Let f(x)+2f((1)/(x))=x^2+5 and 2g(x)-3g((1)/(x))=x,x0.

Question

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

Accepted Answer

$\begin{gathered} f(x)+2f(\frac{1}{x})={x}^{2}+5 \ 2f(\frac{1}{x})+4f(x)=2(\frac{1}{{x}^{2}}+5) \ 3f(x)=\frac{2}{{x}^{2}}-{x}^{2}+5 \ f(x)=\frac{1}{3}(\frac{2}{{x}^{2}}-{x}^{2}+5) \ 2g(x)-3g(\frac{1}{x})=x \ 2g(\frac{1}{x})-3g(x)=\frac{1}{x} \ 4g(x)-6g(\frac{1}{x})=2x \ 6g(\frac{1}{x})-9g(x)=\frac{3}{x} \ -5g(x)=2x+\frac{3}{x} \end{gathered}$

$\begin{gathered} g(x)=-\frac{1}{5}(2x+\frac{3}{x}) \ {\int }_{2}^{1}f(x)dx={\int }_{2}^{1}\frac{1}{3}(\frac{2}{{x}^{2}}-{x}^{2}+5)dx \end{gathered}$

$\begin{gathered} =\frac{1}{3}{[-\frac{2}{x}-\frac{{x}^{3}}{3}+5x]}_{2}^{1} \ =\frac{1}{3}[(-\frac{2}{2}-\frac{8}{3}+10)-(-2-\frac{1}{3}+5)] \ =\frac{1}{3}[-1-\frac{8}{3}+10+2+\frac{1}{3}-5] \ \alpha =\frac{11}{9} \ 2g(x)=x+3g(\frac{1}{2}) \ 2g(\frac{1}{2})=\frac{1}{2}+3g(\frac{1}{2}) \ g(\frac{1}{2})=-\frac{1}{2} \end{gathered}$

$\begin{gathered} 	herefore \beta ={\int }_{2}^{1}g(x)dx \ =\frac{1}{2}{\int }_{2}^{1}(x+3g(\frac{1}{2}))dx \ =\frac{1}{2}{[\frac{{x}^{2}}{2}+3g(\frac{1}{2})x]}_{2}^{1} \ =0 \ 	herefore 9\alpha +\beta =11 \end{gathered}$

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