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

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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

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