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

Karnataka II PUC Mathematics (KSEAB) practice questions

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Question 1Relations and Functions

If f: A → B is a bijection, which statement about the inverse relation f⁻¹ is necessarily true?

  1. f⁻¹ is a function from B onto A Correct answer.
  2. f⁻¹ is a function from A onto B
  3. f⁻¹ is a function from B into A that may not be onto
  4. f⁻¹ is only a relation and need not be a function

Explanation

A bijection is both one-one and onto, so every element of B has exactly one preimage in A. Reversing the ordered pairs therefore yields a function with domain B and codomain A that is itself onto and one-one. Invertibility is exactly the property of being bijective.

Question 2Relations and Functions

The relation R on the set {1, 2, 3} given by R = {(1,1), (2,2), (3,3), (1,2), (2,1)} is

  1. reflexive, symmetric and transitive
  2. reflexive and symmetric but not transitive Correct answer.
  3. symmetric and transitive but not reflexive
  4. reflexive and transitive but not symmetric

Explanation

All three diagonal pairs are present, so R is reflexive, and (1,2) is matched by (2,1), so it is symmetric. But (2,1) and (1,2) belong to R while (2,2) does not, so transitivity fails. Hence R is reflexive and symmetric only.

Question 3Relations and Functions

If f(x) = 3x − 5 and g(x) = x² + 1, then the value of (gof)(2) is

  1. 5
  2. 10
  3. 26
  4. 2 Correct answer.

Explanation

By definition (gof)(x) = g(f(x)), so first compute f(2) = 6 − 5 = 1. Then g(1) = 1² + 1 = 2. Composition applies the inner function first, which is the usual source of error here.

Question 4Relations and Functions

The number of one-one functions from a set with 3 elements to a set with 4 elements is

  1. 81
  2. 24 Correct answer.
  3. 12
  4. 4

Explanation

A one-one function assigns to the three domain elements three distinct images chosen in order from the four codomain elements. The count is the number of injections, 4 × 3 × 2 = 24. This is the falling-factorial formula nPm with n = 4, m = 3.

Question 5Relations and Functions

The domain of the real function f(x) = 1/√(x − 3) is

  1. (3, ∞) Correct answer.
  2. R − {3}
  3. [3, ∞)
  4. (−∞, 3)

Explanation

The expression under the square root must be positive, and the square root sits in a denominator, so x − 3 > 0 strictly. That gives x > 3, i.e. the open interval (3, ∞). The value x = 3 is excluded because it makes the denominator zero.

Question 6Relations and Functions

Let f: R → R be defined by f(x) = 2x + 7. The inverse function f⁻¹(x) equals

  1. (x − 7)/2 Correct answer.
  2. 7 − 2x
  3. (x + 7)/2
  4. 2x − 7

Explanation

Write y = 2x + 7 and solve for x to obtain x = (y − 7)/2, then interchange the names of the variables. Since f is a linear bijection on R, this formula defines f⁻¹ on all of R. Checking, f((x−7)/2) = x confirms the result.

Question 7Relations and Functions

The relation R on Z defined by aRb if and only if a − b is divisible by 5 is

  1. an equivalence relation Correct answer.
  2. symmetric but not transitive
  3. reflexive but not symmetric
  4. transitive but not reflexive

Explanation

Since a − a = 0 is divisible by 5, R is reflexive; if 5 divides a − b then it divides b − a, so R is symmetric; and if 5 divides a − b and b − c then it divides their sum a − c, so R is transitive. Being reflexive, symmetric and transitive, R is an equivalence relation, namely congruence modulo 5.

Question 8Relations and Functions

If f(x) = x/(x − 1) for x ≠ 1, then (fof)(x) equals

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

Explanation

Substituting f(x) into itself gives [x/(x−1)] ÷ [x/(x−1) − 1]. The denominator simplifies to 1/(x−1), so the quotient is x. Thus f is its own inverse, satisfying fof = identity wherever it is defined.

Question 9Relations and Functions

The range of the function f: R → R defined by f(x) = x² + 4x + 7 is

  1. [7, ∞)
  2. [3, ∞) Correct answer.
  3. [4, ∞)
  4. R

Explanation

Completing the square gives f(x) = (x + 2)² + 3. Since a square is never negative, the least value is 3 at x = −2, and the expression increases without bound. Hence the range is [3, ∞).

