Let \(\mathrm{f}:\mathrm{ℝ}-{0}\to \mathrm{ℝ}\) be a function such that \(\mathrm{f}(\mathrm{x})-6\mathrm{f}\left(\frac{…

\(f(x)-6f\left(\frac{1}{x}\right)=\frac{35}{3x}-\frac{5}{2}......(1)\\ f\left(\frac{1}{x}\right)-6f(x)=\left(\frac{35x}{3}-\frac{5}{2}\right)\\ 6f\left(\frac{1}{x}\right)-36f(x)=\left(\frac{35x}{3}-\frac{5}{2}\right)\times 6...(2)\\ \text{Add (1) and (2),we get}\\ -35f(x)=\frac{35}{3x}-\frac{5}{2}+70x-15\\ -35f(x)=70x+\frac{35}{3x}-\frac{35}{2}\\ f(x)=\frac{1}{2}-2x-\frac{1}{3x}\\ \lim _{x\to 0}(\frac{1}{\alpha x}+\frac{1}{2}-2x-\frac{1}{3x})=\beta \\ =\underset{x\to 0}{\lim (}\left(\frac{1}{\alpha }-\frac{1}{3}\right)\frac{1}{x}+\frac{1}{2}-2x)=\beta \\ \alpha =3\beta =\frac{1}{2}\\ now,\alpha +2\beta =3+1=4\)