JEE Challenger
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MathMedAdv 2010 • Paper 2

Derivative of Inverse Function defined by Integral Equation

Let ff be a real-valued function defined on the interval (−1,1)(-1, 1) such that e−xf(x)=2+∫0xt4+1 dte^{-x} f(x) = 2 + \int_{0}^{x} \sqrt{t^4 + 1} \, dt, for all x∈(−1,1)x \in (-1,1), and let f−1f^{-1} be the inverse function of ff.

Then (f−1)′(2)(f^{-1})'(2) is equal to

Options

A

11

B

13\frac{1}{3}

Correct
C

12\frac{1}{2}

D

1e\frac{1}{e}

Topics & Concepts

Continuity and DifferentiabilityDifferentiationInverse Functions

Step-by-Step Solution

To find the derivative of the inverse function (f−1)′(2)(f^{-1})'(2), we use the standard derivative rule for inverse functions:
(f−1)′(y0)=1f′(x0),where f(x0)=y0(f^{-1})'(y_0) = \frac{1}{f'(x_0)}, \quad \text{where } f(x_0) = y_0

Step 1: Find x0x_0 such that f(x0)=2f(x_0) = 2

The given functional equation is:
e−xf(x)=2+∫0xt4+1 dte^{-x} f(x) = 2 + \int_{0}^{x} \sqrt{t^4 + 1} \, dt

Substituting x=0x = 0 into the equation:
e−0f(0)=2+∫00t4+1 dte^{-0} f(0) = 2 + \int_{0}^{0} \sqrt{t^4 + 1} \, dt
1⋅f(0)=2+0  ⟹  f(0)=21 \cdot f(0) = 2 + 0 \implies f(0) = 2

Thus, x0=0x_0 = 0 corresponds to y0=2y_0 = 2, which gives f−1(2)=0f^{-1}(2) = 0.

Step 2: Find f′(0)f'(0)

Differentiating both sides of the given functional equation with respect to xx using the product rule on the left side and Leibniz's rule for differentiation under the integral sign on the right side:
ddx(e−xf(x))=ddx(2+∫0xt4+1 dt)\frac{d}{dx} \left( e^{-x} f(x) \right) = \frac{d}{dx} \left( 2 + \int_{0}^{x} \sqrt{t^4 + 1} \, dt \right)

−e−xf(x)+e−xf′(x)=x4+1-e^{-x} f(x) + e^{-x} f'(x) = \sqrt{x^4 + 1}

Now, evaluate this derivative equation at x=0x = 0:
−e−0f(0)+e−0f′(0)=04+1-e^{-0} f(0) + e^{-0} f'(0) = \sqrt{0^4 + 1}

Substitute f(0)=2f(0) = 2 into the equation:
−1⋅(2)+1⋅f′(0)=1-1 \cdot (2) + 1 \cdot f'(0) = 1
−2+f′(0)=1  ⟹  f′(0)=3-2 + f'(0) = 1 \implies f'(0) = 3

Step 3: Calculate (f−1)′(2)(f^{-1})'(2)

Using the formula for the derivative of the inverse function:
(f−1)′(2)=1f′(0)=13(f^{-1})'(2) = \frac{1}{f'(0)} = \frac{1}{3}

Hence, the correct option is B.

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