Question
The range of the function $$f\left( x \right) = {\sin ^{ - 1}}\left( {\log \left[ x \right]} \right) + \log \left( {{{\sin }^{ - 1}}\left[ x \right]} \right);\,$$ (where [.] denotes the greatest integer function) is
A.
$$R$$
B.
$$\left[ {1,2} \right)$$
C.
$$\left\{ {\log \frac{\pi }{2}} \right\}$$
D.
$$\left\{ { - \sin 1} \right\}$$
Answer :
$$\left\{ {\log \frac{\pi }{2}} \right\}$$
Solution :
$$\eqalign{
& {\sin ^{ - 1}}\left( {\log \left[ x \right]} \right){\text{is defined if}}\, - 1 \leqslant \log \left[ x \right] \leqslant 1\,{\text{and }}\left[ x \right] > 0 \cr
& \Rightarrow \frac{1}{e} \leqslant \left[ x \right] \leqslant e \cr
& \Rightarrow \left[ x \right] = 1,2 \cr
& \Rightarrow x \in \left[ {1,3} \right) \cr
& {\text{Again}},\,\,\log \left( {{{\sin }^{ - 1}}\left[ x \right]} \right){\text{ is defined if}} \cr
& {\sin ^{ - 1}}\left[ x \right] > 0{\text{ and }} - 1 \leqslant \left[ x \right] \leqslant 1 \cr
& \Rightarrow \left[ x \right] > 0{\text{ and }} - 1 \leqslant \left[ x \right] \leqslant 1 \cr
& \Rightarrow 0 < \left[ x \right] \leqslant 1 \cr
& \Rightarrow x \in \left[ {1,2} \right) \cr
& \therefore {\text{Domain of }}f\left( x \right) = \left[ {1,2} \right) \cr
& {\text{For }}1 \leqslant x < 2,\left[ x \right] = 1 \cr
& \therefore f\left( x \right) = {\sin ^{ - 1}}0 + \log \frac{\pi }{2} = \log \frac{\pi }{2},\forall x \in \left[ {1,2} \right) \cr
& \therefore {\text{Range of }}f\left( x \right) = \left\{ {\log \frac{\pi }{2}} \right\} \cr} $$