CUET UG 2025 — Mathematics Calculus

A function increases where its derivative is positive.

Given $f(x) = x^2e^{-2x}$f(x)=x2e−2x

Using the product rule with $u = x^2$u=x2 and $v = e^{-2x}$v=e−2x:

$$f'(x) = 2x \cdot e^{-2x} + x^2 \cdot (-2e^{-2x})$$f′(x)=2x⋅e−2x+x2⋅(−2e−2x)

$$f'(x) = 2xe^{-2x} - 2x^2e^{-2x}$$f′(x)=2xe−2x−2x2e−2x

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Factoring out $2xe^{-2x}$2xe−2x:

$$f'(x) = 2xe^{-2x}(1 - x)$$f′(x)=2xe−2x(1−x)

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For the function to be increasing:

$$2xe^{-2x}(1 - x) > 0$$2xe−2x(1−x)>0

Since $2 > 0$2>0 and $e^{-2x} > 0$e−2x>0 for all real $x$x, the inequality reduces to:

$$x(1 - x) > 0$$x(1−x)>0

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The product $x(1 - x)$x(1−x) is positive when both factors have the same sign.

Both positive:
$x > 0$x>0 and $1 - x > 0$1−x>0

$x > 0$x>0 and $x < 1$x<1

This gives $0 < x < 1$0<x<1

Both negative:
$x < 0$x<0 and $1 - x < 0$1−x<0

$x < 0$x<0 and $x > 1$x>1

This is impossible.

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The function $f(x) = x^2e^{-2x}$f(x)=x2e−2x is increasing on the interval $(0, 1)$(0,1).