A) 1
B) 1/2
C) 2
D) 3
Correct Answer: A
Solution :
In the expansion of \[{{\left( a{{x}^{2}}+\frac{1}{bx} \right)}^{11}}\], the general term is \[{{T}_{r+1}}={{\,}^{11}}{{C}_{r}}{{(a{{x}^{2}})}^{11-r}}{{\left( \frac{1}{bx} \right)}^{r}}={{\,}^{\,11}}{{C}_{r}}{{a}^{11-r}}\frac{1}{{{b}^{r}}}{{x}^{22-3r}}\] For x7, we must have 22 - 3r = 7 Þ r = 5, and the coefficient of x7 =\[^{11}{{C}_{5}}.{{a}^{11-5}}\frac{1}{{{b}^{5}}}={{\,}^{11}}{{C}_{5}}\frac{{{a}^{6}}}{{{b}^{5}}}\] Similarly, in the expansion of \[{{\left( ax-\frac{1}{b{{x}^{2}}} \right)}^{11}},\] the general term is \[{{T}_{r+1}}={{\,}^{11}}{{C}_{r}}{{(-1)}^{r}}\frac{{{a}^{11-r}}}{{{b}^{r}}}.{{x}^{11-3r}}\] For x-7 we must have, 11 - 3r = -7 Þ r = 6, and the coefficient of \[{{x}^{-7}}\]is \[^{11}{{C}_{6}}\frac{{{a}^{5}}}{{{b}^{6}}}={{\,}^{11}}{{C}_{5}}\frac{{{a}^{5}}}{{{b}^{6}}}\]. As given, \[^{11}{{C}_{5}}\frac{{{a}^{6}}}{{{b}^{5}}}={{\,}^{11}}{{C}_{5}}\frac{{{a}^{5}}}{{{b}^{6}}}\Rightarrow ab=1\].
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