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Class 12 Math Practice Test Online

Practise choosing, justifying, and checking methods across board-style numericals before time pressure turns small setup errors into lost marks.

Class 12Math

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Reviewed by MVR · Senior Secondary Maths Faculty, GV Academy · 18+ years teaching Class 11-12 Math, reviewed for curriculum & exam alignment.Curriculum reviewed for alignment with the current CBSE & NCERT Class 12 Maths framework, covering relations and inverse trigonometry through probability.

Last reviewed: August 2026

📘 NCERT curriculum alignmentThis practice test follows NCERT Class 12 Mathematics, India's national curriculum board.▶

Topics covered

Relations and Functions, and Inverse Trigonometric Functions
Matrices and Determinants
Continuity and Differentiability, and Application of Derivatives
Integrals and Application of Integrals
Differential Equations
Vector Algebra and Three Dimensional Geometry
Linear Programming
Probability

View official NCERT textbooks →

⚡ Quick quiz

Test yourself in under a minute: Class 12 Math Practice Test

Check calculus applications, vector geometry, linear programming, and probability, the core Class 12 Maths skills.

1. Find the determinant of the matrix [[5, 3], [2, 4]].
2. Show that f(x) = |x - 3| is continuous at x = 3 but not differentiable there.
3. Find the local maximum and local minimum values of f(x) = x^3 - 3x^2 - 9x + 5.
4. Find the area bounded by the curve y = x^2, the x-axis, and the lines x = 0 and x = 3.
5. Find a × b (the cross product) if a = i + j + k and b = 2i - j + 3k.
6. For the constraints x ≥ 0, y ≥ 0, x + 2y ≤ 8, and 3x + 2y ≤ 12, find all corner points of the feasible region.
7. Bag A contains 4 red and 2 black balls, and Bag B contains 3 red and 3 black balls. A bag is chosen at random and a ball is drawn from it, which turns out to be red. Using Bayes' theorem, find the probability that the ball was drawn from Bag A.

🎓 Where this leads

Class 12 Maths → JEE / engineering, data science & applied mathematics careers

🧭 Bridge question

Can you explain why a critical point is a maximum or minimum using the second derivative, not just state the final answer?

Questions below are grouped by topic, pick the Math topics you want to practice, or swap in your own worksheet, notebook, or textbook questions any time.

Relations, Functions & Inverse Trigonometry

Covers reflexive/symmetric/transitive relations, one-one and onto functions, and inverse trigonometric values and identities.

📋 Quick referenceRelations & Inverse Trig Quick Rules▶

Reflexive: every element relates to itself. Symmetric: if (a,b) is in R, so is (b,a). Transitive: if (a,b) and (b,c) are in R, so is (a,c). Check all three separately.
A function is one-one if different inputs always give different outputs, and onto if every value in the codomain is actually reached by some input.
For inverse trig values outside the principal range, first evaluate the trig function itself, then find the angle within the principal range that gives the same value.

Test Each Property With a Counter-Example First

For Students

For reflexive/symmetric/transitive questions, try to find one counter-example for each property before concluding it holds, a single missing pair is enough to disprove it.
For inverse trig questions, always double check which principal range applies (sin⁻¹: [-π/2,π/2], cos⁻¹: [0,π], tan⁻¹: (-π/2,π/2)) before picking the final angle.

For Teachers

Grading note: a student who correctly computes an inverse trig value but picks an angle outside the function's principal range should not get full marks, since respecting the range is part of what's being tested.

