Let f(x) and g(x) satisfies the functional equation 2g(x) + 3g(1/x)=x and 2f(x) + 3f(1/x) = x^2+5.

Question

Let \(f(x)\) and g(x) satisfies the functional equation \(2 g(x)+3 g\left(\frac{1}{x}\right)=x\) and \(2 f(x)+3 f\left(\frac{1}{x}\right)=x^{2}+5.\)

If \(\alpha=\int_\limits{1}^{2} f(x) d x\) and \(\beta=\int_\limits{1}^{2} g(x) d x\) then \((9 \alpha+\beta)\) is equal to

(1) \(\frac{27+6 \ln _{2}}{10}\)

(2) \(\frac{27-6 \ln _{2}}{10}\)

(3) \(\frac{3}{5} \ln _{2}\)

(4) \(\frac{3}{5} \ln _{2}+\frac{7}{30}\)  

Answer 1

Correct option is: (1) \(\frac{27+6 \ln _{2}}{10}\)  

\(\alpha=\int_\limits{1}^{2} f(x) d x=\frac{11}{30^{2}} \text { and } \beta=\int_\limits{1}^{2} g(x) d x=\frac{3}{5} \ln x-\left.\frac{x^{2}}{5}\right|_{1} ^{2} \)

\(=\left(\frac{3}{5} \ln _{2}-\frac{4}{5}\right)-\left(\frac{-1}{5}\right)=\frac{3}{5}\left(\ln _{2}-1\right)\)