Applications of Derivatives for Competitive Exam Success

Applications of Derivatives

Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the critical points. (2022)

A. (0, 0)
B. (1, 5)
C. (2, 0)
D. (3, 3)

Solution

Set f'(x) = 0. f'(x) = 6x^2 - 18x + 12 = 0. Critical points are x = 2.

Correct Answer: C — (2, 0)

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Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the intervals of increase. (2022)

A. (-∞, 0)
B. (0, 3)
C. (3, ∞)
D. (0, 2)

Solution

f'(x) = 6x^2 - 18x + 12. The critical points are x = 1 and x = 2. Test intervals show increase in (0, 3).

Correct Answer: B — (0, 3)

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Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the local maxima. (2023) 2023

A. (1, 5)
B. (2, 6)
C. (3, 3)
D. (0, 0)

Solution

Find f'(x) = 6x^2 - 18x + 12, set to 0. Local maxima at x = 2 gives f(2) = 6.

Correct Answer: B — (2, 6)

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Q. For the function f(x) = 3x^2 - 12x + 7, find the minimum value. (2022)

A. -5
B. -4
C. -3
D. -2

Solution

The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 7 = -5.

Correct Answer: B — -4

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Q. For the function f(x) = 3x^2 - 12x + 7, find the x-coordinate of the vertex. (2022)

A. 1
B. 2
C. 3
D. 4

Solution

The x-coordinate of the vertex is given by x = -b/(2a) = 12/(2*3) = 2.

Correct Answer: B — 2

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Q. For the function f(x) = 3x^2 - 12x + 9, find the coordinates of the vertex. (2020)

A. (2, 3)
B. (3, 0)
C. (1, 1)
D. (0, 9)

Solution

The vertex is at x = -b/(2a) = 2. f(2) = 3.

Correct Answer: A — (2, 3)

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Q. For the function f(x) = 3x^2 - 12x + 9, find the vertex. (2021)

A. (2, 3)
B. (3, 0)
C. (0, 9)
D. (1, 6)

Solution

The vertex occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

Correct Answer: A — (2, 3)

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Q. For the function f(x) = 3x^2 - 12x + 9, find the x-coordinate of the vertex. (2021)

A. 1
B. 2
C. 3
D. 4

Solution

The vertex x-coordinate is found using -b/(2a) = 12/(2*3) = 2.

Correct Answer: B — 2

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Q. For the function f(x) = x^2 + 2x, find the local maximum. (2022)

A. -1
B. 0
C. 1
D. 2

Solution

f'(x) = 2x + 2. Setting f'(x) = 0 gives x = -1. f(-1) = 1.

Correct Answer: A — -1

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Q. For the function f(x) = x^3 - 6x^2 + 9x, find the local minima. (2022)

A. (1, 4)
B. (2, 1)
C. (3, 0)
D. (0, 0)

Solution

f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f(2) = 1 is a local minimum.

Correct Answer: B — (2, 1)

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Q. If f(x) = x^3 - 3x^2 + 4, find the critical points. (2022)

A. 1, 2
B. 0, 3
C. 2, 4
D. 1, 3

Solution

f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2.

Correct Answer: A — 1, 2

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Q. If the cost function is C(x) = 3x^2 + 12x + 5, find the minimum cost. (2020)

A. 5
B. 8
C. 12
D. 10

Solution

The minimum cost occurs at x = -b/(2a) = -12/(2*3) = -2. C(-2) = 3(-2)^2 + 12(-2) + 5 = 8.

Correct Answer: B — 8

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Q. If the cost function is C(x) = 3x^2 + 12x + 5, find the minimum cost. (2020) 2020

A. 5
B. 8
C. 12
D. 10

Solution

The minimum cost occurs at x = -b/(2a) = -12/(2*3) = -2. C(-2) = 3(-2)^2 + 12(-2) + 5 = 8.

Correct Answer: B — 8

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Q. If the cost function is C(x) = 5x^2 + 20x + 100, find the minimum cost. (2020)

A. 100
B. 120
C. 140
D. 160

Solution

The minimum cost occurs at x = -b/(2a) = -20/(2*5) = -2. C(-2) = 5(-2)^2 + 20(-2) + 100 = 120.

Correct Answer: B — 120

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Q. If the revenue function is R(x) = 100x - 2x^2, find the number of units that maximizes revenue. (2021)

A. 25
B. 50
C. 75
D. 100

Solution

Max revenue occurs at x = -b/(2a) = 100/(2*2) = 25.

Correct Answer: B — 50

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Q. If the revenue function is R(x) = 20x - 0.5x^2, find the quantity that maximizes revenue. (2021)

A. 10
B. 20
C. 15
D. 25

Solution

R'(x) = 20 - x = 0 gives x = 20. This maximizes revenue.

Correct Answer: B — 20

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Q. If the revenue function is R(x) = 50x - 0.5x^2, find the number of units that maximizes revenue. (2023)

A. 25
B. 50
C. 30
D. 40

Solution

Max revenue occurs at x = -b/(2a) = -50/(2*-0.5) = 50.

Correct Answer: A — 25

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Q. What is the derivative of f(x) = 2x^3 - 9x^2 + 12x? (2021)

A. 6x^2 - 18x + 12
B. 6x^2 - 18x
C. 6x^2 + 18x
D. 6x^2 - 12

Solution

f'(x) = 6x^2 - 18x + 12.

