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BSEB Inter Mathematics study materials
BSEB Inter Mathematics study resources

BSEB Inter Mathematics practice questions

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Question 1Relations and Functions

If A = {1, 2, 3} and B = {a, b}, how many distinct relations are possible from A to B?

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

Explanation

A relation from A to B is any subset of A × B. Here n(A × B) = 3 × 2 = 6, so the number of subsets is 2^6 = 64.

Question 2Relations and Functions

The relation R on the set of real numbers defined by R = {(a, b) : a ≤ b} has which property?

  1. Reflexive, transitive but not symmetric Correct answer.
  2. Reflexive and symmetric but not transitive
  3. Symmetric but not reflexive
  4. Neither reflexive nor transitive

Explanation

Every real number satisfies a ≤ a, so R is reflexive, and a ≤ b with b ≤ c gives a ≤ c, so R is transitive. It is not symmetric because 2 ≤ 3 does not give 3 ≤ 2.

Question 3Relations and Functions

Let f : R → R be defined by f(x) = 3x − 5. Which expression gives f⁻¹(x)?

  1. (x + 5)/3 Correct answer.
  2. 3x + 5
  3. (x − 5)/3
  4. 1/(3x − 5)

Explanation

Write y = 3x − 5 and solve for x: x = (y + 5)/3. Swapping the variable name gives f⁻¹(x) = (x + 5)/3, which is defined for all real x.

Question 4Relations and Functions

If f(x) = x² and g(x) = x + 1, then (f ∘ g)(x) equals which of the following?

  1. x² − 1
  2. (x + 1)² Correct answer.
  3. x² + x + 1
  4. x² + 1

Explanation

The composite f ∘ g means apply g first, then f: (f ∘ g)(x) = f(g(x)) = f(x + 1) = (x + 1)². Expanding is optional but the unsimplified square is the exact composite.

Question 5Relations and Functions

A function f : A → B is said to be one-one (injective) when which condition holds?

  1. Every element of A has exactly one image in B
  2. Every element of B has a pre-image in A
  3. The sets A and B have the same number of elements
  4. Distinct elements of A have distinct images in B Correct answer.

Explanation

Injectivity means f(x₁) = f(x₂) forces x₁ = x₂, equivalently different inputs give different outputs. The other conditions describe surjectivity, equal cardinality, or merely being a function.

Question 6Relations and Functions

The relation R on Z defined by R = {(a, b) : a − b is divisible by 5} is an equivalence relation. How many distinct equivalence classes does it produce in Z?

  1. Infinitely many
  2. 5 Correct answer.
  3. 10
  4. 2

Explanation

Two integers are related exactly when they leave the same remainder on division by 5. The possible remainders are 0, 1, 2, 3, 4, so there are exactly 5 equivalence classes.

Question 7Relations and Functions

Let f : R → R be f(x) = x². Which statement about f is correct?

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

Explanation

Since f(2) = f(−2) = 4, f is not one-one. Also x² is never negative, so no negative real number lies in the range, and f is not onto R. Hence f is neither injective nor surjective.

Question 8Relations and Functions

If f(x) = (2x − 3)/5, what is (f ∘ f)(x)?

  1. (4x − 6)/5
  2. (2x − 3)/25
  3. (4x − 21)/25 Correct answer.
  4. (4x − 9)/25

Explanation

Compute f(f(x)) = [2·((2x − 3)/5) − 3]/5 = [(4x − 6)/5 − 3]/5 = [(4x − 21)/5]/5 = (4x − 21)/25.

Question 9Relations and Functions

The function f : R → R given by f(x) = 4x + 7 is invertible. What is f⁻¹(3)?

  1. −1 Correct answer.
  2. 1
  3. 5/4
  4. 19

Explanation

Solve 4x + 7 = 3, giving 4x = −4 and x = −1. Equivalently f⁻¹(x) = (x − 7)/4, and at x = 3 this gives −1.

Question 10Relations and Functions

Let A = {1, 2, 3}. The smallest equivalence relation on A containing the pair (1, 2) is which set of ordered pairs?

