For the function \(f(x) = \sin x + \cos x\), \(x \in [0, 2\pi]\):
(A) The critical points of f in \((0, 2\pi)\) are \(x = \displaystyle\frac{\pi}{4}\) and \(x = \displaystyle\frac{5\pi}{4}\)
(B) The absolute maximum value of f is \(\sqrt{2}\)
(C) f is strictly decreasing in \(\left(\displaystyle\frac{\pi}{4}, \frac{5\pi}{4}\right)\)
(D) The absolute minimum value of f is \(-1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (C) only
Differentiating:
\(f'(x) = \cos x - \sin x\)
f’(x) = 0 when \(\tan x = 1\), that is \(x = \frac{\pi}{4}\) or \(x = \frac{5\pi}{4}\). Comparing f at the critical points and the end points:
\( \displaystyle f\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \)
\( \displaystyle f\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} \)
\(f(0) = 1,\ f(2\pi) = 1\)
Between \(\frac{\pi}{4}\) and \(\frac{5\pi}{4}\), \(\sin x \gt \cos x\), so f’(x) < 0 there.
So the critical points are \(\frac{\pi}{4}\) and \(\frac{5\pi}{4}\), the maximum is \(\sqrt{2}\), the minimum is \(-\sqrt{2}\), and f’(x) < 0 on \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\).
Checking the statements
(A) The critical points of f in \((0, 2\pi)\) are \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\) — Correct. These are the only solutions of \(\tan x = 1\) there.
(B) The absolute maximum value of f is \(\sqrt{2}\) — Correct. \(f\left(\frac{\pi}{4}\right) = \sqrt{2}\), larger than f(0) = 1.
(C) f is strictly decreasing in \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\) — Correct. \(\cos x - \sin x \lt 0\) throughout that interval.
(D) The absolute minimum value of f is \(-1\) — Incorrect. \(f\left(\frac{5\pi}{4}\right) = -\sqrt{2}\), which is less than −1.