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. For the equation x² + 6x + k = 0 to have real roots, what must be the minimum value of k? (2023)

A. -9
B. -6
C. -12
D. -15

Solution

The discriminant must be non-negative: 6² - 4*1*k ≥ 0, thus k ≤ 9.

Correct Answer: A — -9

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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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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. For the matrix E = [[1, 2, 3], [0, 1, 4], [5, 6, 0]], find det(E). (2021)

A. -24
B. 24
C. 0
D. 12

Solution

Using the determinant formula, det(E) = 1*(1*0 - 4*6) - 2*(0*0 - 4*5) + 3*(0*6 - 1*5) = 1*(-24) - 2*(-20) - 15 = -24 + 40 - 15 = 1.

Correct Answer: A — -24

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Q. For the matrix E = [[1, 2, 3], [0, 1, 4], [5, 6, 0]], find the determinant. (2023)

A. -24
B. 24
C. 0
D. 12

Solution

Determinant of E = 1(1*0 - 4*6) - 2(0*0 - 4*5) + 3(0*6 - 1*5) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1.

Correct Answer: A — -24

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Q. For the matrix E = [[1, 2], [2, 4]], what is the determinant? (2021)

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

Solution

Determinant of E = (1*4) - (2*2) = 4 - 4 = 0.

Correct Answer: A — 0

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Q. For the quadratic equation x² + 2x + k = 0 to have real roots, what is the condition on k? (2021)

A. k ≥ 1
B. k ≤ 1
C. k > 1
D. k < 1

Solution

The discriminant must be non-negative: 2² - 4*1*k ≥ 0, which simplifies to k ≤ 1.

Correct Answer: B — k ≤ 1

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Q. For the quadratic equation x² + 6x + k = 0 to have no real roots, what must be the value of k? (2021)

A. k < 9
B. k > 9
C. k = 9
D. k ≤ 9

Solution

The discriminant must be less than zero: 6² - 4*1*k < 0, which simplifies to k > 9.

Correct Answer: B — k > 9

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Q. For the quadratic equation x² + 6x + k = 0 to have real roots, what is the minimum value of k? (2021)

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

Solution

The discriminant must be non-negative: 6² - 4*1*k ≥ 0, thus k ≤ 9. The minimum value of k is -9.

Correct Answer: A — -9

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Q. For which value of k does the equation x² - kx + 9 = 0 have no real roots? (2021)

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

Solution

The discriminant must be less than zero: k² - 4*1*9 < 0, thus k² < 36, so k > 6 or k < -6.

Correct Answer: A — 6

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Q. For which value of k does the equation x² - kx + 9 = 0 have roots that are both positive? (2023)

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

Solution

For both roots to be positive, k must be greater than 0 and k² > 4*9 = 36, thus k > 6.

Correct Answer: B — 8

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Q. For which value of m does the equation x² + mx + 9 = 0 have roots that are both negative? (2021)

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

Solution

For both roots to be negative, m must be greater than 0 and m² < 36. Thus, m must be in the range (-6, 0). The suitable value is -4.

Correct Answer: B — -4

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Q. For which value of m does the equation x² - mx + 9 = 0 have roots 3 and 3? (2023)

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

Solution

The sum of the roots is 3 + 3 = 6, hence m = 6.

Correct Answer: A — 6

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Q. For which value of p does the equation x² + px + 4 = 0 have roots 2 and -2? (2022)

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

Solution

Using the sum of roots: 2 + (-2) = -p, hence p = 0.

Correct Answer: C — -4

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Q. For which value of p does the equation x² + px + 4 = 0 have roots that are both negative? (2022)

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

Solution

For both roots to be negative, p must be greater than 0 and p² > 16. Thus, p < -4.

Correct Answer: C — -4

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Q. For which value of p does the equation x² + px + 9 = 0 have roots that are both negative? (2021)

A. -6
B. -4
C. -3
D. -2

Solution

For both roots to be negative, p must be positive and p² > 4*9. Thus, p > 6, so p = -4 is valid.

Correct Answer: B — -4

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Q. For which value of p does the equation x² - px + 9 = 0 have roots 3 and 3? (2021)

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

Solution

The sum of the roots is 3 + 3 = 6, hence p = 6.

Correct Answer: A — 6

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Q. How many different 4-digit PINs can be formed using the digits 0-9 if digits cannot be repeated?

A. 5040
B. 10000
C. 9000
D. 7200

Solution

The first digit can be any of 10 digits, the second can be any of 9, the third can be any of 8, and the fourth can be any of 7. Total = 10 * 9 * 8 * 7 = 5040.

Correct Answer: A — 5040

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Q. How many different 4-digit PINs can be formed using the digits 0-9 if repetition is allowed? (2020)

A. 10000
B. 1000
C. 100
D. 1000

Solution

Each digit can be any of the 10 digits (0-9), so the total number of PINs is 10^4 = 10000.

Correct Answer: A — 10000

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Mathematics

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