Let [5 f(x)+4 f((1)/(x))=(1)/(x)+3, x 0].

Question

Let $$5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0$$. Then $$18 \int_\limits{1}^{2} f(x) d x$$ is equal to :

Accepted Answer

We have,

$$
5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0
$$ ..........(i)

On replacing $x$ by $\frac{1}{x}$ in (i), we get

$$
5 f\left(\frac{1}{x}\right)+4 f(x)=x+3
$$ ..........(ii)

Now, using Eq. (i) $\times 5-$ (ii) $\times 4$, we get

$$
\begin{aligned}
& 25 f(x)-16 f(x)  =\left(\frac{5}{x}+15\right)-(4 x+12) \\
&\Rightarrow 9 f(x)  =\frac{5}{x}-4 x+3 \\
&\Rightarrow f(x)  =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii)
\end{aligned}
$$

$$
\begin{aligned}
\therefore \quad & 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x  \text { [Using Eq. (iii)] }\\
= & 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x
\end{aligned}
$$

$$
\begin{aligned}
& =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\
& =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\
& =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\
& =10 \log _e 2-6
\end{aligned}
$$

Solutions · 1