A) \[\pm \,\,2\]
B) \[\pm \,\,3\]
C) \[\pm \,\,4\]
D) None of these
Correct Answer: A
Solution :
Given, \[{{\left( \sqrt{x}\,-\frac{C}{{{x}^{2}}} \right)}^{10}}\] General term is \[{{T}_{r+1}}{{=}^{10(}}{{C}_{r}}{{(\sqrt{x})}^{10-r}}{{\left( -\frac{C}{{{x}^{2}}} \right)}^{r}}\] \[{{=}^{10}}{{C}_{r}}\,{{x}^{(5-r/2)}}{{(-C)}^{r}}{{x}^{-2r}}\] \[{{=}^{10\,}}{{C}_{r}}\,{{x}^{(5-r/2-2r)}}\,{{(-C)}^{r}}\] \[{{=}^{10\,}}{{C}_{r}}\,{{x}^{\left( 5-\frac{5r}{2} \right)}}\,{{(-C)}^{r}}\] For the constant term, Put \[5-\frac{5r}{2}=0\] \[\Rightarrow \] \[(r=2)\] \[\Rightarrow \] \[{{T}_{2+1}}{{=}^{10}}{{C}_{2}}\,\,{{x}^{(5-5)}}\,{{(-C)}^{2}}{{=}^{10}}{{C}_{2}}.{{C}^{2}}\] Given, \[{{T}_{3}}=180{{=}^{10}}{{C}_{2}}.{{C}^{2}}\] \[\Rightarrow \] \[{{C}^{2}}.\frac{10\times 9}{2}=180\] \[\Rightarrow \] \[{{C}^{2}}=4\] \[\Rightarrow \] \[C=\pm 2\]
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