Calculus

Integrals

Medium

common

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(x)+f(y)=f\left(\dfrac{x+y}{1-xy}\right)$ , Where $\mathbb{R}$ is set of real numbers then $\alpha+\int_{-\alpha}^{\alpha} f(x)\,dx$ is equal to:

$\alpha$

$2\alpha$

$\alpha - 1$

✅ Correct Option: 1

The functional equation is the addition formula for $\tan^{-1}$ , so $f(x) = k\tan^{-1}x$ . Putting $y = -x$ gives $f(x) + f(-x) = f(0) = 0$ , so $f$ is odd.

An odd function integrated over the symmetric interval $[-\alpha,\alpha]$ gives $0$ .

Hence the expression equals $\alpha + 0 = \alpha$ .

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Let f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R be a function such that f(x)+f(y)=f(x+y1−xy)f(x)+f(y)=f\left(\dfrac{x+y}{1-xy}\right)f(x)+f(y)=f(1−xyx+y​), Where R\mathbb{R}R is set of real numbers then α+∫−ααf(x) dx\alpha+\int_{-\alpha}^{\alpha} f(x)\,dxα+∫−αα​f(x)dx is equal to:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(x)+f(y)=f\left(\dfrac{x+y}{1-xy}\right)$ , Where $\mathbb{R}$ is set of real numbers then $\alpha+\int_{-\alpha}^{\alpha} f(x)\,dx$ is equal to: