Q. If $\lbrack\sin^{- 1}{\cos^{- 1}{\sin^{- 1}{x\rbrack = 1,}}}$ where [.] denotes the greatest integer function, then $x$ belongs to the interval
A) [tan sin $\cos 1$, tan sin cos sin 1]
B) (tan sin $\cos 1$, tan sin$\ \cos{\sin 1}$)
C) $\lbrack - 1,1\rbrack$
D) [sin $\cos{\tan 1}$, $\sin{\cos{\sin{\tan 1}}}$]
Explanation
(a)
We have, $1 \leq \sin^{- 1}{\cos^{- 1}{\sin^{- 1}{\tan^{- 1}{x \leq \frac{\pi}{2}}}}}$
$\Rightarrow \sin{1 \leq \cos^{- 1}{\sin^{- 1}{\tan^{- 1}{x \leq 1}}}}$
$\Rightarrow \cos{\sin{1 \geq \sin^{- 1}{\tan^{- 1}{x \geq \cos 1}}}}$
$\Rightarrow \sin{\cos{\sin{1 \geq \tan^{- 1}{x \geq \sin{\cos 1}}}}}$
$\Rightarrow \tan{\sin{\cos{\sin 1 \geq x \geq \tan{\sin{\cos 1}}}}}$
$\therefore x \in \lbrack\tan{\sin{\cos{1,\tan{\sin{\cos{\sin{1\rbrack}}}}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.