For the function \(f(x) = \sin x + \cos x, x \in [0,\pi]\), which one of the following is correct?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Absolute maximum value is \(\sqrt{2}\)
On a closed interval, the absolute extreme values of a continuous function occur at the critical points or the end points.
\(f'(x) = \cos x - \sin x = 0\)
\(\tan x = 1\)
\(\displaystyle x = \frac{\pi}{4} \quad (\text{the only solution in } [0, \pi])\)
Values at the critical point and the end points:
\(f(0) = 0 + 1 = 1\)
\(\displaystyle f\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}\)
\(f(\pi) = 0 - 1 = -1\)
So the absolute maximum value is \(\sqrt{2}\), at \(x = \frac{\pi}{4}\), and the absolute minimum value is −1, at \(x = \pi\).
Why the other options are incorrect
Option (1) Point of absolute maxima is \(\pi\) — Incorrect. \(f(\pi) = -1\) is the smallest value: \(\pi\) is the point of absolute minima.
Option (3) Absolute minimum value is 0 — Incorrect. The absolute minimum value is \(f(\pi) = -1\).
Option (4) Point of absolute maxima is 0 — Incorrect. \(f(0) = 1 \lt \sqrt{2}\); the point of absolute maxima is \(\frac{\pi}{4}\).