A) \[\left( -4,\,\sqrt{2} \right)\cup \left( \sqrt{2},4 \right)\]
B) \[\left( -4,\,-2\sqrt{2} \right)\,\cup \,\left( 2\sqrt{2},\,4 \right)\]
C) \[\left( -4,\,\sqrt{2} \right)\cup \left( 2\sqrt{2},\,4 \right)\]
D) \[\left( -4,\,4 \right)\]
Correct Answer: B
Solution :
\[\sin \left( \frac{\alpha }{2} \right)=\frac{2}{a}\] \[\alpha \in \left( \frac{\pi }{3},\,\frac{\pi }{2} \right)\] \[\Rightarrow \,\,\frac{\alpha }{2}\in \left( \frac{\pi }{6},\,\frac{\pi }{4} \right)\] \[\Rightarrow \,\,\frac{2}{a}\in \,\left( \frac{1}{2},\,\frac{1}{\sqrt{2}} \right)\] \[\Rightarrow \,\,a\in \,\left( \frac{1}{2},\,\frac{1}{\sqrt{2}} \right)\] If (a, 0) is on left side of origin then \[-\frac{2}{a}\in \left( \frac{1}{2},\,\frac{1}{\sqrt{2}} \right)\] \[\Rightarrow \,\,a\in \left( -4,\,-2\sqrt{2} \right)\] Hence, \[a\in \,\left( -4,\,-2\sqrt{2} \right)\cup \,\left( 2\sqrt{2},\,4 \right)\]You need to login to perform this action.
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