If `f (x) {{:((1 - cos 8 x )/(x^(2)) "," x ge 0 ),(lambda ", " x lt 0 ) :}` is continous at
x = 0 than `lambda` = ?

Options

A. 32
B. 16
C. 64
D. 8

Answer

To determine the value of \( \lambda \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit (LHL) and right-hand limit (RHL) at \( x = 0 \) are equal, and both equal to \( f(0) \).

### Step-by-Step Solution:

1. **Identify the function**:
The function is defined as:
\[ f(x) = \begin{cases} \frac{1 - \cos(8x)}{x^2} & \text{if } x \geq 0 \\ \lambda & \text{if } x < 0 \end{cases} \]

2. **Calculate \( f(0) \)**:
Since \( f(x) \) is defined for \( x \geq 0 \) as \( \frac{1 - \cos(8x)}{x^2} \), we need to find \( f(0) \):
\[ f(0) = \lim_{x \to 0^+} \frac{1 - \cos(8x)}{x^2} \]

3. **Evaluate the limit**:
We can use the limit identity \( \lim_{x \to 0} \frac{1 - \cos(kx)}{x^2} = \frac{k^2}{2} \) for small \( x \). Here, \( k = 8 \):
\[ \lim_{x \to 0} \frac{1 - \cos(8x)}{x^2} = \frac{8^2}{2} = \frac{64}{2} = 32 \]

4. **Set up the continuity condition**:
For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ \text{LHL} = \text{RHL} = f(0) \]
Thus, we have:
\[ \lambda = 32 \]

5. **Conclusion**:
Therefore, the value of \( \lambda \) that makes the function continuous at \( x = 0 \) is:
\[ \lambda = 32 \]