The function \(f(x) = \sin x + \cos x\), \(0 \le x \le 2\pi\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → strictly increasing in \(\left(\displaystyle\frac{5\pi}{4}, 2\pi\right]\)
The function increases where \(f'(x) \gt 0\) and decreases where \(f'(x) \lt 0\).
\(f'(x) = \cos x - \sin x\)
\(f'(x) = 0\) when \(\tan x = 1\), that is at
\( \displaystyle x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4} \)
These two points split \([0, 2\pi]\) into three parts. Checking the sign of f′(x) at one point of each part:
\(x = 0:\ f'(0) = 1 - 0 = 1 \gt 0\), increasing on \(\left[0, \frac{\pi}{4}\right)\)
\(x = \pi:\ f'(\pi) = -1 - 0 = -1 \lt 0\), decreasing on \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\)
\(x = \frac{3\pi}{2}:\ f'\left(\frac{3\pi}{2}\right) = 0 - (-1) = 1 \gt 0\), increasing on \(\left(\frac{5\pi}{4}, 2\pi\right]\)
So f is strictly increasing on \(\left(\frac{5\pi}{4}, 2\pi\right]\).
Why the other options are incorrect
Option (1) strictly decreasing in \(\left[0, \displaystyle\frac{\pi}{4}\right)\) — Incorrect. Here \(f'(x) \gt 0\), so f is increasing on this interval.
Option (2) strictly increasing in \(\left(\displaystyle\frac{\pi}{4}, \frac{5\pi}{4}\right)\) — Incorrect. Here \(f'(x) \lt 0\), so f is decreasing on this interval.
Option (3) strictly decreasing in \(\left(\displaystyle\frac{5\pi}{4}, 2\pi\right]\) — Incorrect. Here \(f'(x) \gt 0\), so f is increasing, not decreasing.
NCERT (2026-27 edition):
Increasing and Decreasing Functions
