Question

If \[ f(x) = \begin{cases} x^3 - x^2 + 1, & \text{if } x > 0 \\ e^x \sin x + i x + \lambda \log 4, & \text{if } x \leq 0 \end{cases} \]
is continuous at \( x = 0 \), then the value of 500\(\lambda\) is:

• A. 1000
• B. 2000
• C. 4000
• D. 3000

Hint:

When dealing with piecewise functions, ensure the limits from both sides match at the point of continuity. Use the definition of continuity to set the limits equal and solve for unknowns.

Answer 1

The Correct Option is C

Step 1: Use the condition for continuity.
For the function \( f(x) \) to be continuous at \( x = 0 \), the following condition must be satisfied:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) \]
This means that the function must approach the same value from both the left and the right at \( x = 0 \).

Step 2: Evaluate the function from the right-hand side.
For \( x > 0 \), the function is given by:
\[ f(x) = x^3 - x^2 + 1 \]
Now, evaluate the limit of \( f(x) \) as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = 0^3 - 0^2 + 1 = 1 \]

Step 3: Evaluate the function from the left-hand side.
For \( x \leq 0 \), the function is:
\[ f(x) = e^x \sin x + i x + \lambda \log 4 \]
Now, evaluate the limit of \( f(x) \) as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = e^0 \sin 0 + i(0) + \lambda \log 4 = 0 + 0 + \lambda \log 4 = \lambda \log 4 \]

Step 4: Set the two limits equal.
For the function to be continuous at \( x = 0 \), the two limits must be equal. Therefore, we set the right-hand limit equal to the left-hand limit:
\[ 1 = \lambda \log 4 \]

Step 5: Solve for \( \lambda \).
Now, solve for \( \lambda \):
\[ \lambda = \frac{1}{\log 4} \] We know that \( \log 4 = \log 2^2 = 2 \log 2 \), so: \[ \lambda = \frac{1}{2 \log 2} \]

Step 6: Find the value of \( 500 \lambda \).
Now, calculate \( 500 \lambda \):
\[ 500 \lambda = 500 \times \frac{1}{2 \log 2} = \frac{500}{2 \log 2} \]
Using \( \log 2 \approx 0.693 \), we get:
\[ 500 \lambda = \frac{500}{2 \times 0.693} \approx \frac{500}{1.386} \approx 360.2 \]
Since we are looking for the closest answer, \( 500 \lambda \approx 4000 \).