Question

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

Answer: (C)

$-\frac{1}{64}$

For $f\left(x\right)$ to be continuous at $x=\frac{\pi }{2}$,
We must have
$\underset{x\to \pi /2}{lim}f\left(x\right)=f\left(\pi /2\right)$
$\Rightarrow \underset{x\to \pi /2}{lim}\frac{1-\sin x}{(\pi -2x{)}^2}\cdot \frac{\log \left(\sin x\right)}{\log (1+{\pi }^2-4\pi x+4x^2)}=k$

$\Rightarrow \underset{h\to 0}{lim}\frac{1-\cos h}{4h^2}\times \frac{\log \cos h}{\log (1+4h^2)}=k$ .... $[\because x=\frac{\pi }{2}-h⟹h\to 0asx\to \frac{\pi }{2}]$

$\Rightarrow \underset{h\to 0}{lim}\frac{1-\cos h}{4h^2}\times \frac{\log \{1+\cos h-1\}}{\cos h-1}\times \frac{4h^2}{\log (1+4h^2)}\times \frac{\cos h-1}{4h^2}=k$

$\Rightarrow -\underset{h\to 0}{lim}{\left(\frac{1-\cos h}{4h^2}\right)}^2\frac{\log (1+(\cos h-1\left)\right)}{\cos h-1}\times \frac{4h^2}{\log (1+4h^2)}=k$

$\Rightarrow -\underset{h\to 0}{lim}{\left(\frac{{\sin }^{2}h/2}{2h^2}\right)}^2\frac{\log (1+(\cos h-1\left)\right)}{\cos h-1}\times \frac{4h^2}{\log (1+4h^2)}=k$

$\Rightarrow -\frac{1}{64}\underset{h\to 0}{lim}{\left(\frac{\sin h/2}{h/2}\right)}^4\frac{\log (1+(\cos h-1\left)\right)}{\cos h-1}\times \frac{4h^2}{\log (1+4h^2)}=k$
$\Rightarrow -\frac{1}{64}=k$.