# NCERT Solutions for Class 11 Maths Chapter 8

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem (द्विपद प्रमेय) Exercise 8.1, Exercise 8.2 or Miscellaneous Exercise to view online or download in PDF format free for session 2020-21.

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 2020-21 following the latest CBSE Syllabus 2020-2021.## NCERT Solutions for Class 11 Maths Chapter 8

Class: | 11 |

Subject: | Maths |

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 2020-2021. NCERT Solutions 2020-21 are based on latest CBSE Curriculum 2020-2021 for CBSE Board, UP Board, Uttarakhand, Bihar Board, etc. who are following NCERT Books 2020-21 for their study.

### 11th Maths Chapter 8 Solutions in English Medium

### 11th Maths Chapter 8 Solutions in PDF

#### 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

(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.

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

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