Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{1-\sin x}{(\pi -2x{)}^2}\cdot \frac{\log \sin x}{\log (1+{\pi }^2-4\pi x+4x^2)},& x\ne \frac{\pi }{2}\\ k,& x=\frac{\pi }{2}\end{array}$ is continuous at $x=\frac{\pi }{2}$, then $k$ is equal to

• A. $-\frac{1}{16}$
• B. $-\frac{1}{32}$
• C. $-\frac{1}{64}$
• D. $-\frac{1}{28}$

Answer 1

The correct option is C $-\frac{1}{64}$

For $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$, we must have
$\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=f\left(\frac{\pi }{2}\right)$
$\Rightarrow \underset{x\to \frac{\pi }{2}}{lim}\frac{1-\sin x}{(\pi -2x{)}^2}\cdot \frac{\log \sin x}{\log (1+{\pi }^2-4\pi x+4x^2)}=k$
$\Rightarrow \underset{h\to 0}{lim}\frac{1-\cos h}{4h^2}\cdot \frac{\log \cos h}{\log (1+4h^2)}=k$
$\Rightarrow \underset{h\to 0}{lim}\frac{1-\cos h}{4h^2}\times \frac{\log (1+\cos h-1)}{\cos h-1}\times \frac{4h^2}{\log (1+4h^2)}\times \frac{\cos h-1}{4h^2}=k$
$\Rightarrow -\underset{h\to 0}{lim}{\left(\frac{1-\cos h}{4h^2}\right)}^2\times \frac{\log (1+(\cos h-1\left)\right)}{\cos h-1}\times \frac{4h^2}{\log (1+4h^2)}=k$
$\Rightarrow -\frac{1}{64}\underset{h\to 0}{lim}{\left(\frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}\right)}^4\frac{\log (1+(\cos h-1\left)\right)}{\cos h-1}\times \frac{4h^2}{\log (1+4h^2)}=k$
$\Rightarrow -\frac{1}{64}=k\Rightarrow k=-\frac{1}{64}$.