Question

Find the value of the function at $x=2$ denoted by $f\left( 2 \right)$ where the function is governed by the functional equation $af\left( x \right)+bf\left( \dfrac{1}{x} \right)=x-1$ subjected to the conditions $x\ne 0$ and $a\ne b$.

Answer 1

Hint: Substitute $x=2$ and $x=\dfrac{1}{2}$ in the given functional equation. The functional equation will be converted to a linear equation and you will obtain a pair of linear equations with two coefficients $a$ and $b$. Proceed by the method of elimination to eliminate the unneeded unknown in the system of equations to express $f\left( 2 \right)$ in terms of $a$ and $b$.

Complete step-by-step solution:
The given functional equation is,
$af\left( x \right)+bf\left( \dfrac{1}{x} \right)=x-1…………........\left( 1 \right)$ \[\]
Substituting $x=2$ in the above equation (1)\[\]
$af\left( 2 \right)+bf\left( \dfrac{1}{2} \right)=2-1=1................(2)$
Substituting $x=\dfrac{1}{2}$ in the above equation (1)
$af\left( \dfrac{1}{2} \right)+bf\left( \dfrac{1}{\dfrac{1}{2}} \right)=af\left( \dfrac{1}{2} \right)+bf\left( 2 \right)=\dfrac{1}{2}-1=-\dfrac{1}{2}..............(3)$
Now the obtained equations (2) and (3) are a pair of linear equations with coefficients $a$ and $b$. The unknowns in the obtained equation are $f\left( 2 \right)$ and $f\left( \dfrac{1}{2} \right)$ . We proceed to eliminate $f\left( \dfrac{1}{2} \right)$ from the system of linear of equations.\[\]
Let us multiply $a$ with equation (3)
$\begin{align}
  & a\cdot \text{equation}\left( 3 \right)=a\cdot af\left( 2 \right)+a\cdot bf\left( \dfrac{1}{2} \right)=a\cdot 1 \\
 & \Rightarrow {{a}^{2}}f\left( 2 \right)+abf\left( \dfrac{1}{2} \right)=a………...(4) \\
\end{align}$ \[\]
Similarly let us multiply $b$ with equation (3)
$\begin{align}
  & b\cdot \text{equation}(4)=b\cdot af\left( \dfrac{1}{2} \right)+b\cdot bf\left( 2 \right)=-\dfrac{1}{2}\cdot b \\
 & \Rightarrow abf\left( \dfrac{1}{2} \right)+{{b}^{2}}f\left( 2 \right)=\dfrac{-b}{2}.....(5) \\
\end{align}$\[\]
Subtracting equation (5) from equation (4).
Equation (5)- Equation (6)=
$\begin{align}
  & {{a}^{2}}f\left( 2 \right)+abf\left( \dfrac{1}{2} \right)-abf\left( \dfrac{1}{2} \right)-{{b}^{2}}f\left( 2 \right)=a+\dfrac{b}{2} \\
 & \Rightarrow {{a}^{2}}f\left( 2 \right)-{{b}^{2}}f\left( 2 \right)=\dfrac{2a+b}{2} \\
 & \Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right)f\left( 2 \right)=\dfrac{2a+b}{2} \\
 & \Rightarrow f\left( 2 \right)=\dfrac{2a+b}{2\left( {{a}^{2}}-{{b}^{2}} \right)} \\
\end{align}$
So the value of $f\left( 2 \right)$ is found to be $\dfrac{2a+b}{2\left( {{a}^{2}}-{{b}^{2}} \right)}$. We can check that the $f\left( 2 \right)$ is not defined when $a=b$. That is why the question already mentions the favourable condition.

Note: The question combines the concept of linear and functional equations. While solving functional equations the key is proper substitution which is in this case 2 and $\dfrac{1}{2}$. If we cannot find the right substitution then we cannot transform the functional equation to simple linear equations. So we need to be careful while substituting.