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

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

Answer 1

The correct option is D $\frac{4}{\pi }$ and 0

${\int }_0^{1}f\left(x\right)dx=\frac{2A}{\pi }\Rightarrow f^{\prime }\left(x\right)=\frac{A\pi }{2}\cos \frac{\pi x}{2}\Rightarrow f^{\prime }\left(\frac{1}{2}\right)=\frac{A\pi }{2}\cos \frac{\pi }{4}=\frac{A\pi }{2\sqrt{2}}$
Now
$f^{\prime }\left(\frac{1}{2}\right)=\sqrt{2}\Rightarrow \frac{A\pi }{2\sqrt{2}}=\sqrt{2}\Rightarrow A=\frac{4}{\pi }$
And ${\int }_0^{1}f\left(x\right)dx=\frac{2A}{\pi }\Rightarrow {\int }_0^1\left(A\sin (\frac{\pi x}{2}+B)\right)dx=\frac{2A}{\pi }\Rightarrow {[-\frac{2A}{\pi }\cos \frac{\pi x}{2}]}_0^1=\frac{2A}{\pi }\Rightarrow B+\frac{2A}{\pi }=\frac{2A}{\pi }\Rightarrow B=0$