A) \[\left( \frac{1}{2},\frac{1}{\sqrt{2}} \right)\]
B) \[\left( \frac{1}{\sqrt{2}},1 \right)\]
C) \[(0,1)\]
D) \[\left( 0,\frac{\sqrt{3}}{2} \right)\]
Correct Answer: B
Solution :
[b] Given \[{{\sin }^{-1}}x>{{\cos }^{-1}}x\]where \[x\in (0,1)\] \[\Rightarrow {{\sin }^{-1}}x>\frac{\pi }{2}-{{\sin }^{-1}}x\Rightarrow 2{{\sin }^{-1}}x>\frac{\pi }{2}\] \[\Rightarrow {{\sin }^{-1}}x>\frac{\pi }{4}\Rightarrow x>\sin \frac{\pi }{4}\Rightarrow x>\frac{1}{\sqrt{2}}\] Maximum value of \[{{\sin }^{-1}}x\] is \[\frac{\pi }{2}\] So, maximum value of x is 1, so, \[x\in \left( \frac{1}{\sqrt{2}},1 \right).\]
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