1. Let A = {1, 2, 3} and R = {(1,1), (2,2), (3,3), (1,2), (2,3)} be a relation defined on A. Check whether R is reflexive, symmetric, and transitive.
▶Reveal answer
Reflexive: yes; Symmetric: no; Transitive: no.
R is reflexive because (1,1), (2,2), (3,3) are all in R. It is not symmetric because (1,2) is in R but (2,1) is not. It is not transitive because (1,2) and (2,3) are in R but (1,3) is not.
2. Show that the function f: R → R defined by f(x) = 2x + 3 is both one-one and onto.
▶Reveal answer
f is one-one and onto (bijective).
One-one: if f(x1) = f(x2), then 2x1 + 3 = 2x2 + 3, giving x1 = x2. Onto: for any y in R, choosing x = (y - 3)/2 gives f(x) = y, so every real y has a pre-image. Hence f is bijective.
3. Find the value of cos^-1(cos(7π/6)).
▶Reveal answer
5π/6
Since 7π/6 lies outside the principal range [0, π] of cos^-1, compute cos(7π/6) = cos(π + π/6) = -cos(π/6) = -√3/2. Then cos^-1(-√3/2) = 5π/6, since 5π/6 is in [0, π] and cos(5π/6) = -√3/2.
4. Using the formula tan^-1 x + tan^-1 y = tan^-1((x + y)/(1 - xy)), find the value of tan^-1(1/2) + tan^-1(1/3).
▶Reveal answer
π/4
Here xy = (1/2)(1/3) = 1/6 < 1, so the formula applies directly: (x + y)/(1 - xy) = (5/6)/(5/6) = 1. So the sum equals tan^-1(1) = π/4.

Matrices & Determinants

Covers matrix operations, transpose, determinant calculation, area via determinants, and solving linear systems using the matrix method.

📋 Quick referenceMatrices & Determinants Quick Rules▶

Matrix addition/subtraction combines entries in the same position; matrix multiplication (AB) multiplies rows of A by columns of B, and requires A's column count to match B's row count.
For a 2x2 matrix [[a,b],[c,d]], determinant = ad - bc.
To solve AX = B by the matrix method, find A⁻¹ (using the determinant and adjugate), then X = A⁻¹B, there is no such thing as 'dividing' by a matrix.

Check the Matrix Is Actually Invertible First

For Students

Before using the matrix method to solve a system, calculate the determinant first, if it's zero, the matrix has no inverse and this method can't be used at all.
When multiplying matrices, write out each row-by-column calculation as its own small sum before combining, rushing this step is the most common source of arithmetic slips.

For Teachers

Common mistake: students confuse matrix multiplication with entry-by-entry multiplication (treating AB like adding/subtracting), especially when the matrices happen to be the same size.

1. If A = [[2, -1], [3, 4]] and B = [[1, 2], [0, -3]], find the matrix 2A - B.
▶Reveal answer
[[3, -4], [6, 11]]
2A = [[4, -2], [6, 8]]. Subtracting B entrywise: [[4-1, -2-2], [6-0, 8-(-3)]] = [[3, -4], [6, 11]].
2. If A = [[1, 2], [3, 4]] and B = [[2, 0], [1, 3]], find the matrix product AB.
▶Reveal answer
[[4, 6], [10, 12]]
Row 1 of AB: (1*2+2*1, 1*0+2*3) = (4, 6). Row 2 of AB: (3*2+4*1, 3*0+4*3) = (10, 12). So AB = [[4, 6], [10, 12]].
3. If A = [[1, 2, 3], [4, 5, 6]], find A^T (the transpose of A) and state its order.
▶Reveal answer
A^T = [[1, 4], [2, 5], [3, 6]], order 3×2.
The transpose is formed by converting rows of A into columns. A is 2×3, so A^T is 3×2, with rows (1,4), (2,5), (3,6).
4. Using determinants, find the area of the triangle whose vertices are (1, 0), (6, 0), and (4, 3).
▶Reveal answer
7.5 square units
Area = (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)| = (1/2)|1(0-3) + 6(3-0) + 4(0-0)| = (1/2)|-3+18+0| = (1/2)(15) = 7.5.
5. Solve the following system of equations using the matrix method: x + 2y = 5 and 3x + y = 5.
▶Reveal answer
x = 1, y = 2
Writing as AX = B with A = [[1,2],[3,1]], det(A) = 1*1 - 2*3 = -5. A^-1 = (1/-5)[[1,-2],[-3,1]]. Then X = A^-1 B = [[-1+2],[3-1]] = [[1],[2]], so x=1, y=2, which checks: 1+2(2)=5 and 3(1)+2=5.

