Let \( f:\mathrm{R}\to\mathrm{R} \) be a function such that \( f(x)+f(y)=f\left(\displaystyle\frac{x+y}{1-xy}\right) \), where \( \mathrm{R} \) is set of real numbers then \( \alpha+\int_{-\alpha}^{\alpha}f(x)\,dx \) is equal to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \alpha \)
Putting x = y = 0 in the given relation,
\( \displaystyle f(0)+f(0)=f(0) \)
\( \displaystyle \Rightarrow\ f(0)=0 \)
Putting y = −x, so that \( \frac{x+y}{1-xy}=0 \),
\( \displaystyle f(x)+f(-x)=f(0) \)
\( \displaystyle =0 \)
\( \displaystyle f(-x)=-f(x) \)
So f is an odd function, and its integral over the symmetric interval vanishes:
\( \displaystyle \int_{-\alpha}^{\alpha}f(x)\,dx=0 \)
\( \displaystyle \alpha+\int_{-\alpha}^{\alpha}f(x)\,dx=\alpha \)