A) \[(3, - 1)\]
B) \[\left( 4,2-\sqrt{2} \right)\]
C) \[\left( 2+\sqrt{2},2-\sqrt{2} \right)\]
D) \[\left( 2+\sqrt{2},-1 \right)\]
Correct Answer: C
Solution :
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 \[2-\sqrt{{{1}^{2}}+{{1}^{2}}}=2-\sqrt{2}.\]
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