Question

The inequality log₂x < sin^-1(sin5) hold if

(A) x ∈(0, 2^{5 - 2π})

(B) x ∈(2^{5 - 2π}, ∞)

(C) x ∈(2^{2π - 5}, ∞)

(D) None of these

Answer 1

Correct option (A) x ∈(0, 2^{5 - 2π})

Explanation:

sin^-1(sin5) = sin^-1[sin(-2π + 5)] log₂x < sin^-1(sin5)

⇒ log₂x < sin^-1[sin(-2π + 5)]

⇒ log₂x < - 2π + 5 = 5 - 2π

⇒ x < 2^5-2π

Also, x ≠ 0 is positive.

Therefore, required value of x belongs to x ∈(0, 2^{5 - 2π}).