$f\left(x\right) = \left\{\begin{array}{ll}\frac{\sin ( a + 1)x + \sin x}{x}& if x <\hspace{0.17em}0\\ \frac{{(x + b{x}^2)}^{{\displaystyle 1/2}} - x^{1/2}}{b x^{3/2}}& if x> 0\end{array}\right.$ and f(0) = c is continuous at x = 0 then

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\sin \left(\mathrm{a}+1\right)+\sin \mathrm{x} }{\mathrm{x}} \mathrm{if} \mathrm{x}<0\\ \mathrm{c} \mathrm{if} \mathrm{x}=0\\ \frac{\sqrt{\mathrm{x}+{\mathrm{bx}}^2}-\sqrt{\mathrm{x}}}{{\mathrm{bx}}^{3/2}} \mathrm{if} \mathrm{x}>0\end{array}\right.$

at  x = 0

$$\begin{gathered} \Rightarrow \mathrm{LHL} = \mathrm{RHL}=\mathrm{f}\left(0\right) \\ \underset{\mathrm{x}\to 0^{-}}{\lim }\frac{\sin \left(\mathrm{a}+1\right)\mathrm{x}}{\mathrm{x}}+\frac{\sin \mathrm{x}}{\mathrm{x}}=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{\sqrt{\mathrm{x}}\left(\sqrt{1+\mathrm{bx}}-1\right)}{\mathrm{bx}.\sqrt{x}} =\mathrm{b} \\ \Rightarrow a+1+1=\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{\left(1+\mathrm{bx}\right)-1}{\mathrm{bx}\left(\sqrt{1+\mathrm{bx}}+1\right)}=\frac{1}{2} \\ \therefore a+2=\frac{1}{2},\mathrm{b}\ne 0 \\ \Rightarrow a=\frac{-3}{2},c=\frac{1}{2} \end{gathered}$$