Question 10Relations and Functions

Let A = {1, 2, 3}. The number of equivalence relations on A is

  1. 5 Correct answer.
  2. 8
  3. 9
  4. 3

Explanation

Equivalence relations on a finite set correspond exactly to partitions of that set. The partitions of a three-element set are: one block, three partitions into a pair plus a singleton, and three singletons, giving 1 + 3 + 1 = 5. These counts are the Bell numbers.

Question 11Relations and Functions

If f: R → R is given by f(x) = x³, then the value of f⁻¹(27) + f⁻¹(−8) is

  1. 5
  2. −1
  3. 3
  4. 1 Correct answer.

Explanation

The cube function is a bijection on R with inverse the cube root. So f⁻¹(27) = 3 and f⁻¹(−8) = −2, whose sum is 1. Note that the cube root of a negative number is negative, which is essential here.

Question 12Relations and Functions

The function f: R → R defined by f(x) = |x| is

  1. neither one-one nor onto Correct answer.
  2. one-one but not onto
  3. onto but not one-one
  4. both one-one and onto

Explanation

Since f(2) = f(−2) = 2 while 2 ≠ −2, the function is not one-one. Also |x| ≥ 0 for every real x, so no negative real number is an image and f is not onto. Therefore it is neither injective nor surjective.

Question 13Relations and Functions

If f(x) = x² − 3x + 2, then the value of f(f(2)) is

  1. 0 Correct answer.
  2. 12
  3. 2
  4. 6

Explanation

First f(2) = 4 − 6 + 2 = 0. Then f(0) = 0 − 0 + 2 = 2? Careful: substituting gives f(0) = 2, so the answer is 2 only if we recompute correctly. Rechecking: f(0) = 0² − 3(0) + 2 = 2, hence f(f(2)) = 2.

Question 14Relations and Functions

Let f: A → B and g: B → C be functions. If gof is injective, which conclusion is valid?

  1. f must be injective Correct answer.
  2. g must be surjective
  3. f must be surjective
  4. g must be injective

Explanation

If f(a₁) = f(a₂), then applying g gives (gof)(a₁) = (gof)(a₂), and injectivity of gof forces a₁ = a₂. So f is injective, while nothing can be concluded about g on elements outside the image of f. Injectivity of a composition therefore passes to the inner function only.

Question 15Relations and Functions

The function f: [0, ∞) → [0, ∞) defined by f(x) = x² + 2x is

  1. neither one-one nor onto
  2. a bijection Correct answer.
  3. onto but not one-one
  4. one-one but not onto

Explanation

On [0, ∞), f is strictly increasing because both terms increase, so it is one-one. Completing the square gives f(x) = (x + 1)² − 1, which rises continuously from 0 to ∞, so every value in [0, ∞) is attained. Hence f is a bijection on the stated domain and codomain.

Question 16Relations and Functions

Let R be a relation on a set A. If R is symmetric and transitive, and for every a in A there exists b in A with aRb, then R is

  1. antisymmetric
  2. reflexive Correct answer.
  3. irreflexive
  4. an equivalence relation

Explanation

Given a, choose b with aRb. Symmetry gives bRa, and transitivity applied to aRb and bRa yields aRa, so R is reflexive. Together with the given symmetry and transitivity, R is therefore an equivalence relation, and reflexivity is the property derived here.

Question 17Inverse Trigonometric Functions

The principal value branch of sin⁻¹ x has range

  1. [0, π]
  2. [-π, π]
  3. [-π/2, π/2] Correct answer.
  4. (-π/2, π/2)

Explanation

For sin⁻¹ to be a function, the sine is restricted to the monotonic interval [-π/2, π/2], and this closed interval is taken as the principal value range. The endpoints are included because sin⁻¹(1) = π/2 and sin⁻¹(-1) = -π/2 are both defined.

Question 18Inverse Trigonometric Functions

The value of cos⁻¹(cos 5π/6) is

  1. 5π/6 Correct answer.
  2. 7π/6
  3. π/6
  4. -5π/6

Explanation

The angle 5π/6 already lies in the principal range [0, π] of cos⁻¹, so cos⁻¹(cos 5π/6) simply returns 5π/6. The identity cos⁻¹(cos x) = x holds exactly when x is inside that principal interval.

Question 19Inverse Trigonometric Functions

The principal value of tan⁻¹(-1) is

  1. -π/4 Correct answer.
  2. π/4
  3. 3π/4
  4. 5π/4

Explanation

The principal range of tan⁻¹ is (-π/2, π/2), and tan⁻¹ is an odd function, so tan⁻¹(-1) = -tan⁻¹(1) = -π/4. This value lies inside the open principal interval.

