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

The Correct Option is B

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 \( \alpha^2 + \beta^2 \):

\[ \alpha^2 + \beta^2 = 4 + 8 = 12 \]