For the function \(f(x) = 2\sin x - \cos 2x\), \(0 \le x \le 2\pi\):
(A) \(f'(x) = 2\cos x\,(1 + 2\sin x)\).
(B) The absolute maximum value of f is 3.
(C) The absolute minimum value of f is −3.
(D) f attains its absolute minimum at \(x = \displaystyle\frac{7\pi}{6}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
Differentiating and using \(\sin 2x = 2\sin x\cos x\):
\( f'(x) = 2\cos x + 2\sin 2x \)
\( = 2\cos x\,(1 + 2\sin x) \)
f′(x) = 0 when cos x = 0 or \(\sin x = -\frac{1}{2}\):
\( \displaystyle x = \frac{\pi}{2}, \ \frac{3\pi}{2}, \ \frac{7\pi}{6}, \ \frac{11\pi}{6} \)
The values at these points and at the end points are
\( \displaystyle f\left(\frac{\pi}{2}\right) = 2 - (-1) = 3 \)
\( \displaystyle f\left(\frac{3\pi}{2}\right) = -2 - (-1) = -1 \)
\( \displaystyle f\left(\frac{7\pi}{6}\right) = f\left(\frac{11\pi}{6}\right) = -1 - \frac{1}{2} = -\frac{3}{2} \)
\( f(0) = f(2\pi) = 0 - 1 = -1 \)
So the absolute maximum value is 3 (at \(x = \frac{\pi}{2}\)) and the absolute minimum value is \(-\frac{3}{2}\) (at \(x = \frac{7\pi}{6}\) and \(x = \frac{11\pi}{6}\)).
Checking the statements
(A) \(f'(x) = 2\cos x\,(1 + 2\sin x)\). — Correct. \(2\cos x + 2\sin 2x = 2\cos x + 4\sin x\cos x\).
(B) The absolute maximum value of f is 3. — Correct. \(f\left(\frac{\pi}{2}\right) = 3\) is the greatest of the values.
(C) The absolute minimum value of f is −3. — Incorrect. The least value is \(-\frac{3}{2}\); the value −3 is never reached.
(D) f attains its absolute minimum at \(x = \displaystyle\frac{7\pi}{6}\). — Correct. \(f\left(\frac{7\pi}{6}\right) = -\frac{3}{2}\), the least value.