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(f⁻¹)′(c) for f(x)=∫₀ˣ√(1+t⁴)dt | JEE

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

Question
Suppose f(x)=∫0x1+t4 dtf(x)=\displaystyle\int_{0}^{x}\sqrt{1+t^{4}}\,dt for all real xx and let f(1)=cf(1)=c. Then the value of (f−1)′(c)\big(f^{-1}\big)'(c) is
A12\dfrac{1}{\sqrt2}correct
B32\dfrac{\sqrt3}{2}
C12\dfrac12
DNone of these
Solution
Step 1: Differentiate ff using the Fundamental Theorem of Calculus:
f′(x)=1+x4.f'(x)=\sqrt{1+x^{4}}.
 Since f′(x)>0f'(x)>0, ff is strictly increasing and therefore invertible. Step 2: Let g=f−1g=f^{-1}, so g(f(x))=xg(f(x))=x. Differentiating both sides:
g′(f(x)) f′(x)=1 ⇒ g′(f(x))=1f′(x).g'(f(x))\,f'(x)=1\ \Rightarrow\ g'(f(x))=\dfrac{1}{f'(x)}.
 Step 3: At x=1x=1, f(1)=cf(1)=c, hence
(f−1)′(c)=g′(c)=1f′(1)=11+14=12.\big(f^{-1}\big)'(c)=g'(c)=\dfrac{1}{f'(1)}=\dfrac{1}{\sqrt{1+1^{4}}}=\dfrac{1}{\sqrt2}.
 Step 4: So (f−1)′(c)=12.\big(f^{-1}\big)'(c)=\dfrac{1}{\sqrt2}. Correct answer: (1)
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