The value of `k(k gt 0)` for which the function `f(x)=(e^x-1)^4/(sin(x^2//k^2) log {1+(x^2//2)}), x ne 0` , f(0) =8 may be continuous function is

Options

A. 1
B. 4
C. 2
D. 3

Answer

To find the value of \( k \) for which the function

\[ f(x) = \frac{(e^x - 1)^4}{\sin\left(\frac{x^2}{k^2}\right) \log\left(1 + \frac{x^2}{2}\right)}, \quad x \neq 0 \]

is continuous at \( x = 0 \) with \( f(0) = 8 \), we need to ensure that

\[ \lim_{x \to 0} f(x) = f(0). \]

### Step 1: Find the limit as \( x \to 0 \)

We start by calculating the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(e^x - 1)^4}{\sin\left(\frac{x^2}{k^2}\right) \log\left(1 + \frac{x^2}{2}\right)}. \]

### Step 2: Apply limit properties

Using the known limits:

1. \( \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \) implies \( \lim_{x \to 0} (e^x - 1) = x \).
2. \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) implies \( \sin\left(\frac{x^2}{k^2}\right) \sim \frac{x^2}{k^2} \) as \( x \to 0 \).
3. \( \lim_{x \to 0} \frac{\log(1 + x)}{x} = 1 \) implies \( \log\left(1 + \frac{x^2}{2}\right) \sim \frac{x^2}{2} \).

### Step 3: Substitute limits into the expression

Thus, we can rewrite:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(x^4)^4}{\left(\frac{x^2}{k^2}\right) \left(\frac{x^2}{2}\right)} = \lim_{x \to 0} \frac{x^4}{\frac{x^4}{2k^2}} = \lim_{x \to 0} \frac{2k^2}{1} = 2k^2. \]

### Step 4: Set the limit equal to \( f(0) \)

Since we know \( f(0) = 8 \), we set:

\[ 2k^2 = 8. \]

### Step 5: Solve for \( k \)

Dividing both sides by 2 gives:

\[ k^2 = 4. \]

Taking the square root of both sides, we find:

\[ k = 2 \quad \text{or} \quad k = -2. \]

### Step 6: Consider the condition \( k > 0 \)

Since we are given that \( k > 0 \), we discard \( k = -2 \) and conclude:

\[ k = 2. \]

### Final Answer

Thus, the value of \( k \) for which the function is continuous is:

\[ \boxed{2}. \]