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If the sum of all the coefficients of $\left(\alpha x^2-2 x+1\right)^{2019}$ is equal to the sum of all the coefficients of $(x-\alpha y)^{2019}$, then $\alpha=$
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The correct answer is:
1
The sum of the coefficient of $\left(\alpha x^2-2 x+1\right)^{2019}$ is $(\alpha-1)^{2019}$. (On putting $\chi=1$ )
and similarly the sum of the coefficients of $(x-\alpha y)^{2019}$ is $(1-\alpha)^{2019}$ (on putting $x=y=1$ )
Now according to the question,
$$
\begin{aligned}
& (\alpha-1)^{2019}=(1-\alpha)^{2019} \\
& \Rightarrow \alpha-1=1-\alpha \Rightarrow \alpha=1
\end{aligned}
$$
and similarly the sum of the coefficients of $(x-\alpha y)^{2019}$ is $(1-\alpha)^{2019}$ (on putting $x=y=1$ )
Now according to the question,
$$
\begin{aligned}
& (\alpha-1)^{2019}=(1-\alpha)^{2019} \\
& \Rightarrow \alpha-1=1-\alpha \Rightarrow \alpha=1
\end{aligned}
$$
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