A) \[\pm 3\]
B) \[\pm 6\]
C) \[\pm 5\]
D) \[\pm 4\]
Correct Answer: A
Solution :
[a] Given expansion is \[{{\left( \sqrt{x}+\frac{k}{{{x}^{2}}} \right)}^{10}}\] \[{{(r+1)}_{th}}term,\,\,{{T}_{r+1}}={{\,}^{10}}{{C}_{r}}{{(\sqrt{x})}^{10-r}}{{\left( \frac{k}{{{x}^{2}}} \right)}^{r}}\] \[\Rightarrow {{T}_{r+1}}={{\,}^{10}}{{C}_{r}}{{x}^{5-r/2}}.{{(k)}^{r}}.{{x}^{-2r}}\] \[\therefore \,\,\,\,{{T}_{r+1}}={{\,}^{10}}{{C}_{r}}{{x}^{(10-5r)/2}}{{(k)}^{r}}\] Since, \[{{T}_{r+1}}\] is independent of x \[\therefore \,\frac{10-5r}{2}=0\Rightarrow r=2\therefore \,405={{\,}^{10}}{{C}_{2}}{{(k)}^{2}}\] \[405=45\times {{k}^{2}}\Rightarrow {{k}^{2}}=9\Rightarrow k=\pm 3\]
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