Let (a, b) ⊂(0,2 π) be the largest interval for which sin ⁻¹(sin θ)-cos ⁻¹(sin θ)>0, θ ∈(0,2 π), holds.
If α x²+β x+sin ⁻¹(x²-6 x+10)+cos ⁻¹(x²-6 x+10)=0 and α-β=b-a, then α is equal to :

Solution

Maths sir · 4 steps

Use sin⁻¹ t + cos⁻¹ t = π/2 twice: once for the interval, once for the equation.

1. Find the interval. Put t = sin θ and replace cos⁻¹ t with π/2 - sin⁻¹ t. The inequality becomes 2 sin⁻¹ t > π/2, so sin⁻¹ t > π/4, which means sin θ > 1/√2.
2. Interval length. In (0, 2π), sin θ > 1/√2 exactly for θ ∈ (π/4, 3π/4). So b - a = π/2, which gives α - β = π/2.
3. Pin down x. The argument x² - 6x + 10 = (x-3)² + 1 is at least 1, but sin⁻¹ and cos⁻¹ need it to be at most 1. So it must equal 1, which forces x = 3, and the two inverse terms add to π/2.

   9α + 3β + π/2 = 0, β = α - π/2

4. So. Substituting, 9α + 3α - 3π/2 + π/2 = 0, so 12α = π and α = π/12, option (d).

The idea to keep. When sin⁻¹ and cos⁻¹ of the same thing appear together, replace their sum by π/2. Always check the domain [-1, 1]: it can pin x to one value.

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