Question

If x and α are real, then the inequation log₂^x + log²_x + 2cos α ≤ 0

(A) Has no solution

(B) Has exactly two solutions

(C) Is satisfied for any real α and any real x in (0, 1)

(D) Is satisfied for any real α and any real x in (1, ∞)

Answer 1

Correct option  (C) Is satisfied for any real α and any real x in (0, 1)

The equation has meaning if x > 0, x ≠ 1

Hence, domain = (o,1) ∪ (1,∞)

If x ∈(0, 1) then log₂x < 0 and log₂x + log_x2 = log x/log 2 +  log/log x

= sum of a negative number ≤ -2.

In this case any α will satisfy since 2cos α can never be more than 2. 

 Thus, the inequation is satisfied for any x in (0, 1) and for any α. If ∈ (1, ∞) then log₂x > 0. So

The inequation cannot be satisfied unless

cosα = −1 and x = 2, that is, log₂x = 1

Option (D) is wrong since in the last case there are infinite solutions.