If `f(x) =(e^(x)-1)^(4)/(sin(x^(2)/lambda^(2))log (1+x^(2)/2)), x ne 0` and f (0) = 8 be a continuous function then `lambda` =

Options

A. 2
B. 1
C. `-1`
D. `-2`

Answer

To solve the problem, we need to find the value of \(\lambda\) such that the function \(f(x) = \frac{(e^x - 1)^4}{\sin\left(\frac{x^2}{\lambda^2}\right) \log\left(1 + \frac{x^2}{2}\right)}\) is continuous at \(x = 0\) and \(f(0) = 8\).

### Step-by-Step Solution:

1. **Understanding Continuity**:
For \(f(x)\) to be continuous at \(x = 0\), we need:
\[ f(0) = \lim_{x \to 0} f(x) \]
Given that \(f(0) = 8\), we need to compute \(\lim_{x \to 0} f(x)\).

2. **Substituting into the Limit**:
We can express the limit as:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(e^x - 1)^4}{\sin\left(\frac{x^2}{\lambda^2}\right) \log\left(1 + \frac{x^2}{2}\right)} \]

3. **Using Taylor Series Expansions**:
- For small \(x\), we have:
\[ e^x - 1 \approx x \quad \text{(as } x \to 0\text{)} \]
Therefore,
\[ (e^x - 1)^4 \approx x^4 \]
- For \(\sin\left(\frac{x^2}{\lambda^2}\right)\):
\[ \sin\left(\frac{x^2}{\lambda^2}\right) \approx \frac{x^2}{\lambda^2} \quad \text{(as } x \to 0\text{)} \]
- For \(\log\left(1 + \frac{x^2}{2}\right)\):
\[ \log\left(1 + \frac{x^2}{2}\right) \approx \frac{x^2}{2} \quad \text{(as } x \to 0\text{)} \]

4. **Substituting the Approximations**:
Now we substitute these approximations into the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x^4}{\left(\frac{x^2}{\lambda^2}\right) \left(\frac{x^2}{2}\right)} = \lim_{x \to 0} \frac{x^4}{\frac{x^4}{2\lambda^2}} = \lim_{x \to 0} 2\lambda^2 \]

5. **Setting the Limit Equal to 8**:
Since we need this limit to equal \(8\):
\[ 2\lambda^2 = 8 \]

6. **Solving for \(\lambda\)**:
Dividing both sides by \(2\):
\[ \lambda^2 = 4 \]
Taking the square root:
\[ \lambda = \pm 2 \]

### Final Answer:
Thus, the values of \(\lambda\) are \(2\) and \(-2\).