Verified Solution • 2025–26 Session moderate

QJEEM23Ma3403: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3403 Solution: Let (a, b) be the largest interval for which , holds. If and ,…

Let (a, b) ⊂(0,2π)\subset(0,2 \pi) be the largest interval for which \sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)>0, \theta \in(0,2 \pi) , holds.

If αx2+βx+sin⁡−1(x2−6x+10)+cos⁡−1(x2−6x+10)=0\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0 and α−β=b−a\alpha-\beta=b-a , then α\alpha is equal to :

Multiple Choice Options

  • A π16\frac{\pi}{16}
  • B π48\frac{\pi}{48}
  • C π8\frac{\pi}{8}
  • D π12\frac{\pi}{12}

Step-by-Step Verified Solution

\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right)>0

\Rightarrow \sin ^{-1} \sin \theta>\frac{\pi}{4}

\Rightarrow \sin \theta>\frac{1}{\sqrt{2}}

So, θ∈(π4,3π4)\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)

θ∈(π4,3π4)=(a,b)\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=(\mathrm{a}, \mathrm{b})

b−a=π2=α−βb-a=\frac{\pi}{2}=\alpha-\beta

⇒β=α−π2\Rightarrow \beta=\alpha-\frac{\pi}{2}

⇒αx2+βx+sin⁡−1[(x−3)2+1]+cos⁡−1[(x−3)2+1]=0\Rightarrow \alpha x^{2}+\beta \mathrm{x}+\sin ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]+\cos ^{-1}\left[(\mathrm{x}-3)^{2}+1\right]=0

x=3,9α+3β+π2+0=0x=3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0

amp;⇒9α+3(α−π2)+π2=0amp;⇒12α−π=0amp;α=π12\begin{aligned}& \Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0 \\\\& \Rightarrow 12 \alpha-\pi=0 \\\\& \alpha=\frac{\pi}{12}\end{aligned}

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