Let \( f:\mathrm{R}\to\mathrm{R} \) be defined by \( f(x)=2x+\cos x \) (where \( \mathrm{R} \) is set of real numbers), then which of following statements are TRUE?
(A) \( f(x) \) is an increasing function on \( \mathrm{R} \)
(B) \( f(x) \) is a decreasing function on \( \mathrm{R} \)
(C) \( f(x) \) has no maxima or minima on \( \mathrm{R} \)
(D) \( f(x) \) has maximum at \( x=0 \)
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
Since \( -1\le\sin x\le 1 \),
\( \displaystyle f'(x)=2-\sin x \)
\( \displaystyle f'(x)\ge 1\gt 0 \)
So \( f'(x)\gt 0 \) for every real x: f is strictly increasing and \( f' \) is never zero.
Checking the statements
(A) \( f(x) \) is an increasing function on \( \mathrm{R} \) — Correct. \( f'(x)\ge 1\gt 0 \) for every x.
(B) \( f(x) \) is a decreasing function on \( \mathrm{R} \) — Incorrect. \( f'(x) \) is never negative.
(C) \( f(x) \) has no maxima or minima on \( \mathrm{R} \) — Correct. \( f'(x) \) is never zero, so there is no turning point.
(D) \( f(x) \) has maximum at \( x=0 \) — Incorrect. \( f'(0)=2 \), not 0.