Q1408: Find k so that $f(x)=\begin{cases}\frac{x^{2}-2x-3}{x+1},&x\...

Class CBSE Class 12 Mathematics Continuity and Differentiability Q #1408

KNOWLEDGE BASED

REMEMBER

3 Marks 2025 AISSCE(Board Exam) SA

Find k so that $f(x)=\begin{cases}\frac{x^{2}-2x-3}{x+1},&x\ne-1\\ k,&x=-1\end{cases}$ is continuous at $x=-1$.

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AI Tutor Explanation

Detailed Solution

Step 1: Understand the condition for continuity

For $f(x)$ to be continuous at $x = -1$, we must have $\lim_{x \to -1} f(x) = f(-1)$. In this case, $f(-1) = k$. So, we need to find the limit of $f(x)$ as $x$ approaches $-1$ and set it equal to $k$.

Step 2: Evaluate the limit

We need to find $\lim_{x \to -1} \frac{x^2 - 2x - 3}{x + 1}$. We can factor the numerator:

Step 3: Factor the numerator

$x^2 - 2x - 3 = (x - 3)(x + 1)$. So, we have $\lim_{x \to -1} \frac{(x - 3)(x + 1)}{x + 1}$.

Step 4: Simplify the expression

Since $x \ne -1$, we can cancel the $(x + 1)$ terms: $\lim_{x \to -1} (x - 3)$.

Step 5: Evaluate the limit

Now, we can substitute $x = -1$ into the simplified expression: $(-1 - 3) = -4$. Therefore, $\lim_{x \to -1} f(x) = -4$.

Step 6: Find the value of k

For continuity, we need $k = \lim_{x \to -1} f(x)$, so $k = -4$.

Final Answer: -4

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because the student needs to recall the definition of continuity and apply it to find the value of $k$.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a procedure (evaluating limits and applying the definition of continuity) to solve the problem.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of the definition of continuity of a function at a point, a core concept in the syllabus.