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Test Yourself on Every Mathematics Chapter (Class 12)
Below, you'll find one test for each Mathematics chapter in Class 12, matched to the CBSE 2026-27 marking scheme. Pick any chapter to start right away - no login required, and you can attempt it again and again.
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Quick Practice - Class 12 Mathematics (NCERT Core) Select any chapter below to test your problem-solving, calculus, algebra, vectors, and probability skills with 5 high-yield multiple-choice questions, instant scoring, and verified step-by-step solutions. Q1.Let set A = {1, 2, 3}. A relation R on A is defined as R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3)}. The relation R is: Answer: (a) Reflexive but neither symmetric nor transitive. Since (1, 1), (2, 2), (3, 3) belong to R, it is reflexive. It is not symmetric because (2, 3) belongs to R but (3, 2) does not belong to R. It is not transitive because (1, 2) and (2, 3) belong to R, but (1, 3) does not belong to R.
Q2.Let f: R → R be defined by f(x) = x4. Then the function f is: Answer: (d) Neither one-one nor onto. f(-1) = f(1) = 1, so it is not one-one. The range of x4 is [0, infinity), which does not equal the codomain R (negative numbers have no pre-images), so it is not onto.
Q3.If a set A has 3 elements and set B has 2 elements, what is the total number of relations that can be defined from A to B? Answer: (b) 64 (26). Number of elements in A × B = n(A) × n(B) = 3 × 2 = 6. Total number of possible relations is the total number of subsets of A × B, which equals 26 = 64.
Q4.Let f: R → R be defined as f(x) = 3x - 4. The function f is: Answer: (a) One-one and onto (Bijective). If f(x1) = f(x2), then 3x1 - 4 = 3x2 - 4, which implies x1 = x2 (one-one). For any y in R, setting y = 3x - 4 gives x = (y + 4)/3 in R (onto). Thus, f is bijective.
Q5.The total number of equivalence relations that can be defined on the set S = {1, 2, 3} is: Answer: (c) 5. The number of equivalence relations on a set with n elements equals the Bell Number Bn. For n = 3, B3 = 5, corresponding to the 5 partitions of a 3-element set.
Q1.What is the Principal Value Branch (Range) of the function y = cos-1(x)? Answer: (b) [0, π]. The principal value branch for cos-1(x) with domain [-1, 1] is [0, π]. For sin-1(x), the principal range is [-π/2, π/2].
Q2.What is the principal value of cos-1(-1/2)? Answer: (c) 2π/3. Using the property cos-1(-x) = π - cos-1(x): cos-1(-1/2) = π - cos-1(1/2) = π - π/3 = 2π/3, which lies in the principal branch [0, π].
Q3.What is the value of sin(π/3 - sin-1(-1/2))? Answer: (a) 1. sin-1(-1/2) = -sin-1(1/2) = -π/6. Therefore: sin(π/3 - (-π/6)) = sin(π/3 + π/6) = sin(π/2) = 1.
Q4.The principal value of tan-1(tan(3π/4)) is: Answer: (d) -π/4. The range of tan-1(x) is (-π/2, π/2). Since 3π/4 is not in (-π/2, π/2), we rewrite tan(3π/4) = tan(π - π/4) = -tan(π/4) = tan(-π/4). Thus, the value is -π/4.
Q5.What is the domain of the function f(x) = sin-1(2x - 1)? Answer: (b) [0, 1]. For sin-1(θ), the argument must satisfy -1 ≤ θ ≤ 1. Here: -1 ≤ 2x - 1 ≤ 1 implies 0 ≤ 2x ≤ 2, which gives 0 ≤ x ≤ 1. Hence, the domain is [0, 1].
Q1.If A and B are symmetric matrices of the same order, then the matrix (AB - BA) is always: Answer: (a) A Skew-Symmetric Matrix. Taking the transpose: (AB - BA)T = (AB)T - (BA)T = BTAT - ATBT. Since AT = A and BT = B, this gives BA - AB = -(AB - BA). Therefore, (AB - BA) is skew-symmetric.
Q2.What is optical/total number of all possible matrices of order 3 × 3 with each entry either 0 or 1? Answer: (c) 512 (29). A 3 × 3 matrix contains 3 × 3 = 9 element positions. Since each position can independently be filled by either 0 or 1 (2 choices), total possible matrices = 29 = 512.
