Let f be a function such that 3f(x)+2f((m)/(19x))=5x,x 0, where m= _i=1^9(i)^2.

Question

Let $f$ be a function such that $3f(x)+2f(\frac{m}{19x})=5x,x
e 0$, where $m=\sum _{i=1}^{9}(i{)}^{2}$. Then $f(5)-f(2)$ is equal to

Accepted Answer

Given $m=\sum _{9}^{i=1}(i{)}^{2}={1}^{2}+{2}^{2}+\cdots 9$

We know, ${1}^{2}+{2}^{2}+{3}^{2}+\cdots {n}^{2}=\frac{n(n+1)(2n+1)}{6}$

so, $m=\frac{9	imes 10	imes 19}{6}=3	imes 5	imes 19=285$

Given equation

$3f(x)+2f(\frac{m}{19x})=5x,x
e 0$

put $m=285,3f(x)+2f(\frac{285}{19x})=5x$

$3f(x)+2f(\frac{15}{x})=5x-(1)$

put $x=\frac{15}{x}$

$\begin{gathered} 3f(\frac{15}{x})+2f(\frac{15}{15/x})=5	imes \frac{15}{x} \ 3f(\frac{15}{x})+2f(x)=\frac{75}{x}-(2) \end{gathered}$

from (1) $f(\frac{15}{x})=\frac{5x-3f(x)}{2}$ substitute in equation (2)

$\begin{gathered} 3[\frac{5x-3f(x)}{2}]+2f(x)=\frac{75}{x} \ 15x-9f(x)+4f(x)=\frac{150}{x} \ 5f(x)=15x-\frac{150}{x} \ f(x)=3x-\frac{30}{x} \end{gathered}$

to find $f(5)-f(2)$

$\begin{gathered} f(5)-f(2)=(3(5)-\frac{30}{5})-(3	imes 2-\frac{30}{2}) \ =(15-6)-(6-15) \end{gathered}$

= $9-(-9)=18$

Solutions · 1