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Karnataka II PUC Basic Mathematics (KSEAB) study materials
Karnataka II PUC Basic Mathematics (KSEAB) study resources

Karnataka II PUC Basic Mathematics (KSEAB) practice questions

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Question 1Sets, Relations and Functions

The set A is defined in set-builder form as A = {x : x is a vowel in the English alphabet}. Which of the following gives A in roster form?

  1. {a, e, i, o, u, y}
  2. {a, b, c, d, e}
  3. {a, e, i, o, u} Correct answer.
  4. {a, e, i, o}

Explanation

The vowels of the English alphabet are exactly a, e, i, o and u, so the roster form lists these five letters inside braces. The letter y is often called a semi-vowel but is not treated as a vowel in this standard classification, and b, c, d are consonants. Hence A = {a, e, i, o, u}.

Question 2Sets, Relations and Functions

The set A is given by A = {2, 4, 6, 8, 10}. How many elements does the power set P(A) contain?

  1. 25
  2. 32 Correct answer.
  3. 5
  4. 10

Explanation

A set with n elements has a power set with 2^n elements, because each element is either included in or excluded from a subset. Here n(A) = 5, so the power set has 2^5 = 32 elements. These range from the empty set to A itself.

Question 3Sets, Relations and Functions

A = {1, 2, 3} and B = {a, b}. Which of the following relations from A to B is a function?

  1. {(1, b), (2, a), (3, b)} Correct answer.
  2. {(1, a), (2, b), (2, a)}
  3. {(1, a), (1, b), (2, b), (3, a)}
  4. {(1, a), (1, b), (2, a)}

Explanation

A relation from A to B is a function when every element of A appears exactly once as a first coordinate. In {(1, b), (2, a), (3, b)} each of 1, 2 and 3 occurs once, so the rule assigns a unique image to every element of A. Repeating a first coordinate, as in the other options, violates the definition.

Question 4Sets, Relations and Functions

In a survey of students, n(A) = 20, n(B) = 15 and n(A ∩ B) = 8. What is n(A ∪ B)?

  1. 27 Correct answer.
  2. 12
  3. 35
  4. 43

Explanation

The addition principle for two sets states that n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting gives 20 + 15 − 8 = 27. The subtraction is needed because the 8 students counted in both sets have been counted twice.

Question 5Sets, Relations and Functions

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}. Which set equals (A ∩ B)′?

  1. {1, 2, 3, 6, 7, 8, 9, 10} Correct answer.
  2. {1, 2, 3, 6, 7}
  3. {4, 5}
  4. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Explanation

First find the intersection: A ∩ B = {4, 5}, the elements common to both sets. The complement of this set in U consists of all elements of U that are not 4 or 5, namely {1, 2, 3, 6, 7, 8, 9, 10}. Note that the complement is taken in the universal set U, not in A ∪ B.

Question 6Sets, Relations and Functions

For any two sets A and B, the set difference A − B is equal to which of the following?

  1. A′ ∩ B
  2. A ∩ B′ Correct answer.
  3. A ∪ B′
  4. B ∩ A′

Explanation

A − B consists of those elements of A that do not belong to B. An element is outside B exactly when it lies in B′, so the condition is membership of both A and B′, giving A − B = A ∩ B′. This is one of the standard identities used to convert set differences into intersections and complements.

Question 7Sets, Relations and Functions

On A = {1, 2, 3} the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is defined. Which description of R is correct?

  1. reflexive and symmetric only
  2. reflexive only Correct answer.
  3. reflexive and transitive only
  4. symmetric and transitive only

Explanation

All three pairs (1, 1), (2, 2) and (3, 3) are present, so R is reflexive. It is not symmetric because (1, 2) ∈ R while (2, 1) ∉ R, and it is not transitive because (1, 2) and (2, 3) belong to R but (1, 3) does not. Hence only reflexivity holds.

Question 8Sets, Relations and Functions

A = {p, q} and B = {1, 2, 3}. How many distinct relations from A to B are there?

