⚡ Quick quiz

Test yourself in under a minute: IIT JEE Maths Practice Test

Ten questions pulled from calculus, algebra, vectors and coordinate geometry test how fast you can spot the right technique, not just whether you know the formula.

1. 1. Evaluate lim(x→0) (1 − cosx)/x².

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2. 2. If the sum of the first n terms of an AP is Sn = 4n² + n, find the 10th term.

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3. 3. If a = i + j and b = j + k, find |a × b|.

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4. 4. Find the equation of the directrix of the parabola y² = 12x.

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5. 5. If sinθ = 1/2 and θ is acute, find cos2θ.

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6. 6. A die is rolled twice. Find the probability that the sum of the two numbers is 8.

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Calculus

Covers indeterminate-form limits, two-sided continuity and differentiability checks, tangent-line and related-rate/optimization applications, definite integration via the King's-rule substitution, an exact-derivative indefinite integral, and area under a curve alongside a linear first-order differential equation.

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

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   ▶Reveal answer

   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.

2. 2. Evaluate lim(x→0) (tanx − sinx)/x³.

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   ▶Reveal answer

   1/2🔊

   Writing tanx−sinx = sinx(1−cosx)/cosx and substituting the small-angle approximations sinx≈x and 1−cosx≈x²/2 with cosx→1 gives tanx−sinx ≈ x·(x²/2) = x³/2. Dividing by x³ leaves the limit 1/2.

3. 3. 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.

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   ▶Reveal answer

   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.

4. 4. Show whether f(x) = x|x| is differentiable at x = 0, and find f'(0) if it exists.

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   ▶Reveal answer

   Yes, f is differentiable at x=0, and f'(0)=0🔊

   Since f(x)=x² for x≥0 and f(x)=−x² for x<0, both one-sided derivatives at 0 equal lim(h→0) |h|·h/h = lim|h| = 0. Because the left-hand and right-hand derivatives agree at 0, f is differentiable there with f'(0)=0.

5. 5. Find the equation of the tangent to the curve y = x³ − 3x + 2 at the point (2, 4).

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   ▶Reveal answer

   y = 9x − 14🔊

   Differentiating gives y'=3x²−3, which at x=2 equals 3(4)−3=9, the slope of the tangent. Using the point-slope form y−4=9(x−2) and simplifying gives y=9x−14.

6. 6. A 10 m long ladder is sliding down a vertical wall, with its foot moving away from the wall at 2 m/s. Find the rate at which the top of the ladder is sliding down when the foot is 6 m from the wall.

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   ▶Reveal answer

   1.5 m/s🔊

   Let x and y be the foot's distance from the wall and the top's height, so x²+y²=100. When x=6, y=8 (a 6-8-10 triangle). Differentiating gives 2x(dx/dt)+2y(dy/dt)=0, so dy/dt=−x(dx/dt)/y=−(6)(2)/8=−1.5, meaning the top slides down at 1.5 m/s.

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

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   ▶Reveal answer

   (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.

8. 8. Evaluate ∫ (x cosx − sinx)/x² dx.

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   ▶Reveal answer

   sinx/x + C🔊

   By the quotient rule, d/dx(sinx/x) = (x·cosx − sinx·1)/x², which is exactly the integrand. So the integral is simply an application of recognising the derivative already worked out, giving sinx/x + C.

9. 9. Find the area enclosed between the curves y = x and y = x² for x between 0 and 1.

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   ▶Reveal answer

   1/6 square units🔊

   On [0,1], the line y=x lies above the parabola y=x², so the area is ∫₀¹(x−x²)dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.

10. 10. Solve the differential equation dy/dx + y = eˣ.

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    ▶Reveal answer

    y = eˣ/2 + Ce⁻ˣ🔊

    This is linear with integrating factor e^(∫1 dx)=eˣ. Multiplying through gives d/dx(y·eˣ)=e^(2x), and integrating gives y·eˣ = e^(2x)/2 + C, so y = eˣ/2 + Ce⁻ˣ.

11. 11. A farmer wants to fence a rectangular field of area 200 m² using an existing wall as one side. Find the dimensions that minimize the total length of fencing needed for the remaining three sides.

