Applications of Derivatives for Competitive Exam Success

Applications of Derivatives

Q. Determine the point where the function f(x) = 4x - x^2 has a maximum. (2022)

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

Solution

The maximum occurs at x = 2, found by setting f'(x) = 4 - 2x = 0. f(2) = 4(2) - (2^2) = 4.

Correct Answer: B — (2, 4)

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Q. Find the critical points of f(x) = x^4 - 8x^2 + 16. (2021)

A. (0, 16)
B. (2, 0)
C. (4, 0)
D. (1, 15)

Solution

f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 0 is a critical point.

Correct Answer: B — (2, 0)

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Q. Find the critical points of the function f(x) = x^4 - 8x^2 + 16. (2019)

A. (0, 16)
B. (2, 0)
C. (4, 0)
D. (1, 9)

Solution

Setting f'(x) = 4x^3 - 16x = 0 gives x = 0, ±2. Evaluating f(2) = 0 shows (2, 0) is a critical point.

Correct Answer: B — (2, 0)

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Q. Find the dimensions of a box with a square base that maximizes volume given a surface area of 600 sq. units. (2020)

A. 10, 10
B. 15, 15
C. 12, 12
D. 20, 20

Solution

Let x be the side of the base and h the height. The surface area constraint gives 2x^2 + 4xh = 600. Max volume occurs at x = 12.

Correct Answer: C — 12, 12

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Q. Find the dimensions of a rectangle with a fixed area of 50 m^2 that minimizes the perimeter. (2021)

A. 5, 10
B. 7, 7.14
C. 8, 6.25
D. 10, 5

Solution

For a fixed area, the minimum perimeter occurs when the rectangle is a square. Thus, dimensions are approximately 7 m by 7.14 m.

Correct Answer: B — 7, 7.14

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Q. Find the dimensions of a rectangle with a fixed area of 50 square units that minimizes the perimeter. (2022) 2022

A. 5, 10
B. 7, 7.14
C. 10, 5
D. 8, 6.25

Solution

For minimum perimeter, the rectangle should be a square. Thus, side = sqrt(50) ≈ 7.07.

Correct Answer: B — 7, 7.14

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Q. Find the dimensions of a rectangle with a fixed area of 50 square units that minimizes the perimeter. (2020)

A. 5, 10
B. 7, 7
C. 10, 5
D. 8, 6.25

Solution

For a fixed area, the perimeter is minimized when the rectangle is a square. Thus, side = √50.

Correct Answer: B — 7, 7

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Q. Find the local maxima of f(x) = -x^2 + 4x + 1. (2020)

A. 1
B. 5
C. 9
D. 7

Solution

The maximum occurs at x = -b/(2a) = -4/(2*-1) = 2. f(2) = -2^2 + 4(2) + 1 = 5.

Correct Answer: B — 5

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Q. Find the local maxima of f(x) = -x^3 + 3x^2 + 1. (2020)

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

Solution

f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f(2) = 5.

Correct Answer: B — (1, 3)

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Q. Find the local maximum of f(x) = -x^3 + 3x^2 + 4. (2020)

A. 4
B. 5
C. 6
D. 3

Solution

Set f'(x) = 0 to find critical points. The local maximum occurs at x = 2. f(2) = 5.

Correct Answer: B — 5

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Q. Find the maximum area of a triangle with a base of 10 m and height varying. (2020)

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. Find 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 = 1/2 * base * height. Max area occurs when height is maximized at 10 units, giving Area = 50.

Correct Answer: B — 50

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Q. Find the maximum area of a triangle with a base of 10 units and height as a function of x. (2022)

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

Solution

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

Correct Answer: B — 50

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Q. Find the maximum area of a triangle with a fixed perimeter of 30 cm. (2022)

A. 75 cm²
B. 100 cm²
C. 50 cm²
D. 60 cm²

Solution

For maximum area, the triangle should be equilateral. Area = (sqrt(3)/4) * (10)^2 = 75 cm².

