Cracking JEE maths comes down to one thing: solving a large number of varied problems until the methods become second nature. The trouble with past papers alone is that you run out of them, and you start remembering answers instead of working them out. Maths Daily Helper generates unlimited fresh JEE-style questions across every topic in the syllabus, so you always have something new to attempt, and every question comes with a complete worked solution so you can check your method, not just your final answer.
What the JEE maths syllabus covers
Your practice here spans the full JEE Main and Advanced maths syllabus:
- Calculus: limits, continuity, differentiation, applications of derivatives, definite and indefinite integration, differential equations, area under curves.
- Coordinate geometry: straight lines, circles, parabola, ellipse, hyperbola.
- Algebra: quadratic equations, sequences and series, binomial theorem, permutations and combinations, complex numbers, matrices and determinants.
- Trigonometry: identities, equations, inverse functions, heights and distances.
- Vectors and 3D geometry.
- Probability and statistics.
Try a few JEE-style problems
Attempt each one on paper first, then reveal the working.
Evaluate the limit: lim (x → 0) (1 − cos x) / x².
1. Use the identity 1 − cos x = 2 sin²(x/2), so the expression becomes 2 sin²(x/2) / x².
2. Rewrite as (1/2) · [ sin(x/2) / (x/2) ]², since x² = 4 · (x/2)².
3. As x → 0, sin(x/2)/(x/2) → 1, so the bracket → 1.
Answer: 1/2.
If z is a complex number satisfying |z − 3| = 2, find the maximum value of |z|.
1. |z − 3| = 2 describes a circle in the Argand plane, centred at 3 (i.e. the point (3, 0)) with radius 2.
2. |z| is the distance from the origin to a point z on that circle.
3. The maximum distance from the origin to a point on the circle is the distance to the centre plus the radius: 3 + 2.
Answer: 5.
Find the equation of the tangent to the parabola y² = 8x at the point (2, 4).
1. For y² = 4ax, here 4a = 8, so a = 2. The tangent at (x₁, y₁) is y·y₁ = 2a(x + x₁).
2. Substitute y₁ = 4, x₁ = 2, 2a = 4: 4y = 4(x + 2).
3. Simplify: y = x + 2, or x − y + 2 = 0.
Answer: x − y + 2 = 0.
Evaluate ∫ x·eˣ dx.
1. Use integration by parts: ∫ u dv = uv − ∫ v du. Let u = x and dv = eˣ dx.
2. Then du = dx and v = eˣ.
3. So ∫ x·eˣ dx = x·eˣ − ∫ eˣ dx = x·eˣ − eˣ + C.
Answer: eˣ(x − 1) + C.
If the roots of x² − px + 12 = 0 differ by 1, find the positive value of p.
1. Let the roots be α and α + 1. Sum = 2α + 1 = p; product = α(α + 1) = 12.
2. Solve α² + α − 12 = 0, which factors as (α + 4)(α − 3) = 0, so α = 3 or α = −4.
3. Taking α = 3 gives p = 2(3) + 1 = 7.
Answer: p = 7.
Two dice are rolled. What is the probability that the sum is 7?
1. Total outcomes when rolling two dice: 6 × 6 = 36.
2. Favourable outcomes summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), that is 6 outcomes.
3. Probability = 6/36 = 1/6.
Answer: 1/6.
Solve for x in [0, 2π): 2 sin x = 1.
1. sin x = 1/2.
2. In [0, 2π), sin x = 1/2 at x = π/6 and x = 5π/6 (first and second quadrants).
Answer: x = π/6 or 5π/6.
Find the angle between the vectors a = i + j and b = j + k.
1. Dot product: a·b = (1)(0) + (1)(1) + (0)(1) = 1.
2. Magnitudes: |a| = √2, |b| = √2, so |a||b| = 2.
3. cos θ = (a·b)/(|a||b|) = 1/2, so θ = 60°.
Answer: 60° (π/3).
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Open Maths Daily Helper →How to use daily practice for JEE
Consistency beats cramming. Fifteen to twenty focused problems a day, spread across topics, keeps every method fresh through the long JEE preparation cycle. Use the hint feature when you are stuck rather than jumping straight to the solution, so you train the habit of finding the next step yourself. Reserve full worked solutions for checking and for the problems that genuinely defeat you. Over weeks, this turns the standard JEE techniques, substitution in integrals, parametric forms of conics, argument and modulus tricks for complex numbers, into reflexes.