NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem (Dwipad Pramey) Exercise 8.1, Exercise 8.2 or Miscellaneous Exercise to view online or download in PDF format free for session 2022-23. UP Board Students can download UP Board solutions for class 11 Maths chapter 8 here in Hindi Medium. Join the discussion Forum to ask your doubts and share your knowledge with your friends. NCERT Solutions Offline Apps are based on new NCERT Books 2022-23 following the latest CBSE Syllabus 2022-2023.
NCERT Solutions for Class 11 Maths Chapter 8
11th Maths Chapter 8 Solutions in English Medium
11th Maths Chapter 8 Solutions in PDF
|Chapter 8:||Binomial Theorem|
11th Maths Chapter 8 Solutions
NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem all exercises including miscellaneous exercise are given below to download in PDF form updated for new session 2022-2023. NCERT Solutions 2022-23 are based on latest CBSE Curriculum 2022-2023 for CBSE Board, UP Board, Uttarakhand, Bihar Board, etc. who are following NCERT Books 2022-23 for their study.
Class 11 Maths Chapter 8 Solution in Hindi Videos
Class 11 Maths Chapter 8 Miscellaneous Exercise Solution
How to find Middle Term?
The Middle Term: In the expansion of (a + b)^n, the total number of terms are (n + 1). The middle term in the expansion of (a + b)^n depend on n.
1. When n is even: Let n = 2m, where m is positive integer. The total number of terms will be 2m + 1. Hence, the middle term of the expansion (a + b)^n will be 1/2[(2m + 1) + 1], i. e. when n is even then (m + 1)th term or (n/2 + 1)th will be the middle term.
2. When n is odd: Let n = 2m + 1, where m is a positive integer. In the expansion of (a + b)^n the total number of terms will be (m + 2). The middle term in the expansion of (a + b)^n will be (m + 1)th and (m + 2)th term or (n + 1)/2th and ((n + 3)/2th term.
Important Terms on Binomial Theorem
1. Binomial Expression: Any expression containing two terms combined by + or – is called Binomial expression. For example: x + 3, 2x + y, x – 4y, 4 – 100x, y – 4, etc.
2. In the expansion of (a + b)^n, the coefficient of first term = coefficient of last term, coefficient of second term = coefficient of second term from last. Thus we get that in the expansion of (a + b)^n, the terms from first term and from the last term at equal distance have the same coefficients.
3. The General Term: The term (r + 1) is called the general term of the expansion (a + b)^n because we can get different terms from this term by giving different values to r. This general term is denoted by Tr+1.
Important Questions on 11th Maths Chapter 8
Using Binomial theorem,indicate which is larger (1.1)¹⁰⁰⁰⁰ or 1000.
By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000 can be obtained as
= (1 + 0.1)^10000 = C(10000, 0) + C(10000, 1) (10000)^1 (0.1)^1 + Other positive terms.
= 1 + 10000×0.1 + Other positive terms.
= 1001 + Other positive terms. > 1000
Hence,(1.1)^10000 > 1000.
Find an approximation of (0.99)^5 using the first three terms of its expansion.
99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 0.99 = 1 – 0.01
= C(5, 0) (-0.01)^0 + C(5, 1) (-0.01)^1 + C(5, 2) (-0.01)^2 + ⋯
= 1 – 5(0.01) + 10(0.0001) + ⋯ ≈ 1 – 0.05 + 0.001 = 0.951
Find the expansion of (3x^2 – 2ax + 3a^2 )^3 using binomial theorem.
Here,(3x^2-2ax+3a^2 )^3=[(3x^2-2ax)-3a^2 ]^3
Using Binomial theorem, the expression [(3x^2-2ax)-3a^2 ]^3 can be expressed as
C(3, 0) (3x^2-2ax)^3 + C(3, 1) (3x^2-2ax)^2 (3a^2) + C(3, 2)(3x^2-2ax)^1 (3a^2)^2 + C(3, 3)(3a^2)^3
= (3x^2-2ax)^3+3(9x^4+4a^2 x^2-12ax^3)(3a^2) + 3(3x^2-2ax)(9a^4) + (27a^6)^
= (3x^2-2ax)^3+(81a^2 x^4+36a^4 x^2-108a^3 x^3)+(81a^4 x^2-54a^5 x)+〖27a^6〗^ …(1)
Again applying Binomial theorem, we have (3x^2-2ax)^3 = C(3, 0)(3x^2 )^3 + C(3, 1)(3x^2 )^2 (-2ax)^1 + C(3, 2)(3x^2)(-2ax)^2 + C(3, 3)(-2ax)^3
= 27x^6-3×18ax^5+3×12a^2 x^4-8a^3 x^3
= 27x^6-54ax^5+36a^2 x^4-8a^3 x^3
Now putting the value of (3x^2-2ax)^3 in (1),we have (3x^2-2ax+3a^2 )^3
= [27x^6-54ax^5+36a^2 x^4-8a^3 x^3 ]+(81a^2 x^4+36a^4 x^2-108a^3 x^3 )+(81a^4 x^2-54a^5 x)+〖27a^6〗
= 27x^6-54ax^5+117a^2 x^4-116a^3 x^3+117a^4 x^2-54a^5 x+27a^6
How much time, teachers needed to complete chapter 8 of class 11th Maths?
Teachers need a maximum of 1 week to complete chapter 8 of class 11th Maths if they properly give 1 hour per day to this chapter. This time also depends on student’s efficiency, capability, working speed, etc because no students can have the same working speed, efficiency, capability, etc.
Which questions of chapter 8 of class 11th Maths have a complete chance to come in the exams?
Chapter 8 of grade 11th Maths has 36 questions and 17 illustrations. Every year questions are asked from chapter 8 in the exams. All the questions of this chapter are significant and can come in the exams. But the most important questions of this chapter that have more chance to come in the terminal exam are questions 2, 4, 13, 14 of the first exercise (Ex 8.1), questions 2, 4, 6, 8, 10, 11 of the second exercise (Ex 8.2), questions 1, 3, 5, 7, 8, 9, of the Miscellaneous exercise on chapter 8, and examples 2, 3, 5, 6, 11, 12, 13, 14, 15, 17.
Which topics teacher teaches under chapter 8 Binomial Theorem of class 11th Maths?
Under chapter 8 Binomial Theorem of class 11th Maths teacher teaches the following topics:
1. Binomial Theorem for positive integral indices.
2. Pascal’s Triangle.
4. Some special cases.
5. General and Middle Terms.
How to get practice material based on Chapter 8 of class 11 Maths NCERT?
NCERT Exemplar book provides a lots of solved examples and practice questions. These questions are sufficient for the preparation of exams if a student has completed the NCERT Textbook thoroughly.