Solutions of class 11 NCERT maths exercise 8.1 of chapter 8 Binomial Theorem - Future Study Point

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

ex.8.1 of chapter binomial theorem

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

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ex.8.1 of chapter binomial theorem

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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