Question

Find the value of k if f(x) is continuous at x = π/2, where
$f\left(x\right)=\left\{\begin{array}{ll}\frac{k\cos x}{\mathrm{\pi }-2x},& x\ne \mathrm{\pi }/2\\ 3,& x=\mathrm{\pi }/2\end{array}\right.$

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{k\cos x}{\mathrm{\pi }-2x},x\ne \frac{\mathrm{\pi }}{2}\\ 3,x=\frac{\mathrm{\pi }}{2}\end{array}\right.$

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, then
$\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

⇒ $\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{k\cos x}{\mathrm{\pi }-2x}=3$ ...(1)

Putting $\frac{\mathrm{\pi }}{2}-x=h$, we get
$\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{k\cos x}{\mathrm{\pi }-2x}=$$\underset{h\to 0}{\lim }\frac{k\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}{\mathrm{\pi }-2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}$

From (1), we have

$\underset{h\to 0}{\lim }\frac{k\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}{\mathrm{\pi }-2\left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)}=3$
$\Rightarrow \underset{h\to 0}{\lim }\frac{k\sin h}{2h}=3$
$\Rightarrow \underset{h\to 0}{\lim }\frac{k\sin h}{h}=6$
$\Rightarrow k\underset{h\to 0}{\lim }\frac{\sin h}{h}=6$
$\Rightarrow k\times 1=6$
$\Rightarrow k=6$

Hence, for $k=6$ , f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$.