If the coefficients of x^{7} in (x^{2}+frac{1 | vedclass

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(A) For the expansion $(x^{2}+\frac{1}{bx})^{11}$,the general term is $T_{r+1} = {}^{11}C_{r}(x^{2})^{11-r}(\frac{1}{bx})^{r} = {}^{11}C_{r} \cdot b^{

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(A) For the expansion $(x^{2}+\frac{1}{bx})^{11}$,the general term is $T_{r+1} = {}^{11}C_{r}(x^{2})^{11-r}(\frac{1}{bx})^{r} = {}^{11}C_{r} \cdot b^{-r} \cdot x^{22-3r}$.Setting $22-3r = 7$,we get $3r = 15$,so $r = 5$.The coefficient of $x^{7}$ is ${}^{11}C_{5} \cdot b^{-5}$.For the expansion $(x-\frac{1}{bx^{2}})^{11}$,the general term is $T_{r+1} = {}^{11}C_{r}(x)^{11-r}(-\frac{1}{bx^{2}})^{r} = {}^{11}C_{r} \cdot (-1)^{r} \cdot b^{-r} \cdot x^{11-3r}$.Setting $11-3r = -7$,we get $3r = 18$,so $r = 6$.The coefficient of $x^{-7}$ is ${}^{11}C_{6} \cdot (-1)^{6} \cdot b^{-6} = {}^{11}C_{6} \cdot b^{-6}$.Equating the coefficients: ${}^{11}C_{5} \cdot b^{-5} = {}^{11}C_{6} \cdot b^{-6}$.Since ${}^{11}C_{5} = {}^{11}C_{6}$,we have $\frac{1}{b^{5}} = \frac{1}{b^{6}}$.Thus,$b = 1$.

If the coefficients of $x^{7}$ in $(x^{2}+\frac{1}{bx})^{11}$ and $x^{-7}$ in $(x-\frac{1}{bx^{2}})^{11}$,$b \neq 0$,are equal,then the value of $b$ is equal to:

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