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g″/g² for inverse of f(x)=∫(1+t³)^{−1/2}dt | JEE

JEE Maths question with a full step-by-step solution.

Question

f(x)=\displaystyle\int_{0}^{x}\big(1+t^{3}\big)^{-1/2}\,dt

and

g(x)

is the inverse of

f

, then the value of

\dfrac{g''(x)}{g^{2}(x)}

\dfrac32

correct

\dfrac23

\dfrac13

\dfrac12

Solution

Step 1: Differentiate

f

by the Fundamental Theorem of Calculus:

f'(x)=\big(1+x^{3}\big)^{-1/2}.

Step 2: Use the inverse relationship

f(g(x))=x

. Differentiating,

f'(g(x))\,g'(x)=1

, so

g'(x)=\dfrac{1}{f'(g(x))}=\big(1+g^{3}\big)^{1/2}\quad\Rightarrow\quad \big(g'\big)^{2}=1+g^{3}.

Step 3: Differentiate

\big(g'\big)^{2}=1+g^{3}

with respect to

x

2g'\,g''=3g^{2}\,g'.

Step 4: Cancel

g'

(non-zero):

g''=\dfrac{3}{2}g^{2}\ \Rightarrow\ \dfrac{g''(x)}{g^{2}(x)}=\dfrac32.

Correct answer: (1)

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