a =\[- \frac{3}{2}\], b ∈ R − {0}, c = \[\frac{1}{2}\]
The given function can be rewritten as
\[f\left( x \right) = \begin{cases}\frac{\sin \left( a + 1 \right) x + x \sin x}{x}, \text{ for} x < 0 \\ c ,\text{ for } x = 0 \\ \frac{\sqrt{x + b x^2} - \sqrt{x}}{b x^\frac{3}{2}} , \text{ for } x > 0\end{cases}\]
\[\Rightarrow f\left( x \right) = \begin{cases}\frac{\sin \left( a + 1 \right)x + \sin x}{x}, \text{ for } x < 0 \\ c , \text{ for } x = 0 \\ \frac{\sqrt{1 + bx} - 1}{bx} , \text{ for } x > 0\end{cases}\]
We have
(LHL at x = 0) = \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right)\]
\[= \lim_{h \to 0} \left[ \frac{- \sin \left( a + 1 \right)h - \sin \left( - h \right)}{h} \right] = \lim_{h \to 0} \left[ \frac{- \sin \left( a + 1 \right)h}{h} - \frac{\sin h}{h} \right]\]
\[= - \left( a + 1 \right) \lim_{h \to 0} \left[ \frac{\sin \left( a + 1 \right)h}{\left( a + 1 \right)h} \right] - \lim_{h \to 0} \frac{\sin h}{h} = - a - 1\]
(RHL at x = 0) = \[\lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right) = \lim_{h \to 0} f\left( h \right)\]
\[\lim_{h \to 0} \left( \frac{\sqrt{1 + bh} - 1}{bh} \right) = \lim_{h \to 0} \left( \frac{bh}{bh\left( \sqrt{1 + bh} + 1 \right)} \right) = \lim_{h \to 0} \left( \frac{1}{\left( \sqrt{1 + bh} + 1 \right)} \right) = \frac{1}{2}\]
Also, \[f\left( 0 \right) = c\]
If \[f\left( x \right)\] is continuous at x = 0, then
\[\lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right) = f\left( 0 \right)\]
\[\Rightarrow - a - 1 = \frac{1}{2} = c\]
\[\Rightarrow - a - 1 = \frac{1}{2} \text{ and } c = \frac{1}{2}\]
\[\Rightarrow a = \frac{- 3}{2}\] \[, c = \frac{1}{2}\]
Now,
\[\frac{\sqrt{1 + bx} - 1}{bx}\] exists only if \[bx \neq 0 \Rightarrow b \neq 0\]
Thus, \[b \in R - \left\{ 0 \right\}\].
