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ExamsJEE AdvancedMaths

Find the complete solution set of the inequality arcsin(sin 5) > x² - 4x.

  1. x in (2 - sqrt(9 - 2pi), 2 + sqrt(9 - 2pi))
  2. x in (2 - sqrt(9 + 2pi), 2 + sqrt(9 + 2pi))
  3. x in (2 - sqrt(13 - 2pi), 2 + sqrt(13 - 2pi))
  4. x in [2 - sqrt(9 - 2pi), 2 + sqrt(9 - 2pi)]

Correct answer: x in (2 - sqrt(9 - 2pi), 2 + sqrt(9 - 2pi))

Solution

5 rad lies in (3pi/2, 2pi) approx (4.712, 6.283), so arcsin(sin 5) = 5 - 2pi (which is about -1.283, within [-pi/2, pi/2]). Inequality: 5 - 2pi > x² - 4x => x² - 4x - (5 - 2pi) < 0 => x² - 4x + (2pi - 5) < 0. Roots: x = [4 +/- sqrt(16 - 4(2pi - 5))]/2 = 2 +/- sqrt(4 - (2pi - 5)) = 2 +/- sqrt(9 - 2pi). So x in (2 - sqrt(9 - 2pi), 2 + sqrt(9 - 2pi)).

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