The function f(x) = x^2 e^-x is strictly increasing in the interval

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

By the product rule:

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

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

    \(x \lt 0\): \(x \lt 0,\ 2 - x \gt 0\), so \(f'(x) \lt 0\)
    \(0 \lt x \lt 2\): both factors positive, \(f'(x) \gt 0\)
    \(x \gt 2\): \(x \gt 0,\ 2 - x \lt 0\), so \(f'(x) \lt 0\)

So f is strictly increasing in (0, 2) and strictly decreasing in \((-\infty, 0)\) and in \((2, \infty)\).