# Solve this following

Question:

Let $f: \mathrm{R} \rightarrow \mathrm{R}$ be a function defined as

$f(x)=\left\{\begin{array}{cc}\frac{\sin (a+1) x+\sin 2 x}{2 x} & , \text { if } x<0 \\ b & , \text { if } x=0 \\ \frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}} & , \text { if } x>0\end{array}\right.$

If $f$ is continuous at $\mathrm{x}=0$, then the value of $a+b$ is equal to :

1. $-\frac{5}{2}$

2. $-2$

3. $-3$

4. $-\frac{3}{2}$

Correct Option: 4,

Solution:

$f(x)$ is continuous at $x=0$

$\lim _{x \rightarrow 0^{+}} f(x)=f(0)=\lim _{x \rightarrow 0^{-}} f(x)$ ..........(1)

$f(0)=b$    .......(2)

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

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

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{x+b x^{3}}-\sqrt{x}}{b x^{5 / 2}}$

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

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

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

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

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

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