CBSE Class 12 Mathematics Relations and Functions Q #2138
COMPETENCY BASED
APPLY
2 Marks Generic-Curriculum VSA
Check whether $f : R - \{3\} \to R$ defined as $f(x) = \frac{x - 2}{x - 3}$ is onto or not. [Delhi 2026]
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Detailed Solution

Step 1: Understand the condition for Onto function

A function $f: A \to B$ is onto (surjective) if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $f(x) = y$.

Step 2: Set up the equation

Let $y = f(x)$. We have: $$y = \frac{x - 2}{x - 3}$$

Step 3: Solve for x in terms of y

Multiply both sides by $(x - 3)$: $$y(x - 3) = x - 2$$ $$xy - 3y = x - 2$$ $$xy - x = 3y - 2$$ $$x(y - 1) = 3y - 2$$ $$x = \frac{3y - 2}{y - 1}$$

Step 4: Analyze the range

For $f$ to be onto, $x$ must exist in the domain $R - \{3\}$ for every $y$ in the codomain $R$. Looking at the expression $x = \frac{3y - 2}{y - 1}$, we see that $x$ is undefined when $y = 1$. Since $y = 1$ is in the codomain $R$ but there is no $x$ such that $f(x) = 1$, the function is not onto.

Final Answer: The function is not onto.

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must apply the algebraic definition of surjectivity to a specific rational function.
Knowledge Dimension: PROCEDURAL
Justification: The student follows a standard algorithmic process of inverting the function to determine the range and checking for domain constraints.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the fundamental understanding of Relations and Functions (Chapter 1), specifically the mapping properties of rational functions.
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