Question

If $f(x) = \begin{cases} \frac{1-sin\,x}{(\pi - 2x)^2} \cdot \frac{log\,sin\, x}{log(1+ \pi^2 -4\pi x +4x^2)} ,& x\ne \frac{\pi}{2}\\[2ex] k& ,x = \frac{\pi}{2} \end{cases}$
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)

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