Question

If $f\left(x\right)=A\sin \left(\frac{\pi x}{2}\right)+B,f^{\prime }\left(\frac{1}{2}\right)=\sqrt{2}$ and ${\int }_0^{1}f\left(x\right)dx=\frac{2A}{\pi }$, then $A$ and $B$ are

• A. $\frac{\pi }{2},\frac{\pi }{2}$
• B. $\frac{2}{\pi },\frac{3}{\pi }$
• C. $0,-4\pi$
• D. $\frac{4}{\pi },0$

Answer 1

Answer: (D)

$\frac{4}{\pi },0$

$f\left(x\right)=A\sin \left(\frac{\pi x}{2}\right)+B$

$\Rightarrow f^{\prime }\left(x\right)=\frac{A\pi }{2}\cos \left(\frac{\pi x}{2}\right)$

$\Rightarrow f^{\prime }\left(\frac{1}{2}\right)=\frac{A\pi }{2}\cos \frac{\pi }{4}=\frac{A\pi }{2\sqrt{2}}$

$\Rightarrow \frac{A\pi }{2\sqrt{2}}=\sqrt{2}\{\because f^{\prime }\left(\frac{1}{2}\right)=\sqrt{2}\}$

$\Rightarrow A=\frac{4}{\pi }$

Now, ${\int }_0^{1}f\left(x\right)dx=\frac{2A}{\pi }$

$\Rightarrow {\int }_0^{1}A\sin \left(\frac{\pi x}{2}\right)+B\cdot dx=\frac{2A}{\pi }$

$\Rightarrow {[\frac{-2A}{\pi }\cos \left(\frac{\pi x}{2}\right)+Bx]}_0^1=\frac{2A}{\pi }$

$\Rightarrow B+\frac{2A}{\pi }=\frac{2A}{\pi }$

$\therefore B=0$.