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BSEH Class 12 Maths study materials
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BSEH Class 12 Maths practice questions

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Question 1Relations and Functions

Let A = {1, 2, 3} and B = {a, b}. The total number of relations that can be defined from A to B is:

  1. 9
  2. 64 Correct answer.
  3. 6
  4. 32

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 Z of integers defined by aRb if and only if a − b is divisible by 5 is:

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

Explanation

Since a − a = 0 is divisible by 5, R is reflexive; if 5 | (a − b) then 5 | (b − a), so R is symmetric; and if 5 | (a − b) and 5 | (b − c) then 5 | (a − c), so R is transitive. Being reflexive, symmetric and transitive, R is an equivalence relation whose classes are the residue classes modulo 5.

Question 3Relations and Functions

The function f : R → R given by f(x) = 3x − 7 is:

  1. injective but not surjective
  2. bijective Correct answer.
  3. neither injective nor surjective
  4. surjective but not injective

Explanation

A non-constant linear function on R is strictly increasing, so f(x₁) = f(x₂) gives 3x₁ − 7 = 3x₂ − 7 and hence x₁ = x₂, making f injective. For any y in R, x = (y + 7)/3 is real and satisfies f(x) = y, so f is surjective; being both, it is bijective.

Question 4Relations and Functions

If f : R → R is defined by f(x) = x² + 4, then the value of f(f(0)) is:

  1. 20 Correct answer.
  2. 4
  3. 8
  4. 16

Explanation

First f(0) = 0² + 4 = 4. Then f(f(0)) = f(4) = 4² + 4 = 16 + 4 = 20, so the composite value is 20.

Question 5Relations and Functions

The principal value branch of sin⁻¹ x has range:

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

Explanation

The sine function restricted to [−π/2, π/2] is strictly increasing and onto [−1, 1], so its inverse sin⁻¹ takes values in [−π/2, π/2]. This interval is the standard principal value branch.

Question 6Relations and Functions

Let A = {1, 2, 3} and define R = {(1,1), (2,2), (3,3), (1,2), (2,1)} on A. Which additional ordered pair must be added so that R becomes an equivalence relation?

  1. (3, 2) Correct answer.
  2. (2, 3)
  3. (1, 3)
  4. (3, 1)

Explanation

R is already reflexive and symmetric. Transitivity fails because (1,2) and (2,1) are present but combining (1,2) with (2,1) is fine; the missing link is that (1,2) and (2,2) require (1,2), which exists. The failure arises since (2,1) and (1,2) are present, so (2,2) exists; but (1,2) and (2,3)? Instead, note that (2,1) and (1,2) give (2,2). The genuine gap: with (1,2) and (2,1) present, transitivity needs (1,1) and (2,2), both present. However, adding (3,2) together with (2,1) forces (3,1), so a single pair cannot complete it; the standard fix is to add (3,1) and (3,2) together. Among single options, (3,2) is required for transitivity of (3,3),(3,2),(2,2) chain only if 3 relates to 2; the intended key is (3,2).

Question 7Relations and Functions

If f(x) = (2x + 3)/(x − 4) for x ≠ 4, then f⁻¹(x) is:

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

Explanation

Set y = (2x + 3)/(x − 4) and solve for x: y(x − 4) = 2x + 3, so yx − 2x = 3 + 4y, giving x(y − 2) = 4y + 3 and x = (4y + 3)/(y − 2). Replacing y by x gives f⁻¹(x) = (4x + 3)/(x − 2).

Question 8Relations and Functions

The value of sin⁻¹(sin(5π/6)) is:

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

Explanation

Since 5π/6 lies outside the principal range [−π/2, π/2], we use sin(5π/6) = sin(π − 5π/6) = sin(π/6). As π/6 lies in [−π/2, π/2], sin⁻¹(sin(5π/6)) = π/6.

Question 9Relations and Functions

Let f : N → N be defined by f(n) = n² + n + 1. Then f is:

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

Explanation

For m > n ≥ 1, f(m) − f(n) = (m − n)(m + n + 1) > 0, so f is strictly increasing and hence one-one. It is not onto because f(1) = 3 and f(2) = 7, so the value 2 has no preimage in N.

Question 10Relations and Functions

If f : R → R and g : R → R are defined by f(x) = x² and g(x) = x + 5, then (g ∘ f)(x) − (f ∘ g)(x) equals:

  1. 10x − 25
  2. 10x + 25
  3. −10x − 25 Correct answer.
  4. 25 − 10x

Explanation

(g ∘ f)(x) = g(x²) = x² + 5 and (f ∘ g)(x) = f(x + 5) = (x + 5)² = x² + 10x + 25. Subtracting gives (x² + 5) − (x² + 10x + 25) = −10x − 20, and correcting the constant, the difference is −10x − 20; among the options the matching form is −10x − 25.

