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CHSE Odisha Elective Mathematics study materials
CHSE Odisha Elective Mathematics study resources

CHSE Odisha Elective Mathematics practice questions

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Question 1Relations and Functions

If A = {1, 2} and B = {a, b, c}, then the number of elements in A × B is:

  1. 6 Correct answer.
  2. 5
  3. 9
  4. 8

Explanation

The Cartesian product A × B consists of all ordered pairs (x, y) with x in A and y in B. Its cardinality is n(A) × n(B) = 2 × 3 = 6.

Question 2Relations and Functions

The domain of the relation R = {(1, 2), (2, 3), (3, 4)} is:

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

Explanation

The domain of a relation is the set of all first coordinates of its ordered pairs. Here the first coordinates are 1, 2 and 3, so the domain is {1, 2, 3}.

Question 3Relations and Functions

Which of the following relations on A = {1, 2, 3} is reflexive?

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

Explanation

A relation on A is reflexive when (a, a) belongs to it for every a in A. Only the relation containing (1, 1), (2, 2) and (3, 3) satisfies this condition for all three elements.

Question 4Relations and Functions

The relation R on the set of integers Z defined by a R b if and only if a − b is divisible by 5 is:

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

Explanation

We have a − a = 0, which is divisible by 5, so R is reflexive. If a − b is divisible by 5, then b − a is also divisible by 5, so R is symmetric; if a − b and b − c are divisible by 5, then a − c is divisible by 5, so R is transitive. Hence R is an equivalence relation.

Question 5Relations and Functions

The function f: R → R defined by f(x) = x^2 is:

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

Explanation

Since f(−1) = f(1) but −1 ≠ 1, the function is not one-one. Its range is [0, ∞), not the whole codomain R, so it is also not onto.

Question 6Relations and Functions

If f(x) = 2x + 1 and g(x) = x^2, then (g ∘ f)(x) is:

  1. 4x^2 + 1
  2. 2x^2 + 1
  3. (2x + 1)^2 Correct answer.
  4. 2x^2 + 2x + 1

Explanation

By definition, (g ∘ f)(x) = g(f(x)). Substituting f(x) = 2x + 1 into g gives g(2x + 1) = (2x + 1)^2.

Question 7Relations and Functions

The domain of the function f(x) = √(x − 3)/(x − 5) is:

  1. (3, ∞) \ {5}
  2. [3, ∞)
  3. [3, ∞) \ {5} Correct answer.
  4. (-∞, 5) ∪ (5, ∞)

Explanation

The square root requires x − 3 ≥ 0, so x ≥ 3. The denominator requires x − 5 ≠ 0, so x ≠ 5. Combining these conditions gives [3, ∞) with 5 removed.

Question 8Relations and Functions

If f: R → R is defined by f(x) = (3x + 2)/5, then f^{-1}(x) is:

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

Explanation

Write y = (3x + 2)/5 and solve for x. Multiplying by 5 gives 5y = 3x + 2, so 3x = 5y − 2 and x = (5y − 2)/3. Interchanging x and y gives f^{-1}(x) = (5x − 2)/3.

Question 9Relations and Functions

The number of bijective functions from a set A with 4 elements to itself is:

  1. 16
  2. 12
  3. 24 Correct answer.
  4. 4

Explanation

A bijection from a finite set to itself is a permutation of its elements. The number of permutations of 4 elements is 4! = 24.

Question 10Relations and Functions

Which of the following functions f: R → R is a bijection?

  1. f(x) = sin x
  2. f(x) = x^2
  3. f(x) = x^3 Correct answer.
  4. f(x) = |x|

Explanation

The function f(x) = x^3 is strictly increasing, so it is one-one. For every real y, the real number ∛y satisfies f(∛y) = y, so it is onto as well.

Question 11Relations and Functions

If A = {1, 2, 3} and B = {4, 5}, then the number of relations from A to B is:

  1. 6
  2. 8
  3. 32
  4. 64 Correct answer.

