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

Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem are created here for the purpose of helping the students in their preparation for the exam and in completing their worksheets. All these NCERT solutions of class 11 maths are prepared by an expert in maths who has huge experience in the field of maths teaching. These NCERT solutions of class 11 maths are the most valuable study material for all CBSE students, therefore they have to be thorough with the knowledge about each exercise of each chapter with proper understanding which is needed them to solve unsolved questions in sample papers, previous year question papers, and then in CBSE board exam.

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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)is given as following

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

∴(1-2x)= 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 + nC1an-1b1+ nC2an-2b2+…. nCna0bn

(2/x – x/2) = 5C0(2/x)5(x/2)05C1(2/x)4(x/2)1+5C2(2/x)3(x/2)25C3(2/x)2(x/2)3+5C4(2/x)1(x/2)45C5(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 + nC1an-1b1+ nC2an-2b2+…. nCna0bn

(2x – 3)6= 6C0(2x)630 – 6C1(2x)531+ 6C2(2x)4326C3(2x)333 +6C4a(2x)2346C5(2x)35+6C636

=64x-576x5 +2160x-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 +nC1an-1b1+ nC2an-2b2+……..nCna0bn

(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 +nC1an-1b1+ nC2an-2b2+……..nCna0bn

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

x6 + 6x5(1/x)+ 15x4(1/x)2 +20x3(1/x)³+15x²(1/x)4+6x(1/x)+ (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 = nC0anb0nC1an-1b1+ nC2an-2b2-……..nCna0bn

(96)³=(100- 4)3= 3C0 (100)33C1(100)24+ 3C2(100)423C3

= 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 +nC1an-1b1+ nC2an-2b2+……..nCna0bn

(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 +nC1an-1b1+ nC2an-2b2+……..nCna0bn

(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 = nC0anb0nC1an-1b1+ nC2an-2b2-……..nCna0bn

(99)5=(100-1)5= 5C0 (100)55C1(100)41+ 5C2(100)³1² –5C3(100)²1³+5C4(100)145C515

=(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 +nC1an-1b1+ nC2an-2b2+……..+nCna0bn

Expansion of other important expressions

(a -b)n = nC0anb0nC1an-1b1+ nC2an-2b2-……..nCna0bn

(1+x)n = 1 +nC1x+ nC2x2+nC3x3……..+xn

(1-x)n = 1 –nC1x+ nC2x2nC3x3+……xn

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

Tr+1 = nCran-rbr

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