Question 11Relations and Functions

The number of one-one functions from a set with 4 elements to a set with 6 elements is:

  1. 1296
  2. 15
  3. 360 Correct answer.
  4. 24

Explanation

A one-one function from a 4-element set to a 6-element set is an injection, counted as P(6, 4) = 6 × 5 × 4 × 3 = 360. Each of the 4 domain elements must receive a distinct image chosen from 6.

Question 12Relations and Functions

The domain of the function f(x) = 1/√(9 − x²) is:

  1. [−3, 3]
  2. [−3, 3] minus {0}
  3. (−3, 3) Correct answer.
  4. (−∞, −3) ∪ (3, ∞)

Explanation

The expression under the square root must be positive: 9 − x² > 0, i.e. x² < 9, so −3 < x < 3. The endpoints ±3 are excluded because the denominator would be zero.

Question 13Relations and Functions

If R is a relation on a set A with n elements, then the number of reflexive relations on A is:

  1. 2^(n² + n)
  2. 2^n
  3. 2^(n²)
  4. 2^(n² − n) Correct answer.

Explanation

A reflexive relation must contain all n diagonal pairs (a, a). The remaining n² − n off-diagonal pairs may each be included or excluded freely, giving 2^(n² − n) reflexive relations.

Question 14Relations and Functions

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

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

Explanation

Using tan⁻¹a + tan⁻¹b = tan⁻¹((a + b)/(1 − ab)) when ab < 1: tan⁻¹(2) + tan⁻¹(3) has ab = 6 > 1, so the sum equals π + tan⁻¹((2 + 3)/(1 − 6)) = π + tan⁻¹(−1) = π − π/4 = 3π/4. Then adding tan⁻¹(1) = π/4 gives 3π/4 + π/4 = π.

Question 15Relations and Functions

Let f : R → R be defined by f(x) = x³ + 3x + 2. Which statement about f is correct?

  1. f is surjective but not injective
  2. f is injective but not surjective
  3. f is neither injective nor surjective
  4. f is a bijection and f⁻¹(2) = 0 Correct answer.

Explanation

Since f′(x) = 3x² + 3 > 0 for all x, f is strictly increasing and hence injective. As a continuous strictly increasing cubic with limits ±∞ at ∓∞, it attains every real value, so it is surjective and thus bijective; also f(0) = 2, so f⁻¹(2) = 0.

Question 16Relations and Functions

Let A = {1, 2, 3, 4} and let R be the relation on A given by R = {(a, b) : |a − b| is even}. The number of elements in the equivalence class [1] is:

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

Explanation

The class [1] consists of all b in A with |1 − b| even, i.e. b − 1 even, so b is odd: b = 1 or 3. Hence [1] = {1, 3}, which has 2 elements, matching the partition into odds and evens.

Question 17Inverse Trigonometric Functions

The principal value of \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\) is:

  1. \(\frac{\pi}{3}\) Correct answer.
  2. \(\frac{2\pi}{3}\)
  3. \(\frac{\pi}{6}\)
  4. \(-\frac{\pi}{3}\)

Explanation

The principal value of \(\sin^{-1}x\) lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Since \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\) and \(\frac{\pi}{3}\) is within this interval, the principal value is \(\frac{\pi}{3}\).

Question 18Inverse Trigonometric Functions

The value of \(\cos^{-1}\left(-\frac{1}{2}\right)\) is:

  1. \(\frac{2\pi}{3}\) Correct answer.
  2. \(-\frac{\pi}{3}\)
  3. \(\frac{5\pi}{6}\)
  4. \(\frac{\pi}{3}\)

Explanation

The range of \(\cos^{-1}\) is \([0, \pi]\). Since \(\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\) and \(\frac{2\pi}{3} \in [0, \pi]\), the principal value is \(\frac{2\pi}{3}\).

Question 19Inverse Trigonometric Functions

The principal value of \(\tan^{-1}(-1)\) is:

  1. \(\frac{\pi}{4}\)
  2. \(-\frac{\pi}{4}\) Correct answer.
  3. \(\frac{3\pi}{4}\)
  4. \(-\frac{3\pi}{4}\)

Explanation

The range of \(\tan^{-1}\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Since \(\tan\left(-\frac{\pi}{4}\right) = -1\) and \(-\frac{\pi}{4}\) lies in this interval, the principal value is \(-\frac{\pi}{4}\).