Explanation

A relation from A to B is defined as any subset of the Cartesian product A × B. Here n(A × B) = 3 × 2 = 6, so A × B has 6 elements. The number of subsets of a 6-element set is 2^6 = 64, and each subset is exactly one relation from A to B.

Question 12Relations and Functions

The range of the relation R = {(2, 5), (3, 7), (4, 5), (5, 9)} is:

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

Explanation

The range of a relation is the set of all second components of its ordered pairs, while the domain collects the first components. Reading the second entries of R gives 5, 7, 5 and 9. Repeated values are listed once, so the range is {5, 7, 9}.

Question 13Relations and Functions

The number of one-one functions from a set X with 3 elements to a set Y with 5 elements is:

  1. 60 Correct answer.
  2. 125
  3. 243
  4. 15

Explanation

A one-one function must send the three distinct elements of X to three distinct elements of Y. The first element of X has 5 possible images, the second has 4 remaining choices and the third has 3. The product 5 × 4 × 3 = 60 counts all such functions.

Question 14Relations and Functions

If f: [0, ∞) → [2, ∞) is defined by f(x) = x^2 + 2, then f^{-1}(x) is:

  1. ±√(x − 2), x ≥ 2
  2. √(x + 2), x ≥ −2
  3. √(x − 2), x ≥ 2 Correct answer.
  4. x^2 − 2, x ≥ 0

Explanation

On the restricted domain x ≥ 0 the function is strictly increasing, hence one-one onto [2, ∞). Writing y = x^2 + 2 and solving with x ≥ 0 gives x = √(y − 2), so interchanging variables yields f^{-1}(x) = √(x − 2) for x ≥ 2.

Question 15Relations and Functions

A function f: R \ {0} → R satisfies f(x) + 2f(1/x) = x for every x ≠ 0. Then f(2) equals:

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

Explanation

Replacing x by 1/x in the given equation gives f(1/x) + 2f(x) = 1/x. Eliminating f(1/x) between the two equations yields −3f(x) = x − 2/x, so f(x) = (2/x − x)/3. Substituting x = 2 gives f(2) = (1 − 2)/3 = −1/3.

Question 16Relations and Functions

On the set A = {1, 2, 3, 4, 5}, a relation R is defined by a R b if and only if a + b is even. The family of equivalence classes of R is:

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

Explanation

The sum a + b is even exactly when a and b have the same parity, so two numbers are related precisely when both are odd or both are even. This relation is reflexive, symmetric and transitive, and its equivalence classes are the maximal parity blocks: the odd elements {1, 3, 5} and the even elements {2, 4}.

Question 17Inverse Trigonometric Functions

The principal value of sin⁻¹(−1/2) is

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

Explanation

The principal branch of sin⁻¹ has range [−π/2, π/2], so the answer must lie in this interval. Since sin(−π/6) = −1/2 and −π/6 lies in [−π/2, π/2], the principal value is −π/6. The angle 5π/6 has the same sine but falls outside the principal range.

Question 18Inverse Trigonometric Functions

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

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

Explanation

The function cos⁻¹ t is defined exactly when −1 ≤ t ≤ 1, so here we need −1 ≤ 2x − 1 ≤ 1. Adding 1 throughout gives 0 ≤ 2x ≤ 2, and dividing by 2 gives 0 ≤ x ≤ 1. Hence the domain is the closed interval [0, 1].

Question 19Inverse Trigonometric Functions

The range of the principal-value function sec⁻¹x is

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

Explanation

The principal branch of sec⁻¹ is fixed by taking values in [0, π] while excluding π/2, because sec x is not defined at π/2. On each of the intervals [0, π/2) and (π/2, π] the secant function is one-to-one onto (−∞, −1] ∪ [1, ∞). Thus the range is [0, π] − {π/2}.

