Question

If \[f\left( x \right) = \dfrac{{5x + 3}}{{6x + a}}\] and f(f(x)) = x then the value of a is:
A. -5
B. 5
C. 6
D. -6

Answer 1

Hint: A function contains a codomain, often known as a range, and a domain that are used to find certain values of the function f(x) for all the values of x that are defined in the function. The values of x for which the denominator of the function becomes zero are not defined for the function. We will find the values of a and x which will satisfy the equation f(f(x)) = x, from which we can also determine the value of a.

Formula Used: f(f(x)) or fof(x) is found by replacing every x in f(x) by f(x).

Complete step-by-step solution:
We have been given a function \[f\left( x \right) = \dfrac{{5x + 3}}{{6x + a}}\]. We will now find the value of f(f(x)) and equate it to x as,
\[
  f\left( {f\left( x \right)} \right) = x \\
   \Rightarrow \dfrac{{5\left( {f\left( x \right)} \right) + 3}}{{6\left( {f\left( x \right)} \right) + a}} = x \\
   \Rightarrow 5\left( {\dfrac{{5x + 3}}{{6x + a}}} \right) + 3 = x\left( {6\left( {\dfrac{{5x + 3}}{{6x + a}}} \right) + a} \right) \\
   \Rightarrow \dfrac{{5\left( {5x + 3} \right) + 3\left( {6x + a} \right)}}{{\left( {6x + a} \right)}} = \dfrac{{6x\left( {5x + 3} \right) + ax\left( {6x + a} \right)}}{{\left( {6x + a} \right)}}
 \]
For denominator equal to zero, the function is not defined, so,
\[
  6x + a \ne 0 \\
   \Rightarrow a \ne - 6x
 \]

Cancelling the denominator on both sides, as they are equal, we get,
\[
  5\left( {5x + 3} \right) + 3\left( {6x + a} \right) = 6x\left( {5x + 3} \right) + ax\left( {6x + a} \right) \\
   \Rightarrow 25x + 15 + 18x + 3a = 30{x^2} + 18x + 6a{x^2} + {a^2}x \\
   \Rightarrow 6\left( {a + 5} \right){x^2} + \left( {a + 5} \right)\left( {a - 5} \right)x - 3\left( {a + 5} \right) = 0 \\
   \Rightarrow \left( {a + 5} \right)\left( {6{x^2} + \left( {a - 5} \right)x - 3} \right) = 0
 \]
So, either $a + 5 = 0$ or \[6{x^2} + \left( {a - 5} \right)x - 3 = 0\]. So, from the first equation $a = -5.$

So, option A, $a = -5$ is the required solution.

Additional Information: f(f(x)) of fof(x) is known as a function inside the same function. The functions like fog(x) or gof(x) are a few examples of a function inside another function.

Note: When we factorize any equation, make sure that the denominator is not equal to zero, otherwise, the equation is not defined for that value. Instead of solving the function with x as the variables, try to factorize and take out the common terms, to check whether we can obtain the required solution or not and if not, only then try to find the solution with variable x in it.