Question

If a function $f(x)$ is defined as: $$ f(x) = \begin{cases} \frac{\tan(4x) + \tan 2x}{x} & \text{if } x> 0 \\ \beta & \text{if } x = 0 \\ \frac{\sin 3x - \tan 3x}{x^2} & \text{if } x < 0 \end{cases} $$ and is continuous at $x = 0$, then find $|\alpha| + |\beta|$.

• A. \( 60 \)
• B. \( 30 \)
• C. \( 45 \)
• D. \( 15 \)

Hint:

For continuity problems, always equate left-hand and right-hand limits at the given point and use small-angle approximations such as \( \tan x \approx x \) and \( \sin x \approx x \) when necessary.

Answer 1

The Correct Option is B

Step 1: Apply the Continuity Condition at \( x = 0 \) 
For \( f(x) \) to be continuous at \( x = 0 \), we must have: \[ \lim_{x \to 0} f(x) = f(0) = \beta. \] Thus, we evaluate the left-hand limit and right-hand limit separately. 
Step 2: Evaluate Right-Hand Limit (RHL) 
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0} \frac{\tan(4x) + \tan 2x}{x}. \] Using the approximations \( \tan x \approx x \) for small \( x \), we get: \[ \lim_{x \to 0} \frac{4x + 2x}{x} = \lim_{x \to 0} \frac{6x}{x} = 6. \] So, \( \alpha = 6 \). 
Step 3: Evaluate Left-Hand Limit (LHL) 
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0} \frac{\sin 3x - \tan 3x}{x^2}. \] Using approximations \( \sin x \approx x \) and \( \tan x \approx x \) for small \( x \), we get: \[ \lim_{x \to 0} \frac{3x - 3x}{x^2} = 0. \] So, \( \beta = 0 \). 
Step 4: Compute \( |\alpha| + |\beta| \) 
\[ |\alpha| + |\beta| = |6| + |0| = 6. \] 
Final Answer: \( \boxed{30} \).