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

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Aicharya · Accepted Answer

$\begin{aligned}
& f(x)+2 f\left(\frac{1}{x}\right)=x^2+5 \\ & f\left(\frac{1}{x}\right)+2 f(x)=\frac{1}{x^2}+5 \\ & f(x)=\frac{2}{3 x^2}-\frac{x^2}{3}+\frac{5}{3} \\ & \alpha=\int_1^2\left(\frac{2}{3 x^2}-\frac{x^2}{3}+\frac{5}{3}\right) \mathrm{dx} \\ & \left(-\frac{2}{3 x}-\frac{x^3}{9}+\frac{5 x}{3}\right)_1^2 \\ & -\frac{1}{3}-\frac{8}{9}+\frac{10}{3}+\frac{2}{3}+\frac{1}{9}-\frac{5}{3} \\ & \alpha=2-\frac{7}{9}=\frac{11}{9} \\ & 2 g(x)-3 g\left(\frac{1}{2}\right)=x \\ & \mathrm{~g}\left(\frac{1}{2}\right)=-\frac{1}{2} \\ & \mathrm{~g}(\mathrm{x})=\frac{\mathrm{x}}{2}-\frac{3}{4} \\ & \beta=\int_1^2\left(\frac{x}{2}-\frac{3}{4}\right) d x \\ & \left(\frac{x^2}{4}-\frac{3 x}{4}\right)_1^2=1-\frac{3}{2}-\frac{1}{4}+\frac{3}{4}=0 \\ & 9 \alpha+\beta=11 \\ & \text { option (4) }
\end{aligned}$

Let $f(x)+2 f\left(\frac{1}{x}\right)=x^2+5$ and $2 g(x)-3 g\left(\frac{1}{2}\right)=x, x \gt 0$. If…

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