Step 1: Understanding the Question:
For a function to be continuous at \( x = 0 \), the left-hand limit (LHL), right-hand limit (RHL), and the functional value at \( x = 0 \) must be equal.
Step 2: Key Formula or Approach:
We evaluate \( \lim_{x \to 0^+} f(x) \) and \( \lim_{x \to 0^-} f(x) \) and equate them.
Step 3: Detailed Explanation:
RHL (\( x \to 0^+ \)):
\[ \lim_{x \to 0^+} \frac{8^x - 4^x - 2^x + 1}{x^2} = \lim_{x \to 0^+} \frac{4^x(2^x - 1) - 1(2^x - 1)}{x^2} \]
\[ = \lim_{x \to 0^+} \frac{(4^x - 1)(2^x - 1)}{x^2} = \lim_{x \to 0^+} \left( \frac{4^x - 1}{x} \right) \left( \frac{2^x - 1}{x} \right) \]
Using the standard limit \( \lim_{h \to 0} \frac{a^h - 1}{h} = \ln a \):
RHL = \( \ln 4 \cdot \ln 2 \).
LHL (\( x \to 0^- \)):
\[ f(0) = \lim_{x \to 0^-} (\text{e}^x \sin x + ix + \lambda \ln 4) \]
Substituting \( x = 0 \):
LHL = \( \text{e}^0 \sin 0 + i(0) + \lambda \ln 4 = \lambda \ln 4 \).
Equating LHL and RHL:
\( \lambda \ln 4 = \ln 4 \cdot \ln 2 \)
\( \lambda = \ln 2 \).
Now, we need the value of \( 500\text{e}^{\lambda} \):
\( \text{e}^{\lambda} = \text{e}^{\ln 2} = 2 \).
Value = \( 500 \times 2 = 1000 \).
Step 4: Final Answer:
The value is 1000.