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

sin−1⁡(sin⁡θ)−cos−1⁡(sin⁡θ)>0 for θ∈(0,2π)sin−1⁡(sin⁡θ)−π2−sin−1⁡(sin⁡θ)>0⇒sin−1⁡(sin⁡θ)>π412<sin⁡θ≤1⇒π4<θ<3π4a=π4,b=3π4,b−a=π2αx2+βx+sin−1⁡(x−3)2+1+cos−1⁡(x−3)2+1=0is true only for x = 39α+3β+π2+0=0⇒3α+β=−π6….(1)α−β=b−a⇒α−β=π2 ...(2) 4α=π2−π6=π3⇒α=π12

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

sin−1⁡(sin⁡θ)−cos−1⁡(sin⁡θ)>0 for θ∈(0,2π)sin−1⁡(sin⁡θ)−π2−sin−1⁡(sin⁡θ)>0⇒sin−1⁡(sin⁡θ)>π412<sin⁡θ≤1⇒π4<θ<3π4a=π4,b=3π4,b−a=π2αx2+βx+sin−1⁡(x−3)2+1+cos−1⁡(x−3)2+1=0is true only for x = 39α+3β+π2+0=0⇒3α+β=−π6….(1)α−β=b−a⇒α−β=π2 ...(2) 4α=π2−π6=π3⇒α=π12

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 ${αx}^{2} + βx + \sin^{- 1} (x^{2} - 6 x + 10) + \cos^{- 1} (x^{2} - 6 x + 10) = 0 and α - β = b - a, the α$ is equal to :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{π}{48}$

$\frac{π}{16}$

$\frac{π}{12}$

$\frac{π}{8}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \sin^{- 1} (\sin θ) - \cos^{- 1} (\sin θ) > 0 for θ \in (0, 2 π) \\ \sin^{- 1} (\sin θ) - (\frac{π}{2} - \sin^{- 1} (\sin θ)) > 0 \\ \Rightarrow \sin^{- 1} (\sin θ) > \frac{π}{4} \\ \frac{1}{\sqrt{2}} < \sin θ \leq 1 \Rightarrow \frac{π}{4} < θ < \frac{3 π}{4} \\ a = \frac{π}{4}, b = \frac{3 π}{4}, b - a = \frac{π}{2} \end{array}$
${αx}^{2} + βx + \sin^{- 1} ((x - 3)^{2} + 1) + (\cos^{- 1} (x - 3)^{2} + 1) = 0$
is true only for x = 3
$\begin{array}{l} 9 α + 3 β + \frac{π}{2} + 0 = 0 \Rightarrow 3 α + β = \frac{- π}{6} \dots . (1) \\ α - β = b - a \Rightarrow α - β = \frac{π}{2} . . . (2) \end{array}$
$4 α = \frac{π}{2} - \frac{π}{6} = \frac{π}{3} \Rightarrow α = \frac{π}{12}$