Continuity, Differentiability & Applications of Derivatives

Covers checking continuity and differentiability, differentiation rules (power, product, chain), rates of change, increasing/decreasing intervals, and maxima/minima.

📋 Quick referenceContinuity & Derivatives Quick Rules▶

A function is continuous at a point if the left-hand limit, right-hand limit, and the function's value there are all equal.
A function can be continuous at a point without being differentiable there (like |x| at x=0), but differentiable always implies continuous.
f is increasing where f'(x) > 0 and decreasing where f'(x) < 0. At a local max, f' changes from + to -; at a local min, from - to +.
For related-rates problems, differentiate the connecting formula with respect to time first, then substitute the given numerical values.

Check Differentiability by Comparing Left and Right Derivatives

For Students

To check differentiability at a sharp point (like |x - a|), calculate the left-hand and right-hand derivatives separately, they must be equal for the function to be differentiable there.
For maxima/minima, always confirm a critical point is genuinely a max or min using the second derivative test (or a sign change in f'), don't assume every critical point is one or the other.

For Teachers

Grading note: 'differentiable everywhere except at sharp corners' is a common but insufficiently justified answer, students should show the actual left/right derivative calculation, not just point at the corner visually.

1. Find dy/dx if y = 3x^4 - 5x^3 + 2x - 7.
▶Reveal answer
dy/dx = 12x^3 - 15x^2 + 2
Differentiate term by term using the power rule: d/dx(3x^4)=12x^3, d/dx(-5x^3)=-15x^2, d/dx(2x)=2, d/dx(-7)=0. Adding gives 12x^3 - 15x^2 + 2.
2. Find dy/dx if y = sin(3x^2 + 1), using the chain rule.
▶Reveal answer
dy/dx = 6x cos(3x^2 + 1)
Let u = 3x^2 + 1, so y = sin(u). By the chain rule, dy/dx = cos(u) * du/dx = cos(3x^2+1) * 6x = 6x cos(3x^2+1).
3. The radius of a circle is increasing at a rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
▶Reveal answer
60π cm^2/s
Area A = πr^2, so dA/dt = 2πr (dr/dt). Substituting r=10 and dr/dt=3: dA/dt = 2π(10)(3) = 60π cm^2/s.
4. Find the interval in which the function f(x) = x^2 - 4x + 6 is increasing.
▶Reveal answer
(2, ∞)
f'(x) = 2x - 4. f is increasing where f'(x) > 0, i.e., 2x - 4 > 0, giving x > 2, so f increases on the interval (2, ∞).

Integrals & Their Applications

Covers standard integrals, integration by substitution, definite integrals, and area under a curve.

📋 Quick referenceIntegrals Quick Rules▶

Standard integrals: ∫xⁿdx = xⁿ⁺¹/(n+1) + C (n≠-1), ∫(1/x)dx = ln|x| + C, ∫sec²x dx = tan x + C.
For substitution, spot a function and its derivative appearing together in the integrand, let u equal the inner function, then rewrite entirely in terms of u.
Area under y = f(x) between x = a and x = b (above the x-axis) = the definite integral ∫ₐᵇ f(x) dx.

Differentiate Your Answer to Check It

For Students

After evaluating any indefinite integral, differentiate your answer, if it doesn't give back the original function, a step went wrong.
For substitution integrals, always convert the limits (for a definite integral) or rewrite dx fully in terms of du, don't leave a mix of x and u in the same expression.

For Teachers

Common mistake: forgetting the '+C' on indefinite integrals, and for area problems, forgetting to check whether the curve dips below the x-axis (which would need the area taken as an absolute value in that stretch).

