Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem

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NCERT solutions of class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

Class 11 -Physics

Class 11-Chemistry

Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem

Q1.Expand the expression (1-2x)⁵

Ans. We are given the expression (1-2x)⁵

The expansion of the expression (1-a)ⁿ  is given as following

(1-a)ⁿ = 1+ⁿC₁a + ⁿC₂a² + ⁿC₃a³ +….. ⁿC_naⁿ

∴(1-2x)⁵ = 1-⁵C₁ 2x+ ⁵C₂(2x)² – ⁵C₃(2x)³ + ⁵C₄(2x)⁴ – ⁵C₅(2x)⁵

= 1 – 10x +40x²-80x³ +80x⁴ -32x⁵

Q2.Expand the expression

(2/x – x/2)⁵

Ans. The given expression is (2/x – x/2)⁵

The expansion of the expression (a – b)ⁿ is given as

(a – b)ⁿ = ⁿC₀aⁿb⁰ + ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²+…. ⁿC_na⁰bⁿ

(2/x – x/2)⁵  = ⁵C₀(2/x)⁵(x/2)⁰ – ⁵C₁(2/x)⁴(x/2)¹+⁵C₂(2/x)³(x/2)²– ⁵C₃(2/x)²(x/2)³+⁵C₄(2/x)¹(x/2)⁴–⁵C₅(2/x)⁰(x/2)⁵

Q3. Expand the expression

(2x – 3)⁶

Ans. The expansion of the expression (a – b)ⁿ is given as

(a – b)ⁿ = ⁿC₀aⁿb⁰ + ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²+…. ⁿC_na⁰bⁿ

(2x – 3)⁶= ⁶C₀(2x)⁶3⁰ – ⁶C₁(2x)⁵3¹+ ⁶C₂(2x)⁴3²– ⁶C₃(2x)³3³ +⁶C₄a(2x)²3⁴–⁶C₅(2x)3⁵+⁶C₆3⁶

=64x⁶ -576x⁵ +2160x⁴ -4320x³ +4860x²-2916x+729

Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem

Q4. Expand the expression

(x/3 -1/x)⁵

Ans. The given expression is

(x/3 -1/x)⁵

Applying binomial theorem

(a + b)ⁿ = ⁿC₀aⁿb⁰ +ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²+……..ⁿC_na⁰bⁿ

(x/3 +1/x)⁵= ⁵C₀(x/3)⁵ + ⁵C₁(x/3)⁴(1/x)+ ⁵C₂(x/3)³(1/x)² +⁵C₃(x/3)²(1/x)³+⁵C₄(x/3)(1/x)⁴+⁵C₅(1/x)⁵

= x⁵/243 + 5(x⁴/81)(1/x) +10(x³/27)(1/x²) +10(x²/9)(1/x³) +5(x/3)(1/x⁴)+1/x⁵

= x⁵/243 +5x³/81 + 10x/27 +10/9x + 5/3x³ +1/x⁵

Q5. Expand the expression

(x+ 1/x)⁶

Ans. Applying the binomial theorem

(a + b)ⁿ = ⁿC₀aⁿb⁰ +ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²+……..ⁿC_na⁰bⁿ

(x+ 1/x)⁶= ⁶C₀ x⁶ + ⁶C₁x⁵(1/x)+ ⁶C₂x⁴(1/x)² +⁶C₃x³(1/x)³+⁶C₄x²(1/x)⁴+⁶C₅x(1/x)⁵ + (1/x)⁶

=  x⁶ + 6x⁵(1/x)+ 15x⁴(1/x)² +20x³(1/x)³+15x²(1/x)⁴+6x(1/x)⁵ + (1/x)⁶

= x⁶ + 6x⁴+ 15x² +20+15/x²+6/x⁴ + (1/x)⁶

Q6. Using binomial theorem evaluate (96)³

Ans.Writting 96 in the following form

96 = 100 – 4

Applying the binomial theorem

(a – b)ⁿ = ⁿC₀aⁿb⁰ –ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²-……..ⁿC_na⁰bⁿ

