Limits and Derivatives Case Study Questions Class 11 Maths Chapter 12

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Hello students, we are providing case study questions for class 11 Maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 11 Maths. In this article, you will find case study questions for cbse class 11 Maths chapter 12 Limits and Derivatives.

Chapter	Limits and Derivatives
Type of Questions	Case Study Questions
Nature of Questions	Competency Based Questions
Board	CBSE
Class	11
Subject	Maths
Useful for	Class 12 Studying Students
Answers provided	Yes
Difficulty level	Mentioned
Important Link	Class 11 Maths Chapterwise Case Study

Table of Contents

Case Study Questions on Limits and Derivatives

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Case Study:
An environmental scientist is tracking the concentration of a pollutant in a lake over time. The concentration (in mg/L) at any time \( t \) (in days) is modeled by the function \( f(t) = \frac{2t^2 + 3t – 5}{t – 1} \), for \( t > 1 \). To predict long-term impacts and short-term sensitivity, she analyzes limits and rates of change at various points.

Questions:

Find \( \displaystyle \lim_{t \to 2} f(t) \).
Find \( \displaystyle \lim_{t \to 1^+} f(t) \). What type of discontinuity (if any) does the function have at \( t=1 \)?
Find the derivative \( f'(t) \) and evaluate it at \( t = 3 \).
Explain what the limit and derivative concepts tell us about the pollutant’s concentration and its change over time.

Solutions:

1. \( \displaystyle \lim_{t \to 2} f(t) \):

Direct substitution: \[ f(2) = \frac{2 \cdot (2)^2 + 3 \cdot 2 – 5}{2 – 1} = \frac{2 \cdot 4 + 6 – 5}{1} = \frac{8 + 6 – 5}{1} = \frac{9}{1} = \boxed{9} \]

2. \( \displaystyle \lim_{t \to 1^+} f(t) \) and discontinuity type:

As \( t \to 1 \): \[ f(t) = \frac{2t^2 + 3t – 5}{t – 1} \] Numerator at \( t=1 \): \[ 2(1)^2 + 3(1) – 5 = 2 + 3 – 5 = 0 \] So, the function is in indeterminate form \( \frac{0}{0} \). Factor numerator: \[ 2t^2 + 3t – 5 = (2t – 2)(t + 5) \] But \( 2t^2 + 3t – 5 = (2t – 5)(t + 1) \), check: \[ (2t – 5)(t + 1) = 2t^2 + 2t – 5t – 5 = 2t^2 – 3t – 5 \] So, actual factorization: \( 2t^2 + 3t – 5 = (2t – 5)(t + 1) + 6t \) – but for limit, let’s use L’Hospital’s Rule: Derivative of numerator: \( 4t + 3 \) Derivative of denominator: \( 1 \) So, \[ \lim_{t \to 1} f(t) = \lim_{t \to 1} \frac{4t + 3}{1} = 4 \cdot 1 + 3 = \boxed{7} \] Because \( f(1) \) is not defined (division by zero), but the limit exists, it’s a **removable discontinuity** at \( t = 1 \).

3. Derivative \( f'(t) \) and its value at \( t = 3 \):

Use quotient rule: \[ f(t) = \frac{2t^2 + 3t – 5}{t – 1} \] \[ f'(t) = \frac{(4t + 3)(t – 1) – (2t^2 + 3t – 5)(1)}{(t – 1)^2} \] Expand numerator: \[ (4t + 3)(t – 1) = 4t^2 – 4t + 3t – 3 = 4t^2 – t – 3 \] \[ 4t^2 – t – 3 – (2t^2 + 3t – 5) = 4t^2 – t – 3 – 2t^2 – 3t + 5 = (4t^2 – 2t^2) + (-t – 3t) + (-3 + 5) = 2t^2 – 4t + 2 \] So, \[ f'(t) = \frac{2t^2 – 4t + 2}{(t – 1)^2} \] At \( t = 3 \): \[ f'(3) = \frac{2 \cdot 9 – 4 \cdot 3 + 2}{(3 – 1)^2} = \frac{18 – 12 + 2}{4} = \frac{8}{4} = \boxed{2} \]

4. Explanation/Interpretation:
The limit at \( t = 2 \) gives the pollutant concentration at that exact time. The limit at \( t = 1 \) tells us the trend as time approaches 1 day, but a value isn’t defined due to a (removable) discontinuity—mathematically, the model “skips” \( t = 1 \). The derivative \( f'(3) = 2 \) means that, at day 3, the concentration is increasing at a rate of 2 mg/L per day. Limits and derivatives help us understand both specific values and how rapidly the environment is changing over time.

We hope the given case study questions for Limits and Derivatives Class 11 helps you in your learning.

Also check

Topics from which case study questions may be asked

Introduction to Limits and Derivatives
Intuitive Idea of Limits
Limits of Algebraic and Trigonometric Functions
Meaning and Definition of Derivative
Derivative as a Rate of Change
Derivative of a Function Using First Principle
Derivatives of Standard Functions (Polynomial, Trigonometric, etc.)
Geometrical Interpretation of Derivatives

Frequently Asked Questions (FAQs) on Limits and Derivatives Case Study Questions

Q1: What is a case study question in mathematics?

A1: A case study question in mathematics is a problem or set of problems based on a real-life scenario or application. It requires students to apply their understanding of mathematical concepts to analyze, interpret, and solve the given situation.

Q2: How should students tackle case study questions in exams?

A2: To tackle case study questions effectively, students should:
Read the problem carefully: Understand the scenario and identify the mathematical concepts involved.
Break down the problem: Divide the case study into smaller parts to manage the information better.
Apply relevant formulas and theorems: Use the appropriate mathematical tools to solve each part of the problem.

Q3: Why are case study questions included in the Class 11 Maths curriculum?

A3: Case study questions are included to bridge the gap between theoretical knowledge and practical application. They help students see the relevance of what they are learning and prepare them for real-life situations where they may need to use these mathematical concepts.

Q4: Are there any online resources or tools available for practicing “Limits and Derivatives” case study questions?

A12: We provide case study questions for CBSE Class 11 Maths on our website. Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit Physics Gurukul website. they are having a large collection of case study questions for all classes.