Inverse Trigonometric Functions Question 72
Question: Let $ x\in (0,1). $ The set of all x such that $ {{\sin }^{-1}}x>{{\cos }^{-1}}x, $ is the interval:
Options:
A) $ ( \frac{1}{2},\frac{1}{\sqrt{2}} ) $
B) $ ( \frac{1}{\sqrt{2}},1 ) $
C) $ (0,1) $
D) $ ( 0,\frac{\sqrt{3}}{2} ) $
Show Answer
Answer:
Correct Answer: 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 ( \frac{1}{\sqrt{2}},1 ). $
BETA