(96)³=(100- 4)³= ³C₀ (100)³ – ³C₁(100)²4+ ³C₂(100)4² –³C₃4³

= 100³ -3(100)²4 +3(100)16 -64

= 1000000 – 120000 +4800 -64

= 884736

Q7. Using binomial theorem evaluate (102)⁵

Ans.Writting 102 in the following form

102 = 100 + 2

Applying the binomial theorem

(a +b)ⁿ = ⁿC₀aⁿb⁰ +ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²+……..ⁿC_na⁰bⁿ

(102)⁵=(100+2)⁵= ⁵C₀ (100)⁵ +⁵C₁(100)⁴2+ ⁵C₂(100)³2² +⁵C₃(100)²2³+ ⁵C₄(100)2⁴+⁵C₅2⁵

= (100)⁵ +5(100)⁴2 +10(100)³2² + 10(100)²2³+5(100)2⁴+2⁵

= 10000000000+ 5(100000000)2 + 10(1000000)4+10(10000)8 +5(100)16 +32

=10000000000 + 1000000000+ 40000000 +800000+ 8000+32

=11040808032

Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem

Q8. Using binomial theorem evaluate (101)⁵

Ans.Writting 101 in the following form

101 = 100 + 1

Applying the binomial theorem

(a +b)ⁿ = ⁿC₀aⁿb⁰ +ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²+……..ⁿC_na⁰bⁿ

(101)⁴=(100+1)⁴= ⁴C₀ (100)⁴ +⁴C₁(100)³1+ ⁴C₂(100)²1² +⁴C₃(100)1³+ ⁴C₄1⁴

=(100)⁴ +4(100)³ +6(100)² + 4(100)+1

=100000000 +4000000+60000+400+1

=104060401

Q9. Using binomial theorem evaluate (99)⁵

Ans.Writting 99 in the following form

99 = 100 – 1

Applying the binomial theorem

(a -b)ⁿ = ⁿC₀aⁿb⁰ –ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²-……..ⁿC_na⁰bⁿ

(99)⁵=(100-1)⁵= ⁵C₀ (100)⁵ –⁵C₁(100)⁴1+ ⁵C₂(100)³1² –⁵C₃(100)²1³+⁵C₄(100)1⁴– ⁵C₅1⁵

=(100)⁵ -5(100)⁴ +10(100)³ – 10(100)²+5(100) – 1

=10000000000 -500000000+10000000-100000+500-1

=9509900499

Q10.Using binomial theorem indicate which number is larger (1.1)¹⁰⁰⁰⁰or 1000

Ans. Using the binomial theorem, expanding the term (1.1)¹⁰⁰⁰⁰

Writing 1.1 in the form of  1 +0.1

(1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰

= ¹⁰⁰⁰⁰C₀ 1¹⁰⁰⁰⁰ +¹⁰⁰⁰⁰C₁1⁹⁹⁹0.1+……

= 1 + 10000×0.1 +….

= 1+1000+…… = 1001 +……>1000

Therefore (1.1)¹⁰⁰⁰⁰> 1000

Summary of the  class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem

Class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem is based on the expansion of two terma (a +b)ⁿ which is expanded as following

(a +b)ⁿ = ⁿC₀aⁿb⁰ +ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²+……..+ⁿC_na⁰bⁿ

Expansion of other important expressions

(a -b)ⁿ = ⁿC₀aⁿb⁰ –ⁿC₁a^n-1b¹+ ⁿC₂a^n-2b²-……..ⁿC_na⁰bⁿ

(1+x)ⁿ = 1 +ⁿC₁x+ ⁿC₂x²+ⁿC₃x³……..+xⁿ

(1-x)ⁿ = 1 –ⁿC₁x+ ⁿC₂x²–ⁿC₃x³+……xⁿ

The general term of the expansion (a +b)ⁿ is written as follows

T_r+1 = ⁿC_ra^n-rb^r