A) \[a-b=1\]
B) \[a+b=1\]
C) \[\frac{a}{b}=1\]
D) \[ab=1\]
Correct Answer: D
Solution :
General term in the expansion of \[{{\left[ a{{x}^{2}}+\frac{1}{bx} \right]}^{11}}\]is \[{{T}_{r+1}}{{=}^{11}}{{C}_{r}}{{(a{{x}^{2}})}^{11-r}}{{\left( \frac{1}{bx} \right)}^{r}}\] \[{{=}^{11}}{{C}_{r}}\frac{{{a}^{11-r}}}{{{b}^{r}}}.{{x}^{22-3r}}\] For coefficient of \[{{x}^{7}},22-3r=7\] \[\Rightarrow \] \[r=5\] \[\therefore \] \[{{T}_{6}}{{=}^{11}}{{C}_{5}}\frac{{{a}^{6}}}{{{b}^{5}}}{{x}^{7}}\] Similarly, in the expansion of\[{{\left( ax-\frac{1}{b{{x}^{2}}} \right)}^{11}}\]the coefficient of \[{{x}^{-7}}{{=}^{11}}{{C}_{6}}\frac{{{a}^{5}}}{{{b}^{6}}}\] According to question, \[^{11}{{C}_{5}}\frac{{{a}^{6}}}{{{b}^{5}}}{{=}^{11}}{{C}_{6}}\frac{{{a}^{5}}}{{{b}^{6}}}\] \[\Rightarrow \] \[\frac{{{a}^{6}}}{{{b}^{5}}}=\frac{{{a}^{5}}}{{{b}^{6}}}\] \[\Rightarrow \] \[ab=1\]
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