1. Evaluate the integral ∫ (2/x + 3 sec^2 x) dx.
▶Reveal answer
2 ln|x| + 3 tan x + C
Using the standard formulas ∫(1/x)dx = ln|x| + C and ∫sec^2 x dx = tan x + C term by term: ∫2/x dx = 2ln|x|, and ∫3sec^2x dx = 3tan x. Summing with a constant gives 2 ln|x| + 3 tan x + C.
2. Evaluate ∫ 2x e^(x^2) dx using substitution.
▶Reveal answer
e^(x^2) + C
Let u = x^2, so du = 2x dx. The integral becomes ∫e^u du = e^u + C, which substituting back gives e^(x^2) + C.
3. Evaluate the definite integral ∫ from 1 to 2 of (2x + 3) dx.
▶Reveal answer
6
The antiderivative of 2x+3 is x^2+3x. Evaluating from 1 to 2: (2^2+3*2) - (1^2+3*1) = (4+6) - (1+3) = 10 - 4 = 6.

Differential Equations

Covers order and degree, and solving variable-separable differential equations with an initial condition.

📋 Quick referenceDifferential Equations Quick Rules▶

Order = the highest derivative present. Degree = the power of that highest-order derivative, once the equation is written as a polynomial in derivatives.
For a variable-separable equation, rearrange so all y-terms (with dy) are on one side and all x-terms (with dx) are on the other, then integrate both sides.
Use the given initial condition (a known x, y pair) to solve for the constant of integration, giving the particular solution rather than the general one.

Separate Variables Fully Before Integrating

For Students

Before integrating, double-check that every y term (including inside dy) is fully separated from every x term, a mixed term left on the wrong side gives a wrong integral entirely.
Substitute the initial condition immediately after integrating, before doing any further simplification, so the constant is pinned down early.

For Teachers

Grading note: a general solution left without applying the given initial condition is incomplete, since the question specifically asks for the particular solution.

1. Find the order and degree of the differential equation (d^2y/dx^2)^3 + (dy/dx)^2 + y = 0.
▶Reveal answer
Order = 2, Degree = 3
The order is the highest derivative present, which is d^2y/dx^2, giving order 2. The degree is the power to which this highest-order derivative is raised (after the equation is a polynomial in derivatives), which is 3.
2. Solve the differential equation dy/dx = x/y, given that y = 2 when x = 0.
▶Reveal answer
y^2 = x^2 + 4
Separating variables: y dy = x dx. Integrating both sides: y^2/2 = x^2/2 + C. Using y=2 at x=0: 4/2 = 0 + C, so C=2. Thus y^2/2 = x^2/2 + 2, i.e., y^2 = x^2 + 4.
3. Solve the differential equation dy/dx = 3x^2, given that y = 4 when x = 0.
▶Reveal answer
y = x^3 + 4
Integrating both sides directly: y = ∫3x^2 dx = x^3 + C. Using y=4 at x=0: 4 = 0 + C, so C=4. The particular solution is y = x^3 + 4.

Vector Algebra & 3D Geometry

Covers vector addition, dot and cross products, magnitude, direction cosines, equations of a line in 3D, and distance between points.

📋 Quick referenceVectors & 3D Geometry Quick Rules▶

Dot product a·b = a₁b₁+a₂b₂+a₃b₃; it equals zero exactly when the vectors are perpendicular.
Cross product a×b gives a vector perpendicular to both a and b, found using the 3x3 determinant method with i, j, k in the first row.
Magnitude of a vector = √(sum of squares of its components). Direction cosines = each component divided by the magnitude.
Distance between two points in 3D = √((x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²), the direct 3D extension of the 2D distance formula.

Check Perpendicularity With the Dot Product, Not by Eye

For Students

To check if two vectors are perpendicular, calculate the dot product and confirm it's exactly zero, never assume perpendicularity just from how the vectors look.
For line-equation questions, clearly identify the fixed point and the direction vector separately before writing either the vector or Cartesian form.

For Teachers

Common mistake: mixing up the formulas for dot product (a scalar) and cross product (a vector), especially when checking perpendicularity versus parallelism.

