Question
Let $$x \in \left( {0,1} \right).$$ The set of all $$x$$ such that $${\sin ^{ - 1}}x > {\cos ^{ - 1}}x,$$ is the interval :
A.
$$\left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)$$
B.
$$\left( {\frac{1}{{\sqrt 2 }},1} \right)$$
C.
$$\left( {0,1} \right)$$
D.
$$\left( {0,\frac{{\sqrt 3 }}{2}} \right)$$
Answer :
$$\left( {\frac{1}{{\sqrt 2 }},1} \right)$$
Solution :
$$\eqalign{
& {\text{Given, }}\,{\sin ^{ - 1}}x > {\cos ^{ - 1}}x{\text{ where}}\,x \in \left( {0,1} \right) \cr
& \Rightarrow {\sin ^{ - 1}}x > \frac{\pi }{2} - {\sin ^{ - 1}}x \cr
& \Rightarrow 2\,{\sin ^{ - 1}}x > \frac{\pi }{2} \cr
& \Rightarrow {\sin ^{ - 1}}x > \frac{\pi }{4} \cr
& \Rightarrow x > \sin \frac{\pi }{4} \cr
& \Rightarrow x > \frac{1}{{\sqrt 2 }} \cr} $$
Maximum value of $${\sin ^{ - 1}}x\,\,{\text{is }}\,\frac{\pi }{2}$$
So, maximum value of $$x$$ is $$1.$$ So, $$x \in \left( {\frac{1}{{\sqrt 2 }},1} \right).$$