Let f x = sin ⁡ π 6 sin ⁡ π 2 sin ⁡ x , for all x ∈ R and g x = π 2 sin ⁡ x , for all x ∈ R . Let f o g x denote f g x and g o f x denote g f x . Then which of the following is (are) true?

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TestHub · Accepted Answer

f x = sin ⁡ π 6 sin ⁡ π 2 sin ⁡ x - π 2 ≤ π 2 sin ⁡ x ≤ π 2 - 1 ≤ sin ⁡ π 2 sin ⁡ x ≤ 1 sin ⁡ - π 6 ≤ sin ⁡ π 6 sin ⁡ π 2 sin ⁡ x ≤ sin ⁡ π 6 - 1 2 ≤ sin ⁡ π 6 sin ⁡ π 2 sin ⁡ x ≤ 1 2 f g x = sin ⁡ π 6 sin ⁡ π 2 sin ⁡ π 2 sin ⁡ x - 1 ≤ sin ⁡ π 2 sin ⁡ x ≤ 1 - π 2 ≤ π 2 sin ⁡ π 2 sin ⁡ x ≤ π 2 - 1 ≤ sin ⁡ π 2 sin ⁡ π 2 sin ⁡ x ≤ 1 - π 6 ≤ π 6 sin ⁡ π 2 sin ⁡ π 2 sin ⁡ x ≤ π 6 - 1 2 ≤ sin ⁡ π 6 sin ⁡ π 2 sin ⁡ π 2 sin ⁡ x ≤ 1 2 lim x → 0 f x g x = lim x → 0 ⁡ sin ⁡ π 6 sin π 2 sin x π 2 sin ⁡ x × π 6 sin ⁡ π 2 sin ⁡ x π 6 sin ⁡ π 2 sin ⁡ x = lim x → 0 ⁡ π 6 sin ⁡ π 2 sin ⁡ x π 2 sin ⁡ x = π 6 . Range of g f x is π 2 sin ⁡ - 1 2 , π 2 sin ⁡ 1 2 = - π 2 sin ⁡ 1 2 , π 2 sin ⁡ 1 2 Hence, 1 , does not belong to this range.

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Doubts & Discussion

JEE Advanced (2027)