  1. 9
  2. 6
  3. 64 Correct answer.
  4. 8

Explanation

A relation from A to B is any subset of A × B. Since n(A × B) = 2 × 3 = 6, the number of subsets of A × B is 2^6 = 64, and each subset is a distinct relation. Thus the count is a power of 2, not a product of the set sizes.

Question 9Sets, Relations and Functions

The function f : R → R is defined by f(x) = √(x − 2) / (x − 5). What is the domain of f?

  1. (2, ∞) with x = 5 excluded
  2. [2, ∞) with x = 5 excluded Correct answer.
  3. [2, ∞) with no value excluded
  4. (−∞, 2] with x = 5 excluded

Explanation

The square root √(x − 2) is real only when x − 2 ≥ 0, that is x ≥ 2, and 2 itself is allowed because √0 = 0. The denominator x − 5 must be non-zero, so x = 5 is removed from the interval. The domain is therefore [2, ∞) except x = 5.

Question 10Sets, Relations and Functions

How many onto functions are there from a set with 3 elements to a set with 2 elements?

  1. 8
  2. 9
  3. 12
  4. 6 Correct answer.

Explanation

Every function from a 3-element set to a 2-element set assigns one of the two images to each of the three elements, giving 2^3 = 8 functions. Of these, exactly two are constant, hitting only one of the two elements, and a constant function is not onto. Hence 8 − 2 = 6 functions are onto.

Question 11Sets, Relations and Functions

Let A = {1, 2, 3} and B = {x : x is a positive integer less than 4}. Which statement is correct?

  1. B is a proper subset of A
  2. A is a proper subset of B
  3. A ∩ B = ∅
  4. A = B Correct answer.

Explanation

B = {1, 2, 3} because the positive integers less than 4 are 1, 2 and 3. This is exactly the roster of A, so A and B contain the same elements and are equal sets. Equal sets are not proper subsets of each other, and their intersection is not empty.

Question 12Sets, Relations and Functions

For any two sets A and B, which set is equal to (A ∪ B)′?

  1. A′ − B′
  2. A ∩ B′
  3. A′ ∪ B′
  4. A′ ∩ B′ Correct answer.

Explanation

De Morgan's law for sets states that the complement of a union is the intersection of the complements: (A ∪ B)′ = A′ ∩ B′. This follows because an element outside A ∪ B must be outside both A and B.

Question 13Sets, Relations and Functions

Which set-builder form represents the interval (2, 5]?

  1. {x : 2 ≤ x < 5}
  2. {x : 2 < x ≤ 5} Correct answer.
  3. {x : 2 ≤ x ≤ 5}
  4. {x : 2 < x < 5}

Explanation

The interval (2, 5] includes all real numbers greater than 2 and less than or equal to 5. The round bracket at 2 excludes 2, while the square bracket at 5 includes 5. Hence the set-builder form is {x : 2 < x ≤ 5}.

Question 14Sets, Relations and Functions

If A = {1, 2} and B = {a, b, c}, what is n(A × B)?

  1. 6 Correct answer.
  2. 5
  3. 2
  4. 9

Explanation

For finite sets A and B, n(A × B) = n(A) × n(B). Here n(A) = 2 and n(B) = 3, so the Cartesian product has 2 × 3 = 6 ordered pairs.

Question 15Sets, Relations and Functions

How many functions are there from A = {1, 2, 3} to B = {p, q}?

  1. 9
  2. 4
  3. 8 Correct answer.
  4. 6

Explanation

A function from A to B assigns exactly one element of B to each element of A. Each of the 3 elements of A has 2 independent choices in B, so the number of functions is 2³ = 8.

Question 16Sets, Relations and Functions

The function f : R → R is defined by f(x) = x² + 2. What is the range of f?

  1. y > 2
  2. all real y
  3. y ≥ 2 Correct answer.
  4. y ≥ 0

Explanation

Since x² ≥ 0 for every real x, we have x² + 2 ≥ 2. The value 2 is attained at x = 0, and larger values occur as |x| increases, so the range is y ≥ 2.

