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Q10:
Let $A = R – \{3\}$ and $B = R – \{1\}$. Consider the function $f : A \to B$ defined by $f(x) = \frac{x-2}{x-3}$. Is $f$ one-one and onto? Justify your answer.

Solution :

Step 1: Analyzing Injectivity (One-One Nature)

To determine if $f(x)$ is one-one, we assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in A$ and verify if $x_1 = x_2$ [Definition of Injectivity].

Given: $f(x) = \frac{x-2}{x-3}$

$\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}$

Cross-multiplying: $(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$

Expanding both sides: $x_1x_2 - 3x_1 - 2x_2 + 6 = x_2x_1 - 3x_2 - 2x_1 + 6$

Subtracting $x_1x_2$ and $6$ from both sides: $-3x_1 - 2x_2 = -3x_2 - 2x_1$

Rearranging terms: $x_2 - x_1 = 0 \implies x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$, the function $f$ is one-one.

Step 2: Analyzing Surjectivity (Onto Nature)

To determine if $f$ is onto, we check if for every $y \in B$, there exists an $x \in A$ such that $f(x) = y$ [Definition of Surjectivity].

Let $y = \frac{x-2}{x-3}$. Solving for $x$ in terms of $y$:

$y(x-3) = x-2$

$yx - 3y = x - 2$

$yx - x = 3y - 2$

$x(y - 1) = 3y - 2$

$x = \frac{3y - 2}{y - 1}$

For $x$ to be defined, $y - 1 \neq 0$, which means $y \neq 1$. Since the codomain $B = R - \{1\}$, this condition is satisfied for all $y \in B$. Furthermore, we must ensure $x \neq 3$ so that $x \in A$.

If $x = 3$, then $3 = \frac{3y - 2}{y - 1} \implies 3y - 3 = 3y - 2 \implies -3 = -2$, which is a contradiction. Thus, for every $y \in B$, there exists a unique $x \in A$. The function $f$ is onto.

Step 3: Graphical Visualization

x y x=3 y=1 O

Step 4: Comparative Summary

Property Status Justification
One-one Yes $f(x_1) = f(x_2) \implies x_1 = x_2$
Onto Yes Range equals Codomain

Final Solution: The function $f$ is both one-one and onto (bijective).


More Questions from Class 12 Mathematics Relations and Functions EXERCISE 1.2


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