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

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

f is decreasing on R when \(f'(x) \le 0\) for every real x (being 0 only at single points is allowed).

    \( f'(x) = 3(k + 2)x^{2} - 6kx + 9k \)

If k = −2, then f′(x) = 12x − 18, which is positive for x > 1.5, so k = −2 fails.

Otherwise f′(x) is a quadratic. It is \(\le 0\) for all x exactly when its leading coefficient is negative and its discriminant is \(\le 0\):

    \( k + 2 \lt 0 \), so \( k \lt -2 \)

    \( (6k)^{2} - 4 \cdot 3(k + 2) \cdot 9k \le 0 \)

    \( 36k^{2} - 108k^{2} - 216k \le 0 \)

    \( -72k(k + 3) \le 0 \)

    \( k(k + 3) \ge 0 \), so \( k \le -3 \) or \( k \ge 0 \)

Both conditions together give \(k \le -3\). The end value is included: for k = −3,

    \( f'(x) = -3x^{2} + 18x - 27 = -3(x - 3)^{2} \le 0 \)

which is 0 only at x = 3, so f is still decreasing on R.

Why the other options are incorrect

• Option (1) \((-\infty, -2)\) — Incorrect. It uses only the leading-coefficient condition. For k = −2.5, \(f'(x) = -1.5x^{2} + 15x - 22.5\) and f′(5) = 15 > 0.

• Option (2) \([-3, 0]\) — Incorrect. This reverses the discriminant inequality; for k = −1, f′(x) = 3x² + 6x − 9 is positive for large x.

• Option (4) \((-\infty, -3)\) — Incorrect. It leaves out k = −3, where \(f'(x) = -3(x - 3)^{2} \le 0\) and f is still decreasing.