Let $$f : R \to R$$ be a function defined as
$$$f(x) = \begin{cases} \frac{\sin(a+1)x + \sin 2x}{2x}, & \text{if } x < 0 \\ b, & \text{if } x = 0 \\ \frac{\sqrt{x+bx^3} - \sqrt{x}}{bx^{5/2}}, & \text{if } x > 0 \end{cases}$$$
If $$f$$ is continuous at $$x = 0$$, then the value of $$a + b$$ is equal to :

Given,

$$f(x)=\begin{cases}\dfrac{\sin(a+1)x+\sin2x}{2x},&x<0\\[4pt]b,&x=0\\[4pt]\dfrac{\sqrt{x+bx^3}-\sqrt{x}}{bx^{5/2}},&x>0\end{cases}$$

For continuity at

$$x=0,$$

we require

$$\text{LHL}=\text{RHL}=f(0)=b$$

First find the left-hand limit.

$$\text{LHL}=\lim_{x\to0^-}\frac{\sin(a+1)x+\sin2x}{2x}$$

$$=\frac12\left[\lim_{x\to0}\frac{\sin(a+1)x}{x}+\lim_{x\to0}\frac{\sin2x}{x}\right]$$

Using

$$\lim_{x\to0}\frac{\sin kx}{x}=k,$$

we get

$$\text{LHL}=\frac12\left((a+1)+2\right)$$

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

Now find the right-hand limit.

$$\text{RHL}=\lim_{x\to0^+}\frac{\sqrt{x+bx^3}-\sqrt{x}}{bx^{5/2}}$$

Factor out

$$\sqrt{x}:$$

$$=\lim_{x\to0^+}\frac{\sqrt{x}\left(\sqrt{1+bx^2}-1\right)}{bx^{5/2}}$$

$$=\lim_{x\to0^+}\frac{\sqrt{1+bx^2}-1}{bx^2}$$

Rationalizing,

$$=\lim_{x\to0^+}\frac{(1+bx^2)-1}{bx^2\left(\sqrt{1+bx^2}+1\right)}$$

$$=\lim_{x\to0^+}\frac{1}{\sqrt{1+bx^2}+1}$$

$$=\frac12$$

Therefore,

$$b=\frac12$$

Now continuity gives

$$\frac{a+3}{2}=b=\frac12$$

Hence,

$$a+3=1$$

$$a=-2$$

Therefore,

$$a+b=-2+\frac12$$

$$=-\frac32$$

Hence, the required value is

$$\boxed{-\frac32}$$