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

Question

Let $f(x)+2f\left(\dfrac1x\right)=x^2+5$ and $2g(x)-3g\left(\dfrac1x\right)=x$, $x>0$. If $\alpha=\displaystyle\int_1^2f(x)\,dx$ and $\beta=\displaystyle\int_1^2g(x)\,dx$, then the value of $9\alpha+\beta$ is:

Accepted Answer

From $f(x)+2f(1/x)=x^2+5$ and (replacing $x\to1/x$) $f(1/x)+2f(x) = 1/x^2+5$, solving simultaneously:

$$f(x) = \frac13\left(\frac2{x^2}-x^2+5\right)$$

$$\alpha = \int_1^2 f(x)\,dx = \frac13\left[-\frac2x-\frac{x^3}3+5x\right]_1^2 = \frac{11}{9}$$

From $2g(x)-3g(1/x)=x$ and $2g(1/x)-3g(x)=1/x$: solving gives $g(x)=-\dfrac15\left(2x+\dfrac3x\right)$, and $g(1/2)=-\dfrac12$.

$$\beta = \int_1^2 g(x)\,dx = 0$$

$$9\alpha+\beta = 9\cdot\frac{11}9+0 = 11$$

Solutions · 1