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IIT JEE Maths Practice Test Online

JEE Maths punishes hesitation more than ignorance — most drops happen not because a concept is unknown, but because the algebra under pressure takes too long. This set is built to fix exactly that.

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About this IIT JEE Math practice test

Calculus and Coordinate Geometry alone account for a huge share of the JEE Maths paper, but what actually separates a 99-percentiler from an average scorer is speed on the messier chapters — matrices, permutations, sequences — that most students under-practice because they feel 'easy'. This set deliberately leans into those chapters alongside the heavy hitters, mixing single-correct, integer-type and multi-step problems the way Mains and Advanced actually do. Every solution below shows the full working, not just the final answer, so you can see exactly where a faster method existed. Once this feels comfortable, paste in a previous year's paper and time yourself against the real clock.

IIT JEE Maths Practice Test sample questions

These starter questions help you launch a math mock test quickly. Swap them with your own worksheet, notebook, or textbook questions any time.

  1. 1. Find the domain of f(x) = √(log_{1/2}((x−1)/(x+5))).

  2. 2. If z is a complex number such that |z − 3 + 2i| ≤ 4, find the maximum value of |z|.

  3. 3. If α and β are the roots of x² − 2x + 4 = 0, find the value of α⁷ + β⁷.

  4. 4. Using the Cayley–Hamilton theorem, find A⁻¹ for A = [[1, 2], [2, 1]].

  5. 5. For what value of k does the system x + y + z = 6, x + 2y + 3z = 10, x + 2y + kz = 12 have no solution?

  6. 6. Find the number of ways of distributing 10 identical balls into 4 distinct boxes so that no box is empty.

  7. 7. In how many ways can the letters of the word MISSISSIPPI be arranged so that the four I's do not all come together?

  8. 8. Find the term independent of x in the expansion of (3x²/2 − 1/(3x))⁹.

  9. 9. If the sum of the first n terms of an AP is Sₙ = 3n² + 5n, find the 15th term.

  10. 10. Find the sum to n terms of the series 1·2·3 + 2·3·4 + 3·4·5 + ... .

  11. 11. Evaluate lim(x→0) (eˣ − e⁻ˣ − 2x)/(x − sinx).

  12. 12. Let f(x) = (1 − cos4x)/x² for x < 0, f(0) = a, and f(x) = √x/(√(16+√x) − 4) for x > 0. If f is continuous at x = 0, find a.

  13. 13. If lim(x→∞) [ (x²+1)/(x+1) − ax − b ] = 0, find the value of a + b.

  14. 14. Find the equation of the tangent to the curve y = x³ − 3x + 2 at the point where the curve crosses the y-axis.

  15. 15. A 13 m ladder leans against a wall; its bottom is pulled away from the wall at 2 m/s. How fast is the top sliding down the wall when the bottom is 5 m from the wall?

  16. 16. Find the maximum value of f(x) = x³ − 12x on the interval [−3, 5].

  17. 17. Evaluate ∫₀^{π/4} ln(1 + tanx) dx.

  18. 18. Evaluate ∫ eˣ [ (x ln x + 1)/x ] dx.

  19. 19. Evaluate ∫₀^{π} x sinx/(1 + cos²x) dx.

  20. 20. Solve dy/dx + y/x = x² given y(1) = 1. Find y(2).

  21. 21. Find the equation of the line through the point of intersection of x + 2y − 3 = 0 and 2x − y + 1 = 0, perpendicular to 3x + 4y − 5 = 0.

  22. 22. Find the equation of the circle passing through (0,0), (2,0), and (0,4).

  23. 23. Find the length of the tangent from the point (5,5) to the circle x² + y² − 2x − 4y − 4 = 0.

  24. 24. Find the equation of the parabola with vertex at the origin, axis along the x-axis, passing through the point (2,−4).

  25. 25. Find the shortest distance between the lines (x−1)/2 = (y−2)/3 = (z−3)/4 and (x−2)/3 = (y−4)/4 = (z−5)/5.

  26. 26. Find the equation of the plane passing through the point (1,2,3) and perpendicular to the line joining the points (3,4,1) and (2,−1,5).

  27. 27. If a = 2i + 3j − k and b = i − 2j + 2k, find a unit vector perpendicular to both a and b.

  28. 28. Find λ such that the vectors a = 2i − j + k, b = i + 2j − 3k, and c = 3i + λj + 5k are coplanar.

  29. 29. The mean and variance of 8 observations are 10 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12, 13, find the remaining two observations.

  30. 30. If cosA = 3/5 and cosB = 5/13, where A and B are acute angles, find the value of cos(A − B).

Syllabus & Core Topics

calculuscoordinate geometryvectors and 3Dalgebratrigonometry

Once these thirty feel comfortable, cross-link calculus with coordinate geometry — Advanced loves combining the two in a single question. Drill the King's-rule substitution (x → a−x) until it's automatic, since it resurfaces in definite integrals almost every year.

