Let `f(x)=(1-cos x sqrt(cos 2x))/x^2 , x ne 0` The value of f(0) so that f is a continuous function is

Answer

To find the value of \( f(0) \) such that the function \( f(x) = \frac{1 - \cos(x \sqrt{\cos(2x)})}{x^2} \) is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0 and set it equal to \( f(0) \).

### Step-by-Step Solution:

1. **Identify the Limit**:
We need to find:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1 - \cos(x \sqrt{\cos(2x)})}{x^2} \]

2. **Substituting \( x = 0 \)**:
When we substitute \( x = 0 \):
\[ f(0) = \frac{1 - \cos(0 \cdot \sqrt{\cos(0)})}{0^2} = \frac{1 - \cos(0)}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
This results in an indeterminate form \( \frac{0}{0} \).

3. **Applying L'Hôpital's Rule**:
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{d}{dx}(1 - \cos(x \sqrt{\cos(2x)})) \bigg/ \frac{d}{dx}(x^2) \]

4. **Differentiate the Numerator**:
Using the chain rule and product rule:
\[ \frac{d}{dx}(1 - \cos(x \sqrt{\cos(2x)})) = \sin(x \sqrt{\cos(2x)}) \cdot \frac{d}{dx}(x \sqrt{\cos(2x)}) \]
Now, differentiate \( x \sqrt{\cos(2x)} \):
\[ \frac{d}{dx}(x \sqrt{\cos(2x)}) = \sqrt{\cos(2x)} + x \cdot \frac{1}{2\sqrt{\cos(2x)}} \cdot (-\sin(2x) \cdot 2) = \sqrt{\cos(2x)} - \frac{x \sin(2x)}{\sqrt{\cos(2x)}} \]

5. **Differentiate the Denominator**:
\[ \frac{d}{dx}(x^2) = 2x \]

6. **Setting Up the Limit Again**:
Now, substituting back into the limit:
\[ \lim_{x \to 0} \frac{\sin(x \sqrt{\cos(2x)}) \left( \sqrt{\cos(2x)} - \frac{x \sin(2x)}{\sqrt{\cos(2x)}} \right)}{2x} \]

7. **Evaluating the Limit**:
As \( x \to 0 \):
- \( \sin(x \sqrt{\cos(2x)}) \approx x \sqrt{\cos(2x)} \) (using small angle approximation)
- \( \sqrt{\cos(2x)} \to 1 \)
- \( \sin(2x) \to 2x \)

Thus, we can simplify:
\[ \lim_{x \to 0} \frac{x \cdot 1 \cdot \left(1 - 0\right)}{2x} = \lim_{x \to 0} \frac{x}{2x} = \frac{1}{2} \]

8. **Conclusion**:
Therefore, to make \( f(x) \) continuous at \( x = 0 \), we set:
\[ f(0) = \frac{1}{2} \]

### Final Answer:
The value of \( f(0) \) so that \( f \) is a continuous function is \( \frac{1}{2} \).