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

\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      $x>1$

Since,

\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.