If \(f(x) = \sin x - \cos x,\ x \in [0, 2\pi]\) then
(A) \(f(x)\) is increasing in \(\left(0, \displaystyle\frac{3\pi}{4}\right)\)
(B) \(f(x)\) is decreasing in \(\left(0, \displaystyle\frac{3\pi}{4}\right)\)
(C) \(f(x)\) is decreasing in \(\left(\displaystyle\frac{3\pi}{4}, \frac{7\pi}{4}\right)\)
(D) \(f(x)\) is decreasing in \(\left(\displaystyle\frac{7\pi}{4}, 2\pi\right)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (C) only
The sign of the derivative decides where f increases or decreases:
\(f'(x) = \cos x + \sin x\)
\( \displaystyle = \sqrt{2}\,\sin\left(x + \frac{\pi}{4}\right) \)
For \(0 \lt x \lt 2\pi\), \(x + \frac{\pi}{4}\) runs from \(\frac{\pi}{4}\) to \(\frac{9\pi}{4}\), and the sine is zero at \(x + \frac{\pi}{4} = \pi\) and \(2\pi\):
\( \displaystyle f'(x) = 0 \text{ at } x = \frac{3\pi}{4},\ \frac{7\pi}{4} \)
So \(f'(x) \gt 0\) on \(\left(0, \frac{3\pi}{4}\right)\), \(f'(x) \lt 0\) on \(\left(\frac{3\pi}{4}, \frac{7\pi}{4}\right)\) and \(f'(x) \gt 0\) on \(\left(\frac{7\pi}{4}, 2\pi\right)\).
Checking the statements
(A) \(f(x)\) is increasing in \(\left(0, \frac{3\pi}{4}\right)\) — Correct. For example, \(f'\left(\frac{\pi}{4}\right) = \sqrt{2} \gt 0\).
(B) \(f(x)\) is decreasing in \(\left(0, \frac{3\pi}{4}\right)\) — Incorrect. \(f'(x) \gt 0\) there, so f is increasing.
(C) \(f(x)\) is decreasing in \(\left(\frac{3\pi}{4}, \frac{7\pi}{4}\right)\) — Correct. For example, \(f'(\pi) = -1 \lt 0\).
(D) \(f(x)\) is decreasing in \(\left(\frac{7\pi}{4}, 2\pi\right)\) — Incorrect. For example, \(f'\left(\frac{15\pi}{8}\right) = \sqrt{2}\sin\frac{\pi}{8} \gt 0\).