CUET UG 2025 — Mathematics Calculus

A function is increasing on an interval when the derivative is positive: $f'(x) > 0$f′(x)>0

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Given $f(x) = x^2e^{-x}$f(x)=x2e−x, apply the Product Rule with $u = x^2$u=x2 and $v = e^{-x}$v=e−x:

$$u' = 2x$$u′=2x

$$v' = -e^{-x}$$v′=−e−x

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

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

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

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

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For $f'(x) > 0$f′(x)>0:

$$e^{-x} \cdot x(2 - x) > 0$$e−x⋅x(2−x)>0

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

$$x(2 - x) > 0$$x(2−x)>0

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The critical points occur when $x = 0$x=0 or $2 - x = 0$2−x=0, giving $x = 0$x=0 and $x = 2$x=2.

Testing the sign of $x(2-x)$x(2−x) in each interval:

For $x < 0$x<0: $x$x is negative, $(2-x)$(2−x) is positive, so $x(2-x) < 0$x(2−x)<0

For $0 < x < 2$0<x<2: $x$x is positive, $(2-x)$(2−x) is positive, so $x(2-x) > 0$x(2−x)>0

For $x > 2$x>2: $x$x is positive, $(2-x)$(2−x) is negative, so $x(2-x) < 0$x(2−x)<0

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Therefore $f'(x) > 0$f′(x)>0 when $0 < x < 2$0<x<2.

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