Question

If the function \(f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3},\) \(x \in \mathbb{R} \), is continuous at \( x = 0 \), then \( f(0) \) is equal to:

• A. 2
• B. -2
• C. 4
• D. -4

Answer 1

The Correct Option is D

\( f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3} \) is continuous at \( x = 0 \).

\[ \lim_{x \to 0} \frac{3x - \left(\frac{3x^3}{3}\right) + \dots + \alpha \left(\frac{x - \frac{x^3}{3}}{3}\right) - \beta \left(1 - \frac{(3x)^2}{2} \dots \right)}{x^3} = f(0) \]

Continuing with the limit:

\[ \lim_{x \to 0} \frac{-\beta + x(3 + \alpha) + \frac{9 \beta x^2}{2} + \left(-\frac{27}{3} + \frac{\alpha}{3}\right)x^3 \dots}{x^3} = f(0) \]

For existence:

\[ \beta = 0, \quad 3 + \alpha = 0, \quad -\frac{27}{3} + \frac{\alpha}{3} = f(0) \]

Calculating:

\[ \alpha = -3, \quad -\frac{27}{6} = -\frac{3}{6} = f(0) \] 
\[ f(0) = \frac{-27 + 3}{6} = -4 \]