Let 5f(x)+4f((1)/(x))=(1)/(x)+3,x0.

Question

Let $5f(x)+4f(\frac{1}{x})=\frac{1}{x}+3,x>0$. Then $18\int _{1}^{2}f(x)dx$ is equal to :

Accepted Answer

We have,

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

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

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

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

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

$\begin{gathered} 	herefore \, 18{\int }_{2}^{1}f(x)dx=18{\int }_{2}^{1}\frac{1}{9}(\frac{5}{x}-4x+3)dx \ \ = 2{\int }_{2}^{1}(\frac{5}{x}-4x+3)dx \end{gathered}$

$\begin{gathered} =2{[5{\log }_{e}x-4(\frac{{x}^{2}}{2})+3x]}_{2}^{1} \ =2[(5{\log }_{e}2-2(2{)}^{2}+3(2))-(5\log 1-2(1{)}^{2}+3(1))] \ =2[5{\log }_{e}2-8+6+2-3][\because \log 1=0] \ =10{\log }_{e}2-6 \end{gathered}$

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