JEE Main 2023 — Mathematics Algebra

Given,

The coefficient of ${x}^{9}$x9 in ${(\alpha {x}^{3}+\frac{1}{\beta x})}^{11}$(αx3+βx1​)11and the coefficient of ${x}^{-9}$x−9 in ${(\alpha x-\frac{1}{\beta {x}^{3}})}^{11}$(αx−βx31​)11 are equal,

Now the ${r}^{th}$rth term of ${(\alpha {x}^{3}+\frac{1}{\beta x})}^{11}$(αx3+βx1​)11 is given by $=Cr11\cdot \frac{{\alpha }^{11-r}}{{\beta }^{r}}\times {x}^{3(11-r)-r}$=Cr11⋅βrα11−r​×x3(11−r)−r

So, for ${x}^{9}$x9, $33-4r=9\Rightarrow r=6$33−4r=9⇒r=6

Hence, coefficient of ${x}^{9}$x9 in ${(\alpha {x}^{3}+\frac{1}{\beta x})}^{11}=C611\cdot \frac{{\alpha }^{5}}{{\beta }^{6}}$(αx3+βx1​)11=C611⋅β6α5​

Similarly, for coefficient of ${x}^{-9}$x−9 in ${(\alpha x-\frac{1}{\beta {x}^{3}})}^{11}=-C511\cdot \frac{{\alpha }^{6}}{{\beta }^{5}}$(αx−βx31​)11=−C511⋅β5α6​

Now given both coefficient are equal,

So, $\frac{11}{{C}_{6}}\cdot \frac{{\alpha }^{5}}{{\beta }^{6}}=-\frac{11}{{C}_{5}}\cdot \frac{{\alpha }^{6}}{{\beta }^{5}}$C6​11​⋅β6α5​=−C5​11​⋅β5α6​

$$\Rightarrow \frac{1}{\beta }=-\alpha$$⇒β1​=−α

$$\Rightarrow \alpha \beta =-1$$⇒αβ=−1

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