Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f\left(x\right)$ is continuous at $x=0$

$\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)=\underset{x\to 0^{-}}{\lim }f\left(x\right) ...\left(1\right)$

$f\left(0\right)=b ...\left(2\right)$

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(\frac{\sin (a+1)x}{2x}+\frac{\sin 2x}{2x}\right)$

$=\frac{a+1}{2}+1 ...\left(3\right)$

$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x+b{x}^3}-\sqrt{x}}{b{x}^{5/2}}$

$=\underset{x\to 0^{+}}{\lim }\frac{\left(x+b{x}^3-x\right)}{b{x}^{5/2}\left(\sqrt{x+b{x}^3}+\sqrt{x}\right)}$

$=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b{x}^2}+1\right)}=\frac{1}{2} ...\left(4\right)$

Use $\left(2\right),\left(3\right) \& \left(4\right)$ in $\left(1\right)$

$\frac{1}{2}=b=\frac{a+1}{2}+1$

$\Rightarrow b=\frac{1}{2},a=-2$
$a+b=\frac{-3}{2}$