A) 1
B) \[\sqrt{10}\]
C) 10
D) \[{{10}^{-2/5}}\]
Correct Answer: C
Solution :
[c] Put \[{{\log }_{10}}x=y\], the given expansion becomes \[{{(x+{{x}^{y}})}^{5}}\]. \[{{T}_{3}}={{\,}^{5}}{{C}_{2}}\,.\,\,{{x}^{3}}{{({{x}^{y}})}^{2}}=10{{x}^{3+2y}}={{10}^{6}}(given)\] \[\Rightarrow (3+2y)lo{{g}_{10}}x=5\,{{\log }_{10}}10=5\] \[\Rightarrow (3+2y)y=5\] \[\Rightarrow y=1,-\frac{5}{2}\Rightarrow {{\log }_{10}}x=1\] or \[{{\log }_{10}}x=-\frac{5}{2}\] \[\therefore \,x=10\] or \[x={{(10)}^{-5/2}}\]
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