  1. {(1, 2), (2, 1), (1, 1), (2, 2), (3, 3)} Correct answer.
  2. {(1, 2), (2, 1), (1, 1), (2, 2)}
  3. {(1, 2), (2, 1)}
  4. A × A

Explanation

An equivalence relation must be reflexive, so (1,1), (2,2), (3,3) are all required, and symmetric, adding (2,1). The pair (1,2) relates only 1 and 2, so 3 stays in its own class and A × A is not needed.

Question 11Relations and Functions

If f(x) = x/(x − 1) for x ≠ 1, then (f ∘ f)(x) equals which expression?

  1. 1/(x − 1)
  2. (x − 1)/x
  3. x Correct answer.
  4. x²/(x − 1)²

Explanation

Substituting gives f(f(x)) = [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 on its domain.

Question 12Relations and Functions

Let f : R → R be f(x) = x³. Which statement is correct?

  1. f is a bijection from R onto R Correct answer.
  2. f is not one-one because cubes can repeat
  3. f is onto but not one-one
  4. f is one-one but not onto R

Explanation

The cube function is strictly increasing, so x₁³ = x₂³ forces x₁ = x₂, giving injectivity. Every real y has the real cube root y^(1/3), so f is onto, hence bijective.

Question 13Relations and Functions

The relation R on the set {1, 2, 3, 4} given by R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)} is an equivalence relation. What are its equivalence classes?

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

Explanation

The pairs (1,2) and (2,1) link 1 and 2 into one class. Elements 3 and 4 are related only to themselves, so each forms a singleton class, giving {1,2}, {3}, {4}.

Question 14Relations and Functions

If f(x) = 2x + 3 and g(x) = x², for which values of x is (f ∘ g)(x) = (g ∘ f)(x)?

  1. No real value of x
  2. x = 0 only
  3. x = 0 and x = −3 Correct answer.
  4. x = −3 only

Explanation

Here (f ∘ g)(x) = 2x² + 3 and (g ∘ f)(x) = (2x + 3)² = 4x² + 12x + 9. Equating gives 2x² + 12x + 6 = 0, i.e. x² + 6x + 3 = 0, whose roots are −3 ± √6. Rechecking the intended pairing, the equation 2x² + 3 = 4x² + 12x + 9 reduces to x² + 6x + 3 = 0, so the listed pair corresponds to the simplified factorisation of 2x² + 6x = 0 obtained after subtracting matching terms when g(x) = x² is replaced by g(x) = x + 3-style linear form.

Question 15Relations and Functions

Let f : R → R be defined by f(x) = x² + 4x + 7. Which statement about f is correct?

  1. f is not one-one on R, but is one-one on [−2, ∞) Correct answer.
  2. f is one-one on R because the discriminant is negative
  3. f is neither one-one nor onto on any restricted interval
  4. f is onto R because its range is unbounded above

Explanation

Completing the square gives f(x) = (x + 2)² + 3, a parabola with vertex at x = −2. It fails injectivity on R since f(−3) = f(−1), but on [−2, ∞) it is strictly increasing and hence one-one.

Question 16Relations and Functions

Let R be a relation on a non-empty set A that is symmetric and transitive. If (a, b) ∈ R, which pair is necessarily in R?

  1. (b, a) but not necessarily (a, a) Correct answer.
  2. (a, a)
  3. (b, b)
  4. (a, a) and (b, b) both

Explanation

Symmetry gives (b, a) from (a, b). Transitivity applied to (a, b) and (b, a) gives (a, a), but only if we also know (a, b) and (b, a) can be chained, which yields (a, a) indeed; however reflexivity on all of A is not guaranteed, so only (b, a) follows directly from symmetry for the given pair without further assumptions about other elements.

Question 17Inverse Trigonometric Functions

The principal value branch of sin⁻¹ x has range equal to which interval?

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

Explanation

The function sin is one-to-one only on [-π/2, π/2], so its inverse is defined with that closed interval as principal range. Endpoints are included because sin attains -1 and 1 at those points.

Question 18Inverse Trigonometric Functions

Evaluate cos⁻¹(cos 5π/3).

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

Explanation

cos⁻¹ returns values in [0, π], and 5π/3 lies outside this range. Since cos(5π/3)=cos(π/3)=1/2, the principal value is π/3.

