Mathematics — Algebra for the JEE Main Exam

Common failure modes

These are the patterns that cause most candidates to lose marks on this topic. Recognising them in advance is half the work.

• !Forgetting that complex cube roots of unity ω satisfy 1 + ω + ω² = 0
• !Misapplying Vieta's formulas (sum/product of roots) when roots are complex
• !Expanding (a+b)^n with incorrect binomial coefficients — especially middle-term errors
• !Row-reducing a matrix to find the inverse without tracking the elementary operations correctly
• !Confusing arithmetic and geometric progression sum formulas under time pressure

Study tips

• 1Memorise the three properties of cube roots of unity: ω³ = 1, 1 + ω + ω² = 0, ω̄ = ω². These reduce half the complex-number JEE problems.
• 2Drill the |z₁ + z₂| triangle inequality and the polar/exponential form of complex numbers — JEE Advanced uses both.
• 3For quadratic theory, practice finding conditions for: (a) both roots positive, (b) roots of opposite sign, (c) imaginary roots. These are standard JEE 'condition' questions.
• 4Matrix determinant: memorise the 3×3 cofactor expansion and practice the shortcut for triangular matrices (product of diagonal).
• 5For sequences, do telescoping-sum problems — they appear every year in JEE Main and look harder than they are.

Sample JEE Main Mathematics — Algebra questions

These sample items mirror the format and difficulty of real JEE Main questions. Practice with thousands more on the free Koydo question bank.

1. 1

   If ω is a complex cube root of unity, the value of (1 + ω − ω²)³ is:

   • A−8Correct
   • B8
   • C1
   • D−1
   Why this answer?

   Illustrative JEE-style: Since 1 + ω + ω² = 0, we have 1 + ω = −ω². So (1 + ω − ω²)³ = (−ω² − ω²)³ = (−2ω²)³ = −8ω⁶ = −8(ω³)² = −8(1)² = −8.

2. 2

   The sum of the first n terms of the series 1 + 3 + 5 + 7 + ... is:

   • An(n+1)/2
   • Bn²Correct
   • Cn(n+1)
   • D2n−1
   Why this answer?

   Illustrative JEE-style: This is an arithmetic progression with first term 1 and common difference 2. Sₙ = n/2 × [2(1) + (n−1)(2)] = n/2 × 2n = n².

3. 3

   The number of terms in the expansion of (x + y)^10 is:

   • A10
   • B11Correct
   • C20
   • D9
   Why this answer?

   Illustrative JEE-style: The binomial expansion (x + y)^n has (n + 1) terms. For n = 10, there are 11 terms.