If A = R - {3} and B = R - {1}. Let f: A → B such that f(x) = (x-2)/(x-3) for all x ∈ A. Then show that f is bijective.

 


Given: 

$\hspace{0.3cm}A=\mathbb{R}-\{3\}$

$\hspace{0.3cm}B=\mathbb{R}-\{1\}$

$\hspace{0.3cm}f:A\rightarrow B$ such that $f(x)$ $=$ $\Large{\frac{x-2}{x-3}}$

To Prove:

$\hspace{0.3cm}$ (i) $f(x)$ is one-one.

$\hspace{0.3cm}$ (ii) $f(x)$ is onto.

Proof:

Condition for One-One:

$\hspace{0.8cm}$If $f(x_1) = f(x_2)$ then 

$\hspace{1.8cm}x_{1}=x_{2}\hspace{0.3cm}\forall \hspace{0.2cm}x_1, x_2 \in A$

$\hspace{0.4cm}$ OR $\hspace{0.3cm}$If $x_{1} \neq x_{2}$ then 

$\hspace{1.1cm}f(x_{1})\neq f(x_{2})\hspace{0.3cm}\forall \hspace{0.2cm}x_1, x_2 \in A$

(i) Let $x_1, x_2 \in A$ be any element such that:

$\hspace{2.5cm}f(x_1)=f(x_2)$

$\hspace{2.25cm}\Large{\frac{x_1-2}{x_1-3}}$ $=$ $\Large{\frac{x_2-2}{x_2-3}}$

$\hspace{0.2cm}(x_{1}-2)(x_{2}-3)=(x_{1}-3)(x_{2}-2)$

$\hspace{0.2cm}x_{1}x_{2}-3x_{1}-2x_{2}+6$

$\hspace{3.7cm}=x_{1}x_{2}-2x_{1}-3x_{2}+6$

$\hspace{1.2cm}-3x_{1}-2x_{2}=-2x_{1}-3x_{2}$

$\hspace{1.2cm}-3x_{1}+2x_{1}=+2x_{2}-3x_{2}$

$\hspace{2.7cm}-x_{1}=-x_{2}$

$\hspace{3.1cm}x_{1}=x_{2}\hspace{0.3cm}\forall \hspace{0.2cm}x_1, x_2 \in A$

$\hspace{1.5cm}\boxed{\text{So, } f(x) \text{ is one-one.}}$


Condition for Onto:

$\hspace{0.3cm}$For every $y\in B$ there exist $x\in A$ such that

$\hspace{2.7cm}y=f(x)$

$\hspace{0.3cm}$OR $\hspace{0.8cm}$ Range $=$ Codomain.

(ii) Let $y\in B$ be any element

We need to find an element $x\in A$ such that:

$\hspace{2.4cm}y=f(x)$

Finding suitable value of $x$:

$\hspace{2.3cm}y$ $=$ $\Large{\frac{x-2}{x-3}}$

$\hspace{0.78cm}(x-3)y = (x-2)$

$\hspace{0.88cm}xy-3y = x-2$

$\hspace{1.05cm}xy-x = 3y-2$

$\hspace{0.68cm}x(y-1) = 3y-2$

$\hspace{2cm}\boxed{x =\frac{3y-2}{y-1}}\hspace{1cm}(1)$

If $x=3$ then,

$\hspace{2.13cm}3$ $=$ $\Large{\frac{3y-2}{y-1}}$

$\hspace{0.65cm}3(y-1)=(3y-2)$

$\hspace{1cm}3y-3=3y-2$

$\hspace{1.7cm}-3=-2$

which cannot be possible.

So, $x\in \mathbb{R}-\{3\}=A$

Using $(1)$, we get

 $\hspace{0.5cm}f(x)$ $=$ $\Large{\frac{x-2}{x-3}}$

$\hspace{1.6cm}=$ $\Large{\frac{\frac{3y-2}{y-1}-2}{\frac{3y-2}{y-1}-3}}$

$\hspace{1.6cm}=$ $\large{\frac{(3y-2)-2(y-1)}{(3y-2)-3(y-1)}}$

$\hspace{1.6cm}=$ $\large{\frac{3y-2-2y+2}{3y-2-3y+3}}$

$\hspace{1.6cm}=$ $\large{\frac{y}{1}}$

$\hspace{0.5cm}f(x)=y\hspace{0.5cm}\forall\hspace{0.25cm}y\in B$

$\hspace{1cm}\boxed{\text{So, } f(x) \text{ is onto.}}$

Manik Bhola

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