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MHT CET202615 April 2026Evening ShiftMathematicsApplication of DerivativesActual

The maximum value of ( 1 x )^x , x > 0 is

Options

  1. Ae^e
  2. Be^ -e
  3. Ce^ 1/e
  4. D( 1 e )^ 1/e

Correct answer

C. e^ 1/e

Step-by-step solution

Let y = ( 1 x )^x = x^ -x Taking natural logarithm on both sides: y = -x x Differentiating with respect to x : 1 y dy dx = - ( x + x 1 x ) = -( x + 1) dy dx = -x^ -x ( x + 1) For maximum or minimum value, dy dx = 0 -x^ -x ( x + 1) = 0 Since x^ -x 0 for x > 0 , we have x + 1 = 0 x = 1 e For 0 0 and for x > 1 e , dy dx Thus, y attains its maximum value at x = 1 e The maximum value is y_ max = ( 1 1/e )^ 1/e = e^ 1/e Answer: e^ 1/e

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