Q1749: Find whether the function $f(x)=\begin{cases}x-1, & x...

CBSE Class 12 Mathematics Continuity and Differentiability Q #1749

COMPETENCY BASED

APPLY

2 Marks 2026 AISSCE(Board Exam) VSA

Find whether the function $f(x)=\begin{cases}x-1, & x<2 \\ 2x-3, & x\ge 2\end{cases}$ at $x=2$ is differentiable or not.

Prev Finish

AI Tutor Explanation

Detailed Solution

Step 1: Understand the condition for continuity

A function $f(x)$ is continuous at a point $x=c$ if the left-hand limit (LHL), right-hand limit (RHL), and the value of the function $f(c)$ are all equal. Here, the point of interest is $x=2$.

Step 2: Calculate the Left-Hand Limit (LHL)

For $x < 2$, $f(x) = x - 1$. $$ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x - 1) = 2 - 1 = 1 $$

Step 3: Calculate the Right-Hand Limit (RHL)

For $x \geq 2$, $f(x) = 3x - 5$. $$ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 5) = 3(2) - 5 = 6 - 5 = 1 $$

Step 4: Calculate the function value at $x=2$

Using the definition for $x \geq 2$: $$ f(2) = 3(2) - 5 = 1 $$

Step 5: Conclusion

Since $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) = 1$, the function is continuous at $x=2$.

Final Answer: The function is continuous at x = 2.

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit

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

Knowledge Dimension: PROCEDURAL

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

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This question tests the fundamental understanding of continuity in calculus, which is a core competency for the Continuity and Differentiability chapter.