The number of points in the interval `[-sqrt(13) sqrt(13)]` at which f(x)`=sin x^2 +cos x^2 ` attains its maximum value is

Options

A. 2
B. 8
C. 0
D. 4

Explanation

To solve the problem, we need to find the number of points in the interval \([- \sqrt{13}, \sqrt{13}]\) at which the function \(f(x) = \sin(x^2) + \cos(x^2)\) attains its maximum value.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function given is:
\[ f(x) = \sin(x^2) + \cos(x^2) \]
We know that the maximum value of \(\sin\) and \(\cos\) functions is \(1\). Therefore, the maximum value of \(f(x)\) can be calculated.

2. **Using Trigonometric Identity**:
We can rewrite the function using a trigonometric identity:
\[ f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin(x^2) + \frac{1}{\sqrt{2}} \cos(x^2) \right) = \sqrt{2} \cos\left(x^2 - \frac{\pi}{4}\right) \]
This means that the maximum value of \(f(x)\) is \(\sqrt{2}\) when:
\[ \cos\left(x^2 - \frac{\pi}{4}\right) = 1 \]

3. **Finding Maximum Points**:
The cosine function equals \(1\) at:
\[ x^2 - \frac{\pi}{4} = 2n\pi \quad \text{for integers } n \]
Rearranging gives:
\[ x^2 = 2n\pi + \frac{\pi}{4} \]

4. **Finding Values of \(x\)**:
Taking the square root:
\[ x = \pm \sqrt{2n\pi + \frac{\pi}{4}} \]

5. **Determining Valid \(n\)**:
We need to find values of \(n\) such that \(x\) lies within the interval \([- \sqrt{13}, \sqrt{13}]\):
\[ -\sqrt{13} \leq \sqrt{2n\pi + \frac{\pi}{4}} \leq \sqrt{13} \]
Squaring the inequality:
\[ 2n\pi + \frac{\pi}{4} \leq 13 \]
This simplifies to:
\[ 2n\pi \leq 13 - \frac{\pi}{4} \]
\[ n \leq \frac{13 - \frac{\pi}{4}}{2\pi} \]

6. **Calculating Maximum \(n\)**:
We can calculate the approximate value of \(n\):
\[ n \leq \frac{13 - 0.785}{6.283} \approx \frac{12.215}{6.283} \approx 1.943 \]
Hence, \(n\) can take values \(0\) and \(1\).

7. **Finding Points for Each \(n\)**:
- For \(n = 0\):
\[ x = \pm \sqrt{\frac{\pi}{4}} = \pm \frac{\sqrt{\pi}}{2} \]
- For \(n = 1\):
\[ x = \pm \sqrt{2\pi + \frac{\pi}{4}} = \pm \sqrt{\frac{9\pi}{4}} = \pm \frac{3\sqrt{\pi}}{2} \]

8. **Checking Validity**:
We need to check if \(\pm \frac{3\sqrt{\pi}}{2}\) is within \([- \sqrt{13}, \sqrt{13}]\):
- Since \(\sqrt{13} \approx 3.605\) and \(\sqrt{\pi} \approx 1.772\),
- \(\frac{3\sqrt{\pi}}{2} \approx 2.646\) which is valid.

9. **Counting Points**:
From both \(n = 0\) and \(n = 1\), we have:
- From \(n = 0\): \(2\) points (\(\pm \frac{\sqrt{\pi}}{2}\))
- From \(n = 1\): \(2\) points (\(\pm \frac{3\sqrt{\pi}}{2}\))

Thus, the total number of points where \(f(x)\) attains its maximum value in the interval \([- \sqrt{13}, \sqrt{13}]\) is \(4\).

### Final Answer:
The number of points in the interval \([- \sqrt{13}, \sqrt{13}]\) at which \(f(x)\) attains its maximum value is **4**.