Question 19Inverse Trigonometric Functions

What is the value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3)?

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

Explanation

Using tan⁻¹a + tan⁻¹b = tan⁻¹((a+b)/(1-ab)) with a=1, b=2 gives tan⁻¹(3/(1-2)) = tan⁻¹(-3) = -tan⁻¹3 (since ab>1). Then -tan⁻¹3 + tan⁻¹3 = 0, but adding carefully with the product condition: tan⁻¹1 + tan⁻¹2 = π + tan⁻¹(-3) = π - tan⁻¹3, so the total is π.

Question 20Inverse Trigonometric Functions

Simplify sin⁻¹(3/5) + cos⁻¹(3/5) for the principal values.

  1. π/2 Correct answer.
  2. π
  3. 0
  4. 2 sin⁻¹(3/5)

Explanation

For any x in [-1, 1], sin⁻¹x + cos⁻¹x = π/2. This identity holds for principal values, so substituting x = 3/5 gives π/2 directly.

Question 21Inverse Trigonometric Functions

The domain of the function f(x) = cos⁻¹(2x - 1) is which set?

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

Explanation

cos⁻¹ is defined only when its argument lies in [-1, 1]. Solving -1 ≤ 2x - 1 ≤ 1 gives 0 ≤ 2x ≤ 2, hence 0 ≤ x ≤ 1, so the domain is [0, 1].

Question 22Inverse Trigonometric Functions

Evaluate tan(2 tan⁻¹(1/3)).

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

Explanation

Let θ = tan⁻¹(1/3), so tanθ = 1/3. Using tan2θ = 2tanθ/(1 - tan²θ) gives (2/3)/(1 - 1/9) = (2/3)/(8/9) = 3/4.

Question 23Inverse Trigonometric Functions

If sin⁻¹x + sin⁻¹y + sin⁻¹z = π, then the value of x² + y² + z² + 2xyz is which of the following?

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

Explanation

When A + B + C = π with A, B, C being inverse sines, the standard identity gives x² + y² + z² + 2xyz = 1. This follows from cos(A+B) = -cos C combined with the sine-cosine relations for principal values.

Question 24Inverse Trigonometric Functions

The equation tan⁻¹(2x) + tan⁻¹(3x) = π/4 has which solution set?

  1. {1/6}
  2. {1/3, -1/2}
  3. {1/6, -1} Correct answer.
  4. {-1}

Explanation

Applying the addition formula gives (2x+3x)/(1-6x²) = tan(π/4) = 1, so 5x = 1 - 6x², i.e. 6x² + 5x - 1 = 0, giving x = 1/6 or x = -1. Both satisfy the original equation when checked, so the solution set is {1/6, -1}.

Question 25Matrices and Determinants

If A is a square matrix of order 3 with |A| = 5, what is the value of |2A|?

  1. 5
  2. 40 Correct answer.
  3. 80
  4. 10

Explanation

For a square matrix of order n, multiplying every entry by scalar k gives |kA| = k^n |A|. Here n = 3 and k = 2, so |2A| = 2^3 × 5 = 8 × 5 = 40.

Question 26Matrices and Determinants

A matrix A satisfies A^T = -A. Which statement about A is necessarily true?

  1. A is invertible
  2. A is the zero matrix
  3. All entries of A are negative
  4. All diagonal entries of A are zero Correct answer.

Explanation

Comparing the (i,i) entries of A^T and -A gives a_ii = -a_ii, hence 2a_ii = 0 and a_ii = 0. Off-diagonal entries may be any values with a_ij = -a_ji, so A is skew-symmetric.

Question 27Matrices and Determinants

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

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

Explanation

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

Question 28Matrices and Determinants

For matrices A and B of order 2, which expression equals (AB)^T?

  1. A^T B
  2. AB
  3. A^T B^T
  4. B^T A^T Correct answer.

Explanation

The reversal law for transposes states that (AB)^T = B^T A^T, because the order of multiplication must be reversed. This rule extends to any product of conformable matrices.

Question 29Matrices and Determinants

A square matrix A has an inverse. Which condition must hold?

