If f(x) satisfies the functional equation f(x)+6f(1/x)=35/3x-7/2, x∈R-{0} and if

Question

If f(x) satisfies the functional equation \(f(x)+6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{7}{2}, x \in R-\{0\}\ \text{and if} \lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)\) exist finitely and is equal to \(\beta,\) then \(( \alpha-2 \beta )\) is

Answer 1

Answer is "4"   

\(f(x)+6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{7}{2}\)  

\( f\left(\frac{1}{x}\right)+6 f(x)=\frac{35 x}{3}-\frac{7}{2} \)

\(36 f(x)+6 f\left(\frac{1}{x}\right)=70 x-21 \)

\(35 f(x)=(70 x-21)-\left[\frac{35}{3 x}-\frac{7}{2}\right] \)

\(35 f(x)=70 x-\frac{35}{3 x}-\frac{35}{2} \)

\(\Rightarrow f(x)=2 x-\frac{1}{3 x}-\frac{1}{2} \)

\(\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+2 x-\frac{1}{3 x}-\frac{1}{2}\right)=\beta\) 

Limit exist finitely if 

\(\alpha=3 \Rightarrow \beta=-\frac{1}{2} \)

\((\alpha-2 \beta)=3-2\left(-\frac{1}{2}\right)=4\)