A) \[\{\theta :\pi /2\le \theta \le 5\pi /6\}\]
B) \[\{\theta :\pi \le \theta \le 3\pi /2\}\]
C) \[\{\theta :\pi /2\le \theta \le 5\pi /6\}\cup \{\theta :\pi \,\le \theta \le 3\pi /2\}\]
D) None of these
Correct Answer: C
Solution :
It is given that\[2{{\cos }^{2}}\theta +\sin \theta \le 2\] \[\Rightarrow \] \[2(1-{{\sin }^{2}}\theta )+\sin \theta \le 2\] \[\Rightarrow \] \[2{{\sin }^{2}}\theta -\sin \theta \ge 0\] \[\Rightarrow \] \[\sin \theta (2\sin \theta -1)\ge 0\] \[\Rightarrow \] \[\sin \theta \ge 0\]and\[2\sin \theta -1\ge 0\] or \[\sin \theta \le 0\]and\[2\sin \theta -1\le 0\] Case I When \[\sin \theta \ge 0\] and \[2\sin \theta -1\ge 0\] Now, \[\sin \theta \ge 0\]and\[2\sin \theta -1\ge 0\] \[\Rightarrow \]\[\sin \theta \ge 0\]and\[\sin \theta \ge \frac{1}{2}\] \[\Rightarrow \]\[\frac{\pi }{6}\le \theta \le \frac{5\pi }{6}\] \[\therefore \] \[A\cap B=\left\{ \theta :\frac{\pi }{2}\le \theta \le \frac{5\pi }{6} \right\}\] Case II When \[\sin \theta \le 0\] and \[2\sin \theta -1\le 0\] Then, \[\sin \theta \le 0\]and \[\sin \theta \le \frac{1}{2}\] \[\Rightarrow \] \[\sin \theta \le 0\] \[\Rightarrow \] \[\theta \le \theta \le 2\pi \] \[\therefore \] \[A\cap B=(\theta :\pi \le \theta \le 3\pi /2)\] Thus, \[A\cap B=\{\theta :\pi /2\le \theta \le 5\pi /6\}\] \[\cup \{\theta :\pi \le \theta \le 3\pi /2\}\]So, correct option is [c].
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