Why this practice page is useful

  • Use this IIT JEE Maths page to drill the high-weightage topics that decide your Mains percentile and Advanced rank.

  • The starter set mixes single-correct, multi-correct and integer-type patterns so practice feels close to the real paper.

  • Replace the starter with previous-year JEE questions to generate a fresh similarly-styled mock with worked answers.

Answer key & quick explanations

Short answers for the sample questions above. Use this to self-check before generating a fresh AI-built mock test.

  1. 1. Find the domain of f(x) = √(log_{1/2}((x−1)/(x+5))).

    (1, ∞)

    Since log base 1/2 is a decreasing function, the expression under the root is non-negative only when 0 < (x−1)/(x+5) ≤ 1. The first part forces x < −5 or x > 1, while simplifying the second gives x > −5, so the two conditions overlap only for x > 1.

  2. 2. If z is a complex number such that |z − 3 + 2i| ≤ 4, find the maximum value of |z|.

    √13 + 4

    The inequality describes a closed disc of radius 4 centred at (3,−2). The farthest a point z on or inside this disc can be from the origin equals the distance from the origin to the centre plus the radius, i.e., √(3²+2²)+4.

  3. 3. If α and β are the roots of x² − 2x + 4 = 0, find the value of α⁷ + β⁷.

    128

    The roots 1±i√3 have modulus 2 and argument ±60°, so α=2cis60° and β=2cis(−60°). Then α⁷+β⁷ = 2·2⁷cos(7·60°) = 256cos420° = 256cos60° = 128.

  4. 4. Using the Cayley–Hamilton theorem, find A⁻¹ for A = [[1, 2], [2, 1]].

    A⁻¹ = (1/3)[[−1, 2], [2, −1]]

    By Cayley-Hamilton, A satisfies its own characteristic equation A²−2A−3I=0, so A²−2A=3I. Multiplying through by A⁻¹ gives A−2I=3A⁻¹, and rearranging yields A⁻¹=(A−2I)/3.

  5. 5. For what value of k does the system x + y + z = 6, x + 2y + 3z = 10, x + 2y + kz = 12 have no solution?

    k = 3

    The coefficient determinant of the system equals k−3, so the system loses a unique solution exactly when k=3. Substituting k=3 turns the third equation into x+2y+3z=12, which directly contradicts the second equation x+2y+3z=10, confirming no solution rather than infinitely many.

  6. 6. Find the number of ways of distributing 10 identical balls into 4 distinct boxes so that no box is empty.

    84

    Distributing n identical items into r distinct boxes with none empty is a standard stars-and-bars count of C(n−1, r−1). Here n=10 and r=4, giving C(9,3)=84.

  7. 7. In how many ways can the letters of the word MISSISSIPPI be arranged so that the four I's do not all come together?

    33810

    MISSISSIPPI has 11 letters with M once, I four times, S four times and P twice, giving 11!/(4!4!2!)=34650 total arrangements. Gluing all four I's into a single block leaves 8 units to arrange with the repeated S's and P's, giving 8!/(4!2!)=840, so subtracting leaves 34650−840=33810.

  8. 8. Find the term independent of x in the expansion of (3x²/2 − 1/(3x))⁹.

    7/18

    The general term is T_{r+1}=C(9,r)(3x²/2)^{9−r}(−1/(3x))^r, whose power of x works out to 18−3r. Setting this to zero gives r=6, and substituting back gives C(9,6)(3/2)³(1/3)⁶ = 84·(27/8)·(1/729) = 7/18.

  9. 9. If the sum of the first n terms of an AP is Sₙ = 3n² + 5n, find the 15th term.

    92

    The nth term of any series is Tn=Sn−S(n−1). Working this out for Sn=3n²+5n gives Tn=6n+2, so T15=6(15)+2=92.

  10. 10. Find the sum to n terms of the series 1·2·3 + 2·3·4 + 3·4·5 + ... .

    n(n+1)(n+2)(n+3)/4

    The rth term of the series is r(r+1)(r+2), which can be written as one-fourth of the telescoping difference r(r+1)(r+2)(r+3) − (r−1)r(r+1)(r+2). Summing this telescoping form from r=1 to n collapses to n(n+1)(n+2)(n+3)/4, which checks out for n=1 (giving 6) and n=2 (giving 30).

  11. 11. Evaluate lim(x→0) (eˣ − e⁻ˣ − 2x)/(x − sinx).

    2

    Expanding both the numerator and denominator as Taylor series around x=0 gives eˣ−e⁻ˣ−2x ≈ x³/3 and x−sinx ≈ x³/6. Dividing these leading terms gives the limit (x³/3)/(x³/6)=2.

