If the co-efficient of x 9 in α x 3 + 1 β x 11 and the co-efficient of x - 9 in α x - 1 &#946…

Given,

The coefficient of $x^9$ in ${\left(\alpha x^3+\frac{1}{\beta x}\right)}^{11}$and the coefficient of $x^{-9}$ in ${\left(\alpha x-\frac{1}{\beta x^3}\right)}^{11}$ are equal,

Now the $r^{th}$ term of ${\left(\alpha x^3+\frac{1}{\beta x}\right)}^{11}$ is given by $={}^{11}C_r·\frac{{\alpha }^{11-r}}{{\beta }^r}\times x^{3\left(11-r\right)-r}$

So, for $x^9$, $33-4r=9\Rightarrow r=6$

Hence, coefficient of ${\mathrm{x}}^9$ in ${\left(\alpha x^3+\frac{1}{\beta x}\right)}^{11}={}^{11}C_6·\frac{{\alpha }^5}{{\beta }^6}$

Similarly, for coefficient of $x^{-9}$ in ${\left(\alpha x-\frac{1}{\beta x^3}\right)}^{11}=-{}^{11}C_5·\frac{{\alpha }^6}{{\beta }^5}$

Now given both coefficient are equal,

So, $\frac{11}{C_6}·\frac{{\alpha }^5}{{\beta }^6}=-\frac{11}{C_5}·\frac{{\alpha }^6}{{\beta }^5}$

$\Rightarrow \frac{1}{\beta }=-\alpha$

$\Rightarrow \alpha \beta =-1$

$\Rightarrow (\alpha \beta {)}^2=1$