Determine the values of '$a$' and '$b$' such that the following function is continuous at $x = 0$: $f(x) = \begin{cases} \frac{x + \sin x}{\sin(a + 1)x}, & \text{if } -\pi < x < 0 \\ 2, & \text{if } x = 0 \\ 2 \frac{e^{\sin bx} - 1}{bx}, & \text{if } x > 0 \end{cases}$

$a = 0, b \neq 0$

The correct answer is Option (3) → $a = 0, b \neq 0$ ##

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^-} \frac{x + \sin x}{\sin(a + 1)x}$

$= \lim\limits_{x \to 0^-} \frac{1 + \frac{\sin x}{x}}{\frac{\sin(a + 1)x}{(a + 1)x}(a + 1)}$

$= \frac{2}{a + 1}$

$\lim\limits_{x \to 0^+} f(x) = \lim\limits_{x \to 0^+} 2 \frac{e^{\sin bx} - 1}{bx}$

$= \lim\limits_{x \to 0^+} 2 \frac{e^{\sin bx} - 1}{\sin bx} \times \frac{\sin bx}{bx}$

$f(0) = 2$

For the function to be continuous at $0$, we must have

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$

i.e., we must have $\frac{2}{a + 1} = 2$ or $a = 0$; $b$ may be any real number other than $0$.