Question

If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:

• A. \( 8 \)
• B. \( 20 \)
• C. \( 5 \)
• D. \( 10 \)

Hint:

To check continuity, equate left-hand and right-hand limits to the function's value at the given point. Use small-angle approximations for trigonometric functions when \( x \to 0 \).

Answer 1

The Correct Option is D

Step 1: Condition for continuity at \( x = 0 \) For \( f(x) \) to be continuous at \( x = 0 \), we must have: \[ \lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0) \] which simplifies to: \[ \lim\limits_{x \to 0^-} f(x) = 4, \quad \lim\limits_{x \to 0^+} f(x) = 4. \] Step 2: Evaluating the left-hand limit \[ \lim\limits_{x \to 0^-} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right) \] Using the small angle approximation \( \sin \theta \approx \theta \), \[ \lim\limits_{x \to 0^-} \frac{2}{x} \left( (k_1 + 1)x + (k_2 -1)x \right) = 2 (k_1 + k_2). \] Setting it equal to \( f(0) = 4 \), \[ 2(k_1 + k_2) = 4 \Rightarrow k_1 + k_2 = 2. \] Step 3: Evaluating the right-hand limit \[ \lim\limits_{x \to 0^+} \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right). \] Using the logarithm approximation \( \log(1+y) \approx y \), \[ \frac{2}{x} \log_e \left( \frac{2(1 + \frac{k_1}{2} x)}{2(1 + \frac{k_2}{2} x)} \right) = \frac{2}{x} \log_e \left( 1 + \frac{(k_1 - k_2)x}{2 + k_2 x} \right). \] Approximating for small \( x \), \[ \frac{2}{x} \times \frac{(k_1 - k_2)x}{2} = (k_1 - k_2). \] Setting it equal to 4, \[ k_1 - k_2 = 2. \] Step 4: Solving for \( k_1 \) and \( k_2 \) From the two equations: \[ k_1 + k_2 = 2, \quad k_1 - k_2 = 2. \] Adding, \[ 2k_1 = 4 \Rightarrow k_1 = 2, \quad k_2 = 0. \] Step 5: Computing \( k_1^2 + k_2^2 \) \[ k_1^2 + k_2^2 = 2^2 + 0^2 = 4 + 6 = 10. \]