Question

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

• A. $2,0$
• B. $2,3$
• C. $0,-4$
• D. $4,0$

Answer 1

The correct option is A $2,0$

We have given that

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

And,

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

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

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

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

$\therefore B=0$ . . . (1)

$f\left(\frac{1}{2}\right)=\sqrt{2}$

$\therefore A\sin \left(\frac{\pi }{4}\right)+B=\sqrt{2}$

$\therefore \frac{A}{\sqrt{2}}=\sqrt{2}$ . . . from (1)

$\therefore A=2$

$\therefore A=2,B=0$

Hence, the option $\left(A\right)$ is correct.