Question

$f\left(x\right)=\frac{2cosx-sin2x}{{\left(\pi -2x\right)}^2},g\left(x\right)=\frac{e^{-cosx}-1}{8x-4\pi }$ and $h\left(x\right)=\{\begin{array}{ll}f(x)& \forall x<\frac{\pi }{2}\\ g(x)& \forall x>\frac{\pi }{2}\end{array}$
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
$h\frac{{\pi }^{-}}{2}=\underset{h\to 0}{\lim }\frac{2\sin h-\sin 2h}{4h^2}$
$h\frac{{\pi }^{-}}{2}=\underset{h\to 0}{\lim }\frac{2\sin h-2\times \sin h\times \cos h}{4h^2}$
$h\frac{{\pi }^{-}}{2}=\underset{h\to 0}{\lim }\frac{2\sin h\times 2{\sin }^2\frac{h}{2}}{4h^2}=0$
Again, $RHL$ is
$h{\frac{\pi }{2}}^{+}=\underset{h\to 0}{\lim }\frac{e^{\sin h_{-1}}}{8\frac{\pi }{2}+h-4\pi }$
$h{\frac{\pi }{2}}^{+}=\underset{h\to 0}{\lim }\frac{e^{\sin h_{-1}}}{8h\times \frac{\sin h}{h}}=\frac{1}{8}$
$h(x)$ is discontinuous at $x=\frac{\pi }{2}$.
As $LHL\ne RHL$
$\Rightarrow$ The function has an irremovable discontinuity at $x=\frac{\pi }{2}$.