If $$f\left( x \right) = A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B$$      and $$f'\left( {\frac{1}{2}} \right) = \sqrt 2 $$    and $$\int_0^1 {f\left( x \right)dx = \frac{{2A}}{\pi },} $$     then what is the value of $$B\,?$$

Solution :
Given function $$f\left( x \right) = A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B$$
Differentiating w.r.t. $$x$$
$$\eqalign{ & f'\left( x \right) = A\,\cos \left( {\frac{{\pi x}}{2}} \right).\frac{\pi }{2} \cr & f'\left( {\frac{1}{2}} \right) = \sqrt 2 = A\left( {\cos \frac{\pi }{4}} \right)\frac{\pi }{2} = A.\frac{1}{{\sqrt 2 }}.\frac{\pi }{2} \cr & \Rightarrow A = \frac{{\left( {\sqrt 2 \times \sqrt 2 } \right) \times 2}}{\pi } = \frac{4}{\pi } \cr & {\text{Now, }}\int_0^1 {f\left( x \right)dx = \frac{{2A}}{\pi }} \cr & \Rightarrow \int_0^1 {\left\{ {A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B} \right\}dx = \frac{{2 \times 4}}{{{\pi ^2}}}} \cr & \Rightarrow \left[ { - A\,\cos \frac{{\pi x}}{2}.\frac{2}{\pi } + Bx} \right]_0^1 = \frac{8}{{{\pi ^2}}} \cr & \Rightarrow - \frac{4}{\pi }.\frac{2}{\pi }\cos \frac{\pi }{2} + B + \frac{4}{\pi }.\frac{2}{\pi }\cos \,0 = \frac{8}{{{\pi ^2}}} \cr & \Rightarrow B + \frac{8}{{{\pi ^2}}} = \frac{8}{{{\pi ^2}}} \cr & \Rightarrow B = 0 \cr} $$