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 constants $A$ and $B$ are

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

Answer 1

Answer: (D)

$A=\frac{4}{\pi },B=0$

$f\left(x\right)=A\sin \left(\pi x/2\right)+B$

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

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

$⟹\sqrt{2}=\frac{A\pi }{2}\frac{1}{\sqrt{2}}$

Hence, $A=\frac{4}{\pi }$

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

$⟹\frac{2A}{\pi }=\frac{2A}{\pi }+B$

Hence, $B=0$

Hence, answer is option-(D).