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

To find the value of \( k_1^2 + k_2^2 \) for the function \( f(x) \) to be continuous at \( x = 0 \), we equate the left-hand limit, right-hand limit, and the function's value at \( x = 0 \).

1. Left-hand limit (\( x \to 0^- \)):

For \( x<0 \), \( f(x) = \frac{2}{x} \left( \sin((k_1 + 1)x) + \sin((k_2 -1)x) \right) \). Using the limit property \( \lim_{x \to 0} \sin(ax)/x = a \):

\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left(\frac{2}{x} \sin((k_1 + 1)x) + \frac{2}{x} \sin((k_2 - 1)x)\right)\)

\(= 2(k_1 + 1) + 2(k_2 - 1) = 2k_1 + 2k_2\)

1. Right-hand limit (\( x \to 0^+ \)):

For \( x>0 \), \( f(x) = \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right) \). Using the series expansion for the logarithmic function (\( \log(1+u) \approx u \) for small \( u \)):

\(\lim_{x \to 0^+} \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right) = \lim_{x \to 0^+} \frac{2}{x} \log_e \left( 1 + \frac{(k_1 - k_2)x}{2 + k_2 x} \right)\)

As \( x \to 0 \), \( \frac{(k_1 - k_2)x}{2 + k_2 x} \to \frac{(k_1 - k_2)x}{2} \).

\(\lim_{x \to 0^+} \frac{2}{x} \cdot \frac{(k_1 - k_2)x}{2} = k_1 - k_2\)

1. Function value at \( x = 0 \):

The function value at \( x = 0 \) is given as 4.

1. Equating limits and function value:

For continuity at \( x = 0 \), the following must hold:

• \(2k_1 + 2k_2 = 4\) (Left-hand limit equals function value)
• \(k_1 - k_2 = 4\) (Right-hand limit equals function value)

1. Solving for \( k_1 \) and \( k_2 \):

From the system of equations:

• \(k_1 + k_2 = 2\)
• \(k_1 - k_2 = 4\)

Solving these equations yields \( k_1 = 3 \) and \( k_2 = -1 \).

1. Calculate \( k_1^2 + k_2^2 \):

\(k_1^2 + k_2^2 = 3^2 + (-1)^2 = 9 + 1 = 10\)

The value of \( k_1^2 + k_2^2 \) is 10.