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

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

Answers (1)

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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