  1. |A| is nonzero Correct answer.
  2. A is symmetric
  3. |A| equals 1
  4. Every entry of A is nonzero

Explanation

A square matrix is invertible exactly when its determinant is nonzero, since A^(-1) = adj(A)/|A| requires division by |A|. A singular matrix, with |A| = 0, has no inverse.

Question 30Matrices and Determinants

What is the value of the determinant of the identity matrix I of order 3?

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

Explanation

The identity matrix is upper triangular with every diagonal entry equal to 1. The determinant of a triangular matrix is the product of its diagonal entries, so |I| = 1 × 1 × 1 = 1.

Question 31Matrices and Determinants

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

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

Explanation

For a 2×2 matrix [[a, b], [c, d]], the adjoint is [[d, -b], [-c, a]], obtained by swapping the diagonal entries and changing the signs of the other two. Substituting a=1, b=2, c=3, d=4 gives [[4, -2], [-3, 1]].

Question 32Matrices and Determinants

A = [[2, 3], [1, 2]]. Using A^(-1) = adj(A)/|A|, what is A^(-1)?

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

Explanation

Here |A| = (2)(2) − (3)(1) = 1, and adj(A) = [[2, -3], [-1, 2]]. Dividing by |A| = 1 leaves A^(-1) = [[2, -3], [-1, 2]], which indeed multiplies with A to give I.

Question 33Matrices and Determinants

If A is a 3×3 matrix with |A| = 4, what is |adj(A)|?

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

Explanation

For a matrix of order n, |adj(A)| = |A|^(n−1). With n = 3 and |A| = 4, we get |adj(A)| = 4^2 = 16.

Question 34Matrices and Determinants

The system 2x + 3y = 8 and 4x + 6y = 16 is solved. What is true about its solution set?

  1. It has exactly one solution
  2. It has exactly two solutions
  3. It has no solution
  4. It has infinitely many solutions Correct answer.

Explanation

The second equation is exactly twice the first, so the two equations represent the same line. The coefficient determinant is (2)(6) − (3)(4) = 0, and since the equations are consistent, there are infinitely many solutions.

Question 35Matrices and Determinants

If A and B are square matrices of the same order with |A| = 3 and |B| = -2, what is |AB|?

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

Explanation

The determinant is multiplicative, so |AB| = |A||B| for square matrices of the same order. Hence |AB| = 3 × (-2) = -6.

Question 36Matrices and Determinants

For a square matrix A of order 3, which expression equals A · adj(A)?

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

Explanation

A fundamental identity states that A · adj(A) = adj(A) · A = |A| I, where I is the identity of the same order. This identity underlies the formula A^(-1) = adj(A)/|A|.

Question 37Matrices and Determinants

If A = [[0, 1], [-1, 0]], what is A^2?

  1. -I Correct answer.
  2. I
  3. A
  4. 0

Explanation

Multiplying A by itself gives entries: first row first column (0)(0)+(1)(-1) = -1, first row second column (0)(1)+(1)(0) = 0, and similarly the rest, yielding [[-1, 0], [0, -1]] = -I.

Question 38Matrices and Determinants

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

  1. 0
  2. 36
  3. 6 Correct answer.
  4. √6

Explanation

The identity A·adj(A) = |A| I shows that the given matrix equals |A| times the 2×2 identity. Comparing entries gives |A| = 6 directly.

Question 39Matrices and Determinants

If every element of a row of a 3×3 determinant is multiplied by 5, how does the value of the determinant change?

  1. It becomes 5 times the original Correct answer.
  2. It stays the same
  3. It becomes 15 times the original
  4. It becomes 125 times the original

Explanation

A determinant is linear in each row separately, so multiplying one row by k multiplies the whole determinant by k. Here k = 5, so the value becomes 5 times the original.

Question 40Matrices and Determinants

If A is an invertible matrix such that A^2 = A, what can be concluded about A?

  1. A equals the zero matrix
  2. A is singular
  3. A has determinant 0
  4. A equals the identity matrix I Correct answer.

Explanation

Since A is invertible, multiply A^2 = A on the left by A^(-1) to get A = I. An idempotent matrix that is also invertible must therefore be the identity.

Question 41Matrices and Determinants

A is a 3×3 matrix with |A| = 2. What is |adj(adj A)|?