1. If vector a = 2i + 3j - k and vector b = i - j + 4k, find a + b.
▶Reveal answer
3i + 2j + 3k
Add corresponding components: (2+1)i + (3-1)j + (-1+4)k = 3i + 2j + 3k.
2. Find the dot product of a = i + 2j + 3k and b = 3i - j + 2k, and state whether a and b are perpendicular.
▶Reveal answer
a·b = 7; they are not perpendicular.
a·b = (1)(3) + (2)(-1) + (3)(2) = 3 - 2 + 6 = 7. Since the dot product is not zero, the vectors are not perpendicular.
3. Find the magnitude of the vector a = 3i - 4j + 12k.
▶Reveal answer
13
Magnitude = sqrt(3^2 + (-4)^2 + 12^2) = sqrt(9+16+144) = sqrt(169) = 13.
4. Find the direction cosines of the vector 2i - 3j + 6k.
▶Reveal answer
(2/7, -3/7, 6/7)
Magnitude = sqrt(2^2+(-3)^2+6^2) = sqrt(4+9+36) = sqrt(49) = 7. Direction cosines are the components divided by the magnitude: 2/7, -3/7, 6/7.
5. Find the vector equation and the Cartesian equation of the line passing through the point (1, 2, 3) and parallel to the vector 2i - j + 2k.
▶Reveal answer
Vector: r = (i + 2j + 3k) + λ(2i - j + 2k); Cartesian: (x-1)/2 = (y-2)/(-1) = (z-3)/2
A line through point (x0,y0,z0) parallel to direction vector (a,b,c) has vector equation r = (x0 i + y0 j + z0 k) + λ(a i + b j + c k) and Cartesian form (x-x0)/a = (y-y0)/b = (z-z0)/c. Substituting the given point and direction gives the stated equations.
6. Find the distance between the points P(1, -2, 3) and Q(4, 2, -1).
▶Reveal answer
sqrt(41)
Distance = sqrt((4-1)^2 + (2-(-2))^2 + (-1-3)^2) = sqrt(9 + 16 + 16) = sqrt(41).

Linear Programming

Covers formulating constraints, identifying corner points of a feasible region, and finding the optimal value of an objective function.

📋 Quick referenceLinear Programming Quick Rules▶

The optimal value (maximum or minimum) of a linear objective function always occurs at a corner point of the feasible region, never in the middle of it.
A corner point is found where two boundary lines of the constraints intersect, solve the two equations simultaneously.
Evaluate the objective function Z at every corner point and compare, don't assume the largest-looking coordinates automatically give the largest Z.

Evaluate Z at Every Corner Point, Don't Guess

For Students

Before calculating the objective function, verify each proposed corner against every inequality; an intersection outside one constraint is not part of the feasible region.
When two constraint lines need to be solved together for a corner point, treat it as a simple simultaneous equations problem, nothing about linear programming changes how that's solved.

For Teachers

Grading note: a correct maximum Z value reached without showing the value at each corner point should not receive full marks, since checking every corner is the method being tested.

1. A factory makes chairs (x) and tables (y), requiring x ≥ 0, y ≥ 0, and 2 hours per chair plus 4 hours per table, with 40 hours available in total. Write this constraint as a linear inequality.
▶Reveal answer
2x + 4y ≤ 40 (or equivalently x + 2y ≤ 20)
The total time used, 2 hours for each chair plus 4 hours for each table, cannot exceed the 40 hours available, giving 2x + 4y ≤ 40, which simplifies by dividing by 2 to x + 2y ≤ 20.
2. The corner points of the feasible region for a linear programming problem are (0, 0), (5, 0), (3, 4), and (0, 5). If Z = 4x + 3y, find the maximum value of Z.
▶Reveal answer
Maximum Z = 24 at (3, 4)
Evaluate Z at each corner point: (0,0) gives 0; (5,0) gives 20; (3,4) gives 4(3)+3(4)=12+12=24; (0,5) gives 15. The maximum value, 24, occurs at (3,4).

Probability

Covers conditional probability, classical probability with dice/cards, Bayes' theorem, and probability distributions with expected value.

