The function f(x) = x^2/3(x - 5) , x ∈ R , is strictly decreasing in

The correct answer is Option (1) → (0, 2)

Writing \(f(x) = x^{5/3} - 5x^{2/3}\) and differentiating (for x ≠ 0):

    \( \displaystyle f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} \)

    \( \displaystyle = \frac{5(x - 2)}{3x^{1/3}} \)

The derivative is 0 at x = 2 and does not exist at x = 0. These two points split R into three intervals. The cube root \(x^{1/3}\) has the sign of x, so the sign of f′(x) is

    x < 0: numerator and denominator both negative, so f′(x) > 0
    0 < x < 2: numerator negative, denominator positive, so f′(x) < 0
    x > 2: both positive, so f′(x) > 0

So f is strictly decreasing in (0, 2).

Why the other options are incorrect

• Option (2) \((-\infty, 2)\) — Incorrect. It includes negative x, where f′(x) > 0; for example f(−1) = −6 is less than f(0) = 0.

• Option (3) \((2, \infty)\) — Incorrect. For x > 2, f′(x) > 0, so f is increasing there.

• Option (4) \((-\infty, 0)\) — Incorrect. For x < 0, both x − 2 and \(x^{1/3}\) are negative, so f′(x) > 0 and f is increasing there.