JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 23)
If the co-efficient of $$x^9$$ in $${\left( {\alpha {x^3} + {1 \over {\beta x}}} \right)^{11}}$$ and the co-efficient of $$x^{-9}$$ in $${\left( {\alpha x - {1 \over {\beta {x^3}}}} \right)^{11}}$$ are equal, then $$(\alpha\beta)^2$$ is equal to ___________.
Answer
1
Explanation
Coefficient of $\mathrm{x}^{9}$ in $\left(\alpha x^{3}+\frac{1}{\beta x}\right)={ }^{11} C_{6} \cdot \frac{\alpha^{5}}{\beta^{6}}$
$\because$ Both are equal
$\therefore \frac{11}{C_{6}} \cdot \frac{\alpha^{5}}{\beta^{6}}=-\frac{11}{C_{5}} \cdot \frac{\alpha^{6}}{\beta^{5}}$
$\Rightarrow \frac{1}{\beta}=-\alpha$
$\Rightarrow \alpha \beta=-1$
$\Rightarrow(\alpha \beta)^{2}=1$
$\because$ Both are equal
$\therefore \frac{11}{C_{6}} \cdot \frac{\alpha^{5}}{\beta^{6}}=-\frac{11}{C_{5}} \cdot \frac{\alpha^{6}}{\beta^{5}}$
$\Rightarrow \frac{1}{\beta}=-\alpha$
$\Rightarrow \alpha \beta=-1$
$\Rightarrow(\alpha \beta)^{2}=1$
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