The function \( f(x)=3x+\cos 3x \) is (where R is set of real numbers)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Increasing on R
The sign of the derivative decides whether f increases or decreases.
\( \displaystyle f'(x)=3-3\sin 3x \)
\( \displaystyle =3(1-\sin 3x) \)
Since \( \sin 3x\le 1 \) for every x, \( 1-\sin 3x\ge 0 \), so
\( \displaystyle f'(x)\ge 0\ \text{for all}\ x\in\mathrm{R} \)
\( f'(x)=0 \) only at the isolated points where \( \sin 3x=1 \); these do not form an interval, so f never becomes constant or starts to decrease. Hence f is increasing on R.
Why the other options are incorrect
Option (2) Decreasing on R — Incorrect. \( f'(x) \) is never negative, so f cannot decrease anywhere.
Option (3) Neither increasing nor decreasing on R — Incorrect. A function whose derivative is \( \ge 0 \) everywhere (zero only at isolated points) is increasing; the points where \( f'(x)=0 \) do not change this.
Option (4) Strictly decreasing on R — Incorrect. \( f'(x) \) is never negative, so f is certainly not strictly decreasing.