A) 7 : 16
B) 7 : 64
C) 1 : 4
D) 1 : 32
Correct Answer: D
Solution :
\[{{T}_{r+1}}{{=}^{15}}{{C}_{r}}{{({{x}^{2}})}^{15-r}}\cdot {{\left( \frac{2}{x} \right)}^{r}}{{=}^{15}}{{C}_{r}}{{x}^{30-2r}}\cdot {{x}^{-r}}\]\[{{=}^{15}}C\cdot {{x}^{30-3r}}\cdot {{2}^{r}}\] For coefficient of \[{{x}^{15}}\], put : \[30-3r-15\Rightarrow 3r=15\] \[\Rightarrow \] \[r=5\] \[\therefore \] Coefficient of \[{{x}^{15}}{{=}^{15}}{{C}_{5}}\cdot {{2}^{5}}\] For coefficient of independent of \[x\] i.e., \[{{x}^{o}}\], put \[30-3r=0\Rightarrow r=10\] \[\therefore \] Coefficient of \[{{x}^{o}}{{=}^{15}}C{{}_{10}}\cdot {{2}^{10}}\] By condition \[\Rightarrow \] \[\frac{\text{Coefficient of }{{x}^{15}}}{\text{Coefficient of }{{x}^{o}}}=\frac{^{15}{{C}_{6}}\cdot {{2}^{5}}}{^{15}{{C}_{10}}\cdot {{2}^{10}}}\] \[=\frac{^{15}{{C}_{10}}\cdot {{2}^{5}}}{^{15}{{C}_{10}}\cdot {{2}^{10}}}=1 & :32\]
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