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 determine the value of \( k_1^2 + k_2^2 \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit, right-hand limit, and the value of the function at \( x = 0 \) are equal. Let's calculate these values step-by-step.

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

For \( x < 0 \), the function is defined as:

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

For the function to be continuous as \( x \to 0^- \), we use the limit property: \( \lim_{x \to 0} \sin(ax)/x = a \).

Let's apply this property:

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

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

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

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

For \( x > 0 \), the function is defined as:

\(f(x) = \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right)\)

We calculate:

Using the series expansion for logarithmic function: \(\log(1+x) \approx x\) for small \( x\).

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

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

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

1. Value at \( x = 0 \):

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

1. Equating Limits and Function Value:

For continuity at \( x = 0 \), the left-hand limit and right-hand limit must equal the value of the function at \( x = 0 \).

Thus, we have:

• \(2k_1 + 2k_2 = 4\) (from left-hand limit)
• \(k_1 - k_2 = 4\) (from right-hand limit)

1. Solving the Equations:

From the equations:

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

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

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

Therefore, the final answer is \( 10 \).

Answer 2

To ensure that the function \( f(x) \) is continuous at \( x = 0 \), the left-hand limit as \( x \to 0^- \), the right-hand limit as \( x \to 0^+ \), and the value at \( x = 0 \) must all be equal to 4.

Step 1: Calculate the left-hand limit
As \( x \to 0^- \), \( f(x) = \frac{2}{x}(\sin((k_1 + 1)x) + \sin((k_2 - 1)x)) \).
Using the small angle approximation \(\sin(ax) \approx ax\) when \(x\) is near zero:
\( \frac{2}{x}((k_1 + 1)x + (k_2 - 1)x) = 2((k_1 + 1) + (k_2 - 1)) = 2(k_1 + k_2) \).

To be continuous at \( x = 0 \),
\( \lim_{x \to 0^-} f(x) = 4 \) implies \( 2(k_1 + k_2) = 4 \) thus \( k_1 + k_2 = 2 \).

Step 2: Calculate the right-hand limit
As \( x \to 0^+ \), \( f(x) = \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right) \).
Using the first-order expansion \(\log(1 + u) \approx u\) when \(x\) is near zero:
\(\log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right) \approx \log_e \left(1 + \frac{(k_1 - k_2)x}{2} \right) \approx \frac{(k_1-k_2)x}{2}\)
So, \( \frac{2}{x} \cdot \frac{(k_1-k_2)x}{2} = (k_1 - k_2) \).

To be continuous at \( x = 0 \),
\( \lim_{x \to 0^+} f(x) = 4 \) implies \( k_1 - k_2 = 4 \).

Step 3: Solve for \( k_1 \) and \( k_2 \)
We have the equations:
\( k_1 + k_2 = 2 \) (Equation 1)
\( k_1 - k_2 = 4 \) (Equation 2)
Adding these equations:
\( 2k_1 = 6 \) so \( k_1 = 3 \).
Substituting back into Equation 1:
\( 3 + k_2 = 2 \) thus \( k_2 = -1 \).

Step 4: Calculate \( k_1^2 + k_2^2 \)
Using the obtained values:
\( k_1^2 + k_2^2 = 3^2 + (-1)^2 = 9 + 1 = 10 \).

Thus, \( k_1^2 + k_2^2 \) is equal to \( \boxed{10} \).