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The domain of $sin^{-1}\big[log_3(\large\frac{x}{3})\big]$ is

(A) $[1,9]$ (B) $[-11,9]$ (C) $[-9,1]$ (D) $[-9,-1]$
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Toolbox:
  • $log\frac{a}{b}=loga-l$ogb
  • $log_b\large a=\frac{loga}{logb} $ (with any base)
  • $log_a\large a=1$
  • Domain of $sin^{-1}x=[-1,1]$
  • $log1=0$
  • $log(x^n)=nlogx$
Ans: (A) [1,9]
Domain of $sin^{-1}x=[-1,1]$
$\Rightarrow\:-1\leq\:log_3(\frac{x}{3})\leq 1$
$\Rightarrow\:-1\leq (log_3\large x$-$log_3\large3)\leq 1$
$\Rightarrow\:-1\leq (log_3\large x$-1)$\leq1$
$\Rightarrow\:0\leq (log_3\large x$)$\leq2$
$\Rightarrow\:log_3\large 1$$\leq (log_3\large x$)$\leq log_33^2$
$\Rightarrow\:1\leq x\leq 9$
$x\in [1,9]$
answered Apr 29, 2013 by rvidyagovindarajan_1
edited May 17, 2014 by rohanmaheshwari0831_1
 
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