Q1494: Check whether function $f(x)$ defined as \[ f(x)= \be...

CBSE Class 12 Mathematics Continuity and Differentiability Q #1494

COMPETENCY BASED

APPLY

2 Marks 2026 AISSCE(Board Exam) VSA

Check whether function $f(x)$ defined as

\[
f(x)=
\begin{cases}
\dfrac{|x-3|}{2(x-3)}, & x<3, \\[6pt]
\dfrac{x-6}{6}, & x\ge 3
\end{cases}
\]

is continuous at $x=3$ or not?

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AI Tutor Explanation

Detailed Solution

Step 1: Analyze the function definition

The function is defined as $f(x) = \frac{|x-3|}{2(x-3)}$ for $x \neq 3$. We need to examine the behavior of the function as $x$ approaches $3$ from the left and the right.

Step 2: Evaluate Left-Hand Limit (LHL)

As $x \to 3^-$, $x < 3$, which implies $(x-3) < 0$. Therefore, $|x-3| = -(x-3)$. $$ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{-(x-3)}{2(x-3)} = \lim_{x \to 3^-} -\frac{1}{2} = -0.5 $$

Step 3: Evaluate Right-Hand Limit (RHL)

As $x \to 3^+$, $x > 3$, which implies $(x-3) > 0$. Therefore, $|x-3| = (x-3)$. $$ \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{(x-3)}{2(x-3)} = \lim_{x \to 3^+} \frac{1}{2} = 0.5 $$

Step 4: Conclusion

Since the Left-Hand Limit ($-0.5$) is not equal to the Right-Hand Limit ($0.5$), the limit of the function at $x=3$ does not exist. Consequently, the function is discontinuous at $x=3$.

Final Answer: The function is discontinuous at x = 3 because the limit does not exist.

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student must apply the definition of limits and continuity to a piecewise function involving absolute values.

Knowledge Dimension: PROCEDURAL

Justification: The student follows a specific algorithmic process (evaluating LHL and RHL) to determine the continuity of the function.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the conceptual understanding of limits in the chapter 'Continuity and Differentiability' rather than rote memorization.