If $f(x) = \sin x - \cos x$, $x \in [0, 2\pi]$ then

(A) $f(x)$ is increasing in $(0, \frac{3\pi}{4})$

(B) $f(x)$ is decreasing in $(0, \frac{3\pi}{4})$

(C) $f(x)$ is decreasing in $(\frac{3\pi}{4}, \frac{7\pi}{4})$

(D) $f(x)$ is decreasing in $(\frac{7\pi}{4}, 2\pi)$

Choose the correct answer from the options given below:

Given $f(x) = \sin x - \cos x$ where $x \in [0, 2\pi]$

The derivative is:

$f'(x) = \cos x + \sin x$

If $f'(x) > 0$, the function is increasing.

If $f'(x) < 0$, the function is decreasing.

---

Setting $f'(x) = 0$:

$\cos x + \sin x = 0$

$\sin x = -\cos x$

$\tan x = -1$

In the interval $[0, 2\pi]$, this occurs at $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.

---

For the interval $(0, \frac{3\pi}{4})$, test at $x = \frac{\pi}{4}$:

$f'\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right)$

$= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

$= \sqrt{2} > 0$

The function is increasing on $(0, \frac{3\pi}{4})$.

Statement (A) is TRUE.

---

For the interval $(\frac{3\pi}{4}, \frac{7\pi}{4})$, test at $x = \pi$:

$f'(\pi) = \cos(\pi) + \sin(\pi)$

$= -1 + 0$

$= -1 < 0$

The function is decreasing on $(\frac{3\pi}{4}, \frac{7\pi}{4})$.

Statement (C) is TRUE.

---

For the interval $(\frac{7\pi}{4}, 2\pi)$, test at $x = \frac{15\pi}{8}$:

$f'\left(\frac{15\pi}{8}\right) = \cos\left(\frac{15\pi}{8}\right) + \sin\left(\frac{15\pi}{8}\right) > 0$

The function is increasing on $(\frac{7\pi}{4}, 2\pi)$.

Statement (D) is FALSE.

---

Statement (A) is TRUE.

Statement (B) is FALSE.

Statement (C) is TRUE.

Statement (D) is FALSE.

Therefore, the correct answer is (A) and (C) only.