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

(A) f(x) is increasing in $\left(0,\frac{3\pi}{4}\right)$
(B) f(x) is decreasing in $\left(0,\frac{3\pi}{4}\right)$
(C) f(x) is decreasing in $\left(\frac{3\pi}{4},\frac{7\pi}{4}\right)$
(D) f(x) is decreasing in $\left(\frac{7\pi}{4},2\pi \right)$

Choose the correct answer from the options given below:

(A) and (C) only

The correct answer is Option (3) → (A) and (C) only

Let \( f(x) = \sin x - \cos x \)

We compute the derivative:

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

To find critical points, set \( f'(x) = 0 \):

\( \cos x + \sin x = 0 \Rightarrow \tan x = -1 \)

This occurs at \( x = \frac{3\pi}{4} \) and \( x = \frac{7\pi}{4} \) in the interval \( [0, 2\pi] \)

Now analyze the sign of \( f'(x) \):

• For \( x \in (0, \frac{3\pi}{4}) \), take \( x = \frac{\pi}{4} \):

\( f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} > 0 \)

⇒ Increasing

• For \( x \in (\frac{3\pi}{4}, \frac{7\pi}{4}) \), take \( x = \pi \):

\( f'(\pi) = \cos(\pi) + \sin(\pi) = -1 + 0 = -1 < 0 \)

⇒ Decreasing

• For \( x \in (\frac{7\pi}{4}, 2\pi) \), take \( x = \frac{15\pi}{8} \):

\( f'(x) > 0 \) (as both sine and cosine are positive in this range)

⇒ Increasing

Correct options: (A) and (C)