Question

The function \(f(x)=sinx+cosx,0\leq x\leq 2\pi \) is :

• A. strictly decreasing in \([0,\frac{\pi}{4}]\)
• B. strictly Increasing in \([\frac{\pi}{4},\frac{5\pi}{4}]\)
• C. strictly decreasing in \([\frac{5\pi}{4},2\pi]\)
• D. strictly increasing in \([\frac{5\pi}{4},2\pi]\)

Answer 1

The Correct Option is D

Given function: \( f(x) = \sin x + \cos x \)

Derivative: \( f'(x) = \cos x - \sin x \)

To find when the function is strictly increasing, we solve:

\( f'(x) = \cos x - \sin x > 0 \Rightarrow \cos x > \sin x \Rightarrow \tan x < 1 \)

Now analyze:

The inequality \( \tan x < 1 \) holds in the intervals:

• \( x \in \left(0, \frac{\pi}{4}\right) \)
• \( x \in \left(\pi, \frac{5\pi}{4}\right) \)

At \( x = \frac{5\pi}{4} \), \( \tan x = 1 \), so the derivative is zero — not strictly increasing at that point. And for \( x \in \left(\frac{5\pi}{4}, 2\pi\right) \), \( \tan x > 1 \), meaning \( f'(x) < 0 \), so the function is decreasing there.

✅ Final Answer:
The function \( f(x) = \sin x + \cos x \) is strictly increasing on the interval:
\( \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \)