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. Find the limit: lim (x -> 0) (x^2)/(sin(x)) (2023)

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

Solution

As x approaches 0, sin(x) approaches x, thus lim (x -> 0) (x^2/sin(x)) = lim (x -> 0) (x^2/x) = lim (x -> 0) x = 0.

Correct Answer: A — 0

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Q. Find the limit: lim (x -> 1) (x^4 - 1)/(x - 1) (2023)

A. 0
B. 1
C. 4
D. Undefined

Solution

Factoring gives ((x - 1)(x^3 + x^2 + x + 1))/(x - 1). For x ≠ 1, this simplifies to x^3 + x^2 + x + 1. Thus, lim (x -> 1) = 4.

Correct Answer: A — 0

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Q. Find the limit: lim (x -> 2) (x^2 + 3x - 10)/(x - 2) (2021)

A. 0
B. 1
C. 5
D. 7

Solution

Factoring gives (x - 2)(x + 5)/(x - 2). For x ≠ 2, this simplifies to x + 5. Evaluating at x = 2 gives 7.

Correct Answer: D — 7

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Q. Find the limit: lim (x -> 2) (x^2 - 3x + 2)/(x - 2) (2021)

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

Solution

The expression is undefined at x=2. The limit does not exist as the function approaches infinity.

Correct Answer: D — Undefined

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Q. Find the limit: lim (x -> 3) (x^2 - 9)/(x - 3) (2023)

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

Solution

The expression can be factored as ((x - 3)(x + 3))/(x - 3). For x ≠ 3, this simplifies to x + 3. Thus, lim (x -> 3) (x + 3) = 6.

Correct Answer: A — 0

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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 magnitude of the vector A = 3i - 4j. (2020)

A. 5
B. 7
C. 10
D. 12

Solution

|A| = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5.

Correct Answer: A — 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 midpoint of the line segment joining the points (2, 3) and (4, 7). (2022) 2022

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

Solution

Midpoint = ((2+4)/2, (3+7)/2) = (3, 5).

Correct Answer: A — (3, 5)

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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 particular solution of dy/dx = 4y with the initial condition y(0) = 2.

A. y = 2e^(4x)
B. y = e^(4x)
C. y = 4e^(x)
D. y = 2e^(x)

Solution

The general solution is y = Ce^(4x). Using the initial condition y(0) = 2, we find C = 2, thus y = 2e^(4x).

Correct Answer: A — y = 2e^(4x)

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Q. Find the particular solution of dy/dx = 4y, given y(0) = 2.

A. y = 2e^(4x)
B. y = e^(4x)
C. y = 4e^(2x)
D. y = 2e^(x/4)

Solution

The general solution is y = Ce^(4x). Using the initial condition y(0) = 2, we find C = 2, thus y = 2e^(4x).

Correct Answer: A — y = 2e^(4x)

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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 of intersection of the lines y = x + 2 and y = -x + 4. (2023)

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

Solution

Setting x + 2 = -x + 4 gives 2x = 2, so x = 1. Substituting x back gives y = 3. Thus, the point is (1, 3).

Correct Answer: A — (1, 3)

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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 real part of the complex number 4 + 5i. (2023)

A. 4
B. 5
C. 9
D. 0

Solution

The real part of the complex number 4 + 5i is 4.

Correct Answer: A — 4

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Q. Find the roots of the equation 3x² - 12x + 12 = 0. (2021)

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

Solution

Dividing by 3 gives x² - 4x + 4 = 0, which factors to (x - 2)² = 0, hence the root is 2.

Correct Answer: A — 2

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Mathematics

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