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    ▶Reveal answer

    20 m (parallel to the wall) by 10 m, using 40 m of fencing🔊

    Let x be the side parallel to the wall and y the two perpendicular sides, with xy=200 so y=200/x. The fence length is L=x+2y=x+400/x. Setting dL/dx=1−400/x²=0 gives x=20, so y=10, and the minimum fencing needed is 20+2(10)=40 m.

Algebra

Covers matrix inversion via Cayley-Hamilton, a determinant-based consistency test for a linear system, AP/series summation from Sn, complex-number powers via De Moivre's theorem, restricted-arrangement counting, and a binomial coefficient extraction.

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

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   ▶Reveal answer

   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.

2. 2. Using the determinant test, find the value of λ for which the system x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = 12 has no solution.

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   ▶Reveal answer

   λ = 3🔊

   The coefficient determinant |1 1 1; 1 2 3; 1 2 λ| expands to (2λ−6)−(λ−3)+(2−2)=λ−3, which vanishes at λ=3. At λ=3, the third equation becomes identical in its left side to the second equation but with a different constant (12 vs 10), an outright contradiction, so the system has no solution at λ=3.

3. 3. If the sum of the first n terms of a series is Sn = 3n² + 5n, find its 15th term.

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   ▶Reveal answer

   92🔊

   The nth term is Tn=Sn−S(n−1). Expanding Sn=3n²+5n gives Tn=6n+2 after subtracting S(n−1)=3(n−1)²+5(n−1). Substituting n=15 gives T15=6(15)+2=92.

4. 4. Find the sum to infinity of the geometric progression 8, 4, 2, 1, ...

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   ▶Reveal answer

   16🔊

   The common ratio is r=4/8=1/2, and since |r|<1 the sum to infinity is S=a/(1−r)=8/(1−0.5)=8/0.5=16.

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

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   ▶Reveal answer

   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.

6. 6. Solve z³ = −8 for all complex roots z.

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   ▶Reveal answer

   z = −2, 1+i√3, and 1−i√3🔊

   Writing −8 in polar form as 8·cis(180°), the three cube roots are 2·cis(60°+120°k) for k=0,1,2. This gives 2cis60°=1+i√3, 2cis180°=−2, and 2cis300°=1−i√3.

7. 7. How many 4-letter arrangements (without repetition) can be formed from the letters of the word NUMBER such that the word always starts with N?

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   ▶Reveal answer

   60🔊

   Fixing N in the first position leaves 3 more positions to fill from the remaining 5 distinct letters (U, M, B, E, R), which can be arranged in 5P3 = 5×4×3 = 60 ways.

8. 8. In how many ways can a committee of 5 be chosen from 6 men and 4 women so that the committee includes at least 3 women?

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   ▶Reveal answer

   66🔊

   At least 3 women means either 3 women and 2 men, or 4 women and 1 man. The first case gives C(4,3)×C(6,2)=4×15=60 ways, and the second gives C(4,4)×C(6,1)=1×6=6 ways. Adding these gives 60+6=66 total committees.

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

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   ▶Reveal answer

   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.

10. 10. Find the middle term in the expansion of (x + 1/x)⁸.

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    ▶Reveal answer

    70🔊

    Since the exponent 8 is even, the expansion has 9 terms and the middle one is the 5th, T5=T_{4+1}=C(8,4)x^{8−4}(1/x)^4=C(8,4)=70. All powers of x cancel exactly at the middle term.

Vectors & 3D Geometry

Covers vector algebra with dot and cross products, a unit-vector-perpendicular-to-two-vectors problem, coplanarity via the scalar triple product, shortest distance between skew lines, and the equation of a plane from a point and a direction vector.

1. 1. If a = 3i − j + 2k and b = i + 2j − k, find a·b and a×b.

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   ▶Reveal answer

   a·b = −1; a×b = −3i + 5j + 7k🔊

   The dot product is (3)(1)+(−1)(2)+(2)(−1)=3−2−2=−1. Expanding the determinant for the cross product with rows (3,−1,2) and (1,2,−1) gives i[(−1)(−1)−(2)(2)]−j[(3)(−1)−(2)(1)]+k[(3)(2)−(−1)(1)] = −3i+5j+7k.

2. 2. Find a unit vector perpendicular to both a = i + j + k and b = 2i − j + 3k.

