a
Given that $f$ is continuous at $x=0$
$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sin \left( {a + 1} \right)x + 2\sin x}}{x}}&{x < 0}\\
2&{x = 0}\\
{\frac{{\sqrt {1 + bx} 11}}{x}}&{x > 0}
\end{array}} \right.$
Since $f(x)$ is continous at $x=0,$
$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} f\left( 0 \right)$
Thus $R.H.L. = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$
$\mathop {\lim }\limits_{x \to 0} f\left( {0 + h} \right)$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {1 + bh} - 1}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {1 + bh} - 1}}{h} \times \frac{{\sqrt {1 + bh} + 1}}{{\sqrt {1 + bh} + 1}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{1 + bh - 1}}{{h\sqrt {1 + bh} + 1}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{bh}}{{h\sqrt {1 + bh} + 1}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{b}{{\sqrt {1 + bh} + 1}}$
$ = \frac{b}{2}$
Given that $f(0)=2$
$ \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( 0 \right)$
$ \Rightarrow \frac{b}{2} = 2$
$ \Rightarrow b = 4$
Similarly,
$L.H.L = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$
$ = \mathop {\lim }\limits_{x \to 0} f\left( {0 - h} \right)$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 1} \right)\left( {0 - h} \right) + 2\sin \left( {0 - h} \right)}}{{0 - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \left( {a + 1} \right)h - 2\sin \,h}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \left( {a + 1} \right)h}}{{ - h}} + \mathop {\lim }\limits_{h \to 0} \frac{{ - 2\sin \,h}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {a + 1} \right)h}}{h}\frac{{\left( {a + 1} \right)}}{{\left( {a + 1} \right)}} + 2\mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h}}{h}$
$ = a + 1 + 2$ [$\because $ $\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1$]
$\therefore a = - 3$