Question

The function$f(x) = \begin{cases} \frac{ k \cos x }{\pi-2 x }, & \text{if $x \neq \frac{\pi}{2}$} \\ 3, & \text{if $x =\frac{\pi}{2}$} \end{cases}$
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) =
\begin{cases}
\frac{ k \cos x }{\pi-2 x }, & \text{if $x \neq \frac{\pi}{2}$} \\
3, & \text{if $x =\frac{\pi}{2}$}
\end{cases}$
$\therefore LHL =\displaystyle\lim _{ x \rightarrow \frac{\pi^{-}}{2}} f ( x )=\displaystyle\lim _{ x \rightarrow \frac{\pi^{-}}{2}} \frac{ k \cos x }{\pi-2 x }$
Putting $x =\frac{\pi}{2}- h$ as $x \rightarrow \frac{\pi^{-}}{2}$ when $h \rightarrow 0$
$\therefore \displaystyle\lim _{h \rightarrow 0} \frac{k \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=\displaystyle\lim _{h \rightarrow 0} \frac{k \sin h}{2 h}$
$=\displaystyle\lim _{ h \rightarrow 0} \frac{ k }{2} \times \frac{\sin h }{ h }=\frac{ k }{2} \times 1=\frac{ k }{2}\left(\because \displaystyle\lim _{ x \rightarrow 0} \frac{\sin x }{ x }=1\right)$
$RHL =\displaystyle\lim _{ x \rightarrow \frac{\pi^{+}}{2}} f ( x )=\displaystyle\lim _{ x \rightarrow \frac{\pi^{+}}{2}} \frac{ k \cos x }{\pi-2 x }$
Putting $x =\frac{\pi}{2}+ h$ as $x \rightarrow \frac{\pi^{+}}{2}$ when $h \rightarrow 0$
$\therefore \displaystyle\lim _{ h \rightarrow 0} \frac{ k \cos \left(\frac{\pi}{2}+ h \right)}{\pi-2\left(\frac{\pi}{2}+ h \right)}=\displaystyle\lim _{ h \rightarrow 0} \frac{- k \sin h }{-2 h }$
$=\displaystyle\lim _{ h \rightarrow 0} \frac{ k }{2} \times \frac{\sinh }{ h }=\frac{ k }{2} \times 1=\frac{ k }{2}\left(\because \displaystyle\lim _{ x \rightarrow 0} \frac{\sin x }{ x }=1\right)$
Also, $f \left(\frac{\pi}{2}\right)=3$. Since $f ( x )$ is continuous at $x =\frac{\pi}{2}$.
$\therefore LHL = RHL = f \left(\frac{\pi}{2}\right) \Rightarrow \frac{ k }{2}=3 \Rightarrow k =6$