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 :

Question

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 : (A) (π/8) (B) (π/48) (C) (π/16) (D) (π/12)

Tardigrade Admin · Accepted Answer

sin -1 sin θ-((π/2)- sin -1 sin θ)>0 ⇒ sin -1 sin θ>(π/4) ⇒ sin θ>(1/√2) text So, θ ∈((π/4), (3 π/4)) θ ∈((π/4), (3 π/4))=(a, b) ⇒ b-a=(π/2)=α-β ⇒ β=α-(π/2) ⇒ α x2+β x+ sin -1[(x-3)2+1]+ cos -1[(x-3)2+1]=0 x=3,9 α+3 β+(π/2)+0=0 ⇒ 9 α+3(α-(π/2))+(π/2)=0 ⇒ 12 α-π=0 α=(π/12)

Q. Let $(a, b) \subset (0, 2 π)$ be the largest interval for which
$sin^{- 1} (sin θ) - cos^{- 1} (sin θ) > 0, θ \in (0, 2 π)$ , holds. If $α x^{2} + β x + sin^{- 1} (x^{2} - 6 x + 10) + cos^{- 1} (x^{2} - 6 x + 10) = 0$ and $α - β = b - a$ , then $α$ is equal to :

$\frac{π}{8}$

$\frac{π}{48}$

$\frac{π}{16}$

$\frac{π}{12}$

Q. Let (a,b)⊂(0,2π) be the largest interval for which sin−1(sinθ)−cos−1(sinθ)>0,θ∈(0,2π), holds. If αx2+βx+sin−1(x2−6x+10)+cos−1(x2−6x+10)=0 and α−β=b−a, then α is equal to :

8π​

48π​

16π​

12π​

Q. Let $(a, b) \subset (0, 2 π)$ be the largest interval for which
$sin^{- 1} (sin θ) - cos^{- 1} (sin θ) > 0, θ \in (0, 2 π)$ , holds. If $α x^{2} + β x + sin^{- 1} (x^{2} - 6 x + 10) + cos^{- 1} (x^{2} - 6 x + 10) = 0$ and $α - β = b - a$ , then $α$ is equal to :

$\frac{π}{8}$

$\frac{π}{48}$

$\frac{π}{16}$

$\frac{π}{12}$