#### Need solution for RD Sharma maths class 12 chapter Functions exercise 2.1 question 5 sub question (xii)

Injective but not surjective

Given:

$f: Q-\{3\} \rightarrow Q, \text { defined by } f(x)=\frac{2 x+3}{x-3}$

Hint:

Injective (one-one): function means every element in the domain has a distinct image in the  co-domain.

Surjective(Onto):  function means every element in the co-domain has at least one  pre image in the domain of function.

Solution:

Injection:

let x,y be any two elements in the domain$\left ( Q-\{3\} \right )$, such that $f(x)=f(y)$.

\begin{aligned} &f(x)=\frac{(2 x+3)}{(x-3)}, f(y)=\frac{(2 y+3)}{(y-3)} \\ &f(x)=f(y) \\ &\frac{(2 x+3)}{(x-3)}=\frac{(2 y+3)}{(y-3)} \\ &\begin{array}{l} (2 x+3)(y-3)=(2 y+3)(x-3) \\ 2 x y-6 x+3 y-9=2 x y-6 y+3 x-9 \\ 9 x=9 y \\ x=y \end{array} \end{aligned}

So, f  is an injection.

Subjection test:

Let y be any element in the co-domain Q, such that $f(x)=y$ for some element x in ($\left ( Q-\{3\} \right )$ (domain).

\begin{aligned} &f(x)=y\\ &\frac{(2 x+3)}{(x-3)}=y\\ &2 x+3=x y-3 y\\ &2 x-x y=-3 y-3\\ &x(2-y)=-3(y+1)\\ &x=-3(y+1) /(2-y), \text { which is not defined at } y=2 \end{aligned}

So, f  is not  surjective.

Therefore, f  is not  bijective.