Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function defined as \[ f(x) = \begin{cases} \dfrac{\sin\!\big((a+1)x\big) + \sin 2x}{2x}, & \text{if } x < 0, \\[6pt] b, & \text{if } x = 0, \\[6pt] \dfrac{\sqrt{x + b x^3} - \sqrt{x}}{b\, x^{5/2}}, & \text{if } x > 0. \end{cases} \] If \( f \) is continuous at \( x = 0 \), then the value of \( a + b \) is equal to :

• A. -5/2
• B. -3
• C. -2
• D. -3/2

Hint:

Use the standard limit $\lim_{\theta \to 0} \frac{\sin k\theta}{\theta} = k$ and Binomial expansion $(1+x)^n \approx 1+nx$ for small $x$.

Answer 1

The Correct Option is D

Step 1: LHL at $x=0$: $\lim_{x \to 0^-} \frac{\sin(a+1)x + \sin 2x}{2x} = \frac{a+1}{2} + \frac{2}{2} = \frac{a+3}{2}$.
Step 2: RHL at $x=0$: $\lim_{x \to 0^+} \frac{\sqrt{x}(1 + bx^2)^{1/2} - \sqrt{x}}{bx^{5/2}} = \lim_{x \to 0^+} \frac{(1 + \frac{1}{2}bx^2 + ...) - 1}{bx^2} = 1/2$.
Step 3: Since $f(x)$ is continuous, $LHL = RHL = f(0) = b$. $b = 1/2$ and $\frac{a+3}{2} = 1/2 \implies a+3 = 1 \implies a = -2$.
Step 4: $a + b = -2 + 1/2 = -3/2$.