JEE Main 2023 · 1 February · morning · Single correct · +4 / −1
Let S be the set of all solutions of the equation cos ⁻¹(2 x)-2 cos ⁻¹(√(1-x²))=π, x ∈[-1/2, 1/2]. Then Σx ∈ S 2 sin ⁻¹(x²-1) is equal to :
- (a) π-2 sin ⁻¹(√3/4)
- (b) π-sin ⁻¹(√3/4)
- (c) (-2 π)/3
- (d) None
Answer and solution
Answer (d) None
Solution
Maths sir · 4 steps
Check the ranges first: the left side can never reach π, so there is nothing to solve.
- Range of each term. cos⁻¹ always gives a value in [0, π]. So cos⁻¹(2x) ≤ π, and cos⁻¹(√(1-x²)) ≥ 0.
cos⁻¹(2x) - 2 cos⁻¹(√(1-x²)) ≤ π - 0 = π
- When is it equal. The left side equals π only if both extremes happen together. cos⁻¹(2x) = π needs 2x = -1, so x = -1/2. cos⁻¹(√(1-x²)) = 0 needs √(1-x²) = 1, so x = 0.
- No common x. One value of x cannot be both -1/2 and 0. So no x satisfies the equation, and S is the empty set.
- So. A sum over an empty set is 0. None of the values in (a), (b) or (c) equals 0, so the answer is (d).
The idea to keep. Before solving an inverse-trig equation, check the range of each term. If the extreme value needs two different values of x at the same time, the equation has no solution.
Answer from the NTA final answer key. Solution by PrepWiser.