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The interval in which the function given by f(x)=x2e−xf(x) = x^2 e^{-x} is strictly increasing is:

Solution

✅ Correct Option: 4

f′(x)=2xe−x−x2e−x=e−x⋅x(2−x)f'(x) = 2x e^{-x} - x^2 e^{-x} = e^{-x} \cdot x(2-x). Since e−x>0e^{-x} > 0, sign of f′f' depends on x(2−x)x(2-x). This is positive when 0<x<20 < x < 2. So ff is strictly increasing on (0,2)(0, 2).

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