If the function f : ℝ → ℝ defined by f x = sin ( a + 1 ) x + sin x x , x < 0 b , x = 0…

Given:

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

Since, $f\left(x\right)$ is continuous on $R$

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

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

Now,

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

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

$\Rightarrow \underset{x\to 0^{-}}{\lim }\left[\left(\left(a+1\right)\frac{\sin (a+1)x}{\left(a+1\right)x}\right)+\frac{\sin x}{x}\right]=b$

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

$\Rightarrow a+1+1=b$

$\Rightarrow b=a+2 ...\left(i\right)$

And,

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

$\Rightarrow \underset{x\to 0^{+}}{\lim }\left\{\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{{\displaystyle \frac{3}{2}}}}\right\}=b$

$\Rightarrow \underset{x\to 0^{+}}{\lim }\left\{\frac{\sqrt{1+x}-1}{x}\right\}=b$

Using L'Hospital rule, we have

$\underset{x\to 0^{+}}{\lim }\left\{\frac{{\displaystyle \frac{1}{2\sqrt{1+x}}}-0}{1}\right\}=b$

$\Rightarrow b=\frac{1}{2} ...\left(ii\right)$

Putting in $\left(i\right)$, we get $a=\frac{-3}{2}$.

Therefore, $a+b=-1$.