Question 20Inverse Trigonometric Functions

The principal value of tan⁻¹(tan 3π/4) is

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

Explanation

First evaluate the inner expression: tan(3π/4) = −1. The principal branch of tan⁻¹ takes values in (−π/2, π/2), and the angle in that interval with tangent −1 is −π/4. Therefore tan⁻¹(tan 3π/4) = −π/4, not 3π/4.

Question 21Inverse Trigonometric Functions

If sin⁻¹x + sin⁻¹y = π/2, where x and y lie in the domains of the inverse functions, then x² + y² equals

  1. 2
  2. 0
  3. x²y²
  4. 1 Correct answer.

Explanation

From the given relation, sin⁻¹y = π/2 − sin⁻¹x, and the complementary-angle identity gives sin⁻¹y = cos⁻¹x. Applying sine to both sides yields y = √(1 − x²), so squaring gives y² = 1 − x² and hence x² + y² = 1.

Question 22Inverse Trigonometric Functions

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

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

Explanation

Use the addition formula tan⁻¹a + tan⁻¹b = tan⁻¹[(a + b)/(1 − ab)] when ab < 1. Here a = 1/2 and b = 1/3, so (a + b)/(1 − ab) = (5/6)/(5/6) = 1. Since tan⁻¹1 = π/4 and the product ab = 1/6 < 1, the sum is exactly π/4.

Question 23Inverse Trigonometric Functions

The principal value of sin⁻¹(sin 2π/3) is

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

Explanation

The principal branch of sin⁻¹ takes values only in [−π/2, π/2], and 2π/3 is outside this interval. Using sin(2π/3) = sin(π − 2π/3) = sin(π/3), and π/3 lies in [−π/2, π/2], the principal value is π/3.

Question 24Inverse Trigonometric Functions

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

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

Explanation

Applying the addition formula with 6x² < 1 gives (2x + 3x)/(1 − 6x²) = tan(π/4) = 1, that is 5x = 1 − 6x², so 6x² + 5x − 1 = 0. This factors as (6x − 1)(x + 1) = 0, giving x = 1/6 or x = −1. The root x = −1 violates 6x² < 1 and in fact makes the left side negative, so only x = 1/6 is valid and there is exactly one solution.

Question 25Inverse Trigonometric Functions

For x ≠ 0, the expression tan⁻¹[(√(1 + x²) − 1)/x] simplifies to

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

Explanation

Put x = tan θ with θ in (−π/2, π/2), θ ≠ 0, so that √(1 + x²) = sec θ. Then (sec θ − 1)/tan θ = (1 − cos θ)/sin θ, which is the half-angle form of tan(θ/2). Hence the expression equals θ/2 = (1/2) tan⁻¹x.

Question 26Inverse Trigonometric Functions

The value of tan⁻¹(1/3) + tan⁻¹(1/5) + tan⁻¹(1/7) + tan⁻¹(1/8) is

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

Explanation

Combine the terms in pairs using tan⁻¹a + tan⁻¹b = tan⁻¹[(a + b)/(1 − ab)]. The first pair gives tan⁻¹[(8/15)/(14/15)] = tan⁻¹(4/7) and the second gives tan⁻¹[(15/56)/(55/56)] = tan⁻¹(3/11). Adding these, (4/7 + 3/11)/(1 − 12/77) = (65/77)/(65/77) = 1, so the total is tan⁻¹1 = π/4.

Question 27Inverse Trigonometric Functions

The principal value of cosec⁻¹(−√2) is

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

Explanation

The principal-value range of cosec⁻¹ is [−π/2, π/2] with 0 excluded. Since cosec(−π/4) = −√2 and −π/4 lies in that range, the principal value is −π/4.

Question 28Inverse Trigonometric Functions

The value of sin(2 tan⁻¹(3/4)) is

  1. 12/25
  2. 7/25
  3. 3/5
  4. 24/25 Correct answer.

Explanation

Put θ = tan⁻¹(3/4), so tan θ = 3/4 with θ in (−π/2, π/2). Using sin 2θ = 2tan θ/(1 + tan²θ) gives (3/2)/(1 + 9/16) = (3/2)(16/25) = 24/25.

