Question

$f\left(x\right)=\frac{2 cos x - sin 2 x}{\left(\pi - 2 x\right)^{2}},g\left(x\right)=\frac{e^{- cos x} - 1}{8 x - 4 \pi }$ and $h\left(x\right)=\begin{cases} f\left(\right.x\left.\right) & \forall x < \frac{\pi }{2} \\ g\left(\right.x\left.\right) & \forall x>\frac{\pi }{2} \end{cases}$
Then which of the following holds?

• A. is continuous at $x=\pi /2$
• B. has an irremovable discontinuity at $x=\frac{\pi }{2}$
• C. has a removable discontinuity at $x=\frac{\pi }{2}$
• D. $f\left(\frac{\pi ^{+}}{2}\right)=g\left(\frac{\pi ^{-}}{2}\right)$

Answer 1

Answer: (B)

For $h(x)$ to be continuous at $\frac{\pi}{2}$
$h \frac{\pi}{2}^{-}=h \frac{\pi}{2}^{+}=h \frac{\pi}{2}$
Now, LHL is
$h \frac{\pi^{-}}{2}=\displaystyle\lim _{h \rightarrow 0} \frac{2 \sin h-\sin 2 h}{4 h^2}$
$h \frac{\pi^{-}}{2}=\displaystyle\lim _{h \rightarrow 0} \frac{2 \sin h-2 \times \sin h \times \cos h}{4 h^2}$
$h \frac{\pi^{-}}{2}=\displaystyle\lim _{h \rightarrow 0} \frac{2 \sin h \times 2 \sin ^2 \frac{h}{2}}{4 h^2}=0$
Again, $RHL$ is
$h \frac{\pi}{2}^{+}=\displaystyle\lim _{h \rightarrow 0} \frac{e^{\sin h_{-1}}}{8 \frac{\pi}{2}+h-4 \pi} $
$h \frac{\pi}{2}^{+}=\displaystyle\lim _{h \rightarrow 0} \frac{e^{\sin h_{-1}}}{8 h \times \frac{\sin h}{h}}=\frac{1}{8}$
$h(x)$ is discontinuous at $x=\frac{\pi}{2}$.
As $LHL \neq RHL$
$\Rightarrow$ The function has an irremovable discontinuity at $x=\frac{\pi}{2}$.