Question 20Inverse Trigonometric Functions

The domain of the function f(x) = sin⁻¹(2x) is

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

Explanation

sin⁻¹ requires its argument to lie in [-1, 1], so we need -1 ≤ 2x ≤ 1. Dividing through by 2 gives -1/2 ≤ x ≤ 1/2, hence the domain [-1/2, 1/2].

Question 21Inverse Trigonometric Functions

The value of sin⁻¹(1/2) + cos⁻¹(1/2) is

  1. π/2 Correct answer.
  2. π/3
  3. π
  4. π/6

Explanation

For any x in [-1, 1], the identity sin⁻¹ x + cos⁻¹ x = π/2 holds. Substituting x = 1/2 gives π/2 directly, without needing the individual values π/6 and π/3.

Question 22Inverse Trigonometric Functions

The value of tan⁻¹(1/2) + tan⁻¹(1/3) is

  1. π/3
  2. π/4 Correct answer.
  3. π/2
  4. π/6

Explanation

Since the product (1/2)(1/3) = 1/6 is less than 1, the addition formula tan⁻¹ a + tan⁻¹ b = tan⁻¹[(a+b)/(1-ab)] applies directly. Here (a+b)/(1-ab) = (5/6)/(5/6) = 1, so the sum is tan⁻¹(1) = π/4.

Question 23Inverse Trigonometric Functions

The value of sin(cos⁻¹(3/5)) is

  1. 4/5 Correct answer.
  2. 16/25
  3. 3/5
  4. 5/4

Explanation

Let θ = cos⁻¹(3/5), so cos θ = 3/5 with θ in [0, π], giving sin θ ≥ 0. Then sin θ = √(1 - 9/25) = √(16/25) = 4/5.

Question 24Inverse Trigonometric Functions

The value of tan(2 tan⁻¹(1/3)) is

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

Explanation

Using tan 2θ = 2tan θ/(1 - tan²θ) with tan θ = 1/3 gives numerator 2/3 and denominator 1 - 1/9 = 8/9. Dividing, (2/3)·(9/8) = 18/24 = 3/4.

Question 25Inverse Trigonometric Functions

The value of sin⁻¹(sin 2π/3) is

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

Explanation

The angle 2π/3 is outside the principal range [-π/2, π/2] of sin⁻¹, so we must reduce it. Since sin(2π/3) = sin(π/3) = √3/2 and π/3 lies in the principal range, sin⁻¹(sin 2π/3) = π/3.

Question 26Inverse Trigonometric Functions

The value of cos⁻¹(-1/2) is

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

Explanation

The principal range of cos⁻¹ is [0, π], and cos(2π/3) = -1/2 with 2π/3 in that interval. Hence cos⁻¹(-1/2) = 2π/3, which also fits the identity cos⁻¹(-x) = π - cos⁻¹ x.

Question 27Inverse Trigonometric Functions

The value of sin⁻¹(3/5) + sin⁻¹(8/17) + cos⁻¹(84/85) is

  1. 3π/4
  2. π
  3. π/2 Correct answer.
  4. π/4

Explanation

Since cos⁻¹(84/85) = π/2 - sin⁻¹(84/85), the sum becomes π/2 + [sin⁻¹(3/5) + sin⁻¹(8/17) - sin⁻¹(84/85)]. Taking sines of the bracket, sin(A+B) = (3/5)(15/17) + (4/5)(8/17) = 77/85, and with 84/85 for the third angle the bracket vanishes, leaving π/2.

Question 28Inverse Trigonometric Functions

The number of real solutions of the equation tan⁻¹(2x) + tan⁻¹(3x) = π/4 is

  1. 2
  2. 1 Correct answer.
  3. 0
  4. infinitely many

Explanation

Applying the addition formula and requiring the sum to be π/4 gives (2x+3x)/(1-6x²) = 1, i.e. 6x² + 5x - 1 = 0, whose roots are x = 1/6 and x = -1. For x = -1 the product 6x² exceeds 1 and the sum formula needs a π correction, giving a negative sum, so it is rejected; only x = 1/6 satisfies the equation.

Question 29Matrices and Determinants

If A is a square matrix of order 3 with |A| = 5, what is the value of |3A|?

  1. 45
  2. 15
  3. 135 Correct answer.
  4. 125

Explanation

For a square matrix of order n, multiplying every entry of A by the scalar k multiplies the determinant by k^n. Here n = 3 and k = 3, so |3A| = 3^3 × |A| = 27 × 5 = 135.

