If the coefficient of x7 in the expansion of  ax2+1bx11 and the coefficient of x– 7 is the expansion of ax−1bx211are equal, then

If the coefficient of x7 in the expansion of  ${\left(a{x}^{2}+\frac{1}{bx}\right)}^{11}$ and the coefficient of x– 7 is the expansion of ${\left(ax-\frac{1}{b{x}^{2}}\right)}^{11}$are equal, then

1. A

ab = 1

2. B

ab = 11

3. C

ab = 5

4. D

ab = 6

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Solution:

${T}_{r+1},$the term in the expansion ${\left(a{x}^{2}+\frac{1}{bx}\right)}^{11}$is

${T}_{r+1}{=}^{11}{C}_{r}{\left(a{x}^{2}\right)}^{11-r}{\left(\frac{1}{bx}\right)}^{r}{=}^{11}{C}_{r}\frac{{a}^{11-r}}{{b}^{r}}{x}^{22-3r}$

For the coefficient of ${x}^{7},$set

$\therefore$Coefficient of x7 in is

Next, , the  term in the expansion of ${\left(ax-\frac{1}{b{x}^{2}}\right)}^{11}$is

${t}_{r+1}{=}^{11}{C}_{r}\left(ax{\right)}^{11-r}{\left(-\frac{1}{b{x}^{2}}\right)}^{r}$

${=}^{11}{C}_{r}\frac{{a}^{11-r}}{{b}^{r}}\left(-1{\right)}^{r}{x}^{11-3r}$

For the coefficient of ${x}^{-7},$ set $11-3r=-7⇒r=6$

$\therefore$Coefficient of  in the expansion of ${=}^{11}{C}_{6}\frac{{a}^{5}}{{b}^{6}}$

We are given

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