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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Mathematics
Chapter
Differentiation

JEE Advanced 2010 · Paper 2 · Mathematics · Differentiation

Let f be a real-valued function defined on the interval (-1,1) such...

Let ff be a real-valued function defined on the interval (−1,1)(-1,1) such that e−xf(x)=2+∫0xt4+1dt,e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t, \quad for all x∈(−1,1)x \in(-1,1) and let f−1f^{-1} be the inverse function of ff. Then (f−1)′(2)\left(f^{-1}\right)^{\prime}(2) is equal to

Options

  1. A1
  2. B13\frac{1}{3}
  3. C12\frac{1}{2}
  4. D1e\frac{1}{e}

Correct answer

B. 13\frac{1}{3}

Step-by-step solution

We have, e−xf(x)=2+∫0xt4+1dtx∈(−1,1)e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t x \in(-1,1) On differentiating w.r.t. xx, we get e−x(f′(x)−f(x))=x4+1⇒f′(x)=f(x)+x4+1ex∵f−1 is the inverse of f∴f−1(f(x))=x⇒f−1′(f(x))f′(x)=1⇒f−1′(f(x))=1f′(x)⇒f−1′(f(x))=1f(x)+x4+1ex At x=0,f(x)=2f−1′(2)=12+1=13\begin{array}{ll} & e^{-x}\left(f^{\prime}(x)-f(x)\right)=\sqrt{x^4+1} \\ \Rightarrow \quad & f^{\prime}(x)=f(x)+\sqrt{x^4+1} e^x \\ \because & f^{-1} \text{ is the inverse of } f \\ \therefore & f^{-1}(f(x))=x \\ \Rightarrow & f^{-1^{\prime}}(f(x)) f^{\prime}(x)=1 \\ \Rightarrow & f^{-1^{\prime}}(f(x))=\frac{1}{f^{\prime}(x)} \\ \Rightarrow \quad & f^{-1^{\prime}}(f(x))=\frac{1}{f(x)+\sqrt{x^4+1} e^x} \\ \text{ At } \quad & x=0, f(x)=2 \\ & f^{-1^{\prime}}(2)=\frac{1}{2+1}=\frac{1}{3} \end{array}
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