Q3.If matrix A = [[cosα, -sinα], [sinα, cosα]], and A + AT = I, then the value of α in (0, π) is: Answer: (b) π/3. A + AT = [[2cosα, 0], [0, 2cosα]] = [[1, 0], [0, 1]], which implies 2cosα = 1, so cosα = 1/2, giving α = π/3.
Q4.If A is a square matrix such that A2 = A, then the value of (I + A)3 - 7A is equal to: Answer: (a) I. Expanding (I + A)3 = I + 3A + 3A + A = I + 7A (since A2 = A and A3 = A). Subtracting 7A gives (I + 7A) - 7A = I.
Q5.If matrix A is of order 2 × 3 and matrix B is of order 3 × 4, then the order of matrix (AB)T is: Answer: (d) 4 × 2. Matrix product AB has order 2 × 4. Taking the transpose swaps rows and columns, giving order 4 × 2.
Q1.If A is a non-singular square matrix of order 3 × 3 and |A| = 4, then the value of |adj(A)| is: Answer: (b) 16. Formula: |adj(A)| = |A|n-1, where n is the matrix order. For n = 3 and |A| = 4: |adj(A)| = 43-1 = 42 = 16.
Q2.If A is a square matrix of order 3 and |A| = -2, what is the value of |3A|? Answer: (c) -54. For an n × n matrix, |kA| = kn|A|. Here n = 3 and k = 3, so |3A| = 33 × (-2) = 27 × (-2) = -54.
Q3.If the points (2, -3), (λ, -1), and (0, 4) are collinear, the value of λ is: Answer: (a) 10/7. The area of the triangle formed by collinear points is zero, which gives the equation -10 + 3λ + 4λ = 0, solving to 7λ = 10, so λ = 10/7.
Q4.A square matrix A is invertible if and only if A is: Answer: (b) Non-singular (|A| ≠ 0). Since A-1 = (1/|A|) adj(A), the inverse exists if and only if the determinant |A| is non-zero.
Q5.If A is an invertible matrix of order 2, then det(A-1) is equal to: Answer: (d) 1 / det(A). Since A × A-1 = I, taking determinants on both sides gives det(A) × det(A-1) = 1, leading to det(A-1) = 1 / det(A).
Q1.The function f(x) = |x - 2| at the point x = 2 is: Answer: (c) Continuous, but not differentiable. The function is continuous at x = 2, but because it has a sharp corner, the Left Hand Derivative (-1) does not equal the Right Hand Derivative (+1), making it non-differentiable there.
Q2.What is the derivative of sin(x2 + 5) with respect to x? Answer: (b) 2x cos(x2 + 5). Applying the Chain Rule: d/dx [sin(x2 + 5)] = cos(x2 + 5) × d/dx (x2 + 5) = 2x cos(x2 + 5).
Q3.If x = a cos(θ) and y = a sin(θ), then dy/dx is equal to: Answer: (a) -cot(θ). Parametric derivatives: dx/dθ = -a sin(θ) and dy/dθ = a cos(θ), yielding dy/dx = (a cos(θ)) / (-a sin(θ)) = -cot(θ).
Q4.If y = e3 log x, what is the value of dy/dx? Answer: (c) 3x2. Using log properties, y = elog(x^3) = x3. Differentiating with respect to x gives dy/dx = 3x2.
Q5.If f(x) = kx + 1 for x ≤ 5 and 3x - 5 for x > 5 is continuous at x = 5, the value of k is: Answer: (d) 9/5. For continuity at x = 5, 5k + 1 = 3(5) - 5, which evaluates to 5k + 1 = 10, giving 5k = 9 and k = 9/5.
Q1.The rate of change of the area of a circle with respect to its radius r at r = 6 cm is: Answer: (b) 12π cm2/cm. Area A = πr2. dA/dr = 2πr. At r = 6 cm, dA/dr = 2π(6) = 12π cm2/cm.
Q2.The function f(x) = x3 - 3x2 + 4x is strictly increasing on: Answer: (a) R (all real numbers). f'(x) = 3x2 - 6x + 4 = 3(x - 1)2 + 1 ≥ 1 > 0 for all x in R, so the function is strictly increasing everywhere.
