Question

If x > 0 and $\left|\begin{array}{ccc}x& sin\theta & cos\theta \\ -sin\theta & x& 1\\ cos\theta & 1& x\end{array}\right|=0$, then

• A. $x<\sqrt{2}$
• B. $x=\sqrt{2}$
• C. $x>\sqrt{2}$
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $x<\sqrt{2}$
B $x=\sqrt{2}$

$\left|\begin{array}{ccc}x& \sin \theta & \cos \theta \\ -\sin \theta & x& 1\\ \cos \theta & 1& x\end{array}\right|=0$

$\Rightarrow x(x^2-1)-\sin \theta (-x\sin \theta -\cos \theta )+\cos \theta (-\sin \theta -x\cos \theta )=0$

$\Rightarrow x^3-x+x{\sin }^{2}x+\sin \theta \cos \theta -\sin \theta \cos \theta -x{\cos }^2\theta =0$

$\Rightarrow x^3-x+x({\sin }^2\theta -{\cos }^2\theta )=0$

$\Rightarrow x^3-x(1+\cos 2\theta =0)$ $\Rightarrow x(x^2-1-\cos 2\theta )=0$

but $x>0\Rightarrow x^2=1+\cos 2\theta$

$\Rightarrow x=\sqrt{1+\cos 2\theta }$

$-1\le \cos 2\theta \le 1$

$0\le 1+\cos 2\theta \le 2$

$0\le \sqrt{1+\cos 2\theta }\le \sqrt{2}$

$0<x\le \sqrt{2}$ as $x>0$

Hence $Opt:A,\left[B\right]$ are the correct answers