The Binomial Theorem is a crucial topic in mathematics, especially for students preparing for competitive exams like the JEE. Understanding this theorem not only helps in solving complex problems but also lays a strong foundation for higher-level mathematics. In this overview, we will delve into the JEE Binomial Theorem Previous Year Question Paper with Solutions, which serves as an excellent resource for students aiming to master this topic.
Binomial theorem expansion formula allows us to expand expressions of the form (a + b)n into a sum involving terms of the form nCk · a{n-k} · bk. This formula is essential for solving various types of problems in exams. For those looking to study further, there are many resources available, including a binomial theorem PDF that provides detailed explanations and examples.
To aid your preparation, you can find a binomial theorem PYQ JEE Mains PDF that compiles previous years' questions along with solutions. This resource is particularly useful for practicing and understanding the types of questions that frequently appear in exams. Additionally, students can download binomial theorem questions and answers PDF to test their knowledge and improve their problem-solving skills.
For those focusing specifically on the JEE, there are tailored resources like binomial theorem JEE questions PDF, which include questions that align with the exam's format and difficulty level. Looking ahead, students preparing for the JEE Mains 2025 can benefit from reviewing binomial theorem JEE Mains questions 2024, ensuring they are well-prepared for the upcoming challenges.
Question 1: If the total coefficients of all equal forces of x in a product (1 + x + x2 + … + x2n) (1 – x + x2 – x3 +… + x2n) are 61, get the value n.
Solution:
Allow (1 + x + x2 +… + x2n) (1 – x + x2 – x3 +… + x2n) = a0 + ax + a2x2 + a3x3 + ……
Set x = 1 => 2n + 1 = a0 + a1 + a2 + a3 + ……… (i)
Put x = -1 => 2n + 1 = a0 – a1 + a2 – a3 + ……… .. (ii)
Add (i) and (ii), we get
2n + 1 = 61
or n = 30
Question 2: If α and β are coefficients of x4 and x2 respectively in addition (x + √ (x2 – 1))) 6 + (x – √ (x2 – 1) 6), then -ke:
(a) α + β = -30
(b) α− β = −132
(c) α− β = 60
(d) α + β = 60
Solution: (x + √ (x2 – 1)) 6 + (x – √ (x2 – 1) 6
= 2 [6C0 x6 + 6C2x4 (x2 – 1) + 6C4x2 (x2 – 1) 2 + 6C6 (x2 – 1) 3]
= 2 [32x6 – 48x4 + 18x2 – 1]
=> Now, α=- 96 and Β=-36
=> α – β = -132
Question 3: Find the coefficient of x4 in addition to (1 + x + x2) 10.
Solution:
The general term for a given speech is,
{10!} \ {P! Q! R!} x {q + 2r}
Here, q + 2r = 4
At p = 6, q = 4, r = 0, coefficient = 10! / [6! x 4!] = 210
At p = 7, q = 2, r = 1, coefficient = 10! / [7! x2! x 1!] = 360
At p = 8, q = 0, r = 2, coefficient = 10! / [8! x2!] = 45
Therefore, total = 615
Question 4: If x is positive, the first negative word in addition to
(1 + x) 27/5 by
(a) term 5 (b) term 6 (c) term 7 (d) term 8
Answer: (d)
Solution:
We know, T{r + 1} = {n (n-1)… (n-r + 1)} \ {r!} X r
Therefore, n – r + 1 <0
=> 27/5 + 1 <r
or r> 6
Here r = 7
Thus, Tr + 1 <0
So Tr + 1 = T7 + 1 = T8 = term 8 is negative.
Question 5: Deposit number greater than (1 + 0.0001) 10000
(a) 4 (b) 5 (c) 2 (d) 3
Answer: (d)
Solution:
(1 + 0.0001) 10000 similar form (1 + 1 / n) n where n = 10000.
Using binomial extensions, we have
(1 + 0.0001) 10000 = (1 + 1 / n) n
= 1 + n x 1 / n + n (n-1) / 2! x 1 / n2 + [n (n-1) (n-2)] / 3! X 1 / n3 + …… ..
= 1 + 1 + 1/2! (1 – 1 / n) + 1/3! (1 – 1 / n) + (1 – 2 / n) + ……
<1 + 1/1! + 1/2! + 1/3! + ….. + 1 / (9999)!
