Question

If $f\left(x\right)$ = $\frac{\sin \left(\cos x\right)-\cos x}{(\pi -2x{)}^3}$ & $x\ne$ $\frac{\pi }{2}k$ & $x=\frac{\pi }{2}$ is a continuous at $x=\frac{\pi }{2}$, then $k$ is equal to

• A. $0$
• B. $-\frac{1}{6}$
• C. $-\frac{1}{24}$
• D. $-\frac{1}{48}$

Answer 1

The correct option is D $-\frac{1}{48}$

The given function is continus at $x=\frac{\pi }{2}$.

Therefore, $\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=f\left(\frac{\pi }{2}\right)$

$\Rightarrow \underset{x\to \frac{\pi }{2}}{lim}\frac{\sin \left(\cos x\right)-\cos x}{(\pi -2x)}=k$

Let $x-\frac{\pi }{2}=h\Rightarrow \frac{2x-\pi }{2}=h$

$\Rightarrow \frac{\pi -2x}{2}=-h$

$x\to \frac{\pi }{2}\Rightarrow x\ne \frac{\pi }{2}\Rightarrow x-\frac{\pi }{2}\ne 0$

$\Rightarrow h\ne 0\Rightarrow h\to 0$

$k=\underset{h\to 0}{lim}\frac{\sin \left(\cos \right(h+\frac{\pi }{2}\left)\right)-\cos (h+\frac{\pi }{2})}{-8h^3}$

$k=\underset{h\to o}{lim}\frac{-\sin \left(\sin h\right)+\sin h}{-8h^3}$

Applying L'hospital's rule to above equation, we get

$k=-\frac{1}{48}$