Q3.What is the absolute maximum value of the function f(x) = sin x + cos x in the interval [0, π]? Answer: (c) √2. Setting f'(x) = cos x - sin x = 0 gives x = π/4. Evaluating at endpoints and critical points yields a maximum value of √2.
Q4.The point on the curve y = x2 - 2x + 7 where the tangent is parallel to the x-axis is: Answer: (b) (1, 6). Setting dy/dx = 2x - 2 = 0 gives x = 1, resulting in y = 6.
Q5.If two positive numbers x and y have a constant sum x + y = 16, their product xy is maximized when: Answer: (a) x = 8, y = 8. For a fixed sum, the product is maximized when the numbers are equal.
Q1.What is the value of the integral ∫ sec2(x) / √tan(x) dx? Answer: (b) 2√tan(x) + C. Let t = tan(x), dt = sec2(x)dx, leading to ∫ t-1/2dt = 2t1/2 + C.
Q2.What is the value of the definite integral ∫ from -π/2 to π/2 of sin7(x) dx? Answer: (d) 0. Since f(x) = sin7(x) is an odd function over symmetric limits, the integral evaluates to zero.
Q3.What is the value of ∫ ex (sin x + cos x) dx? Answer: (a) ex sin x + C. Using the standard theorem ∫ ex[f(x) + f'(x)]dx = exf(x) + C.
Q4.The value of the definite integral ∫ from 0 to π/2 of √sin(x) / (√sin(x) + √cos(x)) dx is: Answer: (c) π/4. Applying the standard definite integral property yields 2I = π/2, so I = π/4.
Q5.What is the value of ∫ 1 / (x2 + 16) dx? Answer: (b) (1/4) tan-1(x/4) + C. Using the standard formula ∫ 1/(x2 + a2) dx = (1/a) tan-1(x/a) + C with a = 4.
Q1.What is the total area enclosed by the circle x2 + y2 = a2? Answer: (a) πa2. Multiplying the area of one quadrant by 4 gives 4 × (πa2 / 4) = πa2.
Q2.The area bounded by the ellipse x2/a2 + y2/b2 = 1 is equal to: Answer: (c) πab. Integrating one quadrant and multiplying by 4 yields πab.
Q3.What is the area of the region bounded by the parabola y2 = 4x and the line x = 3? Answer: (b) 8√3 sq units. Evaluating 2 × ∫ from 0 to 3 of 2√x dx gives 8√3 sq units.
Q4.The area bounded by the curve y = cos x between x = 0 and x = π and the x-axis is: Answer: (d) 2 sq units. Area takes absolute values above and below the x-axis: 1 + 1 = 2 sq units.
Q5.What is the area bounded by the line y = 2x, the x-axis, and the vertical lines x = 0 and x = 4? Answer: (a) 16 sq units. ∫ from 0 to 4 of 2x dx evaluates to [x2] from 0 to 4, which equals 16.
Q1.What are the Order and Degree of the differential equation (d2y/dx2)3 + (dy/dx)4 + sin(dy/dx) = 0? Answer: (c) Order = 2, Degree is Not Defined. The highest derivative is d2y/dx2 (Order 2). Because of sin(dy/dx), the equation is not a polynomial in derivatives, so its degree is not defined.
Q2.What is the Integrating Factor (I.F.) for the linear differential equation dy/dx + y cot x = 2x + x2 cot x? Answer: (b) sin x. I.F. = e∫ cot x dx = elog(sin x) = sin x.
Q3.What is the general solution of the differential equation dy/dx = ex - y? Answer: (a) ey = ex + C. Separating variables: ey dy = ex dx, integrating both sides gives ey = ex + C.
Q4.The number of arbitrary constants in the particular solution of a differential equation of third order is: Answer: (d) 0. A particular solution has specific values substituted for all constants, leaving zero arbitrary constants.
Q5.Which of the following is a homogeneous differential equation of degree 0? Answer: (b) (x2 - y2) dx + 2xy dy = 0. Every term in the numerator and denominator is of homogeneous degree 2.
Q1.What is the value of λ if vectors a = 2i - j + 2k and b = 3i + λj + k are perpendicular? Answer: (c) 8. Dot product equals zero: 2(3) + (-1)(λ) + 2(1) = 0, giving 8 - λ = 0, so λ = 8.
