Let S 1 is the complete solution set of the inequality c o s - 1 x c o s - 1 x 2 and S 2 is the complete solution set of the inequality c o t - 1 x 2 - 5 c o t - 1 x + 6 0 , then S 1 ∩ S 2 is

Question

Let S 1 is the complete solution set of the inequality c o s - 1 x > c o s - 1 x 2 and S 2 is the complete solution set of the inequality c o t - 1 x 2 - 5 c o t - 1 x + 6 > 0 , then S 1 ∩ S 2 is, - 1,0, c o t 3,0, c o t 2 , 0, - 1 , c o t 2

MARKS App · Accepted Answer

c o s - 1 x > c o s - 1 x 2 ⇒ x x 2 & x ∈ - 1 , 1 { ∵ c o s - 1 x is a decreasing function} ⇒ x x – 1 > 0 & x ∈ - 1,1 ⇒ x ∈ - ∞ , 0 ∪ 1 , ∞ & x ∈ - 1 , 1 ⇒ x ∈ - 1 , 0 … 1 Now c o t - 1 x 2 - 5 c o t - 1 x + 6 > 0 ⇒ c o t - 1 x - 2 c o t - 1 x - 3 > 0 ⇒ c o t - 1 x 2 or x c o t 3 … 2 { ∵ c o t - 1 x is a decreasing function} 1 ∩ 2 ⇒ c o t 2 , 0

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