Mathematics - Functions Question with Solution | TestHub

$\begin{array}{rl}& f(x)+2f\left(\frac{1}{x}\right)=x^2+5\\ & f\left(\frac{1}{x}\right)+2f(x)=\frac{1}{x^2}+5\\ & f(x)=\frac{2}{3x^2}-\frac{x^2}{3}+\frac{5}{3}\\ & \alpha ={\int }_1^2\left(\frac{2}{3x^2}-\frac{x^2}{3}+\frac{5}{3}\right)\mathrm{dx}\\ & {\left(-\frac{2}{3x}-\frac{x^3}{9}+\frac{5x}{3}\right)}_1^2\\ & -\frac{1}{3}-\frac{8}{9}+\frac{10}{3}+\frac{2}{3}+\frac{1}{9}-\frac{5}{3}\\ & \alpha =2-\frac{7}{9}=\frac{11}{9}\\ & 2g(x)-3g\left(\frac{1}{2}\right)=x\\ & \mathrm{g}\left(\frac{1}{2}\right)=-\frac{1}{2}\\ & \mathrm{g}(\mathrm{x})=\frac{\mathrm{x}}{2}-\frac{3}{4}\\ & \beta ={\int }_1^2\left(\frac{x}{2}-\frac{3}{4}\right)dx\\ & {\left(\frac{x^2}{4}-\frac{3x}{4}\right)}_1^2=1-\frac{3}{2}-\frac{1}{4}+\frac{3}{4}=0\\ & 9\alpha +\beta =11\\ & \text{ option (4) }\end{array}$