Question

If $f\left(x\right)=\frac{5x+3}{6x+a}$ and $f\left(f\left(x\right)\right)=x$ then the value of $a$ is

• A. $-5$
• B. $5$
• C. $6$
• D. $-6$

Answer 1

The correct option is A

$-5$

Explanation for the correct option.

Step 1. Find the function $f\left(f\left(x\right)\right)$.

For the function $f\left(x\right)=\frac{5x+3}{6x+a}$, the composition function $f\left(f\left(x\right)\right)$ is given as:

$\begin{array}{rcl}f\left(f\left(x\right)\right)& =& \frac{5f\left(x\right)+3}{6f\left(x\right)+a}\\ & =& \frac{5\times {\displaystyle \frac{5x+3}{6x+a}}+3}{6\times {\displaystyle \frac{5x+3}{6x+a}}+a}\\ & =& \frac{5\left(5x+3\right)+3\left(6x+a\right)}{6\left(5x+3\right)+a\left(6x+a\right)}\\ & =& \frac{25x+15+18x+3a}{30x+18+6ax+a^2}\\ & =& \frac{43x+15+3a}{30x+6ax+18+a^2}\end{array}$

Step 2. Find the value of $a$.

It is given that $f\left(f\left(x\right)\right)=x$, so

$\begin{array}{l}\frac{43x+15+3a}{30x+6ax+18+a^2}\end{array}=x$

Form a quadratic equation in $x$.

$$\begin{gathered} \left(30+6a\right)x^2+\left(18+a^2-43\right)x-\left(3a+15\right)=0 \\ \Rightarrow 6\left(a+5\right)x^2+\left(a^2-25\right)x-3\left(a+5\right)=0 \\ \Rightarrow \left(a+5\right)\left[6x^2+\left(a-5\right)x-3\right]=0 \end{gathered}$$

So, $a+5=0$ and thus $a=-5$.

So the value of $a$ is $-5$.

Hence, the correct option is A.