Question

Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:

पर्याय

• |A| = 0

• |A| ∈ (2, ∞)

• |A| ∈ (2, 4)

• |A| ∈ [2, 4]

Answer 1

|A| ∈ [2, 4]

Explanation:

|A| = 2 + 2 sin² α

As −1 ≤ sin α ≤ 1, ∀ 0 ≤ α ≤ 2π

⇒ 2 ≤ 2 + 2  sin² α ≤ 4

⇒ |A| ∈ [2, 4]