Question

If $f\left(x\right)=Rsin\left(\frac{\pi x}{2}\right)+S\&f^{\prime }\left(\frac{1}{2}\right)=\sqrt{2}$ and ${\int }_0^{1}f\left(x\right)dx=\frac{2R}{\pi },$ then the constants $R$ and $S$ are, respectively.

• A. $\frac{2}{\pi }$ and $\frac{16}{\pi }$
• B. $\frac{2}{\pi }$ and $0$
• C. $\frac{4}{\pi }$ and $0$
• D. $\frac{4}{\pi }$ and $\frac{16}{\pi }$

Answer 1

The correct option is C $\frac{4}{\pi }$ and $0$

$f^{\prime }\left(x\right)=\frac{R\pi }{2}cos\left(\frac{\pi x}{2}\right)$
$\Rightarrow f^{\prime }\left(1/2\right)=\sqrt{2}\Rightarrow \frac{R\pi }{2\sqrt{2}}=\sqrt{2}\Rightarrow R=\frac{4}{\pi }$
Also ${\int }_0^{1}f\left(x\right)dx=\frac{2R}{\pi }\Rightarrow {\left[\frac{-2R}{\pi }cos\frac{\pi x}{2}\right]}_{x=0}^1+S(x{)}_0^1=2R/\pi$
$\Rightarrow S=0$