Question 17Sets, Relations and Functions

If f(x) = 2x + 1 and g(x) = x², what is (f ∘ g)(2)?

  1. 7
  2. 9 Correct answer.
  3. 8
  4. 10

Explanation

First evaluate g(2) = 2² = 4. Then apply f to this value: f(4) = 2(4) + 1 = 9. Therefore (f ∘ g)(2) = 9.

Question 18Sets, Relations and Functions

If A = {1, 2, 3, 4} and B = {3, 4, 5}, which set is A Δ B, where A Δ B = (A − B) ∪ (B − A)?

  1. {1, 2}
  2. {3, 4}
  3. {1, 2, 5} Correct answer.
  4. {1, 2, 3, 4, 5}

Explanation

The symmetric difference A Δ B consists of elements that belong to exactly one of A and B. A − B = {1, 2} and B − A = {5}, so their union is {1, 2, 5}. The common elements 3 and 4 are excluded.

Question 19Sets, Relations and Functions

The function f : R → R is defined by f(x) = 3x − 4. What is f⁻¹(x)?

  1. f⁻¹(x) = (x + 3)/4
  2. f⁻¹(x) = 3x + 4
  3. f⁻¹(x) = (x + 4)/3 Correct answer.
  4. f⁻¹(x) = (x − 4)/3

Explanation

Let y = 3x − 4. Solving for x gives y + 4 = 3x, so x = (y + 4)/3. Interchanging x and y gives f⁻¹(x) = (x + 4)/3.

Question 20Sets, Relations and Functions

How many equivalence relations are there on the set A = {1, 2, 3}?

  1. 6
  2. 5 Correct answer.
  3. 8
  4. 3

Explanation

Equivalence relations on a finite set correspond bijectively to partitions of that set. The set {1, 2, 3} has one partition into one block, three partitions into two blocks, and one partition into three singleton blocks. Hence there are 1 + 3 + 1 = 5 equivalence relations.

Question 21Matrices and Determinants

The order of the matrix [[2, -1, 0], [5, 3, 7]] is

  1. 6 × 1
  2. 2 × 2
  3. 2 × 3 Correct answer.
  4. 3 × 2

Explanation

The order of a matrix is written as (number of rows) × (number of columns). The given matrix has 2 horizontal rows and 3 vertical columns, so its order is 2 × 3. The total number of elements is 2 × 3 = 6, but that count is not the order.

Question 22Matrices and Determinants

A square matrix A is said to be symmetric if

  1. A^T = -A
  2. A^(-1) = A^T
  3. A^2 = A
  4. A^T = A Correct answer.

Explanation

A square matrix is symmetric when it equals its own transpose, that is A^T = A. This forces a_ij = a_ji for every pair of indices, so the entries mirror across the main diagonal.

Question 23Matrices and Determinants

The value of the determinant with first row (3, 5) and second row (2, 4) is

  1. 22
  2. 2 Correct answer.
  3. -2
  4. 12

Explanation

For a 2 × 2 determinant, subtract the product of the secondary diagonal from the product of the main diagonal: (3)(4) - (5)(2) = 12 - 10 = 2.

Question 24Matrices and Determinants

In the matrix A = [[2, 1], [4, 3]], the cofactor of the element a_12 is

  1. 4
  2. 1
  3. 3
  4. -4 Correct answer.

Explanation

The minor M_12 is the determinant left after deleting row 1 and column 2, which leaves the single entry 4, so M_12 = 4. The cofactor is C_12 = (-1)^(1+2) M_12 = -4.

Question 25Matrices and Determinants

For A = [[2, 3], [1, 4]], the adjoint adj A is

  1. [[4, -3], [-1, 2]] Correct answer.
  2. [[-4, 3], [1, -2]]
  3. [[2, -3], [-1, 4]]
  4. [[4, 3], [1, 2]]

Explanation

For a 2 × 2 matrix the adjoint is obtained by interchanging the diagonal entries and changing the signs of the off-diagonal entries. Thus adj A = [[d, -b], [-c, a]] = [[4, -3], [-1, 2]].

