Mathematics - Continuity - Differentiability Question with Solution | TestHub

$\because f\left(x\right)=\left\{\begin{array}{c}\frac{\sin \left(p+1\right)x +sinx}{\begin{array}{c}x\end{array}}\\ q\\ \frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3/2}}\end{array} \mathrm{ }\begin{array}{c}:x<0\\ :x=0\\ :x>0\end{array}\right.$

$\because f$ is continuous at $x=0$

$L.H.L=$

$\underset{h\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}} \frac{\sin \left(p+1\right)\left(-h\right)+\mathrm{s}\mathrm{i}\mathrm{n}(-h)}{-h}=\underset{h\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sin \left(p+1\right)h+\sin h}{h}$

$=\underset{h\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sin \left(p+1\right)h}{h\left(p+1\right)}\left(p+1\right)+\underset{h\to 0}{\lim }\frac{\sin h}{h}=p+1+1$

$=p+2$

$R.H.L= \underset{h\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}} \frac{\sqrt{h^2+h-} \sqrt{h}}{h^{\frac{3}{2}}}\times \frac{\sqrt{h^2+h}+\sqrt{h}}{\sqrt{h^2+h}+\sqrt{h}}=\underset{h\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{h^2+h-h}{h^{\frac{3}{2}} h^{\frac{1}{2}}\left(\sqrt{h+1}+1\right)}$

$\underset{h\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{\sqrt{1+h}+1}=\frac{1}{2}$

$\Rightarrow L\mathit{.}H\mathit{.}L\mathit{.}=R.H.L.=f\left(0\right)\Rightarrow p+2=\frac{1}{2}=q$
$=\begin{array}{c}p=-\frac{3}{2} \& q=\frac{1}{2}\end{array}$