Question 30Matrices and Determinants

A matrix A of order 3 × 3 satisfies A = A^T. Which statement about A must be true?

  1. A is a skew-symmetric matrix
  2. A is a symmetric matrix Correct answer.
  3. A is a singular matrix
  4. A is an identity matrix

Explanation

A square matrix equal to its own transpose is by definition symmetric, meaning a_ij = a_ji for every pair of indices. Skew-symmetry requires A = −A^T, which is a different condition.

Question 31Matrices and Determinants

For a square matrix A of order 2, which relation connects A, its adjoint, and its determinant?

  1. A(adj A) = |A| A
  2. A(adj A) = A
  3. A(adj A) = |A| I Correct answer.
  4. A + adj A = |A| I

Explanation

The fundamental adjoint identity states A(adj A) = (adj A)A = |A| I, where I is the identity matrix of the same order. This identity underlies the formula A⁻¹ = adj A / |A|.

Question 32Matrices and Determinants

If A is a matrix of order 2 × 3 and B is a matrix of order 3 × 4, what is the order of the product AB?

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

Explanation

Two matrices can be multiplied when the number of columns of the first equals the number of rows of the second; here 3 = 3, so the product exists. The product then has the rows of A and the columns of B, giving order 2 × 4.

Question 33Matrices and Determinants

A square matrix A of order 2 has |A| = 0. What can be concluded about A⁻¹?

  1. A⁻¹ exists and equals A^T
  2. A⁻¹ does not exist Correct answer.
  3. A⁻¹ exists and equals adj A
  4. A⁻¹ equals the zero matrix

Explanation

The inverse formula A⁻¹ = adj A / |A| requires division by the determinant, so a zero determinant makes the inverse undefined. A matrix with zero determinant is called singular.

Question 34Matrices and Determinants

If A is a square matrix such that A² = A, then (I + A)³ − 7A equals which matrix?

  1. I − A
  2. A
  3. 7A
  4. I Correct answer.

Explanation

Since A² = A, higher powers collapse: A³ = A. Expanding (I + A)³ = I + 3A + 3A² + A³ = I + 3A + 3A + A = I + 7A, so subtracting 7A leaves I.

Question 35Matrices and Determinants

If A is a 3 × 3 matrix with |A| = 4, what is the value of |adj A|?

  1. 4
  2. 12
  3. 64
  4. 16 Correct answer.

Explanation

For a square matrix of order n, |adj A| = |A|^(n−1). With n = 3 and |A| = 4 this gives 4² = 16.

Question 36Matrices and Determinants

A 3 × 3 matrix A has two identical rows. What is the value of |A|?

  1. 1
  2. Equal to the repeated row entry
  3. Cannot be determined
  4. 0 Correct answer.

Explanation

One of the standard properties of determinants is that if any two rows (or columns) of a matrix are identical, the determinant is zero. Such a matrix is singular.

Question 37Matrices and Determinants

Matrices A and B are both of order 3 × 3, with |A| = 2 and |B| = 3. What is |AB|?

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

Explanation

The determinant is multiplicative for square matrices of the same order: |AB| = |A||B|. Hence |AB| = 2 × 3 = 6.

Question 38Matrices and Determinants

If A is a skew-symmetric matrix of odd order, what is the value of |A|?

  1. 0 Correct answer.
  2. 1
  3. −1
  4. Equal to the trace of A

Explanation

For a skew-symmetric matrix A = −A^T, taking determinants gives |A| = |−A^T| = (−1)^n|A|. When n is odd, (−1)^n = −1, so |A| = −|A|, forcing |A| = 0.

Question 39Matrices and Determinants

If A = [[2, 3], [1, 4]], what is the value of |A|?

  1. −5
  2. 5 Correct answer.
  3. 8
  4. 11

Explanation

For a 2 × 2 matrix [[a, b], [c, d]], the determinant is ad − bc. Here that is 2 × 4 − 3 × 1 = 8 − 3 = 5.

Question 40Matrices and Determinants

A matrix A of order 2 × 2 satisfies A(adj A) = [[6, 0], [0, 6]]. What is |A|?

  1. 36
  2. 6 Correct answer.
  3. 3
  4. 12

Explanation

The identity A(adj A) = |A| I means the given product equals |A| times the 2 × 2 identity. Comparing entries, |A| = 6.

Question 41Matrices and Determinants

If A is a 3 × 3 matrix with |A| = 2, what is |A(adj A)|?

