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Mathematics Question on Application of derivatives

The function f(x)=x2  e−2x,x>0f(x) = x^2 \; e^{-2} x, x > 0. Then the maximum value of f(x)f(x) is

A

1e\frac{1}{e}

B

12e\frac{1}{ 2 e}

C

1e2\frac{1}{e^2}

D

4e4\frac{4}{e^4}

Answer

1e2\frac{1}{e^2}

Explanation

Solution

Given : f(x)=x2  e−2x,x>0f (x) = x^2 \; e^{-2x}, x > 0
f′(x)=x2.e−2x(−2)+e−2x.2xf '(x) = x^2.e^{-2x} (-2) + e^{-2x} .2x
put f′(x)=0  ⇒  2e−2x.x(−x+1)=0f '(x) = 0 \; \Rightarrow \; 2e^{-2x}. x (-x + 1) = 0
⇒\Rightarrow x = 1 or x = 0
f"(x)=(−4x2−6x+1)e−2xf"(x) = (-4x^2 - 6x + 1)e^{-2x}
f"(1)=−9e−2x<0f"(1) = -9e^{-2x} < 0
f"(0)=e−2x>0f"(0) = e^{-2x} > 0
∴\therefore value of f(x) is maximum at x = 1
∵f(x)=x2.e−2x⇒f(1)=e−2=1e2\because f\left(x\right)=x^{2}.e^{-2x}\Rightarrow f\left(1\right) =e^{-2} = \frac{1}{e^{2}}