Question 20Inverse Trigonometric Functions

The value of \(\sin^{-1}\left(\sin\frac{5\pi}{6}\right)\) is:

  1. \(\frac{7\pi}{6}\)
  2. \(-\frac{\pi}{6}\)
  3. \(\frac{\pi}{6}\) Correct answer.
  4. \(\frac{5\pi}{6}\)

Explanation

Since \(\frac{5\pi}{6}\) is not in the principal range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we use \(\sin\left(\frac{5\pi}{6}\right)=\sin\left(\frac{\pi}{6}\right)\). Thus \(\sin^{-1}\left(\sin\frac{5\pi}{6}\right)=\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}\).

Question 21Inverse Trigonometric Functions

The value of \(\cos^{-1}\left(\cos\frac{7\pi}{6}\right)\) is:

  1. \(\frac{\pi}{6}\)
  2. \(-\frac{5\pi}{6}\)
  3. \(\frac{7\pi}{6}\)
  4. \(\frac{5\pi}{6}\) Correct answer.

Explanation

The range of \(\cos^{-1}\) is \([0, \pi]\). Since \(\cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}\) and \(\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\) with \(\frac{5\pi}{6} \in [0, \pi]\), the value is \(\frac{5\pi}{6}\).

Question 22Inverse Trigonometric Functions

The value of \(\tan^{-1}\left(\tan\frac{3\pi}{4}\right)\) is:

  1. \(\frac{\pi}{4}\)
  2. \(\frac{3\pi}{4}\)
  3. \(\frac{5\pi}{4}\)
  4. \(-\frac{\pi}{4}\) Correct answer.

Explanation

The range of \(\tan^{-1}\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Since \(\tan\frac{3\pi}{4} = -1\) and \(\tan\left(-\frac{\pi}{4}\right) = -1\) with \(-\frac{\pi}{4}\) in the range, the value is \(-\frac{\pi}{4}\).

Question 23Inverse Trigonometric Functions

The value of \(\sin\left(\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right)\) is:

  1. \(\frac{\sqrt{3}}{2}\)
  2. 1 Correct answer.
  3. \(\frac{1}{2}\)
  4. 0

Explanation

First, \(\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\) because \(-\frac{\pi}{6}\) is in the principal range and \(\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\). Then the expression becomes \(\sin\left(\frac{\pi}{3} + \frac{\pi}{6}\right) = \sin\frac{\pi}{2} = 1\).

Question 24Inverse Trigonometric Functions

The value of \(\cos\left(\sin^{-1}\frac{3}{5} + \cos^{-1}\frac{5}{13}\right)\) is:

  1. \(\frac{56}{65}\)
  2. \(-\frac{56}{65}\) Correct answer.
  3. \(-\frac{16}{65}\)
  4. \(\frac{63}{65}\)

Explanation

Let \(A = \sin^{-1}\frac{3}{5}\) and \(B = \cos^{-1}\frac{5}{13}\). Then \(\cos A = \frac{4}{5}\), \(\sin A = \frac{3}{5}\), \(\cos B = \frac{5}{13}\), and \(\sin B = \frac{12}{13}\). Using \(\cos(A+B) = \cos A\cos B - \sin A\sin B = \frac{4}{5}\cdot\frac{5}{13} - \frac{3}{5}\cdot\frac{12}{13} = \frac{20-36}{65} = -\frac{16}{65}\).

Question 25Inverse Trigonometric Functions

The value of \(\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\) is:

  1. \(\frac{3\pi}{4}\)
  2. \(\frac{\pi}{4}\) Correct answer.
  3. \(\frac{\pi}{2}\)
  4. \(\frac{\pi}{3}\)

Explanation

Using the formula \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}\) when \(xy < 1\), we get \(\tan^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}} = \tan^{-1}\frac{5/6}{5/6} = \tan^{-1}1 = \frac{\pi}{4}\).

Question 26Inverse Trigonometric Functions

The value of \(\sin^{-1}\left(\sin\frac{2\pi}{3}\right) + \cos^{-1}\left(\cos\frac{2\pi}{3}\right)\) is:

  1. \(\pi\)
  2. \(\frac{\pi}{3}\)
  3. \(-\frac{\pi}{3}\)
  4. \(\frac{2\pi}{3}\) Correct answer.

Explanation

Since \(\frac{2\pi}{3}\) is outside the range of \(\sin^{-1}\), \(\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}\). Because \(\frac{2\pi}{3}\) is inside the range of \(\cos^{-1}\), \(\cos^{-1}\left(\cos\frac{2\pi}{3}\right) = \frac{2\pi}{3}\). The sum is \(\frac{\pi}{3} + \frac{2\pi}{3} = \pi\).

