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

Question

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

Accepted Answer

$$\begin{aligned}
& f(x)+2 f\left(\frac{1}{x}\right)=x^2+5 \
& 2 f\left(\frac{1}{x}\right)+4 f(x)=2\left(\frac{1}{x^2}+5\right) \
& 3 f(x)=\frac{2}{x^2}-x^2+5 \
& f(x)=\frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) \
& 2 g(x)-3 g\left(\frac{1}{x}\right)=x \
& 2 g\left(\frac{1}{x}\right)-3 g(x)=\frac{1}{x} \
& \text { Or } 4 g(x)-6 g\left(\frac{1}{x}\right)=2 x \
& 6 g\left(\frac{1}{x}\right)-9 g(x)=\frac{3}{x} \
& -5 g(x)=2 x+\frac{3}{x}
\end{aligned}$$

$$
\begin{aligned}
& \text { Or } g(x)=-\frac{1}{5}\left(2 x+\frac{3}{x}\right) \
& \int_1^2 f(x) d x=\int_1^2 \frac{1}{3}\left(\frac{2}{x^2}-x^2+5\right) d x
\end{aligned}$$

$$
\begin{aligned}
&\begin{aligned}
& =\frac{1}{3}\left[-\frac{2}{x}-\frac{x^3}{3}+5 x\right]_1^2 \
& =\frac{1}{3}\left[\left(-\frac{2}{2}-\frac{8}{3}+10\right)-\left(-2-\frac{1}{3}+5\right)\right] \
& =\frac{1}{3}\left[-1-\frac{8}{3}+10+2+\frac{1}{3}-5\right] \
& \alpha=\frac{11}{9}
\end{aligned}\
\&\text { Now, } 2 g(x)=x+3 g\left(\frac{1}{2}\right)\
\&\begin{aligned}
& 2 g\left(\frac{1}{2}\right)=\frac{1}{2}+3 g\left(\frac{1}{2}\right) \
& g\left(\frac{1}{2}\right)=-\frac{1}{2}
\end{aligned}
\end{aligned}$$

$$\begin{aligned}
& \therefore \beta=\int_1^2 g(x) d x \
& =\frac{1}{2} \int_1^2\left(x+3 g\left(\frac{1}{2}\right)\right) d x \
& =\frac{1}{2}\left[\frac{x^2}{2}+3 g\left(\frac{1}{2}\right) x\right]_1^2 \
& =0 \
& \therefore 9 \alpha+\beta=11
\end{aligned}$$

Solutions · 1