7.Binomial Theorem
hard

જો $\left(\alpha x^3+\frac{1}{\beta x}\right)^{11}$ માં $x^9$ નો સહગુણક અને $\left(\alpha x-\frac{1}{\beta x^3}\right)^{11}$ માં $x^{-9}$ નો સહગુણક સરખા હોય,તો $(\alpha \beta)^2=........$

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

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