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Q.
The term independent of x in the expansion of $ \left( \frac{2\sqrt{x}}{x}-\frac{1}{2x\sqrt{x}} \right) $ is
J & K CETJ & K CET 2007Binomial Theorem
Solution:
Given expansion is $ {{\left( \frac{2\sqrt{x}}{5}-\frac{1}{2x\sqrt{x}} \right)}^{11}} $
General term $ {{T}_{r+1}}={{(-1)}^{r}}{{\,}^{11}}{{C}_{r}}{{\left( \frac{2\sqrt{x}}{5} \right)}^{11-r}}{{\left( \frac{1}{2{{x}^{3/2}}} \right)}^{r}} $
$ =\frac{{{2}^{11-2r}}}{{{5}^{11-r}}}{{(-1)}^{r}}\,{{\,}^{11}}{{C}_{r}}\,{{x}^{\frac{11-r}{2}-\frac{3r}{2}}} $
For term independent of x put
$ \frac{11-r}{2}-\frac{3r}{2}=0 $
$ \Rightarrow $ $ \frac{11-4r}{2}=0 $
$ \Rightarrow $ $ r=\frac{11}{4}\notin N $
$ \therefore $ There is no term which is independent of x.