Q25 / 50 · Manipal (MET) 2024 · Mathematics
A real valued differentiable function $f$ satisfies $f(x)+f(y)+2 x y=f(x+y) \forall x, y \in R$. Given $f^n(0)=0$, then value of $\int_0^{\frac{\pi}{2}} f(\sin x) d x$ will be
- A$0$
- Correct answerB$\frac{\pi}{4}$
- C$\frac{\pi}{2}$
- D$\pi$
Solution
$f(x)+f(y)+2 x y=f(x+y)...(i)$
differentiating the equation,
$\Rightarrow f^{\prime}(y)+2 x=f^{\prime}(x+y)$
at $y=0$
$f^{\prime}(0)+2 x=f^{\prime}(x)$
$\Rightarrow f^{\prime}(x)=2 x$
Integrating both sides.
$\int f^{\prime}(x)=\int 2 x \cdot d x$
$\Rightarrow f(x)=x^2$
$\therefore I=\int_0^{\frac{\pi}{2}} f(\sin x) d x$
$\Rightarrow I=\int_0^{\frac{\pi}{2}} \sin ^2 x \cdot d x...(ii)$
$\Rightarrow I=\int_0^{\frac{\pi}{2}} \sin ^2\left(\frac{\pi}{2}-x\right) d x=\int_0^{\frac{\pi}{2}} \cos ^2 x \cdot d x \ldots(iii)$
Adding Eqs. (ii) and (iii)
$2l=\int_0^{\frac{\pi}{2}}\left(\sin ^2 x+\cos ^2 x\right) d x=\int_0^{\frac{\pi}{2}} 1 \cdot d x$
$\begin{array}{ll}\Rightarrow & 2l=\frac{\pi}{2} \\ \Rightarrow & l=\frac{\pi}{4}\end{array}$
Hence, the correct answer is option: 'B'.