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:

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

Answer 2

Step 1: What continuity at x = 0 requires
We are given \[ f(x)=\frac{\sin(3x)+\alpha\sin x-\beta\cos(3x)}{x^3},\quad x\in\mathbb{R}. \] For \(f\) to be continuous at \(x=0\), the limit \(\lim_{x\to 0} f(x)\) must exist and be finite. Since the denominator is \(x^3\), the numerator must have a Taylor expansion about \(x=0\) starting at order \(x^3\) (i.e., the constant, \(x\), and \(x^2\) terms must vanish).

Step 2: Taylor expansions up to the needed order
Use the standard expansions (up to the first nonzero cubic or quadratic terms that can affect the limit): \[ \sin(3x)=3x-\frac{(3x)^3}{6}+O(x^5)=3x-\frac{27}{6}x^3+O(x^5)=3x-\frac{9}{2}x^3+O(x^5), \] \[ \sin x=x-\frac{x^3}{6}+O(x^5), \] \[ \cos(3x)=1-\frac{(3x)^2}{2}+O(x^4)=1-\frac{9}{2}x^2+O(x^4). \] Therefore, the numerator \[ N(x)=\sin(3x)+\alpha\sin x-\beta\cos(3x) \] expands to \[ N(x)=\bigl(3x-\tfrac{9}{2}x^3\bigr)+\alpha\bigl(x-\tfrac{x^3}{6}\bigr)-\beta\bigl(1-\tfrac{9}{2}x^2\bigr)+O(x^4). \]

Step 3: Match coefficients to kill constant, linear, and quadratic terms
Collect terms by powers of \(x\):
• Constant term: \(-\beta\). For continuity, this must be \(0\) \(\Rightarrow\) \(\beta=0\).
• Coefficient of \(x\): \(3+\alpha\). This must be \(0\) \(\Rightarrow\) \(\alpha=-3\).
• Coefficient of \(x^2\): comes only from \(-\beta\bigl(-\tfrac{9}{2}x^2\bigr)=\beta\tfrac{9}{2}x^2\). With \(\beta=0\), it is already \(0\).
Thus, the smallest nonzero term in \(N(x)\) will be the cubic term when \(\alpha=-3,\ \beta=0\).

Step 4: Determine the cubic term and the limit
With \(\alpha=-3,\ \beta=0\), the cubic coefficient in \(N(x)\) is \[ \underbrace{-\frac{9}{2}}_{\sin(3x)}\;+\;\underbrace{\Bigl(-3\Bigr)\Bigl(-\frac{1}{6}\Bigr)}_{\alpha\sin x}\;=\;-\frac{9}{2}+\frac{1}{2}=-4. \] Hence \[ N(x)=-4\,x^3+O(x^5)\quad\Rightarrow\quad f(x)=\frac{N(x)}{x^3}\;=\;-4+O(x^2). \] Therefore, \[ \lim_{x\to 0} f(x)=-4. \] Continuity at \(x=0\) then forces \(f(0)=\lim_{x\to 0}f(x)=-4\).

Final answer

-4