Question
If $$f\left( x \right) = A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B,\,\,f'\left( {\frac{1}{2}} \right) = \sqrt 2 $$ and $$\int\limits_0^1 {f\left( x \right)dx = \frac{{2A}}{\pi },} $$ then constant $$A$$ and $$B$$ are-
A.
$$\frac{\pi }{2}{\text{ and }}\frac{\pi }{2}$$
B.
$$\frac{2}{\pi }{\text{ and }}\frac{3}{\pi }$$
C.
$$0{\text{ and }}\frac{{ - 4}}{\pi }$$
D.
$$\frac{4}{\pi }{\text{ and 0}}$$
Answer :
$$\frac{4}{\pi }{\text{ and 0}}$$
Solution :
$$\eqalign{
& f\left( x \right) = A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B \cr
& \Rightarrow f'\left( x \right) = \frac{{A\pi }}{2}\cos \left( {\frac{{\pi x}}{2}} \right) \cr
& \Rightarrow f'\left( {\frac{1}{2}} \right) = \frac{{A\pi }}{2}\cos \frac{\pi }{4} = \sqrt 2 \cr
& \Rightarrow A = \frac{4}{\pi }\,\,{\text{and }}\int_0^1 {f\left( x \right)dx = \frac{{2A}}{\pi }} \cr
& \Rightarrow \int_0^1 {\left[ {A\,\sin \left( {\frac{{\pi x}}{2}} \right) + B} \right]dx = \frac{{2A}}{\pi }} \cr
& \Rightarrow \left| { - \frac{{2A}}{\pi }\cos \left( {\frac{{\pi x}}{2}} \right) + Bx} \right|_0^1 = \frac{{2A}}{\pi } \cr
& \Rightarrow B + \frac{{2A}}{\pi } = \frac{{2A}}{\pi }\,\,\,\,\, \Rightarrow B = 0 \cr} $$