FunctionshardFree

Functions — JEE Maths practice question

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

Question

f(x)+f(y) = f\!\left(\dfrac{x+y}{1-xy}\right)

for all

x, y \in \mathbb{R}

with

xy \ne 1

, and

\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x} = 2

, then the value of

\dfrac{15\,f(\sqrt{3})}{\pi\,f'(-2)}

is.

Solution

Answer: 25

Step 1: Identify

f

The functional equation is that of

\tan^{-1}

, so

f(x) = k\tan^{-1}x

\lim_{x\to 0}\frac{k\tan^{-1}x}{x} = k = 2 \implies f(x) = 2\tan^{-1}x

Step 2: Compute the required values

f(\sqrt{3}) = 2\tan^{-1}\sqrt{3} = \frac{2\pi}{3}

f'(x) = \frac{2}{1+x^2} \implies f'(-2) = \frac{2}{5}

Step 3: Evaluate the expression

\frac{15\,f(\sqrt{3})}{\pi\,f'(-2)} = \frac{15\cdot\frac{2\pi}{3}}{\pi\cdot\frac{2}{5}} = \frac{10\pi}{\frac{2\pi}{5}} = 25

Answer: 25

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