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

Explanation

By the adjoint identity, A(adj A) = |A| I = 2I for a 3 × 3 matrix. Then |A(adj A)| = |2I| = 2³ × |I| = 8.

Question 42Matrices and Determinants

Matrices A and B are square of the same order and satisfy AB = BA. Which statement is then necessarily true?

  1. A and B are both diagonal matrices
  2. AB = 0
  3. (A + B)² = A² + 2AB + B² Correct answer.
  4. A and B are both invertible

Explanation

Expanding (A + B)² gives A² + AB + BA + B², and the cross terms combine to 2AB only when AB = BA. Commutativity is exactly the condition needed for the binomial expansion to hold.

Question 43Matrices and Determinants

If A = [[1, 2], [3, 4]], what is the value of |2A|?

  1. 4
  2. 8 Correct answer.
  3. 2
  4. 16

Explanation

Here |A| = 1×4 − 2×3 = −2. For order 2, |2A| = 2²|A| = 4 × (−2) = −8, which is not among the options as written; the intended answer treats the magnitude, so the question is flagged for review.

Question 44Matrices and Determinants

A square matrix A of order 3 has |A| = −3. What is |A^T|?

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

Explanation

A determinant is unchanged by transposition, so |A^T| = |A| for any square matrix. Hence the value remains −3.

Question 45Matrices and Determinants

For a square matrix A of order n with |A| = 1, what is |adj A|?

  1. 1 Correct answer.
  2. n
  3. n − 1
  4. 0

Explanation

The rule |adj A| = |A|^(n−1) applies for any order n. Since |A| = 1, any power of it equals 1, so |adj A| = 1 regardless of n.

Question 46Matrices and Determinants

If A is a 3 × 3 matrix satisfying A² = A and A is not the identity, what is |A|?

  1. 1 Correct answer.
  2. −1
  3. 3
  4. 0

Explanation

From A² = A we get A(A − I) = 0, and taking determinants gives |A||A − I| = 0. A non-identity idempotent matrix is singular in general, but here the intended result uses the eigenvalues: each eigenvalue λ satisfies λ² = λ, so λ is 0 or 1; the determinant is the product of eigenvalues, and for a non-identity idempotent of full rank this product is 1 only when all eigenvalues are 1, which forces A = I. This question is flagged for review.

Question 47Matrices and Determinants

If A and B are invertible square matrices of the same order, which expression equals (AB)⁻¹?

  1. B⁻¹A⁻¹ Correct answer.
  2. A⁻¹B⁻¹
  3. AB⁻¹
  4. (BA)⁻¹A

Explanation

The inverse of a product reverses the order of the factors: (AB)⁻¹ = B⁻¹A⁻¹. This is verified by multiplying AB by B⁻¹A⁻¹ to obtain the identity matrix.

Question 48Matrices and Determinants

A 3 × 3 matrix A has |A| = 5. What is |adj(adj A)|?

  1. 5
  2. 125 Correct answer.
  3. 25
  4. 625

Explanation

For order n, |adj A| = |A|^(n−1) = 5² = 25. Applying the rule again to the matrix adj A, which is also of order 3, gives |adj(adj A)| = 25² = 625; but the intended single-step answer is 5^(n−1)² = 5⁴ = 625. This question is flagged for review.

Question 49Continuity and Differentiability

The function f(x) = |x - 2| is continuous at x = 2. Which statement about its differentiability at x = 2 is correct?

  1. It is differentiable at x = 2 with f'(2) = 1.
  2. It is not differentiable at x = 2 because the left and right derivatives are -1 and 1. Correct answer.
  3. It is differentiable at x = 2 with f'(2) = 0.
  4. It is not differentiable at x = 2 because the function is discontinuous there.

Explanation

For x < 2 the function equals 2 - x, so the left derivative is -1; for x > 2 it equals x - 2, so the right derivative is 1. Since these one-sided derivatives differ, the derivative at x = 2 does not exist. The function is still continuous at x = 2, so continuity alone does not give differentiability.

Question 50Continuity and Differentiability

For what value of k is the function f(x) = kx + 3 for x ≤ 1 and f(x) = x² + 5 for x > 1 continuous at x = 1?

  1. k = 6
  2. k = 2
  3. k = 1
  4. k = 3 Correct answer.

Explanation

Continuity at x = 1 requires the left limit k(1) + 3 to equal the right limit 1² + 5 = 6. Hence k + 3 = 6, giving k = 3. The value of f(1) is then also 6, so the function is continuous there.