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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
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 \((0, 1)\)
- D \(\left( {0,\frac{{\sqrt 3 }}{2}} \right)\)
Answer & Solution
Correct Answer
(B) \(\left( {\frac{1}{{\sqrt 2 }},1} \right)\)
Step-by-step Solution
Detailed explanation
Given \({\sin ^{ - 1}}x > {\cos ^{ - 1}}x\) where \(x \in \left( {0,1} \right)\) \( \Rightarrow {\sin ^{ - 1}}x > \frac{\pi }{2} - {\sin ^{ - 1}}x\) \( \Rightarrow 2{\sin ^{ - 1}}x > \frac{\pi }{2} \Rightarrow {\sin ^{ - 1}}x > \frac{\pi }{4}\)…
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