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The function f(x)=x2e−2xf(x) = x^2e^{-2x} increases on

Solution

✅ Correct Option: 2

A function increases where its derivative is positive.

Given f(x)=x2e−2xf(x) = x^2e^{-2x}

Using the product rule with u=x2u = x^2 and v=e−2xv = e^{-2x}:

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

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


Factoring out 2xe−2x2xe^{-2x}:

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


For the function to be increasing:

2xe−2x(1−x)>02xe^{-2x}(1 - x) > 0

Since 2>02 > 0 and e−2x>0e^{-2x} > 0 for all real xx, the inequality reduces to:

x(1−x)>0x(1 - x) > 0


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

Both positive:

x>0x > 0 and 1−x>01 - x > 0

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

This gives 0<x<10 < x < 1

Both negative:

x<0x < 0 and 1−x<01 - x < 0

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

This is impossible.


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

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