The set of all values of a for which the function f(x) = x^3 + ax^2 +…

The correct answer is Option (1) → \([-3, 3]\)

f is increasing on R when \(f'(x) \ge 0\) for every real x (being 0 only at single points does not stop f from increasing).

    \(f'(x) = 3x^{2} + 2ax + 3\)

A quadratic with a positive leading coefficient is \(\ge 0\) for all x exactly when its discriminant is \(\le 0\):

    \((2a)^{2} - 4 \cdot 3 \cdot 3 \le 0\)
    \(4a^{2} \le 36\)
    \(a^{2} \le 9\)
    \(-3 \le a \le 3\)

The end values are included. For a = 3:

    \(f'(x) = 3x^{2} + 6x + 3 = 3(x + 1)^{2}\)

This is positive for every x except x = −1, where it is 0, so f still increases on all of R. In the same way, a = −3 gives \(f'(x) = 3(x - 1)^{2}\).

So the set is [−3, 3].

Why the other options are incorrect

• Option (2) \((-3, 3)\) — Incorrect. It leaves out a = ±3, where \(f'(x) = 3(x \pm 1)^{2} \ge 0\) and f is still increasing.

• Option (3) \((-\infty, -3] \cup [3, \infty)\) — Incorrect. For \(|a| \gt 3\) the discriminant is positive, so f′(x) is negative between its two roots; for example a = 4 gives \(f'(-1) = 3 - 8 + 3 = -2 \lt 0\).

• Option (4) \([0, 3]\) — Incorrect. It leaves out negative values such as a = −1, where \(f'(x) = 3x^{2} - 2x + 3\) has discriminant \(4 - 36 \lt 0\), so f is increasing.