Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a function given by
$f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & x < 0 \\\\\alpha, & x = 0, \text{ where } \alpha, \beta \in \mathbb{R}. \\\\\beta \sqrt{1 - \cos x} / x, & x > 0 \end{cases}$
If $f$ is continuous at $x = 0$, then $\alpha^2 + \beta^2$ is equal to:

• A. 48
• B. 12
• C. 3
• D. 6

Answer 1

Answer: (B)

12

Answer 2

To ensure continuity at $x = 0$, we require $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.

Left-hand limit:

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos 2x}{x^2} = 2$

This gives $f(0) = \alpha = 2$.

Right-hand limit:

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\beta \sqrt{1 - \cos x}}{x} = \frac{\beta}{\sqrt{2}} = 2 \implies \beta = 2\sqrt{2}$

Calculating $\alpha^2 + \beta^2$:

$\alpha^2 + \beta^2 = 4 + 8 = 12$