FunctionshardFree

Functions: Value

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

Question
If f(x)+f(y)=f ⁣(x+y1−xy)f(x)+f(y) = f\!\left(\dfrac{x+y}{1-xy}\right) for all x,y∈Rx, y \in \mathbb{R} with xy≠1xy \ne 1, and lim⁡x→0f(x)x=2\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x} = 2, then the value of 15 f(3)π f′(−2)\dfrac{15\,f(\sqrt{3})}{\pi\,f'(-2)} is.
Solution
Answer: 25
Step 1: Identify ff The functional equation is that of tan⁡−1\tan^{-1}, so f(x)=ktan⁡−1xf(x) = k\tan^{-1}x.
lim⁡x→0ktan⁡−1xx=k=2  ⟹  f(x)=2tan⁡−1x\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(3)=2tan⁡−13=2π3f(\sqrt{3}) = 2\tan^{-1}\sqrt{3} = \frac{2\pi}{3}
f′(x)=21+x2  ⟹  f′(−2)=25f'(x) = \frac{2}{1+x^2} \implies f'(-2) = \frac{2}{5}
 Step 3: Evaluate the expression
15 f(3)π f′(−2)=15⋅2π3π⋅25=10π2π5=25\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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