Question

The domain of the function `cos^-1((2sin^-1(1/(4x^2-1)))/π)` is ______.

Options

• `R - {-1/2, 1/2}`

• (–∞, –1] ∪ [1, ∞) ∪ {0}

• `(-∞, (-1)/2) ∪ (1/2, ∞) ∪ {0}`

• `(-∞, (-1)/sqrt(2)) ∪ (1/sqrt(2), ∞) ∪ {0}`

Answer 1

The domain of the function `cos^-1((2sin^-1(1/(4x^2-1)))/π)` is `underlinebb((-∞, (-1)/sqrt(2)) ∪ (1/sqrt(2), ∞) ∪ {0})`.

Explanation:

Let f(x) = `cos^-1((2sin^-1(1/(4x^2-1)))/π)`

As we know the domain of cos^–1y is [–1, 1]

⇒ `-1 ≤ ((2sin^-1(1/(4x^2 - 1)))/π) ≤ 1`

⇒ `-π/2 ≤ sin^-1(1/(4x^2 - 1)) ≤ π/2`

⇒ `-1 ≤  1/(4x^2 - 1) ≤ 1`

So, `1/(4x^2 - 1) ≥ -1 and 1/(4x^2 - 1) ≤ 1`

⇒ `1/(4x^2 - 1) + 1 ≥ 0 and 1/(4x^2 - 1) -1 ≤ 0`

⇒ `(4x^2)/(4x^2 - 1) ≥ 0 and (2 - 4x^2)/(4x^2 - 1) ≤ 0` 

⇒ `x^2/((2x - 1)(2x + 1)) ≥ 0 `

and `((1 - sqrt(2)x)(1 + sqrt(2)x))/((2x - 1)(2x + 1)) ≤ 0`

⇒ `x ∈(-∞, -1/sqrt(2)) ∪ (1/sqrt(2), ∞) ∪ {0}`

and `x ∈(-∞, 1/sqrt(2)) ∪ (-1/2, 1/2) ∪ (1/sqrt(2), ∞)`

⇒ `x ∈ (-∞, -1/sqrt(2)) ∪ (1/sqrt(2), ∞) ∪ {0}`