# NCERT Solutions for Class 11 Maths Chapter 8

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem (द्विपद प्रमेय) Exercise 8.1 or Exercise 8.2 or Miscellaneous Exercise to view online or download  in PDF format free. Join the discussion Forum to ask your doubts.

 Class: 11 Subject: Maths Chapter 8: Binomial Theorem

## NCERT Solutions for Class 11 Maths Chapter 8

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### Class 11 Maths Chapter 8 Binomial Theorem Solutions

Solutions are based on latest CBSE Curriculum 2019-2020 for CBSE Board, UP Board, Uttarakhand, Bihar Board, etc. who are following NCERT Books for their study.

##### 11 Maths Chapter 8 Exercise 8.1 Sols

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Exercise 8.1 is given below. For other questions, please visit to Exercise 8.2 or Miscellaneous Exercise Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 8 Exercise 8.2 Sols

NCERT Solutions for Class 11 Maths Chapter 8  Exercise 8.2 in English Medium is given below. For other questions, please visit to Exercise 8.1 or Miscellaneous Exercise Solutions. Visit to Class 11 Maths main page or Top of the page.

##### 11 Maths Chapter 8 Miscellaneous Exercise Sols

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise 8 all answers with explanations are given below. For other questions, please visit to Exercise 8.1 or Exercise 8.2 Solutions. Visit to Class 11 Maths main page or Top of the page.

Visit to Class 11 Maths main page or Top of the page

#### Important Terms on Binomial Theorem

• 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.
• 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.
• 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.

• 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.

#### 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
(1.1)^10000 = (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
Therefore,(0.99)^5=(1-0.01)^5
= 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

## 1 thought on “NCERT Solutions for Class 11 Maths Chapter 8”

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