A) \[{{R}_{1}}=\{u:-1\le u<1\},{{R}_{2}}=\{v:-\infty <v<0\}\]
B) \[{{R}_{1}}=\{u:-\infty <u<0\},{{R}_{2}}=\{v:-\infty <v<0\}\]
C) \[{{R}_{1}}=\{u:-1<u<1\},{{R}_{2}}=\{v:-\infty <v<0\}\]
D) \[{{R}_{1}}=\{u:-1\le u\le 1\},{{R}_{2}}=\{v:-\infty <v\le 0\}\]
Correct Answer: D
Solution :
[d] we have \[fog(x)=f(g(x))=sin(lo{{g}_{e}}\left| x \right|).\] \[{{\log }_{e}}\left| x \right|\] has range R, for which \[sin({{\log }_{e}}\left| x \right|)\in [-1,1].\] Therefore, \[{{R}_{1}}=\{u:-1\le u\le 1\}.\] Also, \[gof(x)=g(f(x))=lo{{g}_{e}}\left| \sin x \right|.\] \[\because 0\le \left| \sin x \right|\le 1\]or \[-\infty <{{\log }_{e}}\left| \sin x \right|\le 0\] Or \[{{R}_{2}}=\{v:-\infty <v\le 0\}\]
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