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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
If f(x)=∫0x(1+t3)−1/2 dtf(x)=\displaystyle\int_{0}^{x}\big(1+t^{3}\big)^{-1/2}\,dt and g(x)g(x) is the inverse of ff, then the value of g′′(x)g2(x)\dfrac{g''(x)}{g^{2}(x)} is
A32\dfrac32correct
B) 23\dfrac23
C13\dfrac13
D12\dfrac12
Solution
Step 1: Differentiate ff by the Fundamental Theorem of Calculus:
f′(x)=(1+x3)−1/2.f'(x)=\big(1+x^{3}\big)^{-1/2}.
 Step 2: Use the inverse relationship f(g(x))=xf(g(x))=x. Differentiating, f′(g(x)) g′(x)=1f'(g(x))\,g'(x)=1, so
g′(x)=1f′(g(x))=(1+g3)1/2⇒(g′)2=1+g3.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 (g′)2=1+g3\big(g'\big)^{2}=1+g^{3} with respect to xx:
2g′ g′′=3g2 g′.2g'\,g''=3g^{2}\,g'.
 Step 4: Cancel g′g' (non-zero):
g′′=32g2 ⇒ g′′(x)g2(x)=32.g''=\dfrac{3}{2}g^{2}\ \Rightarrow\ \dfrac{g''(x)}{g^{2}(x)}=\dfrac32.
 Correct answer: (1)
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