Mathematics for Competitive Exam Success

Mathematics

3D Geometry Applications of Derivatives Binomial Theorem Circles Complex Numbers Conic Sections Coordinate Geometry Differential Equations Differentiation Integration Limits & Continuity Matrices & Determinants Permutations & Combinations Probability Quadratic Equations Sequences & Series Sets & Relations Straight Lines Trigonometry Vector Algebra

Q. Determine the local minima of f(x) = x^3 - 3x + 2. (2021)

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

Solution

f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = 1. f(1) = 0.

Correct Answer: B — 0

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Q. Determine the local minima of f(x) = x^4 - 4x^2. (2021)

A. -2
B. 0
C. 2
D. 4

Solution

f'(x) = 4x^3 - 8x. Setting f'(x) = 0 gives x = 0, ±2. f(0) = 0.

Correct Answer: B — 0

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Q. Determine 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. 30
D. 40

Solution

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

Correct Answer: B — 50

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Q. Determine the maximum height of the function f(x) = -x^2 + 6x + 5. (2020) 2020

A. 8
B. 10
C. 12
D. 6

Solution

The vertex occurs at x = 3. f(3) = -3^2 + 6*3 + 5 = 8.

Correct Answer: A — 8

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Q. Determine the maximum height of the projectile given by h(t) = -16t^2 + 64t + 80. (2023)

A. 80
B. 64
C. 48
D. 96

Solution

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

Correct Answer: A — 80

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Q. Determine the maximum height of the projectile modeled by h(t) = -16t^2 + 64t + 80. (2020)

A. 80
B. 64
C. 48
D. 96

Solution

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

Correct Answer: A — 80

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Q. Determine the maximum value of f(x) = -x^2 + 6x - 8. (2022)

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

Solution

The maximum occurs at x = 3. f(3) = -3^2 + 6(3) - 8 = 6.

Correct Answer: C — 6

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Q. Determine the minimum value of f(x) = x^2 - 4x + 5. (2021)

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

Solution

The vertex form gives the minimum at x = 2. f(2) = 2^2 - 4(2) + 5 = 1.

Correct Answer: A — 1

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Q. Determine 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. Determine the point of inflection for f(x) = x^4 - 4x^3 + 6. (2023)

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

Solution

f''(x) = 12x^2 - 24x. Setting f''(x) = 0 gives x(12x - 24) = 0, so x = 0 or x = 2. Check f(1) = 3.

Correct Answer: A — (1, 3)

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Q. Determine the point where the function f(x) = 2x^3 - 9x^2 + 12x has a local minimum. (2023)

A. (1, 5)
B. (2, 0)
C. (3, 3)
D. (4, 4)

Solution

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

Correct Answer: C — (3, 3)

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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. Determine the product of the roots of the equation x² + 6x + 8 = 0. (2023)

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

Solution

The product of the roots is given by c/a = 8/1 = 8.

Correct Answer: A — 8

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Q. Determine the product of the roots of the equation x² + 6x + 9 = 0. (2021)

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

Solution

The product of the roots is c/a = 9/1 = 9.

Correct Answer: A — 9

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Q. Determine the roots of the equation x² + 2x - 8 = 0. (2023)

A. -4 and 2
B. 4 and -2
C. 2 and -4
D. 0 and 8

Solution

Factoring gives (x + 4)(x - 2) = 0, hence the roots are -4 and 2.

Correct Answer: A — -4 and 2

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Q. Determine the roots of the equation x² + 6x + 9 = 0. (2023)

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

Solution

This is a perfect square: (x + 3)² = 0, hence the root is x = -3.

Correct Answer: A — -3

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Q. Evaluate the integral ∫ (3x^2 + 2x) dx. (2020)

A. x^3 + x^2 + C
B. x^3 + x^2 + 2C
C. x^3 + x^2 + 1
D. x^3 + 2x + C

Solution

The integral is (3/3)x^3 + (2/2)x^2 + C = x^3 + x^2 + C.

Correct Answer: A — x^3 + x^2 + C

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Q. Evaluate the integral ∫(3x^2 + 2)dx. (2022)

A. x^3 + 2x + C
B. x^3 + 2x^2 + C
C. x^3 + 2x^3 + C
D. 3x^3 + 2x + C

Solution

Integrating term by term, ∫3x^2dx = x^3 and ∫2dx = 2x. Thus, ∫(3x^2 + 2)dx = x^3 + 2x + C.