= 1 + 1/1! + 1/2! + 1/3! + …… .∞
= e <3
Question 6: The coefficients of xp and xq in the expression (1 + x) p + q
(a) Equal
(b) It is equal to the opposite sign
(c) Reconciliation with each other
(d) None of this
Answer: (a)
Solution:
The coefficients of xp and xq in the expression (1 + x) p + q
We know, yp + 1 = p + qCp xp and yq + 1 = p + qCq xq
Coefficient of xp = p + qCp = [p + q]! / P! Q!
and
Coefficient of xq = p + qCq = [p + q]! / P! Q!
Therefore, the coefficients of xp and xq in the expression (1 + x) p + q are equal.
Question 7: If an=√7+(√7+(√7+……
having n radical signs then in the form of true mathematical inventions
(a) <7 for all n ≥ 1
(b) an> 7 for all n ≥ 1
(c) 3 for all n ≥ 1
(d) <4 for all n ≥ 1
Answer: (c)
Solution: an=√7+(√7+(√7+……
=> i = √ (7 + an)
=> an2 – an – 7 = 0
Solving above the quadratic equation, we find
i = [1 ± √29] / 2
But> 0, therefore, is = [1 + √29] / 2> 3
=> i> 3
Question 8: If the sum of the coefficients in the extension (a + b) n is 4096, then the largest coefficient in the extension is
Answer: (b)
Solution:
Binomial Expansion: (x + y) n = C0 xn + C1 xn-1 y + C2 xn-2y2 + …. + Cnyn
setting x=y=1,then
2n = C0 + C1 + C2 +…. + Cn = 4096 = 212
=> n = 12
As n is present even here, so the coefficient of the largest term states
nCn / 2 = 12C12 / 2 = 12C6
= 12/6 x 11/5 x 10/4 x 9/3 x8 / 2 x 7/1
= 924
Question 9: The co-efficient amount of all the unusual term terms in the extension of
(x + √ (x3-1) 5 + (x – √ (x3-1) 5, (x> 1)
(a) 0 (b) 2 (c) 1 (d) -1
Solution:
Allow y = √ (x3-1)
Thus, the given expression is reduced to (x + y) 5 + (x – y) 5
Using binomial extensions, we have
(x + y) 5 + (x – y) 5 = x5 + 5C1 x4y + 5C2 x3y2 + 5C3 x2y3 + 5C4 xy4 + 5C5 y5 + x5 – 5C1 x4y + 5C2 x3y2 – 5C3 x2y3 + 5C5
= 2 (x5 + 10 x3y2 + 5xy4)
= 2 (x5 + 10 x3 (x3 – 1) + 5x (x3 – 1) 2)
= 2 {10x6 + 5x7 + x5 – 10x3 – 10x4 + 5x}
Total coefficient of unconventional x: 2 {5 + 1 – 10 + 5} = 2
Question 10: Extend quote (2x-3) 6 using the binomial theorem.
Solution:
Presentation: (2x-3) 6
Using the binomial theorem, expression (2x-3) 6 can be extended as follows:
(2x-3) 6 = 6C0 (2x) 6 –6C1 (2x) 5 (3) + 6C2 (2x) 4 (3) 2 – 6C3 (2x) 3 (3) 3 + 6C4 (2x) 2 (3) 4 – 6C5 (2x) (3) 5 + 6C6 (3) 6
(2x-3) 6 = 64x6 – 6 (32x5) (3) +15 (16x4) (9) – 20 (8x3) (27) +15 (4x2) (81) – 6 (2x) (243) + 729
(2x-3) 6 = 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729
Thus, the binomial extension of the given expression (2x-3) 6 is 64x6 -576x5 + 2160x4 – 4320x3 + 4860x2 – 2916x + 729.
Some basic introduction to the Binomial Theorem introduction is included in the eighth-grade syllabus as well. The concept of the Binomial Theorem becomes easier to understand once students become familiar with its output.
Binomial Theorem is one of the most important chapters of Algebra in the JEE syllabus. !
Binomial Theorem is a powerful mathematical formula used to expand binomial expressions raised to any power. In JEE, it helps in solving problems related to series expansion, finding coefficients, and simplifying algebraic expressions. It's essential because it provides a shortcut for solving complex problems involving powers of binomials.
Solving previous year JEE Binomial Theorem questions helps you understand the types of problems that frequently appear in the exam. It also allows you to practice applying the formula in different scenarios, improving your problem-solving skills and speed.
You can find JEE Binomial Theorem previous year question papers with solutions in various online resources such as educational websites, reference books, and dedicated platforms like Infinity Learn. These solutions provide detailed explanations and step-by-step methods for solving each question.