`f:R->R` is defined by `f(x)={(cos3x-cosx)/(x^2), x!=0lambda, x=0` and `f` is continuous at `x=0;` then `lambda=`

Options

A. `-2`
B. `-4
C. `-6`
D. `-8`

Explanation

To solve the problem, we need to find the value of \( \lambda \) such that the function \( f(x) \) is continuous at \( x = 0 \). The function is defined as follows:

\[ f(x) = \begin{cases} \frac{\cos(3x) - \cos(x)}{x^2} & \text{if } x \neq 0 \\ \lambda & \text{if } x = 0 \end{cases} \]

### Step 1: Apply the definition of continuity
For \( f(x) \) to be continuous at \( x = 0 \), we need:

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

This means:

\[ \lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2} = \lambda \]

### Step 2: Calculate the limit
We need to evaluate the limit:

\[ \lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2} \]

### Step 3: Use the cosine difference identity
We can use the identity \( \cos a - \cos b = -2 \sin\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right) \):

Let \( a = 3x \) and \( b = x \):

\[ \cos(3x) - \cos(x) = -2 \sin\left(\frac{3x + x}{2}\right) \sin\left(\frac{3x - x}{2}\right) = -2 \sin(2x) \sin(x) \]

### Step 4: Substitute back into the limit
Now substitute this back into the limit:

\[ \lim_{x \to 0} \frac{-2 \sin(2x) \sin(x)}{x^2} \]

### Step 5: Rewrite the limit
We can rewrite the limit as:

\[ \lim_{x \to 0} -2 \cdot \frac{\sin(2x)}{x} \cdot \frac{\sin(x)}{x} \]

### Step 6: Apply the standard limit
Using the standard limit \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \), we can evaluate:

\[ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \]

Thus, we have:

\[ \lim_{x \to 0} -2 \cdot 2 \cdot 1 = -4 \]

### Step 7: Set the limit equal to \( \lambda \)
Now we can set this equal to \( \lambda \):

\[ \lambda = -4 \]

### Final Answer
Thus, the value of \( \lambda \) is:

\[ \boxed{-4} \]