JEE Main 2021 · 26 August · evening · Single correct · +4 / −1
The domain of the function cosec ⁻¹( (1 + x)/x ) is :
- (a) ( - 1, - 1/2 ] ∪ (0,∞ )
- (b) [ - 1/2,0 ) ∪ [1,∞ )
- (c) ( - 1/2,∞ ) - { 0}
- (d) [ - 1/2,∞ ) - { 0}
Answer and solution
Answer (d) [ - 1/2,∞ ) - { 0}
Solution
Maths sir · 4 steps
cosec⁻¹ needs |y| ≥ 1, so square both sides and the inequality turns linear.
- Condition to impose. cosec⁻¹ y is defined only when |y| ≥ 1. Here y = (1+x)/x, so we also need x ≠ 0.
|(1+x)/x| ≥ 1 ⇔ |1+x| ≥ |x| ⇔ (1+x)² ≥ x²
- Solve it. Both sides of |1+x| ≥ |x| are non-negative, so squaring is safe. Expanding gives 1 + 2x ≥ 0, that is, x ≥ -1/2.
- Check the endpoint. At x = -1/2, y = (1/2)/(-1/2) = -1, and cosec⁻¹(-1) = -π/2 exists. So -1/2 is included, and only x = 0 is removed.
- So. The domain is [-1/2, ∞) - {0}, which is option (d).
The idea to keep. For sec⁻¹ or cosec⁻¹ of a fraction, write |y| ≥ 1 as |num| ≥ |den| and square. Then test the endpoints, because y = ± 1 is allowed.
Answer from the NTA final answer key. Solution by PrepWiser.