Question 26Matrices and Determinants

The matrix A = [[k, 2], [8, k]] is singular when k equals

  1. 4 only
  2. 2
  3. 16
  4. ±4 Correct answer.

Explanation

A square matrix is singular exactly when its determinant is zero. Here det A = k^2 - 16, so k^2 = 16, giving k = 4 or k = -4.

Question 27Matrices and Determinants

If two rows of a 3 × 3 determinant are identical, then the value of the determinant is

  1. unchanged
  2. -1
  3. 0 Correct answer.
  4. 1

Explanation

A standard property of determinants is that if any two rows or two columns are identical, the determinant equals zero. Swapping the equal rows must change the sign of the determinant while leaving its value unchanged, which is possible only when the value is zero.

Question 28Matrices and Determinants

If A = [[1, 2], [3, 4]] and B = [[0, 1], [1, 0]], then AB is

  1. [[2, 1], [4, 3]] Correct answer.
  2. [[3, 4], [1, 2]]
  3. [[1, 2], [4, 3]]
  4. [[0, 2], [3, 0]]

Explanation

Multiply rows of A by columns of B: (1)(0) + (2)(1) = 2, (1)(1) + (2)(0) = 1, (3)(0) + (4)(1) = 4, and (3)(1) + (4)(0) = 3. Hence AB = [[2, 1], [4, 3]].

Question 29Matrices and Determinants

The system of equations x + y = 3 and 2x + 2y = 7 has

  1. no solution Correct answer.
  2. exactly two solutions
  3. infinitely many solutions
  4. a unique solution

Explanation

The coefficient determinant is (1)(2) - (1)(2) = 0, so the system has either no solution or infinitely many. Comparing ratios, 1/2 = 1/2 but 3/7 is not equal to 1/2, so the equations are inconsistent and there is no solution.

Question 30Matrices and Determinants

If A is a 3 × 3 matrix with |A| = 5, then |adj A| equals

  1. 1/5
  2. 5
  3. 25 Correct answer.
  4. 125

Explanation

For an n × n matrix, the determinant of the adjoint satisfies |adj A| = |A|^(n-1). With n = 3 and |A| = 5, this gives |adj A| = 5^2 = 25.

Question 31Matrices and Determinants

A matrix A has order 3 × 2, and a matrix B has order 2 × 4. The product AB is defined and has order

  1. 3 × 4 Correct answer.
  2. 6 × 8
  3. 2 × 3
  4. 4 × 2

Explanation

For the product AB to be defined, the number of columns of A must equal the number of rows of B. Here A is 3 × 2 and B is 2 × 4, so the product has order 3 × 4.

Question 32Matrices and Determinants

If A is a matrix of order 2 × 3, then the transpose A^T has order

  1. 3 × 2 Correct answer.
  2. 2 × 3
  3. 6 × 6
  4. 3 × 3

Explanation

The transpose of an m × n matrix has order n × m because rows and columns are interchanged. For a 2 × 3 matrix A, A^T therefore has order 3 × 2.

Question 33Matrices and Determinants

The matrix A = [[0, 4, -1], [-4, 0, 6], [1, -6, 0]] satisfies A^T = -A, so A is

  1. upper triangular
  2. scalar
  3. symmetric
  4. skew-symmetric Correct answer.

Explanation

A matrix is skew-symmetric when A^T = -A, which requires zero diagonal entries and opposite off-diagonal pairs. The given matrix has those properties, so it is skew-symmetric.

Question 34Matrices and Determinants

If A = [[1, 2], [3, 4]] and B = [[2, 0], [1, 3]], then the element in the second row and first column of AB is

  1. 12
  2. 5
  3. 10 Correct answer.
  4. 8

Explanation

The element in row 2, column 1 of AB is the dot product of row 2 of A with column 1 of B. That is (3)(2) + (4)(1) = 10.

