Question

Let S₁ = `{x ∈ R - {1, 2}: ((x + 2)(x^2 + 3x + 5))/(-2 + 3x - x^2) ≥ 0}` and S₂ = {x ∈ R : 3^2x – 3^x+1 – 3^x+2 + 27 ≤ 0}. Then, S₁ ∪ S₂ is equal to ______.

पर्याय

• (–∞, –2] ∪ (1, 2)

• (–∞, –2] ∪ [1, 2]

• (–2, 1] ∪ (2, ∞)

• (–∞, 2]

Answer 1

Let S₁ = `{x ∈ R - {1, 2}: ((x + 2)(x^2 + 3x + 5))/(-2 + 3x - x^2) ≥ 0}` and S₂ = {x ∈ R : 3^2x – 3^x+1 – 3^x+2 + 27 ≤ 0}. Then, S₁ ∪ S₂ is equal to (–∞, –2] ∪ [1, 2].

Explanation:

For, S₁ we have

⇒ `((x + 2)(x^2 + 3x + 5))/(x^2 -  3x + 2) ≤ 0`

⇒ x ∈ (–∞, 2] ∪ (1, 2)

For, S₂ we have

= 3^x(3^x – 3) – 3²(3^x – 3) ≤ 0

For, S₂, x ∈ [1, 2]

⇒ (–∞, –2] ∪ [1, 2]