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Question

Question: If we have a function as \(f\left( x \right)=\dfrac{x+2}{3x-1}\), then \[ff\left( x \right)\] is, ...

If we have a function as f(x)=x+23x1f\left( x \right)=\dfrac{x+2}{3x-1}, then ff(x)ff\left( x \right) is,
A. xx
B. x-x
C. 1x\dfrac{1}{x}
D. 1x-\dfrac{1}{x}
E. 0

Explanation

Solution

Hint: We will be using the concept of functions to solve the problem. We will be using the concepts of compound function to find the value of ff(x)ff\left( x \right) and further simplify the solution.

Complete step-by-step solution-
Now, we have been given that f(x)=x+23x1f\left( x \right)=\dfrac{x+2}{3x-1} and we have to find the value of ff(x)ff\left( x \right).
Now, we know that fg(x)fg\left( x \right) means f(g(x))f\left( g\left( x \right) \right). So, we have to find f(f(x))f\left( f\left( x \right) \right).
Now, for this we have to substitute f(x) for x in f(x). Therefore, we have,
f(f(x))=f(x)+23f(x)1f\left( f\left( x \right) \right)=\dfrac{f\left( x \right)+2}{3f\left( x \right)-1}
Now, we will put f(x)=x+23x1f\left( x \right)=\dfrac{x+2}{3x-1} and further simplify it. So, that we have,
=x+23x1+23(x+2)3x11=\dfrac{\dfrac{x+2}{3x-1}+2}{\dfrac{3\left( x+2 \right)}{3x-1}-1}
Now, we will take (3x1)\left( 3x-1 \right) as LCM in both numerator and denominator,
=x+2+2(3x1)3(x+2)1(3x1)=\dfrac{x+2+2\left( 3x-1 \right)}{3\left( x+2 \right)-1\left( 3x-1 \right)}
Now, we will expand in numerator and denominator and solve them,
=x+2+6x23x+63x+1=\dfrac{x+2+6x-2}{3x+6-3x+1}
On further simplifying we have,
=7x7 =x \begin{aligned} & =\dfrac{7x}{7} \\\ & =x \\\ \end{aligned}
So, on simplifying we have ff(x)=xff\left( x \right)=x.
Hence, the correct option is (A).

Note: To solve these types of questions one must know the basic concepts of compound function that ff(x)=f(f(x))ff\left( x \right)=f\left( f\left( x \right) \right). Also it is important to note that to find f(f(x))f\left( f\left( x \right) \right) we have put f(x)f\left( x \right) in f(x)f\left( x \right).