Q2.What is the projection of vector a = i + 3j + 7k on vector b = 7i - j + 8k? Answer: (a) 60 / √114. Projection = (a · b) / |b| = (7 - 3 + 56) / √(49 + 1 + 64) = 60 / √114.
Q3.If |a| = 10, |b| = 2, and a · b = 12, then magnitude of cross product |a × b| is: Answer: (d) 16. Using |a × b|2 = |a|2|b|2 - (a · b)2 = 400 - 144 = 256, giving |a × b| = 16.
Q4.What is the value of i · (j × k) + j · (i × k) + k · (i × j)? Answer: (b) 1. Evaluates to 1 - 1 + 1 = 1.
Q5.What is the unit vector in the direction of vector a = 2i + 3j + √3k? Answer: (a) (2/4)i + (3/4)j + (√3/4)k. Magnitude is √(4 + 9 + 3) = 4, so dividing vector components by 4 gives the unit vector.
Q1.If a line makes angles 90 degrees, 135 degrees, and 45 degrees with positive x, y, and z axes, its direction cosines are: Answer: (b) 0, -1/√2, 1/√2. Evaluated as cos(90) = 0, cos(135) = -1/√2, cos(45) = 1/√2.
Q2.What is the Shortest Distance between two parallel lines with direction vector b? Answer: (a). Formula for shortest distance between parallel lines uses the cross product with the shared direction vector.
Q3.The coordinates of the foot of the perpendicular drawn from point P(a, b, c) to the y-axis are: Answer: (c) (0, b, 0). On the y-axis, x and z coordinates are zero.
Q4.Two lines with direction ratios (a1, b1, c1) and (a2, b2, c2) are perpendicular if: Answer: (d) a1a2 + b1b2 + c1c2 = 0. The dot product of direction vectors for perpendicular lines is zero.
Q5.What is the distance of the point (3, 4, 5) from the xy-plane? Answer: (b) 5 units. Distance from the xy-plane is given by the absolute value of the z-coordinate, which is |5| = 5 units.
Q1.The optimal value of a linear objective function Z = ax + by in an LPP always occurs at: Answer: (a) Corner points (vertices). By the Fundamental Corner Point Theorem, optimal solutions occur at boundary vertices.
Q2.In an LPP, maximize Z = 3x + 9y with corner points (0, 0), (0, 20), (15, 15), and (0, 10). The maximum value of Z is: Answer: (c) 180. Evaluating Z at (0, 20) gives 180, and at (15, 15) gives 3(15) + 9(15) = 180.
Q3.What is the Feasible Region in a Linear Programming Problem? Answer: (b). The feasible region satisfies all constraints and non-negativity conditions simultaneously.
Q4.Constraints x ≥ 0 and y ≥ 0 restrict the feasible region to which quadrant? Answer: (a) First Quadrant. Real-world variables are non-negative, confining solutions to the first quadrant.
Q5.If the feasible region of an LPP is Unbounded, then a maximum value M exists if and only if: Answer: (d). Standard graphical test for unbounded feasible regions.
Q1.If P(A) = 0.6, P(B) = 0.3, and P(A intersection B) = 0.2, what is P(A | B)? Answer: (b) 2/3. P(A | B) = P(A intersection B) / P(B) = 0.2 / 0.3 = 2/3.
Q2.If A and B are Independent Events with P(A) = 1/4 and P(B) = 1/2, then P(neither A nor B) is: Answer: (a) 3/8. P(A' intersection B') = P(A') × P(B') = (3/4) × (1/2) = 3/8.
Q3.Bayes' Theorem is primarily used in probability theory to determine: Answer: (c). Bayes' Theorem computes conditional posterior probabilities of causes given observed outcomes.
Q4.A random variable X has probability distribution P(X = x) = kx for x in {1, 2, 3, 4}. Constant k equals: Answer: (b) 1/10. Sum of probabilities = k(1 + 2 + 3 + 4) = 10k = 1, so k = 1/10.
Q5.If two events A and B are mutually exclusive, then conditional probability P(A | B) is: Answer: (d) 0. Mutually exclusive events cannot occur together, making the intersection probability zero.
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