MCQ
If $x$ takes non-positive permissible value, then ${\sin ^{ - 1}}x =$
- A${\cos ^{ - 1}}\sqrt {1 - {x^2}} $
- ✓$ - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $
- C${\cos ^{ - 1}}\sqrt {{x^2} - 1} $
- D$\pi - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $
Since $ - 1 \le x \le 0,$ therefore $\frac{{ - \pi }}{2} \le {\sin ^{ - 1}}x \le 0$
and so $\frac{{ - \pi }}{2} \le y \le 0$
We have $\cos y = \sqrt {1 - {{\sin }^2}y} $
$ \Rightarrow \,\,\cos y = \sqrt {1 - {x^2}} $, for $0 \le y \le \pi $ …..$(i)$
Now $ - \frac{\pi }{2} \le y \le 0\,\, \Rightarrow \,\,\frac{\pi }{2} \ge - y \ge 0$
$ \Rightarrow \,\,\cos \,\left( { - y} \right) = \sqrt {1 - {x^2}} $ {from $(i)$}
$ \Rightarrow \,\, - y = {\cos ^{ - 1}}\sqrt {1 - {x^2}} \,\, $
$\Rightarrow \,\,y = - {\cos ^{ - 1}}\sqrt {1 - {x^2}} $.
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$g(\theta)=\sqrt{f(\theta)-1}+\sqrt{f\left(\frac{\pi}{2}-\theta\right)-1}$
where
$f(\theta)=\frac{1}{2}\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right|+\left|\begin{array}{ccc}\sin \pi & \cos \left(\theta+\frac{\pi}{4}\right) & \tan \left(\theta-\frac{\pi}{4}\right) \\ \sin \left(\theta-\frac{\pi}{4}\right) & -\cos \frac{\pi}{2} & \log _e\left(\frac{4}{\pi}\right) \\ \cot \left(\theta+\frac{\pi}{4}\right) & \log _e\left(\frac{\pi}{4}\right) & \tan \pi\end{array}\right|$.
Let $p (x)$ be a quadratic polynomial whose roots are the maximum and minimum values of the function $g(\theta)$, and $p(2)=2-\sqrt{2}$. Then, which of the following is/are TRUE ?
$(A)$ $p \left(\frac{3+\sqrt{2}}{4}\right)<0$
$(B)$ $p \left(\frac{1+3 \sqrt{2}}{4}\right)>0$
$(C)$ $p \left(\frac{5 \sqrt{2}-1}{4}\right)>0$
$(D)$ $p \left(\frac{5-\sqrt{2}}{4}\right)<0$