#### Explain solution RD Sharma Class 12 Chapter Functions Exercise 2.2 Question 12 maths textbook solution.

Answer : $f(x)=x \; \text {and}\; g(x)=\left | x \right |$

where $f:N\rightarrow Z \; \text {and}\; g:Z\rightarrow Z$

Hint : An injective function is a function that maps distinct elements of its domain to distinct element of its co-domain.

Given : $f:N\rightarrow Z \; \text {and}\; g:Z\rightarrow Z$

Solution :

Let     $f(x)=x \; \text {and}\; g(x)=\left | x \right |$

$g(x)=|x|=\{x \quad x \geq 0-x \quad x<0$

Checking $g(x)$ injective (one-one)

$g(1)=\left | 1 \right |=1$

$g(-1)=\left | -1 \right |=1$

Since, different elements 1, -1 have the same image 1

$\therefore$ g is not injective (one -one).

Checking for injective (one-one)

\begin{aligned} &f: N \rightarrow Z \quad \& g: Z \rightarrow Z \\ &f(x)=x \text { and } g(x)=|x| \\ &g \circ f(x)=g(f(x))=|f(x)| \\ &=|x|=\{x, x \geq 0-x, \quad x<0\} \\ &\text { Here } g \circ f(x): N \rightarrow Z \end{aligned}

So, x is always a natural number.

Here $\left | x \right |$ will always be a natural number.

So, $g\; o\; f (x)$ has a unique image

$\therefore$   $g\; o\; f (x)$ is injective.

Hence, $g\; o\; f$ is injective but  g is not injective