Question 35Matrices and Determinants

If A = [[2, 1], [7, 4]], then A^(-1) equals

  1. [[2, -1], [-7, 4]]
  2. [[4, 1], [7, 2]]
  3. [[4, -1], [-7, 2]] Correct answer.
  4. [[4, -1], [7, 2]]

Explanation

For A = [[a, b], [c, d]], the inverse is (1/(ad - bc)) [[d, -b], [-c, a]] when ad - bc ≠ 0. Here ad - bc = (2)(4) - (1)(7) = 1, so A^(-1) = [[4, -1], [-7, 2]].

Question 36Matrices and Determinants

Using Cramer's rule for the system x + 2y = 5 and 3x - y = 1, the value of x is

  1. -7
  2. 1 Correct answer.
  3. 7
  4. -1

Explanation

Cramer's rule gives x = Δx/Δ, where Δ is the coefficient determinant and Δx replaces the x-column by the constants. Here Δ = (1)(-1) - (2)(3) = -7 and Δx = (5)(-1) - (2)(1) = -7, so x = 1.

Question 37Matrices and Determinants

If A is a 2 × 2 matrix with |A| = 3, then |2A| equals

  1. 9
  2. 6
  3. 18
  4. 12 Correct answer.

Explanation

For an n × n matrix A, the determinant of kA is k^n times |A| because each of the n rows is multiplied by k. Here n = 2 and |A| = 3, so |2A| = 2^2 × 3 = 12.

Question 38Matrices and Determinants

The system kx + 2y = 3 and 4x + ky = 5 has a unique solution when

  1. k ≠ ±2√2 Correct answer.
  2. k = 0
  3. k = ±2√2
  4. every real k

Explanation

The system has a unique solution exactly when the coefficient determinant is non-zero. The determinant is k·k - 2·4 = k^2 - 8, so uniqueness holds for every real k with k^2 ≠ 8, that is k ≠ ±2√2.

Question 39Matrices and Determinants

If A = [[1, 2], [3, 4]], then A^2 - 5A - 2I equals

  1. zero matrix Correct answer.
  2. I
  3. 2I
  4. 4I

Explanation

Compute A^2 first and then subtract 5A and 2I entrywise. A^2 = [[7, 10], [15, 22]] and 5A = [[5, 10], [15, 20]], so A^2 - 5A = [[2, 0], [0, 2]] = 2I; subtracting 2I gives the zero matrix.

Question 40Matrices and Determinants

If A is a 3 × 3 matrix with |A| = -2, then |adj(adj A)| equals

  1. -32
  2. 16 Correct answer.
  3. 64
  4. -16

Explanation

For any n × n non-singular matrix A, |adj A| = |A|^(n-1). Applying this twice gives |adj(adj A)| = |adj A|^2 for n = 3, and with |A| = -2 we get |adj A| = (-2)^2 = 4 and the final value 4^2 = 16.

Question 41Permutations, Combinations and Binomial Theorem

A first-year student must select exactly one language elective from 5 available language subjects and exactly one laboratory elective from 4 available laboratory subjects. In how many different ways can the pair of electives be chosen?

  1. 24
  2. 20 Correct answer.
  3. 9
  4. 120

Explanation

Because the two choices are made one after the other, the fundamental principle of multiplication applies and the counts are multiplied: 5 × 4 = 20 distinct pairs. Addition would be correct only if the student had to pick a single elective from either group, not one from each.

Question 42Permutations, Combinations and Binomial Theorem

The value of the permutation 7P3, that is, the number of arrangements of 3 distinct objects selected from 7 distinct objects, is:

  1. 35
  2. 343
  3. 210 Correct answer.
  4. 5040

Explanation

By definition nPr = n!/(n − r)!, so 7P3 = 7!/4! = 7 × 6 × 5 = 210. Arrangement matters here, so we do not divide by 3!; doing so would give the combination 7C3 = 35.