📋 Quick referenceProbability Quick Rules▶

Conditional probability: P(A|B) = P(A∩B) / P(B).
Bayes' theorem finds P(cause|observed result) by combining each cause's own probability with how likely that cause was to produce the observed result.
For a probability distribution, all probabilities must sum to 1, use this to solve for an unknown constant first.
Expected value E(X) = the sum of each outcome multiplied by its own probability.

Write Out the Sample Space Before Counting Favourable Outcomes

For Students

For dice or card probability questions, list out the actual favourable outcomes (not just a formula) at least once, this catches miscounted cases like sums on two dice.
For Bayes' theorem questions, organise the given probabilities into a small table (each cause, its own probability, and its probability of producing the result) before combining them.

For Teachers

Common mistake: in probability distribution questions, students calculate E(X) before actually solving for the unknown constant, carrying an unresolved variable through the whole calculation.

1. If P(A) = 0.6, P(B) = 0.4, and P(A∩B) = 0.2, find P(A|B).
▶Reveal answer
0.5
P(A|B) = P(A∩B) / P(B) = 0.2 / 0.4 = 0.5.
2. Two dice are thrown together. Find the probability that the sum of the numbers on the two dice is 8.
▶Reveal answer
5/36
The favourable outcomes summing to 8 are (2,6), (3,5), (4,4), (5,3), (6,2), 5 outcomes out of 36 total equally likely outcomes, giving 5/36.
3. A random variable X has the probability distribution: P(X = 0) = k, P(X = 1) = 2k, P(X = 2) = 3k, P(X = 3) = 4k. Find the value of k and the expected value E(X).
▶Reveal answer
k = 0.1, E(X) = 2
Since probabilities must sum to 1: k+2k+3k+4k=10k=1, so k=0.1. E(X) = 0(k)+1(2k)+2(3k)+3(4k) = 2k+6k+12k = 20k = 20(0.1) = 2.

Where senior secondary students actually lose ground

Numericals across calculus, vectors, and linear programming are usually failed not because of wrong arithmetic but because the wrong method was chosen. Listing what's given and what's asked before picking a formula prevents most of these errors.
Continuity is often confused with differentiability. A function can be continuous at a sharp point without being differentiable there, always check the left-hand and right-hand derivatives separately.
In linear programming, students often guess which corner point gives the optimal value instead of evaluating the objective function at every corner point systematically.
General solutions to differential equations are frequently left without applying the given initial condition, even though the question specifically asks for the particular solution.

Frequently asked questions

How do I help my child get better at Class 12 Calculus numericals?

Before calculating anything, have them state which rule or method the question actually needs (a derivative rule, a substitution, a specific test for maxima/minima) and why, rather than jumping straight to arithmetic. Most mistakes at this level come from applying the wrong method, not from a calculation error.

Why does my child confuse continuity and differentiability?

Ask the student to compare two checks separately: whether nearby function values approach one value, and whether the slopes from each side agree. A graph with a cusp makes the second check visible.

My child gets the right optimal value in linear programming but loses marks. Why?

Have them label each boundary line before graphing and record whether the feasible side is above or below it; a correctly calculated maximum cannot recover a wrongly shaded region.

Explore more for Class 12

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Move between subjects in the same class to build a fuller revision routine.

Math learning ladder — Higher Secondary (Class 11-12)

See where Class 12 Math fits in the higher secondary (class 11-12) years, and jump straight to the next step.

Class 11
Sets, Complex Numbers & a First Taste of Calculus
Class 10 was about applying known formulas quickly under exam pressure; Class 11 introduces genuinely new territory instead, sets, complex numbers, permutations and combinations, and a first, intuitive introduction to limits and derivatives that Class 12 builds calculus on directly.
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Class 12
Full Calculus, Vector Geometry & Applied Optimisation
Class 11 built algebraic and coordinate foundations; Class 12 turns them into full differential and integral calculus, 3D vector geometry, and applied optimisation (linear programming, probability), the exact mix the board exam and engineering entrance exams weight most heavily.

Once you can say why a method applies before working the numbers, board-exam time pressure stops being the main obstacle.

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