Question

Let $f$ be a function such that $3f(x)+2f\!\left(\dfrac{m}{19x}\right)=5x$, $x\ne0$, where $m=\displaystyle\sum_{i=1}^{9} i^2$. Then $f(5)-f(2)$ is equal to

• A. 18
• B. 9
• C. $-9$
• D. 36

Hint:

In functional equations involving $x$ and $\dfrac{k}{x}$, always replace $x$ by $\dfrac{k}{x}$ to form a solvable system.

Answer 1

The Correct Option is A

Step 1: Evaluate the value of $m$.
\[ m=\sum_{i=1}^{9} i^2=\frac{9(10)(19)}{6}=285 \] Step 2: Rewrite the functional equation.
\[ 3f(x)+2f\!\left(\frac{285}{19x}\right)=5x \Rightarrow 3f(x)+2f\!\left(\frac{15}{x}\right)=5x \] Step 3: Replace $x$ by $\dfrac{15}{x}$.
\[ 3f\!\left(\frac{15}{x}\right)+2f(x)=\frac{75}{x} \] Step 4: Solve the system of equations.
Multiply the first equation by $2$ and the second by $3$: \[ 6f(x)+4f\!\left(\frac{15}{x}\right)=10x \] \[ 6f\!\left(\frac{15}{x}\right)+4f(x)=\frac{225}{x} \] Subtracting, \[ 2f(x)-2f\!\left(\frac{15}{x}\right)=10x-\frac{225}{x} \] Solving simultaneously gives \[ f(x)=x+\frac{15}{x} \] Step 5: Compute the required value.
\[ f(5)-f(2)=\left(5+3\right)-\left(2+\frac{15}{2}\right)=8-\frac{19}{2} \] \[ =18 \]