Correct Answer: A — 6x^2 - 18x + 12

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Q. What is the maximum area of a triangle with a base of 10 cm and height as a function of x? (2020)

A. 25
B. 50
C. 75
D. 100

Solution

Area = 1/2 * base * height = 1/2 * 10 * x. Max area occurs when x = 10, giving Area = 50.

Correct Answer: B — 50

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Q. What is the maximum area of a triangle with a base of 10 cm and height varying with x? (2021)

A. 25
B. 50
C. 75
D. 100

Solution

Area = 1/2 * base * height. Max area occurs when height is maximized, thus Area = 1/2 * 10 * 10 = 50.

Correct Answer: B — 50

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Q. What is the maximum area of a triangle with a base of 10 m and height as a function of x? (2021)

A. 25 m²
B. 50 m²
C. 30 m²
D. 20 m²

Solution

Area = 1/2 * base * height = 1/2 * 10 * x. Max area occurs at x = 10, giving 50 m².

Correct Answer: B — 50 m²

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Q. What is the maximum area of a triangle with a base of 10 units and height as a function of x? (2020)

A. 25
B. 50
C. 75
D. 100

Solution

Area = 1/2 * base * height = 5h. Max area occurs when h = 10, giving area = 50.

Correct Answer: B — 50

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Q. What is the maximum area of a triangle with a base of 10 units and height as a function of the base? (2021)

A. 25
B. 50
C. 30
D. 40

Solution

Area = 0.5 * base * height. Max area occurs when height is 10, giving Area = 0.5 * 10 * 10 = 50.

Correct Answer: B — 50

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Q. What is the maximum height of the projectile modeled by h(t) = -16t^2 + 32t + 48? (2023)

A. 48
B. 64
C. 80
D. 32

Solution

The maximum height occurs at t = -b/(2a) = 1. h(1) = 64.

Correct Answer: B — 64

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Q. What is the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48? (2021)

A. 48
B. 64
C. 80
D. 32

Solution

The maximum height occurs at t = -b/(2a) = -64/(2*-16) = 2. h(2) = -16(2^2) + 64(2) + 48 = 80.

Correct Answer: B — 64

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Q. What is the maximum profit if the profit function is P(x) = -x^2 + 10x - 16? (2021)

A. 4
B. 6
C. 8
D. 10

Solution

The maximum profit occurs at x = -b/(2a) = 10/2 = 5. P(5) = -5^2 + 10*5 - 16 = 9.

Correct Answer: C — 8

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Q. What is the maximum value of f(x) = -x^2 + 4x + 1? (2023)

A. 5
B. 6
C. 7
D. 8

Solution

The vertex occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 5.

Correct Answer: A — 5

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Q. What is the maximum value of f(x) = -x^2 + 6x - 8? (2023)

A. 2
B. 4
C. 6
D. 8

Solution

The vertex occurs at x = 3. f(3) = -9 + 18 - 8 = 1.

Correct Answer: C — 6

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Q. What is the minimum distance from the point (3, 4) to the line 2x + 3y - 6 = 0? (2023)

A. 2
B. 3
C. 1
D. 4

Solution

Using the distance formula from a point to a line, the minimum distance is calculated to be 2 units.

Correct Answer: A — 2

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Q. What is the minimum value of f(x) = 3x^2 - 12x + 12? (2021)

A. 0
B. 3
C. 6
D. 9

Solution

The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 12 = 0.

Correct Answer: B — 3

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Applications of Derivatives MCQ & Objective Questions

The "Applications of Derivatives" is a crucial topic in mathematics that plays a significant role in various school and competitive exams. Understanding this concept not only enhances your problem-solving skills but also helps in scoring better. By practicing MCQs and objective questions, you can solidify your grasp on important questions and improve your exam preparation effectively.

What You Will Practise Here

Understanding the concept of derivatives and their applications in real-life scenarios.
Finding the maxima and minima of functions using derivatives.
Application of derivatives in motion problems and rates of change.
Using derivatives to determine the concavity of functions and points of inflection.
Solving problems related to optimization in various contexts.
Graphical interpretation of derivatives and their significance.
Key formulas and definitions related to derivatives and their applications.

Exam Relevance

The topic of "Applications of Derivatives" is frequently featured in CBSE, State Boards, NEET, and JEE exams. Students can expect questions that assess their understanding of how to apply derivatives in practical situations, such as optimization problems and motion analysis. Common question patterns include multiple-choice questions that require students to identify maximum or minimum values, as well as theoretical questions that test conceptual clarity.

Common Mistakes Students Make

Confusing the concepts of increasing and decreasing functions.
Overlooking the importance of critical points in optimization problems.
Misinterpreting the meaning of concavity and points of inflection.
Neglecting to apply the first and second derivative tests correctly.
Failing to connect the graphical representation of functions with their derivatives.

FAQs

Question: What are the key applications of derivatives in real life?
Answer: Derivatives are used in various fields such as physics for motion analysis, economics for maximizing profit, and engineering for optimizing designs.

Question: How can I improve my understanding of derivatives?
Answer: Regular practice of MCQs and objective questions, along with reviewing key concepts and formulas, can significantly enhance your understanding.

Start solving practice MCQs today to test your understanding of "Applications of Derivatives" and boost your confidence for upcoming exams!

Applications of Derivatives

Applications of Derivatives MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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