If f : R `rarr R ` is defined by
`f(x) = {{:((cos 3x-cosx )/(x^(2)), "for" x ne 0),(lambda, "for" x=0):}`
and if f is continuous at x = 0, then `lambda` is equal to

Options

A. `-2`
B. 1
C. 2
D. 3

Answer

To find the value of \( \lambda \) such that the function \( f(x) \) is continuous at \( x = 0 \), we will follow these steps:

1. **Define the function**: The function is given as:
\[ f(x) = \begin{cases} \frac{\cos(3x) - \cos(x)}{x^2} & \text{for } x \neq 0 \\ \lambda & \text{for } x = 0 \end{cases} \]

2. **Check for continuity at \( x = 0 \)**: For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ \lim_{x \to 0} f(x) = f(0) = \lambda \]

3. **Calculate the limit**:
We need to evaluate:
\[ \lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2} \]
When we substitute \( x = 0 \), both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \). Thus, we can apply L'Hôpital's Rule.

4. **Apply L'Hôpital's Rule**: Differentiate the numerator and the denominator:
- The derivative of the numerator \( \cos(3x) - \cos(x) \) is:
\[ -3\sin(3x) + \sin(x) \]
- The derivative of the denominator \( x^2 \) is:
\[ 2x \]
Therefore, we have:
\[ \lim_{x \to 0} \frac{-3\sin(3x) + \sin(x)}{2x} \]

5. **Evaluate the limit again**: As \( x \to 0 \), we again encounter the \( \frac{0}{0} \) form. We apply L'Hôpital's Rule a second time:
- The derivative of the numerator \( -3\sin(3x) + \sin(x) \) is:
\[ -9\cos(3x) + \cos(x) \]
- The derivative of the denominator \( 2x \) is:
\[ 2 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{-9\cos(3x) + \cos(x)}{2} \]

6. **Evaluate the limit as \( x \to 0 \)**:
\[ = \frac{-9\cos(0) + \cos(0)}{2} = \frac{-9 \cdot 1 + 1}{2} = \frac{-8}{2} = -4 \]

7. **Set the limit equal to \( \lambda \)**:
Since \( \lim_{x \to 0} f(x) = \lambda \), we have:
\[ \lambda = -4 \]

Thus, the value of \( \lambda \) is \( -4 \).