Question

For x\in R-\left\\{ 0,1 \right\\} , let ${{f}_{1}}\left( x \right)=\dfrac{1}{x}$,${{f}_{2}}\left( x \right)=1-x$ and ${{f}_{3}}\left( x \right)=\dfrac{1}{1-x}$ be three given functions. If a function, $J\left( x \right)$ satisfies $\left( f_{2}^{{}^\circ }{{J}^{{}^\circ }}{{f}_{1}} \right)\left( x \right)={{f}_{3}}\left( x \right)$ then $J\left( x \right)$ is equal to
(A) ${{f}_{3}}\left( x \right)$
(B) ${{f}_{1}}\left( x \right)$
(C) ${{f}_{2}}\left( x \right)$
(D) $\dfrac{1}{x}{{f}_{3}}\left( x \right)$

Answer 1

For answering this question we will use the given definitions of the functions ${{f}_{1}}\left( x \right)=\dfrac{1}{x},{{f}_{2}}\left( x \right)=1-x\text{ and }{{f}_{3}}\left( x \right)=\dfrac{1}{1-x}$ and simplify the given expression $\left( f_{2}^{{}^\circ }{{J}^{{}^\circ }}{{f}_{1}} \right)\left( x \right)={{f}_{3}}\left( x \right)$ by deriving the inverse of ${{f}_{2}}$ as $f_{2}^{-1}\left( x \right)=1-x$. And derive the definition of $J\left( x \right)$ and compare them with the functions we have and strive to a conclusion.

Complete step-by-step solution:
Now considering from the question we have ${{f}_{1}}\left( x \right)=\dfrac{1}{x},{{f}_{2}}\left( x \right)=1-x\text{ and }{{f}_{3}}\left( x \right)=\dfrac{1}{1-x}$ we will use these definitions of the function and find the value of $J\left( x \right)$.
We have $\left( f_{2}^{{}^\circ }{{J}^{{}^\circ }}{{f}_{1}} \right)\left( x \right)={{f}_{3}}\left( x \right)$ .
By using the value of ${{f}_{1}}\left( x \right)=\dfrac{1}{x}$ we will simplify this and write it as $\left( f_{2}^{{}^\circ }J \right)\left( \dfrac{1}{x} \right)={{f}_{3}}\left( x \right)$.
By using the value of ${{f}_{3}}=\dfrac{1}{1-x}$ we will simplify this and write it as $\left( f_{2}^{{}^\circ }J \right)\left( \dfrac{1}{x} \right)=\dfrac{1}{1-x}$.
By applying the inverse of ${{f}_{2}}$ on both sides we will have $J\left( \dfrac{1}{x} \right)=f_{2}^{-1}\left( \dfrac{1}{1-x} \right)$ .
We have ${{f}_{2}}\left( x \right)=1-x\text{ }$ by assuming ${{f}_{2}}\left( x \right)=y$ we will have $y=1-x$ then we can say $x=1-y$ that implies $f_{2}^{-1}\left( x \right)=1-x$ .
By using this value $f_{2}^{-1}\left( x \right)=1-x$ we will have it as $J\left( \dfrac{1}{x} \right)=1-\left( \dfrac{1}{1-x} \right)$ .
After simplifying this we will have it as $J\left( \dfrac{1}{x} \right)=\dfrac{1-x-1}{1-x}=\dfrac{-x}{1-x}=\dfrac{x}{x-1}$ .
Let us now simplify it in the form of $\dfrac{1}{x}$ by assuming $p=\dfrac{1}{x}$ we can write it as $J\left( y \right)=\dfrac{\dfrac{1}{y}}{\dfrac{1}{y}-1}=\dfrac{1}{1-y}$ .
Now we can say that $J\left( x \right)=\dfrac{1}{1-x}$ that is the same as ${{f}_{3}}\left( x \right)$.
Hence, option A is correct.

Note: While answering questions of this type we should be careful while performing the simplifications and inverse of a function if in case by committing a mistake and writing the inverse of ${{f}_{2}}$ as $f_{2}^{-1}\left( x \right)=x$ we will have the conclusion as $J\left( x \right)=\dfrac{x}{x-1}$ which is not there in the options.