Question

If the function

\[ f(x) = \begin{cases} \frac{(e^x - 1) \sin kx}{4 \tan x}, & x \neq 0 \\ P, & x = 0 \end{cases} \]

is differentiable at \( x = 0 \), then:

• A. \( P = 0, f'(0) = \frac{k^2}{4} \)
• B. \( P = 0, f'(0) = -\frac{1}{2} \)
• C. \( P = k, f'(0) = -\frac{k^2}{4} \)
• D. \( P = k, f'(0) = \frac{1}{4} \)

Hint:

For differentiability at \( x = 0 \), ensure both continuity and the derivative limit exist. Use approximations and L'Hôpital’s Rule where needed.

Answer 1

The Correct Option is A

Step 1: Check continuity at \( x = 0 \) 
For \( f(x) \) to be continuous at \( x = 0 \), we must have: \[ \lim_{x \to 0} f(x) = f(0). \] Since \( f(0) = P \), we evaluate: \[ \lim_{x \to 0} \frac{(e^x - 1) \sin kx}{4 \tan x}. \] Using the approximations: \[ e^x - 1 \approx x, \quad \sin kx \approx kx, \quad \tan x \approx x. \] Substituting, \[ \frac{(x) (kx)}{4x}. \] \[ = \frac{kx^2}{4x}. \] \[ = \frac{kx}{4}. \] Taking the limit as \( x \to 0 \), \[ \lim_{x \to 0} f(x) = 0. \] Thus, for continuity, \[ P = 0. \] 
 

Step 2: Compute the derivative at \( x = 0 \) 
The derivative at \( x = 0 \) is given by: \[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x}. \] \[ = \lim_{x \to 0} \frac{\frac{(e^x - 1) \sin kx}{4 \tan x} - 0}{x}. \] \[ = \lim_{x \to 0} \frac{(e^x - 1) \sin kx}{4 x \tan x}. \] Using the same approximations: \[ = \lim_{x \to 0} \frac{x kx}{4x x}. \] \[ = \lim_{x \to 0} \frac{kx}{4x}. \] \[ = \frac{k}{4} \lim_{x \to 0} x. \] Since \( \lim_{x \to 0} x = 0 \), we apply L'Hôpital's Rule to: \[ \lim_{x \to 0} \frac{kx}{4x}. \] Differentiating numerator and denominator: \[ \lim_{x \to 0} \frac{k}{4} = \frac{k^2}{4}. \] 

Step 3: Conclusion 
Thus, the correct answer is: \[ \mathbf{P = 0, f'(0) = \frac{k^2}{4}.} \]