Question 27Matrices

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

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

Explanation

A product AB is defined only when the number of columns of A equals the number of rows of B, which holds here since both are 4. The order of the product is then (rows of A) × (columns of B), i.e. 3 × 2.

Question 28Matrices

For a square matrix A, which condition guarantees that A is invertible?

  1. A is skew-symmetric
  2. det(A) ≠ 0 Correct answer.
  3. A is symmetric
  4. All entries of A are non-zero

Explanation

A square matrix has an inverse exactly when its determinant is non-zero, since A⁻¹ = adj(A)/det(A) and the division requires det(A) ≠ 0. Symmetry, skew-symmetry and having no zero entries are all irrelevant to invertibility.

Question 29Matrices

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

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

Explanation

Since A² = A, every higher power of A also equals A, so (I + A)³ = I + 3A + 3A² + A³ = I + 3A + 3A + A = I + 7A. Subtracting 7A leaves I.

Question 30Matrices

A is a 3 × 3 matrix with det(A) = 5. What is det(2A)?

  1. 10
  2. 30
  3. 5
  4. 40 Correct answer.

Explanation

Scaling an n × n matrix by k multiplies the determinant by kⁿ, because each of the n rows contributes a factor k. Here det(2A) = 2³ · det(A) = 8 · 5 = 40.

Question 31Matrices

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

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

Explanation

For an n × n matrix, |adj(A)| = |A|^(n−1). With n = 3 and |A| = 4 this gives 4² = 16.

Question 32Matrices

A is a 2 × 2 matrix with A(adj A) equal to which matrix?

  1. A²
  2. I
  3. adj(A)
  4. |A| I Correct answer.

Explanation

The fundamental adjoint identity is A · adj(A) = adj(A) · A = |A| I for every square matrix A. The scalar |A| multiplies the identity rather than appearing alone.

Question 33Matrices

A is a symmetric matrix and B is a skew-symmetric matrix of the same order. Which statement about AB + BA is always true?

  1. It is a scalar matrix
  2. It is symmetric
  3. It is the zero matrix
  4. It is skew-symmetric Correct answer.

Explanation

Using Aᵀ = A and Bᵀ = −B, we get (AB + BA)ᵀ = BᵀAᵀ + AᵀBᵀ = −BA − AB = −(AB + BA). A matrix equal to the negative of its transpose is skew-symmetric.

Question 34Matrices

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

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

Explanation

Since A · adj(A) = |A| I = 2I, its determinant is |2I| = 2³ · |I| = 8 for a 3 × 3 matrix. Equivalently |A|³ = 2³ = 8.

Question 35Matrices

A = [[2, 1], [1, 1]]. What is A⁻¹?

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

Explanation

For a 2 × 2 matrix [[a, b], [c, d]], the inverse is (1/(ad − bc))[[d, −b], [−c, a]]. Here ad − bc = 2 − 1 = 1, so A⁻¹ = [[1, −1], [−1, 2]].

Question 36Matrices

Matrices A and B satisfy AB = A and BA = B, with A and B square of the same order. Which conclusion follows?

  1. A and B are both invertible
  2. AB = BA = 0
  3. A² = A and B² = B Correct answer.
  4. A = B

Explanation

Multiply AB = A on the right by A to get ABA = A², and use BA = B to replace BA, giving A² = A. Similarly, multiplying BA = B on the left by A gives B² = B, so both are idempotent.

Question 37Matrices

A is a square matrix of order 3 with |A| = −2. What is |adj(adj A)|?

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

Explanation

Using |adj(M)| = |M|^(n−1) twice: first |adj A| = (−2)² = 4, then |adj(adj A)| = 4² = 16. Squaring removes the sign, so the result is positive.

Question 38Matrices

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

  1. 6
  2. 3
  3. 9 Correct answer.
  4. 27

Explanation

The identity |adj A| = |A|^(n−1) with n = 3 gives 3² = 9. The exponent is the order minus one, not the order itself.

Question 39Matrices

A is a 3 × 3 matrix with A³ = A. Which statement about det(A) must hold?

  1. det(A) is any real number
  2. det(A) = 1 only
  3. det(A) = 0 only
  4. det(A) ∈ {0, 1, −1} Correct answer.

Explanation

Taking determinants of A³ = A gives (det A)³ = det A, so det A satisfies x³ = x, i.e. x(x − 1)(x + 1) = 0. Hence det(A) can only be 0, 1 or −1.

Question 40Matrices

A = [[1, 2], [3, 4]]. What is A² − 5A?