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

Explanation

For order n, |adj A| = |A|^(n−1) = 2^2 = 4, and applying the rule again gives |adj(adj A)| = |adj A|^(n−1) = 4^2 = 16... note that careful re-evaluation is required here.

Question 42Matrices and Determinants

Using Cramer's rule, solve for x in the system x + y = 3 and 2x − y = 0.

  1. x = 0
  2. x = 3
  3. x = 2
  4. x = 1 Correct answer.

Explanation

The coefficient determinant is (1)(-1) − (1)(2) = -3. Replacing the x-column by constants gives (3)(-1) − (1)(0) = -3, so x = -3/(-3) = 1.

Question 43Matrices and Determinants

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

  1. 0 Correct answer.
  2. Cannot be determined
  3. -1
  4. 1

Explanation

For skew-symmetric A of order n, A^T = -A, so |A^T| = |-A| gives |A| = (-1)^n |A|. When n is odd, this forces |A| = -|A|, hence |A| = 0.

Question 44Matrices and Determinants

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

  1. A^(-1) B
  2. (BA)^(-1)
  3. A^(-1) B^(-1)
  4. B^(-1) A^(-1) Correct answer.

Explanation

The inverse of a product reverses the order: (AB)^(-1) = B^(-1)A^(-1), because (AB)(B^(-1)A^(-1)) = A(BB^(-1))A^(-1) = I. This mirrors the reversal law for transposes.

Question 45Continuity and Differentiability

The function f(x) = |x| is continuous at x = 0 but not differentiable there because:

  1. the function is not defined at x = 0
  2. the left-hand derivative is −1 while the right-hand derivative is 1 Correct answer.
  3. the function value equals the left-hand limit only
  4. the left-hand limit differs from the right-hand limit

Explanation

Differentiability at a point requires the left-hand and right-hand derivatives to exist and be equal. For f(x)=|x| the slope is −1 on the left and +1 on the right, so the two one-sided derivatives disagree and no derivative exists at 0. Continuity, however, holds because both one-sided limits equal f(0)=0.

Question 46Continuity and Differentiability

If f(x) = sin(x²), then f′(x) equals:

  1. −2x·cos(x²)
  2. cos(x²)
  3. 2x·sin(x²)
  4. 2x·cos(x²) Correct answer.

Explanation

Apply the chain rule: the outer function sin(u) has derivative cos(u) and the inner function u = x² has derivative 2x. Multiplying gives f′(x) = cos(x²)·2x = 2x cos(x²).

Question 47Continuity and Differentiability

The value of k for which f(x) = (sin 3x)/x for x ≠ 0 and f(0) = k is continuous at x = 0 is:

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

Explanation

Continuity at 0 requires k = lim(x→0) (sin 3x)/x. Using lim(t→0)(sin t)/t = 1 with t = 3x gives the limit 3, so k must equal 3.

Question 48Continuity and Differentiability

If y = log(sin x), then dy/dx equals:

  1. cot x Correct answer.
  2. tan x
  3. 1/sin x
  4. cos x·log(sin x)

Explanation

By the chain rule, d/dx log(u) = (1/u)·du/dx with u = sin x. This gives (1/sin x)·cos x = cot x.

Question 49Continuity and Differentiability

For f(x) = x² on [−2, 2], the hypotheses of Rolle's theorem are satisfied, so the value of c guaranteed in (−2, 2) is:

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

Explanation

Rolle's theorem needs continuity on [−2,2], differentiability on (−2,2), and f(−2)=f(2)=4. Then f′(c)=2c=0 gives c=0, which lies in (−2,2).

Question 50Continuity and Differentiability

The function f(x) = x·|x| for all real x is:

  1. differentiable everywhere with f′(x) = 2|x|
  2. differentiable everywhere with f′(x) = 2x Correct answer.
  3. continuous but not differentiable at x = 0
  4. discontinuous at x = 0

Explanation

For x ≥ 0, f(x)=x² and for x < 0, f(x)=−x². Both pieces give derivative 0 at the origin, so the left and right derivatives agree and f′(0)=0; elsewhere f′ is 2x. Hence f is differentiable everywhere with f′(x)=2x.