JEE (Main) · Mathematics, Class 11
Binomial Theorem: JEE Main PYQ, 2020–2025
Binomial Theorem (Mathematics, Class 11) has been asked 145 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.3 questions a paper, 5% of all Mathematics questions. It is the 12th most-asked of 19 Mathematics chapters. The 2025 shifts had 21 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 16in 14 papers |
| 2021 | 36in 26 papers |
| 2022 | 29in 24 papers |
| 2023 | 24in 12 papers |
| 2024 | 19in 18 papers |
| 2025 | 21in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- General and middle term76
- Simple applications67
- Binomial theorem for a positive integral index2
Try it
10 Binomial Theorem questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Mathematics · Binomial Theorem
Which of the following is the simplified form of (x + 1)^4 + (x − 1)^4?
Answer: (A)
(x + 1)^4 = x⁴ + 4x³ + 6x² + 4x + 1 and (x − 1)^4 = x⁴ − 4x³ + 6x² − 4x + 1. Adding them, the odd-power terms 4x³ and 4x cancel, and the rest double. The result is 2x⁴ + 12x² + 2.
Q2 · Mathematics · Binomial Theorem
In the expansion of (2 + x/3)^n, the coefficients of x^7 and x^8 are equal. What is the value of n?
Answer: (A)
The coefficient of x^r is C(n, r)·2^(n−r)·(1/3)^r. Setting the x^7 and x^8 coefficients equal: C(n,7)·2^(n−7)/3^7 = C(n,8)·2^(n−8)/3^8. Dividing, C(n,8)/C(n,7) = 2·3 = 6. Since C(n,8)/C(n,7) = (n − 7)/8, we get n − 7 = 48, so n = 55.
Q3 · Mathematics · Binomial Theorem
Find the term independent of x in the expansion of (1 + x + 2x³)·((3/2)x² − 1/(3x))⁹.
Answer: (A)
In the second factor, the general term is C(9, r)(3/2)^(9−r)(−1/3)^r·x^(18−3r). The factor 1 needs x⁰, so 18 − 3r = 0 and r = 6. This gives C(9,6)(3/2)³(1/3)⁶ = 84·(27/8)·(1/729) = 7/18. The factor x needs x⁻¹, so 18 − 3r = −1, which has no integer solution. The factor 2x³ needs x⁻³, so r = 7. This gives 2·C(9,7)(3/2)²(−1/3)⁷ = 2·36·(9/4)·(−1/2187) = −2/27. The total is 7/18 − 2/27 = 21/54 − 4/54 = 17/54.
Q4 · Mathematics · Binomial Theorem
What is the remainder when 11⁹ is divided by 10?
Answer: (A)
Write 11⁹ = (10 + 1)⁹ = C(9,0)10⁹ + C(9,1)10⁸ + … + C(9,8)10 + C(9,9). Every term except the last has a factor of 10. The last term is C(9,9)·1 = 1. Hence 11⁹ = 10k + 1 and the remainder on division by 10 is 1.
Q5 · Mathematics · Binomial Theorem
What is the coefficient of x^5 in the expansion of (1+x)^5 + (1+x)^6 + (1+x)^7 + (1+x)^8 + (1+x)^9 + (1+x)^10?
Answer: (D)
The coefficient of x^5 in (1+x)^k is C(k,5). Add over k = 5 to 10: 1 + 6 + 21 + 56 + 126 + 252 = 462. This agrees with the identity C(5,5)+C(6,5)+…+C(10,5) = C(11,6) = 462. The value 252 counts only (1+x)^10, and 461 omits the first term.
Q6 · Mathematics · Binomial Theorem
What is the remainder when 6⁸³ + 8⁸³ is divided by 49?
Answer: (A)
Use 8⁸³ = (1 + 7)⁸³ and 6⁸³ = (7 − 1)⁸³. Every term containing 7² or a higher power is divisible by 49, so only the first two terms of each expansion matter. 8⁸³ ≡ 1 + 83·7 = 582 ≡ 43 (mod 49). 6⁸³ = (7 − 1)⁸³ ≡ −1 + 83·7 = 580 ≡ 41 (mod 49), because the exponent 83 is odd and the linear term has sign (−1)⁸² = +1. The sum is 84 ≡ 35 (mod 49). Equivalently, the 1 and −1 cancel and the linear terms add to 2·83·7 = 1162 = 49·23 + 35.
Q7 · Mathematics · Binomial Theorem
What is the constant term (the term independent of x) in the expansion of (x + 1/x)^6 ?
Answer: (A)
The general term is T₍ᵣ₊₁₎ = C(6,r) x^(6−r) (1/x)^r = C(6,r) x^(6−2r). For the constant term, 6 − 2r = 0, so r = 3. The term is C(6,3) = 20.
Q8 · Mathematics · Binomial Theorem
What is the remainder when 7^103 is divided by 25?
Answer: (B)
Write 7^103 = 7·(7^2)^51 = 7·(50 − 1)^51. Expanding, (50 − 1)^51 = 50k + (−1)^51 = 50k − 1 for an integer k. So 7^103 = 350k − 7, and 350k is a multiple of 25. The remainder is −7 + 25 = 18.
Q9 · Mathematics · Binomial Theorem
Let (3 + √5)^7 = I + f, where I is an integer and 0 ≤ f < 1. What is the value of (I + f)(1 − f)?
Answer: (A)
Principle: (a + b)^n + (a − b)^n is an integer, because the odd powers of √5 cancel in the sum. Let g = (3 − √5)^7. Since 0 < 3 − √5 < 1, we have 0 < g < 1. Now I + f + g is an integer, so f + g is an integer in (0, 2). Since f is irrational and non-zero, f + g = 1, so 1 − f = g. Then (I + f)(1 − f) = (3+√5)^7 (3−√5)^7 = (9 − 5)^7 = 4^7 = 16384.
Q10 · Mathematics · Binomial Theorem
How many terms are there in the expansion of (2x + 3y)^12 ?
Answer: (A)
By the binomial theorem, (a + b)^n = Σ C(n, r) a^(n−r) b^r with r running from 0 to n. That is n + 1 terms. The coefficients 2 and 3 do not change the count. For n = 12 there are 13 terms.
Free with an account
All 145 real JEE Main questions from Binomial Theorem.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
Solve them freeQuestions
About Binomial Theorem in JEE Main.
How many questions come from Binomial Theorem in JEE Main?
Binomial Theorem (Mathematics, Class 11) has been asked 145 times in the 111 JEE (Main) shift papers from 2020–2025: about 1.3 questions a paper, 5% of all Mathematics questions. It is the 12th most-asked of 19 Mathematics chapters. The 2025 shifts had 21 from it in all.
Which topics of Binomial Theorem are asked most in JEE Main?
General and middle term (76), Simple applications (67), Binomial theorem for a positive integral index (2), counted over the papers from 2020–2025.
Is Binomial Theorem important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 12 of 19 in Mathematics; use that to decide how much time it gets, not whether to study it.
Where can I solve Binomial Theorem JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Binomial Theorem with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
Your JEE Main 2027 plan, chapter by chapter.
A week-wise plan that gives the heavy chapters their time. Free, from a non-profit trust.