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Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $\mathrm{f}(\mathrm{x})=\frac{2 \mathrm{x}+1}{3}$. If $\alpha$ is an element in the domain of $\mathrm{f}$ whose image is $\frac{1}{\alpha}$, then the sum of all possible values of such $\alpha$ is
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The correct answer is:
$\frac{-1}{2}$
Given that : $f(x)=\frac{2 x+1}{3}$ ...(i)
Also, $f(\alpha)=\frac{1}{\alpha}$
$\begin{aligned} & \Rightarrow \frac{2 \alpha+1}{3}=\frac{1}{\alpha} \\ & \Rightarrow 2 \alpha^2+\alpha=3 \Rightarrow 2 \alpha^2+\alpha-3=0 \\ & \Rightarrow(\alpha-1)(2 \alpha+3)=0 \\ & \Rightarrow \alpha=-\frac{3}{2}, 1\end{aligned}$
Then sum of all possible value of $\alpha=-\frac{3}{2}+1=-\frac{1}{2}$
Also, $f(\alpha)=\frac{1}{\alpha}$
$\begin{aligned} & \Rightarrow \frac{2 \alpha+1}{3}=\frac{1}{\alpha} \\ & \Rightarrow 2 \alpha^2+\alpha=3 \Rightarrow 2 \alpha^2+\alpha-3=0 \\ & \Rightarrow(\alpha-1)(2 \alpha+3)=0 \\ & \Rightarrow \alpha=-\frac{3}{2}, 1\end{aligned}$
Then sum of all possible value of $\alpha=-\frac{3}{2}+1=-\frac{1}{2}$
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