  12. 12. Let f(x) = (1 − cos4x)/x² for x < 0, f(0) = a, and f(x) = √x/(√(16+√x) − 4) for x > 0. If f is continuous at x = 0, find a.

    a = 8

    For x<0, (1−cos4x)/x² = 2sin²(2x)/x² → 8 as x→0 using sinθ/θ→1. For x>0, rationalising √x/(√(16+√x)−4) by multiplying with (√(16+√x)+4) collapses the expression to √(16+√x)+4, which also tends to 8. Continuity at 0 therefore forces a=8.

  13. 13. If lim(x→∞) [ (x²+1)/(x+1) − ax − b ] = 0, find the value of a + b.

    0

    Dividing gives (x²+1)/(x+1) = x−1+2/(x+1), so the bracket becomes (1−a)x+(−1−b)+2/(x+1). For the limit at infinity to vanish, the coefficient of x must be zero, giving a=1, and the remaining constant then forces b=−1, so a+b=0.

  14. 14. Find the equation of the tangent to the curve y = x³ − 3x + 2 at the point where the curve crosses the y-axis.

    3x + y − 2 = 0

    The curve meets the y-axis where x=0, giving the point (0,2). The derivative y'=3x²−3 equals −3 there, so the tangent is y−2=−3(x−0), which rearranges to 3x+y−2=0.

  15. 15. A 13 m ladder leans against a wall; its bottom is pulled away from the wall at 2 m/s. How fast is the top sliding down the wall when the bottom is 5 m from the wall?

    5/6 m/s

    With x and y as the horizontal and vertical distances, x²+y²=13²=169, and at x=5 this gives y=12. Differentiating with respect to time, 2x(dx/dt)+2y(dy/dt)=0, so dy/dt=−(x/y)(dx/dt)=−(5/12)(2)=−5/6, meaning the top slides down at 5/6 m/s.

  16. 16. Find the maximum value of f(x) = x³ − 12x on the interval [−3, 5].

    65

    Setting f'(x)=3x²−12=0 gives critical points x=±2, both inside [−3,5]. Comparing f(−3)=9, f(−2)=16, f(2)=−16 and the endpoint f(5)=65 shows the maximum occurs at the right endpoint, x=5.

  17. 17. Evaluate ∫₀^{π/4} ln(1 + tanx) dx.

    (π/8) ln 2

    Replacing x by π/4−x inside the integral turns tanx into (1−tanx)/(1+tanx), and 1 plus that expression simplifies to 2/(1+tanx). This gives I = (π/4)ln2 − I, so 2I=(π/4)ln2 and I=(π/8)ln2.

  18. 18. Evaluate ∫ eˣ [ (x ln x + 1)/x ] dx.

    eˣ ln x + C

    Writing the integrand as eˣ[ln x + 1/x] matches the pattern ∫eˣ[f(x)+f'(x)]dx with f(x)=ln x. That standard identity integrates directly to eˣf(x)+C, giving eˣ ln x + C.

  19. 19. Evaluate ∫₀^{π} x sinx/(1 + cos²x) dx.

    π²/4

    Replacing x by π−x leaves sinx and cos²x unchanged, so I = π∫₀^π sinx/(1+cos²x)dx − I, giving 2I = π·J where J=∫₀^π sinx/(1+cos²x)dx. Substituting u=cosx converts J into ∫_{−1}^{1} du/(1+u²) = π/2, so 2I = π·(π/2) = π²/2 and I = π²/4.

  20. 20. Solve dy/dx + y/x = x² given y(1) = 1. Find y(2).

    y(2) = 19/8

    This is a linear first-order ODE with integrating factor x, since ∫(1/x)dx=ln x. Multiplying through gives d/dx(xy)=x³, and integrating yields xy=x⁴/4+C; using y(1)=1 fixes C=3/4, so y=x³/4+3/(4x) and y(2)=2+3/8=19/8.

  21. 21. Find the equation of the line through the point of intersection of x + 2y − 3 = 0 and 2x − y + 1 = 0, perpendicular to 3x + 4y − 5 = 0.

    20x − 15y + 17 = 0

    Solving x+2y=3 and 2x−y=−1 simultaneously gives the intersection point (1/5, 7/5). Since 3x+4y−5=0 has slope −3/4, the perpendicular line through this point has slope 4/3, and writing y−7/5=(4/3)(x−1/5) and clearing fractions gives 20x−15y+17=0.

  22. 22. Find the equation of the circle passing through (0,0), (2,0), and (0,4).

    x² + y² − 2x − 4y = 0

    Taking the general circle x²+y²+Dx+Ey+F=0 and substituting (0,0) gives F=0. Substituting (2,0) and (0,4) then give D=−2 and E=−4 respectively, so the circle is x²+y²−2x−4y=0.

  23. 23. Find the length of the tangent from the point (5,5) to the circle x² + y² − 2x − 4y − 4 = 0.

