Mathematics - Continuity - Differentiability Question with Solution | TestHub

We have,

$f\left(x\right)=\left\{\begin{array}{l}a\sin \frac{\pi }{2}\left(x-1\right),\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{for} x\le 0\\ \frac{\tan 2x-\sin 2x}{b{x}^3},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{for} x>0\end{array}\right.$

If function is continuous at $x=0$, then

$\mathrm{LHL}=\mathrm{RHL}=f\left(0\right)$

Now,

$\mathrm{LHL}=\underset{x\to 0^{-}}{\lim }\left[a\sin \frac{\pi }{2}\left(x-1\right)\right]$

$\Rightarrow \mathrm{LHL}=\underset{h\to 0}{\lim }\left[a\sin \frac{\pi }{2}\left(0-h-1\right)\right]$

$\Rightarrow \mathrm{LHL}=\underset{h\to 0}{\lim }\left[a\sin \frac{\pi }{2}\left(-1\right)\right]$

$\Rightarrow \mathrm{LHL}=-a$

Now,

$\mathrm{RHL}=\underset{x\to 0^{+}}{lim} \left(\frac{\tan 2x-\sin 2x}{b{x}^3}\right)$

$\because \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots \dots$

$\because \tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}+\dots .$

$\mathrm{RHL}=\underset{x\to 0^{+}}{lim} \left[\frac{\frac{(2x{)}^3}{3}+\frac{(2x{)}^3}{6}}{b{x}^3}\right]=\frac{{\displaystyle \frac{8}{3}}+{\displaystyle \frac{8}{6}}}{b}=\frac{4}{b}$

And,

$f\left(0\right)=a\sin \left[\frac{\pi }{2}(0-1)\right]=a\sin \left(-\frac{\pi }{2}\right)=-a$

Then,

$\mathrm{RHL}=f\left(0\right)$

$\Rightarrow \frac{4}{b}=-a\Rightarrow -ab=4\Rightarrow 10-ab=14$