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Question

(A) Given that: $f(x) = \frac{2x+1}{3}$ ...$(i)$Also,$f(\alpha) = \frac{1}{\alpha}$.Substituting $\alpha$ in the function definition:$\frac{2\alpha+1}

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$\frac{-1}{2}$

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(A) Given that: $f(x) = \frac{2x+1}{3}$ ...$(i)$Also,$f(\alpha) = \frac{1}{\alpha}$.Substituting $\alpha$ in the function definition:$\frac{2\alpha+1}{3} = \frac{1}{\alpha}$$\Rightarrow 2\alpha^2 + \alpha = 3$$\Rightarrow 2\alpha^2 + \alpha - 3 = 0$Factoring the quadratic equation:$2\alpha^2 + 3\alpha - 2\alpha - 3 = 0$$\Rightarrow \alpha(2\alpha + 3) - 1(2\alpha + 3) = 0$$\Rightarrow (\alpha - 1)(2\alpha + 3) = 0$Thus,the possible values for $\alpha$ are $\alpha = 1$ and $\alpha = -\frac{3}{2}$.The sum of all possible values of $\alpha$ is $1 + (-\frac{3}{2}) = 1 - \frac{3}{2} = -\frac{1}{2}$.

Let $f: R \rightarrow R$ be a function defined by $f(x) = \frac{2x+1}{3}$. If $\alpha$ is an element in the domain of $f$ whose image is $\frac{1}{\alpha}$,then the sum of all possible values of such $\alpha$ is

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