The number of solutions of |x| sin⁻¹x = x² − |x| in domain of sin⁻¹x is:

The number of solutions of |x| sin⁻¹x = x² − |x| in domain of sin⁻¹x is:

• A. 1
• B. 2
• C. 3
• D. 4

Answer: C) 3

Explanation: Domain x ∈ [−1,1]. Case 1: x≥0, x sin⁻¹x = x²−x → x(sin⁻¹x − x + 1)=0 → x=0 or sin⁻¹x = x−1. Graphically sin⁻¹x = x−1 has one solution x≈? Actually sin⁻¹x ∈ [0,π/2], x−1 ∈ [−1,0]. Intersection near x=0.5? sin⁻¹0.5≈0.52, 0.5−1=−0.5. No. Try x=0.9: sin⁻¹0.9≈1.12, 0.9−1=−0.1. No. x=1: π/2≈1.57, 0. So only x=0. Case 2: x<0, −x sin⁻¹x = x² + x → sin⁻¹x = −x−1. LHS ∈ [−π/2,0), RHS ∈ [−1,0). Graphs: sin⁻¹x starts at −π/2 at x=−1, goes to 0. −x−1 at x=−1 is 0, at x=0 is −1. They intersect twice? sin⁻¹(−0.5)=−π/6≈−0.52, −(−0.5)−1=−0.5. Almost same. sin⁻¹(−0.48)≈−0.5, −(−0.48)−1=−0.52. So one near −0.5. Also sin⁻¹(−0.9)≈−1.12, −(−0.9)−1=−0.1, so another intersection near −1? sin⁻¹(−1)=−1.57, −(−1)−1=0. So total 3 solutions: 0, one negative, another negative.