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 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 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 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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Showing 61 to 90 of 104 (4 Pages)

Applications of Derivatives

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