Correct Answer: A — x^3 + 2x + C

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Q. Evaluate the limit: lim (x -> 0) (tan(x)/x) (2023)

A. 0
B. 1
C. ∞
D. Undefined

Solution

Using the limit property lim (x -> 0) (tan(x)/x) = 1, we find that the limit is 1.

Correct Answer: B — 1

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Q. Evaluate the limit: lim (x -> 0) (x - sin(x))/x^3 (2022)

A. 0
B. 1/6
C. 1/3
D. 1/2

Solution

Using the Taylor series expansion for sin(x), we find that lim (x -> 0) (x - sin(x))/x^3 = 1/6.

Correct Answer: B — 1/6

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Q. Evaluate the limit: lim (x -> 0) (x^3)/(sin(x)) (2022)

A. 0
B. 1
C. Infinity
D. Undefined

Solution

As x approaches 0, x^3 approaches 0 and sin(x) approaches 0, thus the limit is 0.

Correct Answer: A — 0

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Q. Evaluate the limit: lim (x -> 3) (x^2 - 9)/(x - 3) (2020)

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

Solution

Factoring gives (x - 3)(x + 3)/(x - 3). Canceling (x - 3) gives lim (x -> 3) (x + 3) = 6.

Correct Answer: B — 6

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Q. Evaluate ∫ (2x + 3) dx. (2022)

A. x^2 + 3x + C
B. x^2 + 3 + C
C. x^2 + 3x + 1
D. 2x^2 + 3 + C

Solution

The integral is (2/2)x^2 + 3x + C = x^2 + 3x + C.

Correct Answer: A — x^2 + 3x + C

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Q. Evaluate ∫ (4x^3 - 2x) dx. (2019)

A. x^4 - x^2 + C
B. x^4 - x^2 + 2C
C. x^4 - x + C
D. 4x^4 - 2x^2 + C

Solution

The integral is (4/4)x^4 - (2/2)x^2 + C = x^4 - x^2 + C.

Correct Answer: A — x^4 - x^2 + C

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Q. Evaluate ∫ (5 - 3x) dx. (2022)

A. 5x - (3/2)x^2 + C
B. 5x - (3/3)x^2 + C
C. 5x - (3/4)x^2 + C
D. 5x - (3/5)x^2 + C

Solution

The integral is 5x - (3/2)x^2 + C.

Correct Answer: A — 5x - (3/2)x^2 + C

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Q. Evaluate ∫(2x^2 + 3x + 1)dx. (2021)

A. (2/3)x^3 + (3/2)x^2 + x + C
B. (2/3)x^3 + (3/2)x + C
C. (2/3)x^3 + (3/2)x^2 + C
D. (2/3)x^3 + 3x + C

Solution

Integrating term by term: ∫2x^2dx = (2/3)x^3, ∫3xdx = (3/2)x^2, and ∫1dx = x. Thus, ∫(2x^2 + 3x + 1)dx = (2/3)x^3 + (3/2)x^2 + x + C.

Correct Answer: A — (2/3)x^3 + (3/2)x^2 + x + C

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Q. Evaluate ∫(5x^4)dx. (2020)

A. (5/5)x^5 + C
B. (1/5)x^5 + C
C. (5/4)x^4 + C
D. (1/4)x^4 + C

Solution

The integral of 5x^4 is (5/5)x^5 + C = x^5 + C.

Correct Answer: A — (5/5)x^5 + C

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Q. Evaluate ∫(6x^2 + 3)dx. (2022)

A. 2x^3 + 3x + C
B. 2x^3 + 3 + C
C. 2x^3 + 3x^2 + C
D. 2x^3 + 3x^3 + C

Solution

Integrating term by term: ∫6x^2dx = 2x^3 and ∫3dx = 3x. Thus, ∫(6x^2 + 3)dx = 2x^3 + 3x + C.

Correct Answer: A — 2x^3 + 3x + C

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Q. Find the area of a triangle with vertices at A(0, 0, 0), B(1, 0, 0), and C(0, 1, 0). (2023)

A. 0.5
B. 1
C. 2
D. 0

Solution

Area = 0.5 * base * height = 0.5 * 1 * 1 = 0.5 square units.

Correct Answer: A — 0.5

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Q. Find the area of the triangle formed by the points (0, 0), (4, 0), and (0, 3). (2022) 2022

A. 6
B. 12
C. 8
D. 10

Solution

Area = 1/2 * base * height = 1/2 * 4 * 3 = 6.

Correct Answer: A — 6

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Mathematics

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