NCERT Solutions for Class 11 Maths Chapter 8 : Binomial Theorem

Q: Q1. What exactly is the Binomial Theorem?

Answer: The technique of algebraically expanding the power of sums of two or more binomials is defined as the binomial theorem. Binomial coefficients are the coefficients of binomial terms during the expansion process. The first sections of these chapters contain appropriate formulations of various facets of the binomial theorem.

Q: Q2. How many exercises are there in the Binomial Theorem Class 11 NCERT Solutions?

Answer: Chapter 8 Class 11 Maths Binomial Theorem comprises of 36 questions divided into 14 simple formula-based sums, 6 intermediate issues, and 16 difficult problems. The miscellaneous exercise is made up of high-order thinking issues that demand learners to use their critical thinking skills. The solutions to the exercise-specific problems are also accessible in PDF format, with the goal of assisting students in performing well in the yearly exam.

Q: Q3. What are the key subtopics in Binomial Theorem Class 11 Maths Chapter 8 that could be tested?

Answer: NCERT Solutions Class 11 Maths Chapter 8 begins with a brief overview of the Binomial Theorem. The NCERT Solutions Class 11 Maths Chapter 8 focuses on two important topics: expression expansion using the binomial theorem and locating a term within an exponential expression. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

Q: Q4. Is it necessary to practice all of the questions in Binomial Theorem Class 11 NCERT Solutions?

Answer: Binomial Theorem Class 11 NCERT Solutions are well-designed to help students study Binomial Theorem, Positive Integral Indices, and Pascal’s Triangle easily. It thoroughly covers all essential principles, allowing students to grasp the concepts. These solutions use examples to clarify the subjects presented so that students may easily relate to the concepts being addressed.

Question 1

Expand

(1 - 2 x)^{5}

Easy

Solution

Verified by Toppr

Answer

Using Binomial Theorem, the expression

(1 - 2 x)^{5}

can be expanded as

(1 - 2 x)^{5} =^{5} C_{0} (1)^{5} -^{5} C_{1} (1)^{4} (2 x) +^{5} C_{2} (1)^{3} (2 x)^{2} -^{5} C_{3} (1)^{2} (2 x)^{3} +^{5} C_{4} (1)^{1} (2 x)^{4} -^{5} C_{5} (2 x)^{5}

= 1 - 5 (2 x) + 10 (4 x^{2}) - 10 (8 x^{3}) + 5 (16 x^{4}) - (32 x^{5})

= 1 - 10 x + 40 x^{2} - 80 x^{3} + 80 x^{4} - 32 x^{5}

Answer

Using Binomial Theorem, the expression

(1 - 2 x)^{5}

can be expanded as

(1 - 2 x)^{5} =^{5} C_{0} (1)^{5} -^{5} C_{1} (1)^{4} (2 x) +^{5} C_{2} (1)^{3} (2 x)^{2} -^{5} C_{3} (1)^{2} (2 x)^{3} +^{5} C_{4} (1)^{1} (2 x)^{4} -^{5} C_{5} (2 x)^{5}

= 1 - 5 (2 x) + 10 (4 x^{2}) - 10 (8 x^{3}) + 5 (16 x^{4}) - (32 x^{5})

= 1 - 10 x + 40 x^{2} - 80 x^{3} + 80 x^{4} - 32 x^{5}

Question 2

Expand the expression

(2 x - 3)^{6}

Easy

Solution

Verified by Toppr

Answer

Using Binomial theorem,

(2 x - 3)^{6} =^{6} C_{0} (2 x)^{6} -^{6} C_{1} (2 x)^{5} (3) +^{6} C_{2} (2 x)^{4} (3)^{2} -^{6} C_{3} (2 x)^{3} (3)^{3} +^{6} C_{4} (2 x)^{2} (3)^{4} -^{6} C_{5} (2 x) (3)^{5} +^{6} C_{6} (3)^{6}

= 64 x^{6} - 6 (32 x^{5}) (3) + 15 (16 x^{4}) (9) - 20 (8 x^{3}) (27) + 15 (4 x^{2}) (81) - 6 (2 x) (243) + 729

= 64 x^{6} - 576 x^{5} + 2160 x^{4} - 4320 x^{3} + 4860 x^{2} - 2916 x + 729

Answer

Using Binomial theorem,

(2 x - 3)^{6} =^{6} C_{0} (2 x)^{6} -^{6} C_{1} (2 x)^{5} (3) +^{6} C_{2} (2 x)^{4} (3)^{2} -^{6} C_{3} (2 x)^{3} (3)^{3} +^{6} C_{4} (2 x)^{2} (3)^{4} -^{6} C_{5} (2 x) (3)^{5} +^{6} C_{6} (3)^{6}

