• # question_answer If $f(\theta )=\left| \begin{matrix} 1 & \cos \theta & 1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \\ \end{matrix} \right|$and A and B are respectively the maximum and the minimum values of f(q), then (A, B) is equal to:   [JEE Main Online Paper ( Held On 12 Apirl  2014 ) A) $(3, - 1)$ B) $\left( 4,2-\sqrt{2} \right)$ C) $\left( 2+\sqrt{2},2-\sqrt{2} \right)$     D) $\left( 2+\sqrt{2},-1 \right)$

Let $f(\theta )=\left| \begin{matrix} 1 & \cos \theta & 1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \\ \end{matrix} \right|$ $=(1+\sin \theta \cos \theta )-\cos \theta (-\sin \theta -\cos \theta )+1(-{{\sin }^{2}}\theta +1)$$=1+\sin \theta \cos \theta +\sin \theta \cos \theta +{{\cos }^{2}}\theta -{{\sin }^{2}}\theta +1$ $=2+2\sin \theta \cos \theta +\cos 2\theta$ $=2+\sin 2\theta +\cos 2\theta$                                           ?(1) Now, maximum value of (1) is $2+\sqrt{{{1}^{2}}+{{1}^{2}}}=2+\sqrt{2}$ and minimum value of (1) is $2-\sqrt{{{1}^{2}}+{{1}^{2}}}=2-\sqrt{2}.$