Question

If $f(x)=\{\begin{array}{ll}\frac{1-sinx}{(\pi -2x{)}^2}\cdot \frac{logsinx}{log(1+{\pi }^2-4\pi x+4x^2)},& x\ne \frac{\pi }{2}\\ k& ,x=\frac{\pi }{2}\end{array}$
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

Answer: (C)

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