The domain of the function cos ⁻¹( (2sin ⁻¹( 1/(4x² - 1) ))/π ) is :

Solution

Maths sir · 4 steps

The outer cos⁻¹ is automatic here, so just make 1/(4x²-1) lie in [-1, 1].

1. Outer function. cos⁻¹ needs its input in [-1, 1]. Since sin⁻¹ t always lies in [-π/2, π/2], the quantity (2 sin⁻¹ t)/π always lies in [-1, 1]. So the outer function adds no new condition.
2. Inner function. sin⁻¹ needs -1 ≤ 1/(4x²-1) ≤ 1, with x ≠ ±1/2. Put u = 4x² - 1. Then |1/u| ≤ 1 means |u| ≥ 1.

   |4x² - 1| ≥ 1 ⇔ 4x² - 1 ≥ 1 or 4x² - 1 ≤ -1

3. Solve both cases. First case: 4x² ≥ 2, so x² ≥ 1/2, which gives |x| ≥ 1/√2. Second case: 4x² ≤ 0, so x = 0. Check x = 0: then u = -1 and 1/u = -1, which is allowed.
4. So. The domain is (-∞, -1/√2] ∪ [1/√2, ∞) ∪ {0}, which is option (d).

The idea to keep. For nested inverse trig functions, check whether the outer range condition is automatic first. For |1/u| ≤ 1, flip it to |u| ≥ 1 and solve both sides.

Answer from the NTA final answer key. Solution by PrepWiser.