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STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 2. inverse trigonometric function1 Mark
MCQ
Let $( a , b ) \subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta) > 0, \theta \in(0,2 \pi)$ holds. If $\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}$ $\left(x^2-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to:
  • A
    $\frac{\pi}{48}$
  • B
    $\frac{\pi}{16}$
  • C
    $\frac{\pi}{8}$
  • ✓
    $\frac{\pi}{12}$
✓

Answer

Correct option: D.
$\frac{\pi}{12}$
d
$\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right) > 0$

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

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

$\text { So, } \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$

$\theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)=( a , b )$

$b - a =\frac{\pi}{2}=\alpha-\beta$

$\Rightarrow \beta=\alpha-\frac{\pi}{2}$

$\Rightarrow \alpha x^2+\beta x+\sin ^{-1}\left[(x-3)^2+1\right]+\cos ^{-1}\left[(x-3)^2+1\right]=0$

$x =3,9 \alpha+3 \beta+\frac{\pi}{2}+0=0$

$\Rightarrow 9 \alpha+3\left(\alpha-\frac{\pi}{2}\right)+\frac{\pi}{2}=0$

$\Rightarrow 12 \alpha-\pi=0$

$\alpha=\frac{\pi}{12}$

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