A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
General term in the expansion of\[{{\left( a{{x}^{2}}+\frac{1}{bx} \right)}^{11}}\] is \[{{T}_{r+1}}{{=}^{11}}{{C}_{r}}{{(a{{x}^{2}})}^{r}}{{\left( \frac{1}{bx} \right)}^{11-r}}\] \[\Rightarrow \] \[{{T}_{r+1}}{{=}^{11}}{{C}_{r}}{{a}^{r}}{{b}^{r-11}}{{x}^{2r-11+r}}\] For the coefficient of\[{{x}^{7}},\] put \[2r-11+r=7\] \[\Rightarrow \] \[r=6\] Thus, \[{{T}_{7}}{{=}^{11}}{{C}_{6}}{{a}^{6}}{{b}^{-5}}\] Now, general term in the expansion of\[{{\left( ax-\frac{1}{b{{x}^{2}}} \right)}^{11}}\]is \[T{{'}_{r+1}}{{=}^{11}}{{C}_{r}}{{(ax)}^{r}}{{\left( -\frac{1}{b{{x}^{2}}} \right)}^{11-r}}\] \[{{=}^{11}}{{C}_{r}}{{a}^{r}}{{(-1)}^{r-11}}{{b}^{r-11}}{{x}^{r-22+2r}}\] For the coefficient of\[{{x}^{-7}},\] put \[r-22+2r=-7\] \[\Rightarrow \] \[3r=15\] \[\Rightarrow \] \[r=5\] Thus, \[T{{'}_{6}}{{=}^{11}}{{C}_{5}}{{a}^{5}}{{b}^{-6}}\] Given, \[{{T}_{7}}=T{{'}_{6}}\] \[\therefore \] \[^{11}{{C}_{6}}{{a}^{6}}{{b}^{-5}}{{=}^{11}}{{C}_{5}}{{a}^{5}}-{{b}^{-6}}\] \[\Rightarrow \] \[ab{{=}^{11}}{{C}_{5}}{{/}^{11}}{{C}_{6}}\] \[{{=}^{11}}{{C}_{5}}{{/}^{11}}{{C}_{(11-6)}}\] \[({{\because }^{n}}{{C}_{r}}{{=}^{n}}{{C}_{n-r}})\] \[\Rightarrow \] \[ab=1\]
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