If S={x ∈ ℝ: sin ⁻¹((x+1)/√(x²+2 x+2))-sin ⁻¹(x/√(x²+1))=π/4}, then Σx ∈ s(sin ((x²+x+5) π/2)-cos ((x²+x+5) π)) is equal to ____________.

Solution

Maths sir · 4 steps

Each sin⁻¹ here is really a tan⁻¹ in disguise, and then the equation collapses.

1. Rewrite as tan inverse. If tan θ = t with θ ∈ (-π/2, π/2), then sin θ = t/√(t²+1), so sin⁻¹ t/√(t²+1) = tan⁻¹ t. Since x²+2x+2 = (x+1)²+1, use this once with t = x+1 and once with t = x.

   tan⁻¹(x+1) - tan⁻¹ x = π/4

2. Take tan. Because x+1 > x, the left side lies in (0, π), and there tan equals 1 only at π/4. So ((x+1)-x)/(1+x(x+1)) = 1, which gives x²+x = 0, that is x = 0 or x = -1. Both work: tan⁻¹ 1 - tan⁻¹ 0 = π/4 and tan⁻¹ 0 - tan⁻¹(-1) = π/4.
3. Evaluate each term. For both roots x²+x = 0, so x²+x+5 = 5. Each term is sin 5π/2 - cos 5π = 1 - (-1) = 2.
4. So. S = {0, -1} has two elements, each contributing 2, so the sum is 2 + 2 = 4.

The common slip: 2. Keeping only the root x = 0 and dropping x = -1 gives 2, not 4. Both roots satisfy the equation, and each adds 2.

The idea to keep. When you see t/√(t²+1) inside sin⁻¹, read it as tan⁻¹ t. And if the final expression contains x²+x, reuse the equation you solved instead of substituting roots one by one.

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