Question

The function$f(x)=\{\begin{array}{ll}\frac{k\cos x}{\pi -2x},& \text{if x=2π}\\ 3,& \text{if x=2π}\end{array}$
is continuous at $x=\frac{\pi }{2}$, when $k$ equals

• A. -6
• B. 6
• C. 5
• D. -5

Answer 1

Answer: (B)

Here,$f(x)=<br/><br/>\{\begin{array}{ll}<br/><br/>\frac{k\cos x}{\pi -2x},& \text{if x=2π}\\ <br/><br/>3,& \text{if x=2π}<br/><br/>\end{array}$
$\therefore LHL=\underset{x\to \frac{{\pi }^{-}}{2}}{\lim }f(x)=\underset{x\to \frac{{\pi }^{-}}{2}}{\lim }\frac{k\cos x}{\pi -2x}$
Putting $x=\frac{\pi }{2}-h$ as $x\to \frac{{\pi }^{-}}{2}$ when $h\to 0$
$\therefore \underset{h\to 0}{\lim }\frac{k\cos \left(\frac{\pi }{2}-h\right)}{\pi -2\left(\frac{\pi }{2}-h\right)}=\underset{h\to 0}{\lim }\frac{k\sin h}{2h}$
$=\underset{h\to 0}{\lim }\frac{k}{2}\times \frac{\sin h}{h}=\frac{k}{2}\times 1=\frac{k}{2}\left(\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right)$
$RHL=\underset{x\to \frac{{\pi }^{+}}{2}}{\lim }f(x)=\underset{x\to \frac{{\pi }^{+}}{2}}{\lim }\frac{k\cos x}{\pi -2x}$
Putting $x=\frac{\pi }{2}+h$ as $x\to \frac{{\pi }^{+}}{2}$ when $h\to 0$
$\therefore \underset{h\to 0}{\lim }\frac{k\cos \left(\frac{\pi }{2}+h\right)}{\pi -2\left(\frac{\pi }{2}+h\right)}=\underset{h\to 0}{\lim }\frac{-k\sin h}{-2h}$
$=\underset{h\to 0}{\lim }\frac{k}{2}\times \frac{\sinh }{h}=\frac{k}{2}\times 1=\frac{k}{2}\left(\because \underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right)$
Also, $f\left(\frac{\pi }{2}\right)=3$. Since $f(x)$ is continuous at $x=\frac{\pi }{2}$.
$\therefore LHL=RHL=f\left(\frac{\pi }{2}\right)\Rightarrow \frac{k}{2}=3\Rightarrow k=6$