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For which values of x does the inequality sin⁻¹(sin 5) > x² − 4x hold?

1. x = 2 − √(9 − 2π)
2. x = 2 + √(9 − 2π)
3. x > 2 + √(9 − 2π)
4. x ∈ (2 − √(9 − 2π), 2 + √(9 − 2π))

Correct answer: x ∈ (2 − √(9 − 2π), 2 + √(9 − 2π))

Solution

Since 5 lies in (3π/2, 2π), sin^-1(sin5)=5-2π. The inequality x^2-4x < 5-2π becomes x^2-4x-(5-2π)<0, i.e. x^2-4x-(9-2π)+4<0, whose roots are 2 ± sqrt(9-2π). So x ∈ (2 - sqrt(9-2π), 2 + sqrt(9-2π)).

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