Class CBSE Class 12 Mathematics Continuity and Differentiability Q #597
KNOWLEDGE BASED
APPLY
1 Marks 2025 AISSCE(Board Exam) MCQ SINGLE
If \(f(x)=\begin{cases}3x-2,&0 \lt x\le 1\\ 2x^{2}+ax,&1\lt x\lt 2\end{cases}\) is continuous for \(x\in(0,2)\), then a is equal to:
(A) -4
(B) \(-\frac{7}{2}\)
(C) -2
(D) -1
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Correct Answer: D
Explanation
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Detailed Solution

Step 1: Understanding Continuity

For the function $f(x)$ to be continuous at $x=1$, the left-hand limit (LHL) and the right-hand limit (RHL) at $x=1$ must be equal, and also equal to the value of the function at that point.

Step 2: Calculate the Left-Hand Limit (LHL) at x=1

The left-hand limit at $x=1$ is given by: $$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x - 2) $$ Substituting $x=1$, we get: $$ \lim_{x \to 1^-} f(x) = 3(1) - 2 = 1 $$

Step 3: Calculate the Right-Hand Limit (RHL) at x=1

The right-hand limit at $x=1$ is given by: $$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x^2 + ax) $$ Substituting $x=1$, we get: $$ \lim_{x \to 1^+} f(x) = 2(1)^2 + a(1) = 2 + a $$

Step 4: Apply the Continuity Condition

For $f(x)$ to be continuous at $x=1$, the LHL must equal the RHL. Therefore: $$ 1 = 2 + a $$

Step 5: Solve for a

Solving for $a$, we get: $$ a = 1 - 2 = -1 $$

Final Answer: -1

AI Suggestion: Option D

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the definition of continuity to find the value of 'a'.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a procedure to determine the value of 'a' by equating the left-hand limit and right-hand limit of the function at a point.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of the concept of continuity of a function, a core topic in the syllabus.
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