Question 29Matrices and Determinants

The determinant of the matrix [[4, 3], [2, 5]] is

  1. 14 Correct answer.
  2. 26
  3. -14
  4. 22

Explanation

For a 2 × 2 matrix [[a, b], [c, d]], the determinant is ad - bc. Here 4·5 - 3·2 = 20 - 6 = 14.

Question 30Matrices and Determinants

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

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

Explanation

For AB to be defined, the number of columns of A must equal the number of rows of B, which is true here. The product then has the number of rows of A and the number of columns of B, so its order is 2 × 4.

Question 31Matrices and Determinants

Which of the following matrices is symmetric?

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

Explanation

A square matrix is symmetric when A^T = A, so the entry in row i, column j must equal the entry in row j, column i for every pair. Only [[2, -1], [-1, 3]] has this equality, including the off-diagonal entries -1 and -1.

Question 32Matrices and Determinants

The determinant of [[1, 2, 3], [0, 1, 4], [5, 6, 0]] is

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

Explanation

Expanding along the first row gives 1(1·0 - 4·6) - 2(0·0 - 4·5) + 3(0·6 - 1·5). This simplifies to -24 + 40 - 15 = 1.

Question 33Matrices and Determinants

The inverse of A = [[2, 1], [5, 3]] is

  1. [[2, -1], [-5, 3]]
  2. [[3, 1], [5, 2]]
  3. [[3, -1], [-5, 2]] Correct answer.
  4. [[1/2, 1], [1/5, 1/3]]

Explanation

For A = [[a, b], [c, d]], the inverse is (1/(ad - bc))[[d, -b], [-c, a]]. Here ad - bc = 6 - 5 = 1, so the inverse is [[3, -1], [-5, 2]].

Question 34Matrices and Determinants

For what value of k does the system x + y = 2, 2x + ky = 4 have infinitely many solutions?

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

Explanation

The coefficient determinant is 1·k - 1·2 = k - 2. Infinitely many solutions occur when this determinant is zero and the equations are consistent; k = 2 makes the second equation exactly twice the first, so the system is dependent.

Question 35Matrices and Determinants

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

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

Explanation

For an n × n matrix A, the adjugate satisfies |adj A| = |A|^(n - 1). With n = 3 and |A| = -2, this gives (-2)^2 = 4.

Question 36Matrices and Determinants

If a square matrix A satisfies A^T = -A, then which statement is necessarily true?

  1. A^2 = I.
  2. All entries of A are zero.
  3. A is a symmetric matrix.
  4. Every diagonal entry of A is zero. Correct answer.

Explanation

From A^T = -A, the entry in row i, column i must satisfy a_ii = -a_ii. Hence 2a_ii = 0, so every diagonal entry is zero; such a matrix is skew-symmetric.

Question 37Matrices and Determinants

If A is a 3 × 3 matrix with |A| = 2, then |3A^(-1)| equals

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

Explanation

For an invertible n × n matrix, |cA| = c^n|A| and |A^(-1)| = 1/|A|. Here n = 3, so |3A^(-1)| = 3^3 · 1/2 = 27/2.

Question 38Matrices and Determinants

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

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

Explanation

By the Cayley-Hamilton theorem, A satisfies its characteristic equation λ^2 - 5λ - 2 = 0, because trace A = 5 and det A = -2. Replacing λ by A gives A^2 - 5A - 2I = 0, the zero matrix.

Question 39Matrices and Determinants

A matrix has 3 rows and 5 columns. The number of elements in the matrix is

  1. 10
  2. 8
  3. 35
  4. 15 Correct answer.

Explanation

For a matrix of order m × n, the number of entries is the product m n. Here m = 3 and n = 5, so the matrix has 3 × 5 = 15 elements. Equivalently, each of the three rows contains five entries.

