The interval in which the function g(x) = x^2e^-x is increasing is:

The correct answer is Option (4) → \((0, 2)\)

A function is increasing where its derivative is positive. By the product rule:

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

    \(g'(x) = x(2 - x)\,e^{-x}\)

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

    \(x \lt 0:\ g'(x) \lt 0\)
\(0 \lt x \lt 2:\ g'(x) \gt 0\)
\(x \gt 2:\ g'(x) \lt 0\)

So g is increasing on (0, 2).

NCERT (2026-27 edition):
Increasing and Decreasing Functions