The number of real solutions of is
- A.zero
- B.one
- C.two
- D.infinite
Show answer and solution
Correct option: C
Solution
For the expressions to be defined, x(x+1) ≥ 0, so x ≤ -1 or x ≥ 0, and x²+x+1 must lie in [0,1] for sin⁻¹ to be defined. Since x²+x+1 = (x+1/2)²+3/4 ≥ 3/4 > 0 always, we need x²+x+1 ≤ 1, i.e., x²+x ≤ 0, giving x ∈ [-1, 0]. Combined with x ≤ -1 or x ≥ 0, the only valid points are x = 0 and x = -1. At both these values, x(x+1) = 0 and x²+x+1 = 1, so the equation becomes tan⁻¹(0) + sin⁻¹(1) = 0 + π/2 = π/2 ✓. Thus there are exactly two real solutions.
[1999]