Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be a function defined as
$f(x)=\left\{\begin{array}{cl}\frac{\sin (a+1) x+\sin 2 x}{2 x} & , \text { if } x<0 \\ b & , \text { if } x=0 \\ \frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}} & , \text { if } x>0\end{array}\right.$
If $f$ is continuous at $x=0$, then the value of $\mathrm{a}+\mathrm{b}$ is equal to:
Correct Option: , 4
$f(x)$ is continuous at $x=0$
$\lim _{x \rightarrow 0^{+}} f(x)=f(0)=\lim _{x \rightarrow 0^{-}} f(x) \ldots(1)$
$f(0)=b \ldots(2)$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin (a+1) x}{2 x}+\frac{\sin 2 x}{2 x}\right)$
$=\frac{a+1}{2}+1 \ldots(3)$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}}$
$=\lim _{x \rightarrow 0^{+}} \frac{\left(x+b x^{3}-x\right)}{b x^{5 / 2}\left(\sqrt{x+b x^{3}}+\sqrt{x}\right)}$
$=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1+b x^{2}}+1\right)}=\frac{1}{2}$.....(4)
Use $(2),(3) \&(4)$ in $(1)$
$\frac{1}{2}=b=\frac{a+1}{2}+1$
$\Rightarrow b=\frac{1}{2}, a=-2$
$a+b=\frac{-3}{2}$