Question

Let a function $f(x)$ satisfy \[ 3f(x)+2f\!\left(\frac{m}{19x}\right)=5x \] where $m=\sum_{i=1}^{9} i^2$. Find $f(5)+f(2)$.

Hint:

For functional equations, always try replacing $x$ by reciprocal expressions to form solvable systems.

Answer 1

Correct Answer: 0

Step 1: Find value of $m$.
\[ m=\frac{9\cdot10\cdot19}{6}=285 \Rightarrow \frac{m}{19x}=\frac{15}{x} \] Step 2: Substitute in functional equation.
\[ 3f(x)+2f\!\left(\frac{15}{x}\right)=5x \quad (1) \] Replace $x$ by $\frac{15}{x}$ in (1): \[ 3f\!\left(\frac{15}{x}\right)+2f(x)=\frac{75}{x} \quad (2) \] Step 3: Solve equations.
Multiply (1) by 3 and (2) by 2 and subtract: \[ 5f(x)=15x-\frac{150}{x} \] \[ f(x)=3x-\frac{30}{x} \] Step 4: Compute required sum.
\[ f(5)=15-6=9,\quad f(2)=6-15=-9 \] \[ f(5)+f(2)=0 \] Final conclusion.
The required value is $0$.