Question

A Function $y=f\left(x\right)$ stasfies $f^{\prime \prime }\left(x\right)=-\frac{1}{x^2}-{\pi }^2\sin \left(\pi x\right)$; $f^{\prime }\left(2\right)=\pi +\frac{1}{2}$ and $f\left(1\right)=0$. The value of $f\left(\frac{1}{2}\right)$ is:

• A. $\ln 2$
• B. $1$
• C. $\frac{\pi }{2}-\ln 2$
• D. $1-\ln 2$

Answer 1

The correct option is D $1-\ln 2$

$f^{\prime \prime }\left(x\right)=-\frac{1}{x^2}-{\pi }^2\sin \left(\pi x\right)$
Integrating w.r.t x on both sides,
$f^{\prime }\left(x\right)=\frac{1}{x}+{\pi }^2\frac{\cos \pi x}{\pi }+C$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}+\pi \cos \pi x+C$
$f^{\prime }\left(2\right)=\frac{1}{2}+\pi +C$
$\pi +\frac{1}{2}=\frac{1}{2}+\pi +C$
$\Rightarrow C=0$
Hence, $f^{\prime }\left(x\right)=\frac{1}{x}+\pi \cos \pi x$
Again , integrating w.r.t. x
$\Rightarrow f\left(x\right)=ln|x|+\sin \pi x+C_1$
Put $x=1$
$0=0+0+C_1$
$\Rightarrow C_1=0$
Hence, $f\left(x\right)=\ln |x|+\sin \pi x$
$f\left(\frac{1}{2}\right)=\ln |\frac{1}{2}|+1$
$\Rightarrow f\left(\frac{1}{2}\right)=-\ln 2+1$
Hence, option 'D' is correct.