#### Provide solution for RD Sharma maths class12 Chapter Functions exercise 2.1 question 3

$f:N \rightarrow N$, given by$f(x)=x^2+x+1$is one-one but not onto.

Hint:

Let x and y be any two elements in the domain (N), such that $f(x)=f(y)$

So,f is one-one.

Given:

$f:N \rightarrow N$,  defined by$f(x)=x^2+x+1$

Let us prove that the given function is one-one.

Solution:

We have$f(x)=x^2+x+1$                                                                                                                 … (i)

Calculate $f(y)=y^2+x+1$                                                                                                              … (ii)

Now equate equation (i) and (ii)

\begin{aligned} &f(x)=f(y) \\ &x^{2}+x+1=y^{2}+y+1 \\ &x^{2}+x-y^{2}-y=0 \end{aligned}

$\begin{array}{r} x^{2}-y^{2}+x-y=0 \\ {\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]} \\ (x+y)(x-y)+x-y=0 \\ (x-y)[(x+y+1)]=0 \end{array}$

From (iii) equation, we can write

$x-y=0, x+y+1\neq 0$

$x+y+1\neq 0$ , because if we substitute$x=1, y= 1$

$1+1+1=3$

Hence, $x+y+1\neq 0$ for any$x\in N$

Where as,

$x-y =0$

$x=y$

So,f is a one-one function.

But, $f(x)=x^3+x+1\geq 3$ for all $x\in N$

So,$f(x)$ doesn’t assume values of1 and 2

$\therefore f$ is not an onto function.

Hence, we proved that, $f(x)=x^2+x+1$ is one-one function not an onto function.