Question

If $f\left( x \right) = \dfrac{x}{{x - 1}}$ then $\dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}}$ =
A.$f\left( {2a} \right)$
B.$f\left( {{a^2}} \right)$
C.$f\left( {a - 1} \right)$
D.$f\left( {a + 1} \right)$

Answer 1

Hint: We will find the value of $f(a)$ and $f(a + 1)$ using the given function defined as$f\left( x \right) = \dfrac{x}{{x - 1}}$. Then, we will put them in the given function $\dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}}$to find the required value and then, we will again use the given function to compare the obtained equation and check which option is correct.

Complete step-by-step answer:
We are given that $f\left( x \right) = \dfrac{x}{{x - 1}}$.
We need to calculate the value of $\dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}}$and then check which option is correct.
Let us first calculate the value of $f(a)$ and $f(a + 1)$ using $f\left( x \right) = \dfrac{x}{{x - 1}}$as:
$ \Rightarrow f(a) = \dfrac{a}{{a - 1}}$ and $f(a + 1) = \dfrac{{a + 1}}{{(a + 1) - 1}} = \dfrac{{a + 1}}{a}$
Putting these values in $\dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}}$to find the required value, we get
$ \Rightarrow \dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}} = \dfrac{{\dfrac{a}{{a - 1}}}}{{\dfrac{{a + 1}}{a}}}$
$ \Rightarrow \dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}} = \dfrac{a}{{a - 1}} \times \dfrac{a}{{a + 1}} = \dfrac{{a(a)}}{{\left( {a - 1} \right)\left( {a + 1} \right)}}$
Using the algebraic identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ to simplify the denominator of the above equation, we get
$ \Rightarrow \dfrac{{f(a)}}{{f(a + 1)}} = \dfrac{{{a^2}}}{{{a^2} - 1}}$
We have obtained the value of $\dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}}$as $\dfrac{{{a^2}}}{{{a^2} - 1}}$ . Comparing it with the definition of $f(x)$ given as $f\left( x \right) = \dfrac{x}{{x - 1}}$, we get
$ \Rightarrow f\left( {{a^2}} \right) = \dfrac{{{a^2}}}{{{a^2} - 1}}$
Therefore, we can say that $\dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}} = f\left( {{a^2}} \right)$
Hence, option (B) is correct.

Note: In this question, you may get confused in the comparison of the obtained value $\dfrac{{{a^2}}}{{{a^2} - 1}}$with the definition of $f(x)$ given in the question. You can also solve for the correct option by simply calculating the given functions in the options defined as $f\left( x \right) = \dfrac{x}{{x - 1}}$ of each option and then you can check which option matches the obtained value of $\dfrac{{f\left( a \right)}}{{f\left( {a + 1} \right)}}$, whichever does, is the correct option. We have used an algebraic identity to simplify the denominator. An algebraic identity can be defined as an equality which holds true for each and every real value of the variables used in the equality.