Let f: R → R be a function defined as f(x) = \ cases ( ( a + 1 ) x...

$f ( x )$ is continuous at $x = 0$

$\lim_{x \to {0}^{+}}f\left(x\right) = f\left(0\right) = \lim_{x \to {0}^{-}}f\left(x\right) . . . \left(1\right)$

$f ( 0 ) = b . . . \left(2\right)$

$\lim_{x \to {0}^{-}}f\left(x\right) = \lim_{x \to {0}^{-}}\left(\dfrac{\sin ( a + 1 ) x}{2x} + \dfrac{\sin 2x}{2x}\right)$

$= \dfrac{a + 1}{2} + 1 . . . \left(3\right)$

$\lim_{x \to {0}^{+}}f\left(x\right) = \lim_{x \to {0}^{+}}\dfrac{\sqrt{x + b{x}^{3}} - \sqrt{x}}{b{x}^{5 / 2}}$

$= \lim_{x \to {0}^{+}}\dfrac{\left(x + b{x}^{3} - x\right)}{b{x}^{5 / 2}\left(\sqrt{x + b{x}^{3}} + \sqrt{x}\right)}$

$= \lim_{x \to {0}^{+}}\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{1 + b{x}^{2}} + 1\right)} = \dfrac{1}{2} . . . \left(4\right)$

Use $( 2 ) , ( 3 ) \text{ and } ( 4 )$ in $( 1 )$

$\dfrac{1}{2} = b = \dfrac{a + 1}{2} + 1$

$⇒ b = \dfrac{1}{2}, a = - 2$
$a + b = \dfrac{- 3}{2}$