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MHT CET202616 April 2026Morning ShiftMathematicsApplication of DerivativesActual

The minimum value of x x in the interval (2, ) is

Options

  1. A0
  2. Be
  3. C1 e
  4. DDoes not exist

Correct answer

D. Does not exist

Step-by-step solution

Let f(x) = x x for x (2, ) . Differentiating with respect to x , we get f'(x) = 1 - x x^2 . Setting f'(x) = 0 gives x = 1 x = e . For x (2, e) , f'(x) > 0 , so f(x) is strictly increasing. For x (e, ) , f'(x) Thus, f(x) attains its maximum value at x = e . To find the minimum value, we check the behavior at the boundaries of the interval. At x = 2 , f(2) = 2 2 > 0 . As x , _ x x x = 0 . Since f(x) > 0 for all x (2, ) and f(x) 0 as x , the infimum of f(x) is 0 . However, there is no finite x (2, ) for which f(x) = 0

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