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

Answer: (D)

-4

Answer 2

$f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}$ 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:

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