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   ▶Reveal answer

   (4i − j − 3k)/√26🔊

   The cross product a×b, expanded from rows (1,1,1) and (2,−1,3), gives i(1·3−1·(−1))−j(1·3−1·2)+k(1·(−1)−1·2) = 4i−j−3k. Its magnitude is √(16+1+9)=√26, so dividing by this gives the unit vector (4i−j−3k)/√26.

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

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   ▶Reveal answer

   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.

4. 4. 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.

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   ▶Reveal answer

   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.

5. 5. Find the equation of the plane passing through the point (1, 2, 3) with normal vector 2i − j + 2k.

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   ▶Reveal answer

   2x − y + 2z = 6🔊

   The plane equation is n·(r−r0)=0, so 2(x−1)−1(y−2)+2(z−3)=0. Expanding gives 2x−2−y+2+2z−6=0, which simplifies to 2x−y+2z=6.

Coordinate Geometry

Covers straight lines through the intersection of two given lines, a circle through three points, the tangent-length formula, and the equation of a parabola (or ellipse/hyperbola) from a vertex condition and one known point.

1. 1. Find the equation of the line through the intersection of x + 2y − 3 = 0 and 2x − y + 1 = 0 that also passes through the point (2, 3).

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   ▶Reveal answer

   8x − 9y + 11 = 0🔊

   Any line through the intersection can be written as (x+2y−3)+λ(2x−y+1)=0. Substituting (2,3) gives (2+6−3)+λ(4−3+1)=5+2λ=0, so λ=−5/2. Substituting back and clearing fractions gives 8x−9y+11=0.

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

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   ▶Reveal answer

   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.

3. 3. Find the length of the tangent from the point (7, 4) to the circle x² + y² − 6x − 8y + 21 = 0.

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   ▶Reveal answer

   2√3🔊

   The tangent length from (x1,y1) to x²+y²+Dx+Ey+F=0 is √(x1²+y1²+Dx1+Ey1+F). Substituting (7,4) gives 49+16−42−32+21=12, and the square root of 12 is 2√3.

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

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   ▶Reveal answer

   y² = 8x🔊

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

5. 5. An ellipse is centred at the origin with semi-major axis 5 along the x-axis and semi-minor axis 3 along the y-axis. Write its standard equation and find its eccentricity.

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   ▶Reveal answer

   x²/25 + y²/9 = 1; eccentricity e = 4/5🔊

   With a=5 and b=3, the standard equation is x²/25+y²/9=1. Eccentricity follows from e=√(1−b²/a²)=√(1−9/25)=√(16/25)=4/5.

Trigonometry, Sets & Relations

Covers a compound-angle trigonometric identity built from two given cosine values, and a sets-and-relations check for whether a given relation is an equivalence relation.

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

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   ▶Reveal answer

   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.

2. 2. If sinA = 8/17 and cosB = 3/5, where A and B are acute angles, find the value of sin(A + B).

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   ▶Reveal answer

   84/85🔊

   Since A and B are acute, cosA=15/17 (from the 8-15-17 triple) and sinB=4/5 (from the 3-4-5 triple). Then sin(A+B)=sinAcosB+cosAsinB=(8/17)(3/5)+(15/17)(4/5)=24/85+60/85=84/85.

3. 3. Let R be the relation on the set of integers Z defined by aRb if and only if (a − b) is divisible by 5. Show that R is an equivalence relation.

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   ▶Reveal answer

   R is reflexive, symmetric, and transitive, so it is an equivalence relation🔊

   R is reflexive since a−a=0 is divisible by 5 for every integer a. It is symmetric because if 5 divides (a−b), it also divides −(a−b)=(b−a). It is transitive because if 5 divides (a−b) and (b−c), their sum (a−c) is also divisible by 5. Satisfying all three properties makes R an equivalence relation.

Probability & Statistics

Covers a reverse mean-variance problem where two observations must be reconstructed from a given mean and variance, and a conditional-probability calculation.

1. 1. 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.

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   ▶Reveal answer

   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.

2. 2. A bag contains 5 red and 3 blue balls. Two balls are drawn one after another without replacement. Given that the first ball drawn is red, find the probability that the second ball drawn is also red.

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   ▶Reveal answer

   4/7🔊

   Since the first ball drawn is red and is not replaced, only 4 red balls remain out of a total of 7 balls left in the bag. So P(second red | first red) = 4/7, a direct application of conditional probability without needing Bayes' theorem here.