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ExamsJEE AdvancedMaths

Find the range of the function f(x) = sqrt(4 - x²) + sqrt(x² - 1).

  1. [sqrt(3), sqrt(7)]
  2. [sqrt(3), sqrt(6)]
  3. [sqrt(2), sqrt(3)]
  4. [sqrt(3), 3]

Correct answer: [sqrt(3), sqrt(6)]

Solution

The domain is 1 <= |x| <= 2. With t = x² in [1,4], analyse g(t) = sqrt(4-t) + sqrt(t-1). The minimum occurs at the endpoints and the maximum at the symmetric point t = 5/2.

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