If the function f ( x ) = ( x 1 ) 2 x ; x > 0 attains the maximum value at x = e 1 then :

Let y = ( x 1 ) 2 x ℓ ny = 2 x ℓ n ( x 1 ) ℓ ny = − 2 x ℓ ln x y 1 d x d y = − 2 ( 1 + ℓ n x ) for x > e 1 f n is decreasing so, e ( π 1 ) 2 π e π > π e

Home
PYQ PracticeFree
JEE Rank Predictor
Answer Key

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minPYQ_2024

MathematicsMediumsingle choice

If the function $f (x) = (\frac{1}{x})^{2 x}; x > 0$ attains the maximum value at $x = \frac{1}{e}$ then :

Options:

Answer:

Solution:

Let $y = (\frac{1}{x})^{2 x}$ $ℓ ny = 2 x ℓ n (\frac{1}{x}) ℓ ny = - 2 x ℓ ln x \frac{1}{y} \frac{d y}{d x} = - 2 (1 + ℓ n x)$ for $x > \frac{1}{e} f^{n}$ is decreasing $so, e < π$ $(\frac{1}{e})^{2 e} > (\frac{1}{π})^{2 π} e^{π} > π^{e}$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

⏱ 2mℹ️ Source: PYQ_2024

Doubts & Discussion

Loading discussions...