= 64 x^{6} - 6 (32 x^{5}) (3) + 15 (16 x^{4}) (9) - 20 (8 x^{3}) (27) + 15 (4 x^{2}) (81) - 6 (2 x) (243) + 729

= 64 x^{6} - 576 x^{5} + 2160 x^{4} - 4320 x^{3} + 4860 x^{2} - 2916 x + 729

Question 3

Expand

(\frac{x}{3} + \frac{1}{x})^{5}

Easy

Solution

Verified by Toppr

Answer

Using Binomial theorem,

(\frac{x}{3} + \frac{1}{x})^{5}

=^{5} C_{0} (\frac{x}{3})^{5} +^{5} C_{1} (\frac{x}{3})^{4} (\frac{1}{x}) +^{5} C_{2} (\frac{x}{3})^{3} (\frac{1}{x})^{2}

+^{5} C_{3} (\frac{x}{3})^{2} (\frac{1}{x})^{3} +^{5} C_{4} (\frac{x}{3}) (\frac{1}{x})^{4} +^{5} C_{5} (\frac{1}{x})^{5}

= \frac{x ^{5}}{2 4 3} + 5 (\frac{x ^{4}}{8 1}) (\frac{1}{x}) + 10 (\frac{x ^{3}}{2 7}) (\frac{1}{x ^{2}}) + 10 (\frac{x ^{2}}{9}) (\frac{1}{x ^{3}}) + 5 (\frac{x}{3}) (\frac{1}{x ^{4}}) + \frac{1}{x ^{5}}

= \frac{x ^{3}}{2 4 3} + \frac{5 x ^{3}}{8 1} + \frac{1 0 x}{2 7} + \frac{1 0}{9 x} + \frac{5}{3 x ^{3}} + \frac{1}{x ^{5}}

Answer

Using Binomial theorem,

(\frac{x}{3} + \frac{1}{x})^{5}

=^{5} C_{0} (\frac{x}{3})^{5} +^{5} C_{1} (\frac{x}{3})^{4} (\frac{1}{x}) +^{5} C_{2} (\frac{x}{3})^{3} (\frac{1}{x})^{2}

+^{5} C_{3} (\frac{x}{3})^{2} (\frac{1}{x})^{3} +^{5} C_{4} (\frac{x}{3}) (\frac{1}{x})^{4} +^{5} C_{5} (\frac{1}{x})^{5}

= \frac{x ^{5}}{2 4 3} + 5 (\frac{x ^{4}}{8 1}) (\frac{1}{x}) + 10 (\frac{x ^{3}}{2 7}) (\frac{1}{x ^{2}}) + 10 (\frac{x ^{2}}{9}) (\frac{1}{x ^{3}}) + 5 (\frac{x}{3}) (\frac{1}{x ^{4}}) + \frac{1}{x ^{5}}

= \frac{x ^{3}}{2 4 3} + \frac{5 x ^{3}}{8 1} + \frac{1 0 x}{2 7} + \frac{1 0}{9 x} + \frac{5}{3 x ^{3}} + \frac{1}{x ^{5}}

Question 4

Expand

(x + \frac{1}{x})^{6}

Easy

Solution

Verified by Toppr

Answer

(x + \frac{1}{x})^{6}

=^{6} C + 0 (x)^{6} +_{1}^{C} (x)^{5} (\frac{1}{x}) +^{6} C_{2} (x)^{4} (\frac{1}{x})^{2} +^{6} C_{3} (x)^{3} (\frac{1}{x})^{3} +

^{6} C_{4} (x)^{2} (\frac{1}{x})^{4} +^{6} C_{5} (x) (\frac{1}{x})^{5} +^{6} C_{6} (\frac{1}{x})^{6}

= x^{6} + 6 (x)^{5} (\frac{1}{x}) + 15 (x)^{4} (\frac{1}{x ^{2}}) + 20 (x)^{3} (\frac{1}{x ^{3}}) + 15 (x)^{2} (\frac{1}{x ^{4}}) + 6 (x) (\frac{1}{x ^{5}}) + \frac{1}{x ^{6}}

= x^{6} + 6 x^{4} + 15 x^{2} + 20 + \frac{1 5}{x ^{2}} + \frac{6}{x ^{4}} + \frac{1}{x ^{6}}

