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

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Aicharya · Accepted Answer

$\begin{aligned}
& f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}...(1) \\ & 6\left(f\left(\frac{1}{x}\right)-6 f(x)=\frac{35 x}{3}-\frac{5}{2}\right) \\ & 6 f\left(\frac{1}{x}\right)-36 f(x)=\frac{210 x}{3}-\frac{30}{2}...(2) \\ & (1)+(2) \\ & -35 f(x)=\frac{35}{3}\left[\frac{1}{x}+6 x\right]-\frac{5}{2}(1+6) \\ & -f(x)=\frac{1}{3}\left(\frac{1}{x}+6 x\right)-\frac{1}{2} \\ & f(x)=-\frac{1}{3 x}-2 x+\frac{1}{2} \\ & \lim _{x \rightarrow 0}\left[\frac{1}{\alpha x}-\frac{1}{3 x}-2 x+\frac{1}{2}\right]=\beta \\ & \Rightarrow \alpha=3 \\ & \beta+2 \beta=3+2 \times \frac{1}{2}=4
\end{aligned}$ .

Let $f: R-\{0\} \rightarrow R$ be a function such that $f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3…

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