JEE Main 2023 — Mathematics Trigonometry

Given, ${\mathrm{sin}}^{-1}(\mathrm{sin}\theta )-{\mathrm{cos}}^{-1}(\mathrm{sin}\theta )>0,\theta \in (0,2\pi )$sin−1(sinθ)−cos−1(sinθ)>0,θ∈(0,2π)

$$\because {\mathrm{sin}}^{-1}(\theta )+{\mathrm{cos}}^{-1}(\theta )=\frac{\pi }{2}$$∵sin−1(θ)+cos−1(θ)=2π​

$$\therefore {\mathrm{sin}}^{-1}(\mathrm{sin}\theta )-(\frac{\pi }{2}-{\mathrm{sin}}^{-1}(\mathrm{sin}\theta ))>0$$∴sin−1(sinθ)−(2π​−sin−1(sinθ))>0

$${\mathrm{sin}}^{-1}(\mathrm{sin}\theta )>\frac{\pi }{4}\Rightarrow \theta \in (\frac{\pi }{4},\frac{3\pi }{4})$$sin−1(sinθ)>4π​⇒θ∈(4π​,43π​)

$$(a,b)=(\frac{\pi }{4},\frac{3\pi }{4})\Rightarrow b-a=\frac{\pi }{2}$$(a,b)=(4π​,43π​)⇒b−a=2π​

Given $b-a=\alpha -\beta$b−a=α−β

$$\therefore \alpha -\beta =\frac{\pi }{2}.....(1)$$∴α−β=2π​.....(1)

Now $\alpha {x}^{2}+\beta x+{\mathrm{sin}}^{-1}({x}^{2}-6x+10)+{\mathrm{cos}}^{-1}({x}^{2}-6x+10)=0$αx2+βx+sin−1(x2−6x+10)+cos−1(x2−6x+10)=0

Now defining ${x}^{2}-6x+10=1+{(x-3)}^{2}\geq 1$x2−6x+10=1+(x−3)2≥1

Hence $x=3$x=3 is the only possible solution

$$9\alpha +3\beta +\frac{\pi }{2}=0\ldots (2)$$9α+3β+2π​=0…(2)

On solving equations $(1)$(1) and $(2)$(2) we get,

$$\alpha =\frac{\pi }{12}$$α=12π​