Answer

(x + \frac{1}{x})^{6}

=^{6} C + 0 (x)^{6} +_{1}^{C} (x)^{5} (\frac{1}{x}) +^{6} C_{2} (x)^{4} (\frac{1}{x})^{2} +^{6} C_{3} (x)^{3} (\frac{1}{x})^{3} +

^{6} C_{4} (x)^{2} (\frac{1}{x})^{4} +^{6} C_{5} (x) (\frac{1}{x})^{5} +^{6} C_{6} (\frac{1}{x})^{6}

= x^{6} + 6 (x)^{5} (\frac{1}{x}) + 15 (x)^{4} (\frac{1}{x ^{2}}) + 20 (x)^{3} (\frac{1}{x ^{3}}) + 15 (x)^{2} (\frac{1}{x ^{4}}) + 6 (x) (\frac{1}{x ^{5}}) + \frac{1}{x ^{6}}

= x^{6} + 6 x^{4} + 15 x^{2} + 20 + \frac{1 5}{x ^{2}} + \frac{6}{x ^{4}} + \frac{1}{x ^{6}}

Question 5

Evaluate

(96)^{3}

Easy

Solution

Verified by Toppr

Answer

(96)^{3} = (100 - 4)^{3}

=^{3} C_{0} (100)^{3} -^{3} C_{1} (100)^{2} (4) +^{3} C_{2} (100) (4)^{2} -^{3} C_{3} (4)^{3}

= (100)^{3} - 3 (100)^{2} (4) + 3 (100) (4)^{2} - (4)^{3}

= 1000000 - 120000 + 4800 - 64

= 884736

Answer

(96)^{3} = (100 - 4)^{3}

=^{3} C_{0} (100)^{3} -^{3} C_{1} (100)^{2} (4) +^{3} C_{2} (100) (4)^{2} -^{3} C_{3} (4)^{3}

= (100)^{3} - 3 (100)^{2} (4) + 3 (100) (4)^{2} - (4)^{3}

= 1000000 - 120000 + 4800 - 64

= 884736

Question 6

Using Binomial Theorem, evaluate

(102)^{5}

Medium

Solution

Verified by Toppr

Answer

(102)^{5} = (100 + 2)^{5}

=^{5} C_{0} (100)^{5} +^{5} C_{1} (100)^{4} (2) +^{5} C_{2} (100)^{3} (2)^{2} +^{5} C_{3} (100)^{2} (2)^{3} +^{5} C_{4} (100) (2)^{4} +^{5} C_{5} (2)^{5}

= (100)^{5} + 5 (100)^{4} (2) + 10 (100)^{3} (2)^{2} + 10 (100)^{2} (2)^{3} + 5 (100) (2)^{4} + (2)^{5}

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

= 11040808032

Answer

(102)^{5} = (100 + 2)^{5}

=^{5} C_{0} (100)^{5} +^{5} C_{1} (100)^{4} (2) +^{5} C_{2} (100)^{3} (2)^{2} +^{5} C_{3} (100)^{2} (2)^{3} +^{5} C_{4} (100) (2)^{4} +^{5} C_{5} (2)^{5}

= (100)^{5} + 5 (100)^{4} (2) + 10 (100)^{3} (2)^{2} + 10 (100)^{2} (2)^{3} + 5 (100) (2)^{4} + (2)^{5}

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

= 11040808032

Question 7

Using Binomial Theorem, evaluate

(101)^{4}

Medium

Solution

Verified by Toppr

Answer

(101)^{4} = (100 + 1)^{4}

=^{4} C_{0} (100)^{4} +^{4} C_{1} (100)^{3} (1) +^{4} C_{2} (100)^{2} (1)^{2} +^{4} C_{3} (100) (1)^{3} +^{4} C_{4} (1)^{4}

= (100)^{4} + 4 (100)^{3} + 6 (100)^{2} + 4 (100) + (1)^{4}

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

Answer

(101)^{4} = (100 + 1)^{4}

=^{4} C_{0} (100)^{4} +^{4} C_{1} (100)^{3} (1) +^{4} C_{2} (100)^{2} (1)^{2} +^{4} C_{3} (100) (1)^{3} +^{4} C_{4} (1)^{4}

= (100)^{4} + 4 (100)^{3} + 6 (100)^{2} + 4 (100) + (1)^{4}

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

Question 8

Using Binomial theorem, evaluate

(99)^{5}

Medium

Solution

Verified by Toppr

Answer

(99)^{5} = (100 - 1)^{5}

=^{5} C_{0} (100)^{5} -^{5} C_{1} (100)^{4} (1) +^{5} C_{2} (100)^{3} (1)^{2} -^{5} C_{3} (100)^{2} (1)^{3} +^{5} C_{4} (100) (1)^{4} -^{5} C_{5} (1)^{5}

