A) [1, 4]
B) [2, 4]
C) [3, 4]
D) [4,\[\infty \]]
Correct Answer: B
Solution :
Given, \[\Delta =\left| \begin{matrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \\ \end{matrix} \right|\] \[=1[1+{{\sin }^{2}}\theta ]-\sin \theta [-\sin \theta +\sin \theta ]\] \[+1[{{\sin }^{2}}\theta +1]\] \[=1+{{\sin }^{2}}\theta -0+{{\sin }^{2}}\theta +1\] \[=2si{{n}^{2}}\theta +2\] Since \[0\le {{\sin }^{2}}\theta \le 1\] \[\Rightarrow \] \[2\le 2+2{{\sin }^{2}}\theta \le 4\] \[\therefore \] \[\Delta \in [2,4]\]
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