\(f(x)=\sin x+\displaystyle\frac{1}{2}\cos 2x\) in \(\left[0, \displaystyle\frac{\pi}{2}\right]\)
(A) \(f'(x)=\cos x-\sin 2x\)
(B) The critical points of the function are \(x=\displaystyle\frac{\pi}{6}\) and \(x=\displaystyle\frac{\pi}{2}\)
(C) The minimum value of the function is 2
(D) The maximum value of the function is \(\displaystyle\frac{3}{4}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
Differentiating, with \(\frac{d}{dx}\left(\frac{1}{2}\cos 2x\right) = -\sin 2x\):
\(f'(x) = \cos x - \sin 2x\)
The critical points are where f′(x) = 0:
\(\cos x - 2\sin x\cos x = 0\)
\(\cos x\,(1 - 2\sin x) = 0\)
\( \displaystyle \cos x = 0 \text{ or } \sin x = \frac{1}{2} \)
In \(\left[0, \frac{\pi}{2}\right]\) this gives
\( \displaystyle x = \frac{\pi}{2} \text{ or } x = \frac{\pi}{6} \)
On a closed interval the absolute maximum and minimum occur at a critical point or an end point:
\( \displaystyle f(0) = 0 + \frac{1}{2}\times 1 = \frac{1}{2} \)
\( \displaystyle f\left(\frac{\pi}{6}\right) = \frac{1}{2} + \frac{1}{2}\times\frac{1}{2} = \frac{3}{4} \)
\( \displaystyle f\left(\frac{\pi}{2}\right) = 1 + \frac{1}{2}\times(-1) = \frac{1}{2} \)
So \(f'(x) = \cos x - \sin 2x\), the critical points are \(\frac{\pi}{6}\) and \(\frac{\pi}{2}\), the minimum value is \(\frac{1}{2}\) and the maximum value is \(\frac{3}{4}\).
Checking the statements
(A) \(f'(x)=\cos x-\sin 2x\) — Correct. The derivative of \(\frac{1}{2}\cos 2x\) is \(-\sin 2x\).
(B) The critical points of the function are \(x=\frac{\pi}{6}\) and \(x=\frac{\pi}{2}\) — Correct. \(\cos x\,(1 - 2\sin x) = 0\) only at these two points of the interval.
(C) The minimum value of the function is 2 — Incorrect. The minimum value is \(\frac{1}{2}\), at x = 0 and \(x = \frac{\pi}{2}\).
(D) The maximum value of the function is \(\frac{3}{4}\) — Correct. \(f\left(\frac{\pi}{6}\right) = \frac{3}{4}\) is the largest of the three values.