Question 43Permutations, Combinations and Binomial Theorem

If nC2 = 45, where n is a positive integer, then the value of n is:

  1. 11
  2. 10 Correct answer.
  3. 9
  4. 45

Explanation

Using nC2 = n(n − 1)/2, we get n(n − 1) = 90, and since 10 × 9 = 90 the solution is n = 10. The other root of the quadratic is negative and is rejected because the number of objects must be positive.

Question 44Permutations, Combinations and Binomial Theorem

The number of distinct arrangements of all the letters of the word INDIA is:

  1. 60 Correct answer.
  2. 120
  3. 20
  4. 24

Explanation

The word has 5 letters in which I appears twice, so the number of distinct arrangements is 5!/2! = 120/2 = 60. Dividing by 2! removes the arrangements that differ only by exchanging the two identical I's.

Question 45Permutations, Combinations and Binomial Theorem

From 5 boys and 4 girls, a committee of 3 students is to be formed containing exactly one girl. The number of such committees is:

  1. 84
  2. 30
  3. 40 Correct answer.
  4. 80

Explanation

Choose the single girl in 4C1 = 4 ways and the two boys in 5C2 = 10 ways, then multiply by the fundamental principle: 4 × 10 = 40. Within a committee the order of members is irrelevant, so combinations rather than permutations are used.

Question 46Permutations, Combinations and Binomial Theorem

In the expansion of (2x + 3)^6, the coefficient of x^3 is:

  1. 1080
  2. 4320 Correct answer.
  3. 8640
  4. 2160

Explanation

The general term is 6Cr (2x)^(6−r) 3^r, whose x-power is x^(6−r). For x^3 we need 6 − r = 3, so r = 3 and the coefficient is 6C3 × 2^3 × 3^3 = 20 × 8 × 27 = 4320.

Question 47Permutations, Combinations and Binomial Theorem

The 5th term in the expansion of (a + b)^8 is:

  1. 28a^4b^4
  2. 56a^3b^5
  3. 70a^4b^4 Correct answer.
  4. 70a^5b^3

Explanation

The (r + 1)th term of (a + b)^n is nCr a^(n−r) b^r, so the 5th term corresponds to r = 4 and equals 8C4 a^4 b^4 = 70a^4b^4. Because the exponent 8 is even, this term is also the middle term of the expansion.

Question 48Permutations, Combinations and Binomial Theorem

The term independent of x in the expansion of (2x^2 − 1/x)^9 is:

  1. 5376
  2. 84
  3. 672 Correct answer.
  4. −672

Explanation

The general term is 9Cr (2x^2)^(9−r) (−1/x)^r, which contains x^(18−3r). Setting 18 − 3r = 0 gives r = 6, so the required term is 9C6 × 2^3 × (−1)^6 = 84 × 8 = 672, which is positive because the sixth power of −1 is +1.

Question 49Permutations, Combinations and Binomial Theorem

In the expansion of (3 + kx)^9, the coefficients of x^2 and x^3 are equal, where k is a non-zero constant. The value of k is:

  1. 3/7
  2. 9/7 Correct answer.
  3. 7/9
  4. 7/3

Explanation

The coefficient of x^2 is 9C2 × 3^7 × k^2 = 36 × 3^7 × k^2 and that of x^3 is 9C3 × 3^6 × k^3 = 84 × 3^6 × k^3. Equating them and cancelling 3^6 k^2 gives 36 × 3 = 84k, that is 108 = 84k, so k = 9/7.

Question 50Permutations, Combinations and Binomial Theorem

In how many ways can 7 students be seated around a circular table, if seatings that differ only by a rotation are regarded as the same arrangement?

  1. 720 Correct answer.
  2. 120
  3. 5040
  4. 360

Explanation

For n distinct objects around a circle the number of arrangements is (n − 1)!, because fixing one student's position removes the n rotational copies of each seating. Here (7 − 1)! = 6! = 720.