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

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$F(x)-6 f(1 / x)=\frac{35}{3 x}-\frac{5}{2}\quad\text{..... (1)}$

$$\begin{aligned} &\begin{aligned} & \text { Replace } x \rightarrow \frac{1}{x} \\ & F(1 / x)-6(x)=\frac{35 x}{3}-\frac{5}{2}\quad\text{..... (2)} \end{aligned}\\ &\text { Using (1) & (2) }\\ &\begin{aligned} & f(x)=-2 x-\frac{1}{3 x}+\frac{1}{2} \\ & B=\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right) \end{aligned} \end{aligned}$$

$$\begin{aligned} &= \lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}-2 x-\frac{1}{3 x}+\frac{1}{2}\right) \\ & \alpha=3, \quad B=1 / 2 \\ & \text { So, } \alpha+2 B=3+1=4 \end{aligned}$$

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

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

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