Class CBSE Class 12 Mathematics Continuity and Differentiability Q #794
KNOWLEDGE BASED
APPLY
1 Marks 2023 MCQ SINGLE
The value of k for which f(x)=\begin{cases}3x+5,&x\ge2\\ kx^{2},&x<2\end{cases} is a continuous function, is :
(A) -\frac{11}{4}
(B) \frac{4}{11}
(C) 11
(D) \frac{11}{4}
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

For the function to be continuous at x=2, the left-hand limit (LHL) must be equal to the right-hand limit (RHL) at x=2.

LHL at x=2: lim (x->2-) f(x) = lim (x->2-) kx2 = k(2)2 = 4k

RHL at x=2: lim (x->2+) f(x) = lim (x->2+) (3x+5) = 3(2) + 5 = 6 + 5 = 11

For continuity, LHL = RHL, so 4k = 11

Solving for k: k = 11/4

Correct Answer: \frac{11}{4}

AI Suggestion: Option D

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concept of continuity to find the value of k. They must set the left-hand limit and right-hand limit equal to each other at x=2 and solve for k.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to determine the value of 'k' that ensures the function is continuous. This involves evaluating limits and solving equations.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of the concept of continuity, which is a core topic in the syllabus.

More from this Chapter

VSA
If $y=x^{\frac{1}{x}}$ then find $\frac{dy}{dx}$ at $x=1$.
MCQ_SINGLE
6. The function $f(x)=|x|$ is
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:
MCQ_SINGLE
The function f(x)=[x], where [x] denotes the greatest integer less than or equal to x, is continuous at
MCQ_SINGLE
The value of k for which function $f(x)=\begin{cases}x^{2},&x\ge0\\ kx,&x
View All Questions