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Explain solution RD Sharma Class 12 Chapter Functions Exercise 2.2 Question 11 maths textbook solution.

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Answer :

 \begin{aligned} &f: N \rightarrow N \text { by } f(x)=x+1 \\ &g: N \rightarrow N \text { by } \\ &g(x)=\{x-1 \text { if } x>1,1 \text { if } x=1 \end{aligned}

Hint : Onto functions means every range has some preimage in the domain of function.

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

Solution :

Let      f:N\rightarrow N \; \text {be}\; f(x)=x+1

and     g:N\rightarrow N \; \text {be}\; g(x)= \left \{ x-1,x>11,x=1 \right.

We will first show that f is not onto

Checking f is not onto

             \begin{aligned} &f: N \rightarrow N \text { be } f(x)=x+1 \\ &y=f(x), \text { where } y \in N \\ &y=x+1 ; x=y-1 \end{aligned}

for        y=1;x=1-1=0

But 0 is not a natural number

\therefore f is not onto.

Finding g\; o\; f :

           \begin{aligned} &f(x)=x+1 \\ &g(x)=\{x-1 \quad x>1 \quad 1 \quad x=1 \\ &f(x)=x+1 ; g(x)=1 \end{aligned}

Since   g(x)=1,g(f(x))=1

So,      g\; o\; f =1

For      x>1


           \begin{aligned} &g(x)=x-1, g(f(x))=f(x)-1 \\ &g \circ f=(x+1)-1 \\ &g \circ f=x \end{aligned}

So,     g \circ f=\{x, x>11, x=1\}

Let      g \circ f=y, \text { where } y \in N

So,     y=\{x, x>11, x=1\}

Here, y is a natural number. As y=x

So, x is also a natural number.

Hence, g\; o\; f is onto.

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