99 Percentile Qs Bank for JEE MainMathematicsInverse Trigonometric Functions

If s i n - 1 x 2 - 2 s i n - 1 x + 1 ≤ 0 (where, . represents the greatest integral part of x ), then

Options

Ax ∈ sin ⁡ 1 , sin ⁡ 2
Bx ∈ - sin ⁡ 1 , sin ⁡ 1
Cx ∈ sin ⁡ 1 , 1
Dx ∈ - sin ⁡ 1 , sin ⁡ 2

Correct answer

C. x ∈ sin ⁡ 1 , 1

Step-by-step solution

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

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