The function f(x) = x^2e^-2x increases on

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

By the product rule:

    \(f'(x) = 2x\,e^{-2x} - 2x^2 e^{-2x}\)
    \(f'(x) = 2x(1 - x)\,e^{-2x}\)

Since \(e^{-2x} \gt 0\) for every x, the sign of f'(x) is the sign of \(x(1 - x)\):

    \(x \lt 0:\ x \lt 0,\ 1 - x \gt 0,\ \text{so } f'(x) \lt 0\)
    \(0 \lt x \lt 1:\ \text{both factors positive, so } f'(x) \gt 0\)
    \(x \gt 1:\ x \gt 0,\ 1 - x \lt 0,\ \text{so } f'(x) \lt 0\)

So f increases on (0, 1).

Why the other options are incorrect

• Option (1) \((-\infty,\infty)\) — Incorrect. f is decreasing for x < 0 and for x > 1.

• Option (3) \((-\infty,0) \cup (1,\infty)\) — Incorrect. These are exactly the intervals where \(f'(x) \lt 0\), so f decreases there.

• Option (4) \((1,\infty)\) — Incorrect. For x > 1, \(1 - x \lt 0\), so f decreases.