Let $f(x)+2 f\left(\frac{1}{x}\right)=x^2+5$ and $2 g(x)-3 g\left(\frac{1}{2}\right)=x, x>0$. If $\alpha=\int_1^2 f(x) \mathrm{d} x$, and $\beta=\int_1^2 g(x) \mathrm{d} x$, then the value of $9 \alpha+\beta$ is :

<p>$$\begin{aligned} & f(x)+2 f\left(\frac{1}{x}\right)=x^2+5 \\ & 2 f\left(\frac{1}{x}\right)+4 f(x)=2\left(\frac{1}{x^2}+5\right) \\ & 3 f(x)=\frac{2}{x^2}-x^2+5 \\ & f(x)=\frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) \\ & 2 g(x)-3 g\left(\frac{1}{x}\right)=x \\ & 2 g\left(\frac{1}{x}\right)-3 g(x)=\frac{1}{x} \\ & \text { Or } 4 g(x)-6 g\left(\frac{1}{x}\right)=2 x \\ & 6 g\left(\frac{1}{x}\right)-9 g(x)=\frac{3}{x} \\ & -5 g(x)=2 x+\frac{3}{x} \end{aligned}$$</p> <p>$$ \begin{aligned} & \text { Or } g(x)=-\frac{1}{5}\left(2 x+\frac{3}{x}\right) \\ & \int_1^2 f(x) d x=\int_1^2 \frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) d x \end{aligned}$$</p> <p>$$ \begin{aligned} &\begin{aligned} & =\frac{1}{3}\left[-\frac{2}{x}-\frac{x^3}{3}+5 x\right]_1^2 \\ & =\frac{1}{3}\left[\left(-\frac{2}{2}-\frac{8}{3}+10\right)-\left(-2-\frac{1}{3}+5\right)\right] \\ & =\frac{1}{3}\left[-1-\frac{8}{3}+10+2+\frac{1}{3}-5\right] \\ & \alpha=\frac{11}{9} \end{aligned}\\ &\text { Now, } 2 g(x)=x+3 g\left(\frac{1}{2}\right)\\ &\begin{aligned} & 2 g\left(\frac{1}{2}\right)=\frac{1}{2}+3 g\left(\frac{1}{2}\right) \\ & g\left(\frac{1}{2}\right)=-\frac{1}{2} \end{aligned} \end{aligned}$$</p> <p>$$\begin{aligned} & \therefore \beta=\int_1^2 g(x) d x \\ & =\frac{1}{2} \int_1^2\left(x+3 g\left(\frac{1}{2}\right)\right) d x \\ & =\frac{1}{2}\left[\frac{x^2}{2}+3 g\left(\frac{1}{2}\right) x\right]_1^2 \\ & =0 \\ & \therefore 9 \alpha+\beta=11 \end{aligned}$$</p>