Question

Find the domain and range of \[f\left( x \right) = {\sin ^{ - 1}}\left( {\log \left[ x \right]} \right) + \log \left( {{{\sin }^{ - 1}}\left[ x \right]} \right)\] where$\left[ . \right]$ denotes greater integer function.

Answer 1

Hint- To find the domain, first find the intersection of the values for which the given function will exist .Choose the positive value as x here is the greater integer function.

Complete step-by-step answer:
Given \[f\left( x \right) = {\sin ^{ - 1}}\left( {\log \left[ x \right]} \right) + \log \left( {{{\sin }^{ - 1}}\left[ x \right]} \right)\]
Here,since ${\sin ^{ - 1}}\left( {\log \left[ x \right]} \right)$ is defined if and only if $\left[ x \right]$ >0 and $ - 1$ ≤$\log \left[ x \right] \leqslant 1$$ \Rightarrow x \in \left[ {1,1} \right)$-- (i) {on solving the inequality }
And $\log \left( {{{\sin }^{ - 1}}\left[ x \right]} \right)$ is defined if and only if $ - 1 \leqslant \left[ x \right] \leqslant 1$ and${\sin ^{ - 1}}\left[ x \right] > 0$ $ \Rightarrow \left[ x \right] = 1$----(ii)
Now,(i)∩(ii) →$x \in \left[ {1,2} \right)$ .This is the domain of the function.Now f(x) is defined only if [x]=1 only.Then range =\[{\sin ^{ - 1}}\left( {\log \left[ 1 \right]} \right) + \log \left( {{{\sin }^{ - 1}}\left[ 1 \right]} \right)\]=${\sin ^{ - 1}}0 + \log \dfrac{\pi }{2}$ $ = \log \dfrac{\pi }{2}$
So the domain is $\left[ {1,2} \right)$ and range is $\log \dfrac{\pi }{2}$ .

Note: Here, the students may mistake the domain as $\left( {1,2} \right)$ or $\left[ {1,2} \right]$ but this is wrong as it changes the meaning here the domain starts with closed interval denoted by[ and end with open interval denoted by ).