need

g'(4)

, so first find

g(4)

.

Since

g=f^{-1}

,

g(4)

is the value of

x

for which

f(x)=4

.

Given

f(x)=3x^3+4e^{x/2}

.

Check

x=0

:

f(0)=3(0)^3+4e^0=0+4=4

.

Hence

g(4)=0

.

Differentiate

f(x)

:

f'(x)=9x^2+4\cdot\frac12 e^{x/2}=9x^2+2e^{x/2}

.

At

x=0

,

f'(0)=9(0)^2+2e^0=2

.

Therefore,

g'(4)=\frac{1}{f'(g(4))}=\frac{1}{f'(0)}=\frac12

.

Answer:

\frac{1}{2}

, i.e. Option D.

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