1\le {{\SIN }^{-1}}{{\cos }^{-1}}{{\sin }^{-1}}{{\tan }^{-1}}x\le \frac{\PI }{2}\] \[\Rightarrow \sin 1\le {{\cos }^{-1}}{{\sin }^{-1}}{{\tan }^{-1}}x\le 1\] \[\Rightarrow \cos \sin 1\ge {{\sin }^{-1}}{{\tan }^{-1}}x\ge \cos 1\] \[\Rightarrow \sin \cos \sin 1\ge {{\tan }^{-1}}x\ge \sin \cos 1\] \[\Rightarrow \tan \sin \cos \sin 1\ge x\ge \tan \,sin\,cos1\] \[\therefore x\in [tan\,sin\,cos1,tan\,sin\,cos\,SIN1]\]

"> 1\le {{\SIN }^{-1}}{{\cos }^{-1}}{{\sin }^{-1}}{{\tan }^{-1}}x\le \frac{\PI }{2}\] \[\Rightarrow \sin 1\le {{\cos }^{-1}}{{\sin }^{-1}}{{\tan }^{-1}}x\le 1\] \[\Rightarrow \cos \sin 1\ge {{\sin }^{-1}}{{\tan }^{-1}}x\ge \cos 1\] \[\Rightarrow \sin \cos \sin 1\ge {{\tan }^{-1}}x\ge \sin \cos 1\] \[\Rightarrow \tan \sin \cos \sin 1\ge x\ge \tan \,sin\,cos1\] \[\therefore x\in [tan\,sin\,cos1,tan\,sin\,cos\,SIN1]\]

">
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If \[[si{{n}^{-1}}co{{s}^{-1}}si{{n}^{-1}}ta{{n}^{-1}}x]=1,\] where \[[.]\] denotes the greatest integer function, then x belongs to the interval

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 1 year ago

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[a] We have, \[1\le {{\SIN }^{-1}}{{\cos }^{-1}}{{\sin }^{-1}}{{\tan }^{-1}}x\le \frac{\PI }{2}\] \[\Rightarrow \sin 1\le {{\cos }^{-1}}{{\sin }^{-1}}{{\tan }^{-1}}x\le 1\] \[\Rightarrow \cos \sin 1\ge {{\sin }^{-1}}{{\tan }^{-1}}x\ge \cos 1\] \[\Rightarrow \sin \cos \sin 1\ge {{\tan }^{-1}}x\ge \sin \cos 1\] \[\Rightarrow \tan \sin \cos \sin 1\ge x\ge \tan \,sin\,cos1\] \[\therefore x\in [tan\,sin\,cos1,tan\,sin\,cos\,SIN1]\]

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Posted on 09 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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