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7.Binomial Theorem
hard
If the co-efficient of $x^9$ in $\left(\alpha x^3+\frac{1}{\beta x}\right)^{11}$ and the co-efficient of $x^{-9}$ in $\left(\alpha x-\frac{1}{\beta x^3}\right)^{11}$ are equal, then $(\alpha \beta)^2$ is equal to $.............$.
A
$2$
B
$4$
C
$1$
D
$6$
(JEE MAIN-2023)
Solution
Coefficient of $x ^9$ in $\left(\alpha x^3+\frac{1}{\beta x}\right)={ }^{11} C_6 \cdot \frac{\alpha^5}{\beta^6}$
$\because$ Both are equal
$\therefore \frac{11}{C_6} \cdot \frac{\alpha^5}{\beta^6}=-\frac{11}{C_5} \cdot \frac{\alpha^6}{\beta^5}$
$\Rightarrow \frac{1}{\beta}=-\alpha$
$\Rightarrow \alpha \beta=-1$
$\Rightarrow(\alpha \beta)^2=1$
Standard 11
Mathematics