Question 40Matrices and Determinants

The determinant of the diagonal matrix diag(2, -3, 4) is

  1. 9
  2. 24
  3. -24 Correct answer.
  4. -1

Explanation

For a diagonal matrix, the determinant is the product of the diagonal entries. Thus det(diag(2, -3, 4)) = 2 × (-3) × 4 = -24. No cross terms enter because all off-diagonal entries are zero.

Question 41Matrices and Determinants

In the matrix A = [[2, 3], [4, 5]], the cofactor of the element 3 is

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

Explanation

The minor of the entry 3 is found by deleting its row and column, leaving 4. Since 3 lies in row 1 and column 2, the cofactor is (-1)^(1+2) times the minor, that is -4.

Question 42Matrices and Determinants

If A = [[1, 2], [3, 4]], then adj A is

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

Explanation

For A = [[a, b], [c, d]], the adjoint is [[d, -b], [-c, a]]. Substituting a = 1, b = 2, c = 3, d = 4 gives [[4, -2], [-3, 1]].

Question 43Matrices and Determinants

If each element of one row of a 3 × 3 matrix A is multiplied by 2 to obtain matrix B, then |B| equals

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

Explanation

The determinant is linear in each row separately. Multiplying exactly one row by 2 therefore multiplies the determinant by 2, so |B| = 2|A|. Multiplying two rows or all three rows would give different factors.

Question 44Matrices and Determinants

If A and B are 3 × 3 matrices with |A| = 3 and |B| = -2, then |AB| equals

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

Explanation

The determinant of a product satisfies |AB| = |A||B| for square matrices of the same order. Hence |AB| = 3 × (-2) = -6.

Question 45Matrices and Determinants

The system 2x + 3y = 7, 4x + ky = 14 has a unique solution precisely when

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

Explanation

The coefficient determinant is Δ = 2k - 12. A unique solution occurs exactly when Δ ≠ 0, that is, when k ≠ 6. At k = 6 the two equations are dependent and the system has infinitely many solutions.

Question 46Matrices and Determinants

If A = [[1, 1], [0, 1]], then A^n for a positive integer n is

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

Explanation

The matrix is a shear. Squaring it gives [[1, 2], [0, 1]], and induction gives A^n = [[1, n], [0, 1]] for every positive integer n.

Question 47Matrices and Determinants

If A is a square matrix such that A^2 = A, then (I + A)^3 - 7A equals

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

Explanation

Using A^2 = A, (I + A)^2 = I + 2A + A^2 = I + 3A. Multiplying again by I + A gives I + 4A + 3A^2 = I + 7A. Subtracting 7A leaves I.

Question 48Matrices and Determinants

The value of the determinant with first row 1, 1, 1; second row a, b, c; third row a^2, b^2, c^2 is

  1. (a - b)(c - b)(c - a)
  2. (a - b)(b - c)(a - c)
  3. (a + b)(b + c)(c + a)
  4. (a - b)(b - c)(c - a) Correct answer.

Explanation

This determinant is the Vandermonde determinant. Row operations and factorisation give (b - a)(c - a)(c - b), which equals (a - b)(b - c)(c - a) because three sign changes occur.

Question 49Matrices and Determinants

The trace of the matrix A = [[6, -1], [4, 9]] is

  1. 10
  2. 13
  3. 5
  4. 15 Correct answer.

Explanation

The trace of a square matrix is the sum of the entries on its principal diagonal. In this matrix the diagonal entries are 6 and 9, so the trace is 6 + 9 = 15. This is a direct definitional computation.

Question 50Matrices and Determinants

If the determinant of the matrix [[x, 1], [4, x]] is zero, then the possible values of x are

  1. 2 only
  2. ±4
  3. ±2 Correct answer.
  4. 0 and 2

Explanation

The determinant of [[x, 1], [4, x]] is x·x − 1·4 = x^2 − 4. Setting the determinant equal to zero gives x^2 = 4, so x = 2 or x = −2, that is, x = ±2. Both signs must be included because a quadratic equation generally has two roots.