A) \[\left[ -\infty ,{{\log }_{e}}\left( {{2}^{\frac{1}{2}}}+{{2}^{\frac{3}{2}}} \right) \right]\]
B) \[\left( -\infty ,{{\log }_{2}}\left( {{2}^{\frac{1}{2}}}-1 \right) \right]\cup \left[ \frac{1}{2},\infty \right)\]
C) \[\left[ {{\log }_{10}}{{2}^{\frac{1}{2}}},{{\log }_{e}}{{2}^{\frac{1}{2}}} \right]\]
D) \[\left( {{2}^{{{\log }_{e}}2}},\infty \right)\]
Correct Answer: B
Solution :
(b): In \[x>0,{{2}^{x}}+{{2}^{x}}\ge {{2}^{\frac{3}{2}}}\Rightarrow {{2}^{x}}\ge {{2}^{\frac{1}{2}}}\Rightarrow x\ge \frac{1}{2}\] And in \[x<0,{{2}^{x}}+{{2}^{-x}}\ge {{2}^{\frac{3}{2}}}\]; Putting \[{{2}^{x}}=m\], we get., \[\Rightarrow m+\frac{1}{m}\ge {{2}^{\frac{3}{2}}}\Rightarrow {{m}^{2}}-{{2}^{\frac{3}{2}}}m+1\ge 0\] \[\Rightarrow \left[ m-\left( {{2}^{\frac{1}{2}}}-1 \right) \right]\left[ m-\left( {{2}^{\frac{1}{2}}}+1 \right) \right]\ge 0\] \[\Rightarrow m\le {{2}^{\frac{1}{2}}}-1\] or \[m\ge {{2}^{\frac{1}{2}}}+1\] but \[m>0\] \[\Rightarrow -\infty <x\le {{\log }_{2}}\left( {{2}^{\frac{1}{2}}}+1 \right)\] Or \[x\ge {{\log }_{2}}\left( {{2}^{\frac{1}{2}}}-1 \right)\] (but not acceptable as x < 0) \[\therefore x\in \left( -\infty ,{{\log }_{2}}\left( {{2}^{\frac{1}{2}}}-1 \right) \right]\cup \left[ \frac{1}{2},\infty \right)\]
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