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If s i n - 1 x 2 + s i n - 1 x - 2 ≤ 0 (where, [ · ] represents the greatest integral part of x ), then the maximum value of x is

Correct answer

1

Step-by-step solution

s i n - 1 x - 1 s i n - 1 x + 2 ≤ 0 ⇒ s i n - 1 x ∈ - 2,1 ⇒ s i n - 1 x ∈ - 2,2 ⇒ - 2 ≤ s i n - 1 x < 2 but s i n, - 1 x ∈ - π 2 ≤ π 2 ∴ - - π 2 ≤ s i n - 1 x ≤ π 2 - s i n π 2 ≤ x ≤ s i n π 2

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