The absolute maximum value of \(f(x) = \sin^{2} x - \cos x\) on the interval \([0, \pi]\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{5}{4}\)
Differentiating:
\( f'(x) = 2\sin x\cos x + \sin x \)
\( = \sin x(2\cos x + 1) \)
In \([0, \pi]\), f′(x) = 0 when \(\sin x = 0\) (x = 0 or \(\pi\)) or \(\cos x = -\frac{1}{2}\) (\(x = \frac{2\pi}{3}\)). Comparing the values at these points and the end points:
\( f(0) = 0 - 1 = -1 \)
\( \displaystyle f\left(\frac{2\pi}{3}\right) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4} \)
\( f(\pi) = 0 + 1 = 1 \)
So the absolute maximum value is \(\frac{5}{4}\).
Why the other options are incorrect
Option (1) 1 — Incorrect. This is f(π), the value at the end point; f(2π/3) is larger.
Option (2) 2 — Incorrect. This adds the separate largest values of \(\sin^{2} x\) and \(-\cos x\), which do not occur at the same x.
Option (4) \(\displaystyle\frac{3}{4}\) — Incorrect. This is only the \(\sin^{2} x\) part at x = 2π/3; the \(-\cos x = \frac{1}{2}\) part is left out.