A) \[2\sqrt{2}\]
B) \[4\sqrt{2}\]
C) \[\frac{1}{8}\]
D) \[\frac{1}{4}\]
Correct Answer: A
Solution :
\[{{\left( 1+{{x}^{{{\log }_{{{2}^{\,x}}}}}} \right)}^{5}}\] \[{{T}_{3}}\,={{\,}^{5}}{{C}_{2}}{{({{x}^{lo{{g}_{{{2}^{x}}}}}})}^{2}}\,\,=\,\,2560\] \[{{x}^{{{\log }_{{{2}^{x}}}}}}=16\] \[{{\log }_{2x}}\,{{\log }_{2x}}\,=\,{{\log }_{2}}16\,=\,4\] \[lo{{g}_{2}}_{X}=\pm \,2\] \[x={{2}^{2}}=4x={{2}^{-2}}=\frac{1}{4}\]
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