Correct Answer: A — 75 cm²

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Q. Find the maximum height of the projectile modeled by h(t) = -16t^2 + 32t + 48. (2020)

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

Solution

The maximum occurs at t = -b/(2a) = -32/(2*-16) = 1. h(1) = 64.

Correct Answer: A — 48

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Q. Find the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 48. (2020)

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

Solution

The maximum 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. Find the maximum value of the function f(x) = -2x^2 + 8x - 3. (2021) 2021

A. 3
B. 8
C. 12
D. 6

Solution

The function is a downward-opening parabola. The maximum occurs at x = -b/(2a) = -8/(2*-2) = 2. f(2) = -2(2^2) + 8(2) - 3 = 8.

Correct Answer: B — 8

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Q. Find the minimum value of f(x) = 4x^2 - 16x + 20. (2022)

A. 4
B. 5
C. 6
D. 7

Solution

The vertex gives the minimum at x = 2. f(2) = 4(2^2) - 16(2) + 20 = 4.

Correct Answer: A — 4

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Q. Find the minimum value of f(x) = x^2 - 4x + 6. (2021)

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

Solution

The vertex form gives the minimum at x = 2. f(2) = 2.

Correct Answer: A — 2

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Q. Find the minimum value of f(x) = x^2 - 4x + 7. (2021)

A. 3
B. 5
C. 4
D. 6

Solution

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4*2 + 7 = 3.

Correct Answer: A — 3

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Q. Find the minimum value of f(x) = x^2 - 4x + 7. (2021) 2021

A. 3
B. 5
C. 4
D. 6

Solution

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4*2 + 7 = 3.

Correct Answer: A — 3

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Q. Find the minimum value of the function f(x) = 2x^2 - 8x + 10. (2022)

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

Solution

The minimum occurs at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 6.

Correct Answer: B — 4

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Q. Find the point of inflection for f(x) = x^3 - 6x^2 + 9x. (2022)

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

Solution

f''(x) = 6x - 12. Setting f''(x) = 0 gives x = 2. f(2) = 3.

Correct Answer: C — (3, 0)

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Q. Find the point on the curve y = x^3 - 3x^2 + 4 that has a horizontal tangent. (2023)

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

Solution

To find horizontal tangents, set the derivative y' = 3x^2 - 6x = 0. This gives x = 0 and x = 2. The point (1, 2) has a horizontal tangent.

Correct Answer: B — (1, 2)

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Q. Find the point on the curve y = x^3 - 3x^2 + 4 where the tangent is horizontal. (2023)

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

Solution

To find horizontal tangents, set dy/dx = 0. dy/dx = 3x^2 - 6x = 0 gives x = 0 and x = 2. At x = 1, y = 2.

Correct Answer: B — (1, 2)

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Q. Find the slope of the tangent line to f(x) = 2x^3 - 3x^2 + 4 at x = 1. (2021)

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

Solution

f'(x) = 6x^2 - 6. f'(1) = 6(1)^2 - 6 = 0.

Correct Answer: B — 2

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Q. Find the slope of the tangent line to f(x) = x^2 + 2x at x = 1. (2022)

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

Solution

f'(x) = 2x + 2. At x = 1, f'(1) = 4.

Correct Answer: A — 2

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Q. For the function f(x) = -x^2 + 4x + 1, find the x-coordinate of the vertex. (2023)

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

Solution

The vertex x-coordinate is given by -b/(2a) = -4/(2*-1) = 2.

Correct Answer: A — 2

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Q. For the function f(x) = -x^2 + 6x, find the x-coordinate of the vertex. (2022)

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

Solution

The vertex x-coordinate is found using x = -b/(2a) = -6/(-2) = 3.

Correct Answer: A — 3

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Q. For the function f(x) = 2x^2 - 8x + 10, find the minimum value. (2022)

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

Solution

The minimum occurs at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 2.

Correct Answer: C — 4

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Showing 31 to 60 of 104 (4 Pages)

Applications of Derivatives

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