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

We are given two functional equations for $$x \gt 0$$:
  $$f(x)+2f\!\left(\dfrac{1}{x}\right)=x^{2}+5$$   $$-(1)$$
  $$2g(x)-3g\!\left(\dfrac{1}{2}\right)=x$$   $$-(2)$$
and we need the value of $$9\alpha+\beta$$ where
  $$\alpha=\int_{1}^{2}f(x)\,dx,\qquad \beta=\int_{1}^{2}g(x)\,dx.$$

Case 1: Determining $$f(x)$$

Write equation $$-(1)$$ for $$x$$ and for $$\dfrac{1}{x}$$.
For $$x$$ we already have
  $$f(x)+2f\!\left(\dfrac{1}{x}\right)=x^{2}+5.$$
Replace $$x$$ by $$\dfrac{1}{x}$$ to get
  $$f\!\left(\dfrac{1}{x}\right)+2f(x)=\dfrac{1}{x^{2}}+5.$$   $$-(3)$$

Let
  $$A=f(x),\qquad B=f\!\left(\dfrac{1}{x}\right).$$
Then $$-(1)$$ and $$-(3)$$ become a linear system

$$\begin{cases} A+2B = x^{2}+5,\\ 2A+B = \dfrac{1}{x^{2}}+5. \end{cases}$$

Solve for $$A$$ and $$B$$.
Multiply the first equation by $$2$$ and subtract the second:

$$2A+4B-(2A+B)=2(x^{2}+5)-\left(\dfrac{1}{x^{2}}+5\right).$$

That gives $$3B=2x^{2}+10-\dfrac{1}{x^{2}}-5,$$ hence
  $$B=\dfrac{2x^{2}+5-\dfrac{1}{x^{2}}}{3}.$$

Substitute $$B$$ back into $$A+2B=x^{2}+5$$:

$$A=x^{2}+5-2B=x^{2}+5-\dfrac{2\!\left(2x^{2}+5-\dfrac{1}{x^{2}}\right)}{3}.$$

Simplify the numerator:
$$3x^{2}+15-\left(4x^{2}+10-\dfrac{2}{x^{2}}\right)=-x^{2}+5+\dfrac{2}{x^{2}}.$$
Therefore

$$f(x)=A=\dfrac{-x^{2}+5+\dfrac{2}{x^{2}}}{3}.$$

Case 2: Evaluating $$\alpha=\displaystyle\int_{1}^{2}f(x)\,dx$$

Write the integrand term-by-term:

$$\alpha=\frac{1}{3}\int_{1}^{2}\!\left(-x^{2}+5+\frac{2}{x^{2}}\right)dx =\frac{1}{3}\Bigg[\int_{1}^{2}-x^{2}\,dx+\int_{1}^{2}5\,dx+\int_{1}^{2}\frac{2}{x^{2}}\,dx\Bigg].$$

Compute each integral:

$$\int_{1}^{2}-x^{2}\,dx=-\left[\frac{x^{3}}{3}\right]_{1}^{2}= -\left(\frac{8}{3}-\frac{1}{3}\right)= -\frac{7}{3},$$
$$\int_{1}^{2}5\,dx=5[x]_{1}^{2}=5(2-1)=5,$$
$$\int_{1}^{2}\frac{2}{x^{2}}\,dx=2\left[-\frac{1}{x}\right]_{1}^{2}=2\!\left(-\frac{1}{2}+1\right)=1.$$

Add the results:
$$-\frac{7}{3}+5+1=-\frac{7}{3}+6=\frac{11}{3}.$$

Hence
$$\alpha=\frac{1}{3}\cdot\frac{11}{3}=\frac{11}{9}.$$

Case 3: Determining $$g(x)$$

From equation $$-(2)$$:
  $$2g(x)-3g\!\left(\dfrac{1}{2}\right)=x \;\;\Longrightarrow\;\; g(x)=\frac{x}{2}+\frac{3}{2}\,g\!\left(\dfrac{1}{2}\right).$$   $$-(4)$$

Put $$x=\dfrac{1}{2}$$ in $$-(2)$$ to find the constant $$g\!\left(\dfrac{1}{2}\right)$$:

$$2g\!\left(\dfrac{1}{2}\right)-3g\!\left(\dfrac{1}{2}\right)=\dfrac{1}{2} \;\;\Longrightarrow\;\; -g\!\left(\dfrac{1}{2}\right)=\dfrac{1}{2} \;\;\Longrightarrow\;\; g\!\left(\dfrac{1}{2}\right)=-\dfrac{1}{2}.$$

Substitute this value into $$-(4)$$ to get an explicit formula:

$$g(x)=\frac{x}{2}+\frac{3}{2}\!\left(-\frac{1}{2}\right) =\frac{x}{2}-\frac{3}{4}.$$

Case 4: Evaluating $$\beta=\displaystyle\int_{1}^{2}g(x)\,dx$$

$$\beta=\int_{1}^{2}\left(\frac{x}{2}-\frac{3}{4}\right)dx =\frac{1}{2}\int_{1}^{2}x\,dx-\frac{3}{4}\int_{1}^{2}dx.$$

Compute the two integrals:

$$\frac{1}{2}\left[\frac{x^{2}}{2}\right]_{1}^{2} =\frac{1}{2}\left(\frac{4}{2}-\frac{1}{2}\right) =\frac{1}{2}\left(\frac{3}{2}\right)=\frac{3}{4},$$
$$\frac{3}{4}[x]_{1}^{2}=\frac{3}{4}(2-1)=\frac{3}{4}.$$

Hence $$\beta=\frac{3}{4}-\frac{3}{4}=0.$$

Case 5: Final computation

$$9\alpha+\beta=9\!\left(\frac{11}{9}\right)+0=11.$$

The required value is $$11$$, which corresponds to Option D.