= (100)^{5} - 5 (100)^{4} + 10 (100)^{3} - 10 (100)^{2} + 5 (100) - 1

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

= 10010000500 - 500100001

= 9509900499

Answer

(99)^{5} = (100 - 1)^{5}

=^{5} C_{0} (100)^{5} -^{5} C_{1} (100)^{4} (1) +^{5} C_{2} (100)^{3} (1)^{2} -^{5} C_{3} (100)^{2} (1)^{3} +^{5} C_{4} (100) (1)^{4} -^{5} C_{5} (1)^{5}

= (100)^{5} - 5 (100)^{4} + 10 (100)^{3} - 10 (100)^{2} + 5 (100) - 1

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

= 10010000500 - 500100001

= 9509900499

Question 9

Find

(a + b)^{4} - (a - b)^{4}

. Hence find

s

if

(3 + 2)^{4} - (3 - 2)^{4} = 40 s

Medium

Solution

Verified by Toppr

Answer

Using Binomial theorem, the expressions

(a + b)^{4}

and

(a - b)^{4}

can be expanded as,

(a + b)^{4} =^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a b^{3} +^{4} C_{4} b^{4})

(a - b)^{4} =^{4} C_{0} a^{4} -^{4} C_{1} a^{3} b -^{4} C_{2} a^{2} b^{2} -^{4} C_{3} a b^{3} +^{4} C_{4} b^{4})

∴ (a + b)^{4} - (a - b)^{4}

=

^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a b^{3} +^{4} C_{4} b^{4} - [^{4} C_{0} a^{4} -^{4} C_{1} a^{3} b -^{4} C_{2} a^{2} b^{2} -^{4} C_{3} a b^{3} +^{4} C_{4} b^{4}]

=

2 (^{4} C_{1} a^{3} b +^{4} C_{3} a b^{3}) = 2 (4 a^{3} b + 4 a b^{3})

=

8 a b (a^{2} + b^{2})

Now by putting

a = 3

and

b = 2

, we obtain

(3 + 2)^{4} - (3 - 2)^{4} = 8 (3) (2) {(3)^{2} + (2)^{2}} = 8 (6) {3 + 2} = 406

Answer

Using Binomial theorem, the expressions

(a + b)^{4}

and

(a - b)^{4}

can be expanded as,

(a + b)^{4} =^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a b^{3} +^{4} C_{4} b^{4})

(a - b)^{4} =^{4} C_{0} a^{4} -^{4} C_{1} a^{3} b -^{4} C_{2} a^{2} b^{2} -^{4} C_{3} a b^{3} +^{4} C_{4} b^{4})

∴ (a + b)^{4} - (a - b)^{4}

=

^{4} C_{0} a^{4} +^{4} C_{1} a^{3} b +^{4} C_{2} a^{2} b^{2} +^{4} C_{3} a b^{3} +^{4} C_{4} b^{4} - [^{4} C_{0} a^{4} -^{4} C_{1} a^{3} b -^{4} C_{2} a^{2} b^{2} -^{4} C_{3} a b^{3} +^{4} C_{4} b^{4}]

=

2 (^{4} C_{1} a^{3} b +^{4} C_{3} a b^{3}) = 2 (4 a^{3} b + 4 a b^{3})

=

8 a b (a^{2} + b^{2})

Now by putting

a = 3

and

b = 2

, we obtain

(3 + 2)^{4} - (3 - 2)^{4} = 8 (3) (2) {(3)^{2} + (2)^{2}} = 8 (6) {3 + 2} = 406

Question 10

Using binomial theorem, indicate which is larger

(1.1)^{10000}

or

1000

.

Easy

Solution

Verified by Toppr

Answer

(1.1)^{10000} = (1 + 0.1)^{10000}

= 1000 0_{C_{0}} + 1000 0_{C_{1}} (1.1)

+ other positive terms

= 1 + 10000 \times 1.1

+ other positive terms

= 1 + 11000

+ other positive terms

> 1000

Hence,

(1.1)^{10000} > 1000

.

Answer

(1.1)^{10000} = (1 + 0.1)^{10000}

= 1000 0_{C_{0}} + 1000 0_{C_{1}} (1.1)

+ other positive terms

= 1 + 10000 \times 1.1

+ other positive terms

= 1 + 11000

+ other positive terms

> 1000

Hence,

(1.1)^{10000} > 1000

.

NCERT Solutions for Class 11 Maths Chapter 8 : Binomial Theorem

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