Class CBSE Class 12 Mathematics Relations and Functions Q #1770
COMPETENCY BASED
APPLY
3 Marks 2026 AISSCE(Board Exam) SA
Let $A=\mathbb{R}-\{3\}$ and $B=\mathbb{R}-\{1\}$. A function $f:A\rightarrow B$ is defined by $f(x)=(\frac{x-2}{x-3})$. Find whether f is one-one and onto.
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Detailed Solution

Step 1: Check for One-One (Injective)

To check if f is one-one, we assume f(x1) = f(x2) for x1, x2 ∈ A. $$ \frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3} $$ Cross-multiplying gives: $$ (x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3) $$ $$ x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6 $$ Simplifying the equation: $$ -3x_1 - 2x_2 = -3x_2 - 2x_1 $$ $$ x_2 = x_1 $$ Since f(x1) = f(x2) implies x1 = x2, the function is one-one.

Step 2: Check for Onto (Surjective)

To check if f is onto, we set y = f(x) and solve for x in terms of y. $$ y = \frac{x - 2}{x - 3} $$ $$ y(x - 3) = x - 2 $$ $$ xy - 3y = x - 2 $$ $$ xy - x = 3y - 2 $$ $$ x(y - 1) = 3y - 2 $$ $$ x = \frac{3y - 2}{y - 1} $$ For every y ∈ B (where y ≠ 1), there exists a corresponding x ∈ A (where x ≠ 3). Since x is defined for all y ≠ 1, the range of f is B. Thus, the function is onto.

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

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the standard algebraic definitions of injectivity and surjectivity to a specific rational function.
Knowledge Dimension: PROCEDURAL
Justification: The student follows a specific algorithmic process (setting f(x1)=f(x2) and solving for x in terms of y) to verify the properties of the function.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This question tests the fundamental understanding of Relations and Functions, specifically the mapping properties required for inverse functions.

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