consider the following statements:
Statement 1: The function \[f:R\to R\] such that $f\left( x \right)={{x}^{3}}$ for all $x\in R$ is one-one.
Statement 2: $f\left( a \right)=f\left( b \right)\Rightarrow a=b$ for all a,b$\in $ R if the function f is one-one.
Which one of the following is correct in respect of above statements?
a)Both the statements are true and Statement 2 is the correct explanation of Statement1.
b)Both the statements are true and Statement 2 is not the correct explanation of Statement1.
c)Statement 1 is true but Statement 2 is false.
d)Statement is true Statement 2 is true.

Hint: One-one function has mapping of distinct elements from domain to co domain i.e. image of each element from domain will be different in the co domain. Mathematically, it is defined as if $f\left( x \right)=f\left( y \right)$ then $x=y$ , where x,y$\in $ R and taken from the domain of function ‘f’.

Complete step-by-step answer:

As we know the definition of a one-one function has mapping of distinct elements of its domain to distinct elements of co-domain. In other words, every element of the function’s co domain is the image of at most one element of its domain.

Mathematically, one-one function is defined as whenever $f\left( x \right)=f\left( y \right)$then $x=y$will always be true or vice-versa is also true.

Hence, we get the definition of one-one function as

If $f\left( x \right)=f\left( y \right)$ then $x=y$ (i)

Statement 1: Here, it is given that $f\left( x \right)={{x}^{3}}$ for \[f:R\to R\], and we need to verify whether $f\left( x \right)$ is one-one or not.

So, let us prove or disprove the statement with the help of a fundamental definition of one-one function from equation (i). So, we get $f\left( x \right)={{x}^{3}}$ and $f\left( y \right)={{y}^{3}}$ So, let $f\left( x \right)=f\left( y \right)$ , and hence, we get

${{x}^{3}}={{y}^{3}}$

${{x}^{3}}-{{y}^{3}}=0$ (ii)

We know the algebraic identity of ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$

So, we get equation (ii) as $\left( x-y \right)\left( {{x}^{2}}+{{y}^{2}}+xy \right)=0$

So, we know that if two numbers in multiplication and their product is 0; then either of the number should be 0 so, we get $x-y=0\Rightarrow {{x}^{2}}+{{y}^{2}}+xy=0$

Ignore ${{x}^{2}}+{{y}^{2}}+xy=0$ as it will not be 0, for any real x or y except $x=y=0$ so, ignore it.

So, $x=y$ is the only possible solution of ${{x}^{3}}={{y}^{3}}$ .

Hence we get that if ${{x}^{3}}={{y}^{3}}$then $x=y$i.e. if $f\left( x \right)=f\left( y \right)$ then $x=y$.

So, $f\left( x \right)={{x}^{3}}$ is one-one function

Hence, Statement 1 is true.

Statement 2:-

Here, it is given that $f\left( a \right)=f\left( b \right)$ if $a=b$ for a,b$\in $ R, if ‘f’ is a one-one function. So, this statement is true as per the equation (i).

Hence, Statement 2 is true as well.

Now, we can observe that statement 1 is proved by using the fundamental definition of one-one function, which is given in statement 2. So, statement 2 is the correct explanation of Statement 1 as well.

So, option (a) is the correct answer.


Note: The relation “$f\left( x \right)=f\left( y \right)$ then $x=y$” can be proved with the help of examples. It is suggesting that two values in the range ($f\left( x \right)$ and $f\left( y \right)$) can only be equal if (x and y) their pre-image in domain are same as per the fundamental definition of one-one that image of all elements in domain are distinct. So, don’t confuse with the given relation, it's only the mathematical representation of the fundamental definition of a one-one function.

One may prove $x=y$ directly from ${{x}^{3}}={{y}^{3}}$which is wrong. We need to observe the other factors of ${{x}^{3}}-{{y}^{3}}=0$ as well. So, factorization at this step is also an important part.
${{x}^{2}}+{{y}^{2}}+xy=0$ can be written as $\left( x-y\omega \right)\left( x-y\omega \right)$ where $\omega $ is the cube root of unit i.e. $\omega =\dfrac{-1+\sqrt{3i}}{2}$ and ${{\omega }^{2}}=\dfrac{-1-\sqrt{3i}}{2}$, it means value of x will never be real for any real value of y. that’s why ${{x}^{2}}+{{y}^{2}}+xy=0$is ignored.