Find the range of $f (x) = {sin}^{- 1} (ln [x]) + ln ({sin}^{- 1} [x]),$ where $[x]$ is the greatest function.

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Solution

${sin}^{- 1} (ln [x])$ is defined if only if
$- 1 \leq ln [x] \leq 1$
$\Rightarrow e^{- 1} \leq [x] \leq e^{1}$
$\Rightarrow [x] = 1, 2$
$ln ({sin}^{- 1} [x])$ is defined if only if
${sin}^{- 1} [x] > 0$ and $- 1 \leq [x] \leq 1 \Rightarrow [x] = 1$
$f (x)$ is defined if $[x] = 1$ only,
Range of $f (x) = {sin}^{- 1} (ln [1]) + ln ({sin}^{- 1} [1])$
$= {sin}^{- 1} (ln 1) + ln ({sin}^{- 1} 1)$
$= {sin}^{- 1} (0) + ln (\frac{π}{2})$
$= ln (\frac{π}{2})$
Hence range is ${ln (\frac{π}{2})}$

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