If the function \( f(x)=x^3-kx \) is increasing for all real value of \( x \), then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( k\le 0 \)
\( \displaystyle f'(x)=3x^2-k \)
f is increasing for all x when \( f'(x)\ge 0 \) for all x. The least value of \( 3x^2 \) is 0 (at x = 0), so we need
\( \displaystyle 0-k\ge 0 \)
\( \displaystyle k\le 0 \)
Why the other options are incorrect
Option (1) \( k\ge 0 \) — Incorrect. The sign is reversed: for k > 0, \( f'(0)=-k\lt 0 \), so f decreases near x = 0.
Option (3) \( k\gt 0 \) — Incorrect. Every k > 0 gives \( f'(0)=-k\lt 0 \).
Option (4) \( k\lt 1 \) — Incorrect. It allows 0 < k < 1; for example, k = 1/2 gives \( f'(0)=-\frac{1}{2}\lt 0 \).