    4

    The length of the tangent from an external point (x1,y1) to a circle x²+y²+Dx+Ey+F=0 is √(x1²+y1²+Dx1+Ey1+F). Substituting (5,5) gives √(25+25−10−20−4)=√16=4.

  24. 24. Find the equation of the parabola with vertex at the origin, axis along the x-axis, passing through the point (2,−4).

    y² = 8x

    A parabola with vertex at the origin and axis along the x-axis has the form y²=4ax. Plugging in the point (2,−4) gives 16=8a, so a=2 and the equation is y²=8x.

  25. 25. Find the shortest distance between the lines (x−1)/2 = (y−2)/3 = (z−3)/4 and (x−2)/3 = (y−4)/4 = (z−5)/5.

    1/√6

    The two lines have direction vectors (2,3,4) and (3,4,5), whose cross product is (−1,2,−1) with magnitude √6. The vector joining the given points on each line is (1,2,2), and its dot product with the cross product is 1, so the shortest distance is 1/√6.

  26. 26. Find the equation of the plane passing through the point (1,2,3) and perpendicular to the line joining the points (3,4,1) and (2,−1,5).

    x + 5y − 4z + 1 = 0

    Since the plane is perpendicular to the line joining (3,4,1) and (2,−1,5), the line's direction vector (−1,−5,4) serves as the plane's normal. Using the point (1,2,3), the plane is −1(x−1)−5(y−2)+4(z−3)=0, which simplifies to x+5y−4z+1=0.

  27. 27. If a = 2i + 3j − k and b = i − 2j + 2k, find a unit vector perpendicular to both a and b.

    (4i − 5j − 7k)/(3√10)

    A vector perpendicular to both a and b is given by their cross product, a×b=(4,−5,−7). Its magnitude is √(16+25+49)=√90=3√10, so dividing by this gives the unit vector (4i−5j−7k)/(3√10).

  28. 28. Find λ such that the vectors a = 2i − j + k, b = i + 2j − 3k, and c = 3i + λj + 5k are coplanar.

    λ = −4

    Three vectors are coplanar exactly when their scalar triple product vanishes. Expanding the determinant with rows (2,−1,1), (1,2,−3) and (3,λ,5) gives 7λ+28=0, so λ=−4.

  29. 29. The mean and variance of 8 observations are 10 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12, 13, find the remaining two observations.

    6 and 14

    With the sum fixed at 80 by the mean, the six known observations sum to 60, so the remaining two must sum to 20. Using the variance to fix Σxi²=874 leaves x²+y²=232 for the unknowns, and solving x+y=20 together with x²+y²=232 gives the pair 6 and 14.

  30. 30. If cosA = 3/5 and cosB = 5/13, where A and B are acute angles, find the value of cos(A − B).

    63/65

    Since A and B are acute, sinA=4/5 and sinB=12/13 follow from the Pythagorean triples 3-4-5 and 5-12-13. Then cos(A−B)=cosAcosB+sinAsinB=(3/5)(5/13)+(4/5)(12/13)=(15+48)/65=63/65.

Curriculum Mapping & Learning Guide

Use this breakdown to identify which skills each question tests and guide post-test review.

Algebra Foundations (Questions 1-10)

Covers function domains, complex-number geometry, matrix inversion via Cayley-Hamilton, a determinant-based consistency test for a linear system, restricted-arrangement counting, a binomial coefficient extraction, and AP/series summation formulas.

Calculus Core (Questions 11-20)

Works through indeterminate-form limits via series expansion, a two-sided continuity check, tangent-line/related-rate/optimization applications, definite integrals solved with the King's-rule substitution, an exact-derivative indefinite integral, and a linear first-order differential equation.

Geometry, Vectors and Applied Topics (Questions 21-30)

Spans 2D coordinate geometry (lines, a circle, tangent length, a parabola), 3D shortest-distance and plane equations, cross-product and scalar-triple-product vector problems, a reverse mean-variance statistics problem, and a compound-angle trigonometry identity.

IIT JEE Math units covered

  1. Chapter 1: Sets, Relations and Functions
  2. Chapter 2: Complex Numbers and Quadratic Equations
  3. Chapter 3: Matrices and Determinants
  4. Chapter 4: Permutations and Combinations
  5. Chapter 5: Binomial Theorem
  6. Chapter 6: Sequences and Series
  7. Chapter 7: Limits, Continuity and Differentiability
  8. Chapter 8: Application of Derivatives
  9. Chapter 9: Indefinite and Definite Integrals
  10. Chapter 10: Differential Equations
  11. Chapter 11: Coordinate Geometry (Straight Lines, Circles, Conics)
  12. Chapter 12: Three Dimensional Geometry
  13. Chapter 13: Vector Algebra
  14. Chapter 14: Statistics
  15. Chapter 15: Trigonometry

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