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(C) Given equation: $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} \quad (1)$ Replacing $x$ with $\frac{1}{x}$ in $(1)$: $f\left(\fr

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$4$

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(C) Given equation: $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2} \quad (1)$ Replacing $x$ with $\frac{1}{x}$ in $(1)$: $f\left(\frac{1}{x}\right) - 6f(x) = \frac{35x}{3} - \frac{5}{2} \quad (2)$ Multiply $(2)$ by $6$: $6f\left(\frac{1}{x}\right) - 36f(x) = 70x - 15 \quad (3)$ Adding $(1)$ and $(3)$: $-35f(x) = \frac{35}{3x} - 5 + 70x - 15$ $-35f(x) = \frac{35}{3x} + 70x - 20$ $f(x) = -\frac{1}{3x} - 2x + \frac{4}{7}$ Given $\lim_{x \rightarrow 0} \left(\frac{1}{\alpha x} + f(x)\right) = \beta$: $\lim_{x \rightarrow 0} \left(\frac{1}{\alpha x} - \frac{1}{3x} - 2x + \frac{4}{7}\right) = \beta$ For the limit to exist,the term $\frac{1}{x}$ must vanish,so $\frac{1}{\alpha} - \frac{1}{3} = 0 \implies \alpha = 3$. Then $\beta = \lim_{x \rightarrow 0} (-2x + \frac{4}{7}) = \frac{4}{7}$. Thus,$\alpha + 2\beta = 3 + 2\left(\frac{4}{7}\right) = 3 + \frac{8}{7} = \frac{29}{7}$. *Correction*: Re-evaluating the system: $f(x) - 6f(1/x) = \frac{35}{3x} - \frac{5}{2}$ $f(1/x) - 6f(x) = \frac{35x}{3} - \frac{5}{2}$ Solving for $f(x)$: $f(x) = -\frac{1}{3x} - 2x + \frac{1}{2}$ $\lim_{x \rightarrow 0} (\frac{1}{\alpha x} - \frac{1}{3x} - 2x + \frac{1}{2}) = \beta \implies \alpha = 3, \beta = 1/2$. $\alpha + 2\beta = 3 + 2(1/2) = 4$.

Let $f : R - \{0\} \rightarrow R$ be a function such that $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$. If $\lim_{x \rightarrow 0} \left(\frac{1}{\alpha x} + f(x)\right) = \beta$,where $\alpha, \beta \in R$,then $\alpha + 2\beta$ is equal to:

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