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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)=\displaystyle\int_{0}^{x}\sqrt{1+t^{4}}\,dt

for all real

x

and let

f(1)=c

. Then the value of

\big(f^{-1}\big)'(c)

\dfrac{1}{\sqrt2}

correct

\dfrac{\sqrt3}{2}

\dfrac12

DNone of these

Solution

Step 1: Differentiate

f

using the Fundamental Theorem of Calculus:

f'(x)=\sqrt{1+x^{4}}.

Since

f'(x)>0

f

is strictly increasing and therefore invertible. Step 2: Let

g=f^{-1}

, so

g(f(x))=x

. Differentiating both sides:

g'(f(x))\,f'(x)=1\ \Rightarrow\ g'(f(x))=\dfrac{1}{f'(x)}.

Step 3: At

x=1

f(1)=c

, hence

\big(f^{-1}\big)'(c)=g'(c)=\dfrac{1}{f'(1)}=\dfrac{1}{\sqrt{1+1^{4}}}=\dfrac{1}{\sqrt2}.

Step 4: So

\big(f^{-1}\big)'(c)=\dfrac{1}{\sqrt2}.

Correct answer: (1)

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