Explanation

(d)
We have, $f(x) = A\sin{\left( \frac{\pi x}{2} \right) + B,\ f^{'}\left( \frac{1}{2} \right) = \sqrt{2}}$
and $\int_{0}^{1}{f(x)dx = \frac{2A}{\pi}}$
$f^{'}(x) = \frac{A\pi}{2}\cos\frac{\pi x}{2}$
$\Rightarrow f^{'}\left( \frac{1}{2} \right) = \frac{A\pi}{2}\cos{\frac{\pi}{4} = \frac{A\pi}{2\sqrt{2}}}$
But $f^{'}\left( \frac{1}{2} \right) = \sqrt{2}$
Therefore, $\frac{A\pi}{2\sqrt{2}} = \sqrt{2}\ \Rightarrow A = \frac{4}{\pi}$
Now, $\int_{0}^{1}{f(x)dx = \frac{2A}{\pi}}$
$\Rightarrow \ \int_{0}^{1}{\left\{ A\sin{\left( \frac{\pi x}{2} \right) + B} \right\} dx = \frac{2\ A}{\pi}}$
$\Rightarrow \ \ \left\lbrack - \frac{2A}{\pi}\cos{\frac{\pi x}{2} + Bx} \right\rbrack_{0}^{1} = \frac{2\ A}{\pi}$
$\Rightarrow \ B + \frac{2\ A}{\pi} = \frac{2\ A}{\pi}\ \ \Rightarrow \ B = 0$