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

Question

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

Aicharya · Accepted Answer

It is given that $f(x)$ is continuous at $x=\frac{\pi}{2}$. $	herefore \quad k=\lim _{x ightarrow \pi / 2} f(x)$ $\begin{aligned} & \Rightarrow \quad k=\lim _{x ightarrow \pi / 2} \frac{\sin (\cos x)-\cos x}{(\pi-2 x)^3} \ & \Rightarrow \quad k=\lim _{x ightarrow \pi / 2} \frac{\sin (\cos x)-\cos x}{\cos ^3 x} 	imes \frac{\sin ^3\left(\frac{\pi}{2}-xight)}{8\left(\frac{\pi}{2}-xight)^3} \ & 	herefore \quad k=-\frac{1}{6} 	imes \frac{1}{8}=-\frac{1}{48}\left[\because \lim _{x ightarrow 0} \frac{\sin x-x}{x^3}=-\frac{1}{6}ight]\end{aligned}$

If $f(x)= \begin{cases}\frac{\sin (\cos x)-\cos x}{(\pi-2 x)^3}, & x \neq \frac{\pi}{2} \\ k, &…

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