JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Analyzing +4 | -1

The function

f\left( x \right) = {x^2}{e^{ - 2x}},\;x > 0

Then, the maximum value of

f\left( x \right)

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it represents the maximum value of the function

f(x) = x^2e^{-2x}

for

x > 0

, found by analyzing its first and second derivatives.

More Questions from MAXIMA & MINIMA

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f(x)=\left(1+b^2\right) x^2+2 b x+1

m(b)

f(x)

b

m(b)

f: R \rightarrow R

f(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N

|2 a x-3|+|a x+1|+|5-a x|=\frac{1}{2}

f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p-9) x

x<0

x>0

p

f\left( x \right)

f\left( x \right) = {\tan ^{ - 1}}x - \frac{1}{2}{\log _e}x

\left[ {1/\sqrt 3 ,\;\sqrt 3 } \right]

f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right),

\frac{k}{e}

\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}