A) \[8th\]
B) \[7th\]
C) \[9th\]
D) \[10th\]
Correct Answer: C
Solution :
The general term in the expansion of\[{{\left( 2{{x}^{2}}-\frac{1}{x} \right)}^{12}}\]is \[{{T}_{r+1}}{{=}^{12}}{{C}_{r}}{{(2{{x}^{2}})}^{12-r}}\cdot {{\left( -\frac{1}{x} \right)}^{r}}\] \[={{(-1)}^{r}}^{12}{{C}_{r}}{{2}^{12-r}}\cdot {{x}^{24-2r-r}}\] \[={{(-1)}^{r}}^{12}{{C}_{r}}\cdot {{2}^{12-r}}\cdot {{x}^{24-3r}}\] The term independent of\[x\], if \[24-3r=0\] \[\Rightarrow \] \[24=3r\] \[\Rightarrow \] \[r=8\] Now, \[r+1=8+1\] \[=9th\]term
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