MDoubt No 000005 - GATE

GATE [CS, 2017, 1 mark]
Question : If $\displaystyle f(x)=R\sin\left(\frac{\pi x}{2}\right)+S, f'\left(\frac{1}{2}\right)=\sqrt{2}$ and $\displaystyle\int\limits_0^1 f(x)dx=\frac{2R}{\pi}$, then the constants $R$ and $S$ are, respectively

(a) $\displaystyle\frac{2}{\pi}$ and $\displaystyle\frac{16}{\pi}$
(b) $\displaystyle\frac{2}{\pi}$ and $0$
(c) $\displaystyle\frac{4}{\pi}$ and $0$
(d) $\displaystyle\frac{4}{\pi}$ and $\displaystyle\frac{16}{\pi}$

Solution : Given $f(x)=R\sin\left(\dfrac{\pi x}{2}\right)+S$
$\implies f'(x)=R\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{\pi x}{2}\right)$
Now $f'\left(\dfrac{1}{2}\right)=\sqrt{2}$
$\implies R\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{\pi}{4}\right)=\sqrt{2}$
$\implies R\left(\dfrac{\pi}{2}\right)\left(\dfrac{1}{\sqrt{2}}\right)=\sqrt{2}$
$\implies R=\dfrac{4}{\pi}$
We have $\displaystyle\int\limits_0^1 f(x)dx=\frac{2R}{\pi}$
$\implies\displaystyle\int\limits_0^1 R\sin\left(\dfrac{\pi x}{2}\right)+S dx=\frac{2R}{\pi}$
$\implies\displaystyle\int\limits_0^1 \dfrac{4}{\pi}\sin\left(\dfrac{\pi x}{2}\right)+S dx=\frac{8}{\pi^2}$
$\implies\dfrac{4}{\pi}\left[\dfrac{\cos\left(\dfrac{\pi x}{2}\right)}{\dfrac{\pi}{2}}+S\right]_0^1=\dfrac{8}{\pi^2}$
$\implies\dfrac{8}{\pi^2}+S=\dfrac{8}{\pi^2}$
$\implies S=0$
CORRECT ANSWER : C