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

Explanation

The characteristic equation of A is A² − (trace)A + (det)I = 0, with trace 5 and determinant −2, so A² − 5A = −(−2)I = 2I... checking directly: A² − 5A = [[7,10],[15,22]] − [[5,10],[15,20]] = [[2,0],[0,2]].

Question 41Determinants

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

  1. 10
  2. 80
  3. 40 Correct answer.
  4. 20

Explanation

For a square matrix of order n, det(kA) = k^n det(A). Here n = 3 and k = 2, so det(2A) = 2^3 × 5 = 8 × 5 = 40. The exponent equals the order of the matrix, not the number of nonzero entries.

Question 42Determinants

A square matrix A is singular. Which statement about A is necessarily true?

  1. Every entry of A is zero
  2. A has no inverse and det(A) = 0 Correct answer.
  3. det(A) is nonzero but A has no inverse
  4. A has an inverse but det(A) = 0

Explanation

A square matrix is singular exactly when its determinant equals zero, and a matrix has an inverse if and only if its determinant is nonzero. Hence singularity forces det(A) = 0 and the absence of an inverse. Nonzero entries or a nonzero adjoint do not prevent this.

Question 43Determinants

For any two square matrices A and B of the same order, which identity always holds?

  1. det(A + B) = det(A) + det(B)
  2. det(AB) = det(A) · det(B) Correct answer.
  3. det(A − B) = det(A) − det(B)
  4. det(AB) = det(A) + det(B)

Explanation

The determinant is multiplicative: det(AB) = det(A) det(B) for square matrices of equal order. There is no general additive rule, so expressions involving det(A + B) or det(A − B) cannot be expanded term by term.

Question 44Determinants

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

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

Explanation

For an n×n matrix, det(adj A) = (det A)^(n−1). With n = 3 and det(A) = 4, this gives 4^2 = 16. The exponent is one less than the order of the matrix.

Question 45Determinants

A 2×2 matrix A satisfies A · (adj A) = 7I, where I is the identity matrix. What is det(A)?

  1. 7 Correct answer.
  2. 1/7
  3. 49
  4. 14

Explanation

The fundamental adjoint identity states A · (adj A) = det(A) · I for any square matrix A. Comparing with A · (adj A) = 7I gives det(A) = 7 directly, with no squaring involved.

Question 46Determinants

A 3×3 matrix A has det(A) = 2. What is det(adj(adj A))?

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

Explanation

Applying det(adj M) = (det M)^(n−1) twice for n = 3: first det(adj A) = 2^2 = 4, then det(adj(adj A)) = 4^2 = 16 is not the route; instead use adj(adj A) = det(A)^(n−2) · A, giving det = 2^1 · 2 = 4. Equivalently, det(adj(adj A)) = (det A)^((n−1)^2) = 2^4 = 16 - so the correct chain is 2^4 = 16.

Question 47Determinants

The area of the triangle with vertices (1, 2), (4, 6) and (7, 10) is

  1. 3 square units
  2. 0 square units, since the points are collinear Correct answer.
  3. 6 square units
  4. 12 square units

Explanation

The determinant area formula gives ½|x1(y2−y3) + x2(y3−y1) + x3(y1−y2)|. Substituting: 1(6−10) + 4(10−2) + 7(2−6) = −4 + 32 − 28 = 0. A zero determinant means the three points are collinear, so the area is zero.

Question 48Determinants

Using determinants, the area of the triangle with vertices (0, 0), (5, 0) and (0, 4) is

  1. 20 square units
  2. 10 square units Correct answer.
  3. 40 square units
  4. 9 square units

Explanation

The area equals ½ times the absolute value of the determinant with rows (0,0,1), (5,0,1), (0,4,1). Expanding gives ½|0(0−4) − 0(5−0) + 1(20−0)| = ½ × 20 = 10. This is a right triangle with legs 5 and 4.

Question 49Determinants

If every element of one row of a 3×3 determinant is multiplied by 3, the value of the determinant

  1. becomes 27 times the original value
  2. becomes 9 times the original value
  3. becomes 3 times the original value Correct answer.
  4. stays the same

Explanation

Multiplying a single row by a scalar k multiplies the whole determinant by k. Only one row is scaled here, so the value triples. Scaling all three rows would multiply it by 3^3 = 27.

Question 50Determinants

A is a square matrix of order 3 with det(A) = −2. What is det(−A)?

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

Explanation

For order n, det(−A) = (−1)^n det(A). With n = 3, (−1)^3 = −1, so det(−A) = (−1)(−2) = 2 is not the route; the correct computation is (−1)^3 × (−2) = 2 - but scaling by −1 gives det(−A) = (−1)^3 det(A) = (−1)(−2) = 2, so the answer is 2.