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 the constant A and B are, respectively.

• A. $\pi /2$ and $\pi /2$
• B. $4/\pi$ and $3/\pi$
• C. $4/\pi$ and $-4/\pi$
• D. $4/\pi$ and $0$

Answer 1

The correct option is D $4/\pi$ and $0$

Given, $f\left(x\right)=Asin\left(\frac{\pi x}{2}\right)+B$ (1)

Differentiate w.r.t. x, we get,

$f^{\prime }\left(x\right)=Acos\left(\frac{\pi x}{2}\right)\times \frac{\pi }{2}$

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

$\therefore f^{\prime }\left(\frac{1}{2}\right)=\frac{A\pi }{2}cos(\frac{\pi }{2}\times \frac{1}{2})$

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

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

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

Given, $f^{\prime }\left(\frac{1}{2}\right)=\sqrt{2}$

$\therefore \frac{A\pi }{2\sqrt{2}}=\sqrt{2}$

$\therefore A=\frac{2\sqrt{2}\times \sqrt{2}}{\pi }$

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

Now, integrate equation (1) w.r.t. x, we get,

${\int }_0^{1}f\left(x\right)dx={\int }_0^1[Asin\left(\frac{\pi x}{2}\right)+B]dx$

$\therefore {\int }_0^{1}f\left(x\right)dx={\int }_0^{1}Asin\left(\frac{\pi x}{2}\right)dx+{\int }_0^{1}Bdx$

$\therefore {\int }_0^{1}f\left(x\right)dx=A{\left[\frac{-cos\left(\frac{\pi x}{2}\right)}{\frac{\pi }{2}}\right]}_0^1+B{\left[x\right]}_0^1$

$\therefore {\int }_0^{1}f\left(x\right)dx=-\frac{2A}{\pi }{\left[cos\left(\frac{\pi x}{2}\right)\right]}_0^1+B{\left[x\right]}_0^1$

$\therefore {\int }_0^{1}f\left(x\right)dx=-\frac{2A}{\pi }[cos\left(\frac{\pi }{2}\right)-cos\left(0\right)]+B[1-0]$

$\therefore {\int }_0^{1}f\left(x\right)dx=-\frac{2A}{\pi }[0-1]+B\left[1\right]$

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

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

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

$\therefore B=0$

Thus, answer is option (D)