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JEE Main202126 Aug 2021Evening ShiftMathematicsApplication of DerivativesActual

The local maximum value of the function, f x = 2 x x 2 , x > 0 , is

Options

  1. A1
  2. B4 e e 4
  3. C( e ) 2 e
  4. D( 2 e ) 1 e

Correct answer

C. ( e ) 2 e

Step-by-step solution

f ' ( x ) = 0 for maximum value Let y = 2 x x 2 ln y = x 2 ln 2 x 1 y y ' = 2 x ln 2 x + x 2 1 2 x × - 2 x 2 y ' = ( x y ) 2 ln 2 x - 1 y ' = 2 x x 2 x 2 ln 2 x - 1 2 ln 2 x = 1 2 x = e 1 2 x = 2 e - 1 2 Then maximum value will be f 2 e - 1 2 = 2 2 e - 1 2 4 e - 1 = e 2 e - 1 = e 2 e

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