The interval in which y = x² e^{-x} is decreasing is . . . . . . .

Correct Answer is : (−∞,0)∪(2,∞)(-∞, 0) (2, ∞)(−∞,0)∪(2,∞)

Correct Answer is : (−∞,0)∪(2,∞)(-∞, 0) (2, ∞)(−∞,0)∪(2,∞)

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GUJCET 2026 Math Paper

Question : 7 of 40

Marks: +1, -0

y = x^2 e^{-x}

is decreasing is . . . . . . .

Solution:

Concept:

For a differentiable function, it decreases where its derivative is negative.

Explanation:

Let

y = x^2 e^{-x}

Differentiate using the product rule:

\frac{dy}{dx} = 2x e^{-x} - x^2 e^{-x} = x e^{-x}(2-x)

For decreasing, require

\frac{dy}{dx} < 0

Since

e^{-x} > 0

for all real

x

, solve

x(2-x) < 0

This gives

x < 0

x > 2

Thus, the decreasing interval is

(-\infty,0) \cup (2,\infty)

Answer:

(-\infty,0) \cup (2,\infty)

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