Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# NCERT Solutions for Class 11 Maths Chapter 7 - Binomial Theorem Miscellaneous Exercise

Last updated date: 13th Aug 2024
Total views: 11.1k
Views today: 0.11k

## NCERT Solutions for Maths Class 11 Binomial Theorem Miscellaneous Exercise - Free PDF Download

Chapter 7 of Class 11 Maths focuses on the Binomial Theorem, which is a useful technique for multiplying formulas with any power. Because it makes complex algebraic calculations easier this theorem has significance for solving difficult mathematical problems. This chapter's Miscellaneous Exercise offers various kinds of difficult tasks that help in the understanding and successful application of the binomial theorem for students.

Table of Content
1. NCERT Solutions for Maths Class 11 Binomial Theorem Miscellaneous Exercise - Free PDF Download
2. Access NCERTSolutions Class 11 Maths Chapter 7 Binomial Theorem
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 7: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 7 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Students can get a thorough understanding of binomial coefficients and the general shape of the binomial expansion by completing these exercises. With the detailed clarifications and step-by-step instructions provided in Class 11 Maths NCERT Solutions, students may confidently solve any related problem and improve their overall mathematical skills. Get the latest CBSE Class 11 Maths Syllabus here.

Competitive Exams after 12th Science

## Access NCERTSolutions Class 11 Maths Chapter 7 Binomial Theorem

### Miscellaneous Exercise

1. If a and b are distinct integers, prove that a-b is a factor of ${a^n} - {b^n}$, whenever n is a positive integer

(hint: ${a^n} = {(a - b + b)^n}$)

Ans: To prove to prove that(a-b) is a factor of (${a^n} - {b^n}$), it must be proved that

${a^n} - {b^n}$= $k(a - b)$, where k is some natural number

It can be written that, $a = a - b + b$

${((a - b) + b)^n}{ = ^n}{C_0}{(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}}b{ + ^n}{C_2}{(a - b)^{n - 2}}{b^2} + \ldots { + ^n}{C_{n - 1}}(a - b){b^{n - 1}}{ + ^n}{C_n}{b^n}$

=${(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}}b{ + ^n}{C_2}{(a - b)^{n - 2}}{b^2} + \ldots { + ^n}{C_{n - 1}}(a - b){b^{n - 1}} + {b^n}$

=${a^n} - {b^n} = (a - b)$$[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}}b{ + ^n}{C_2}{(a - b)^{n - 3}}{b^2} + \ldots { + ^n}{C_{n - 1}}{b^{n - 1}}]$

$\Rightarrow {a^n} - {b^n} = k(a - b)$

Where k =$[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}}b{ + ^n}{C_2}{(a - b)^{n - 3}}{b^2} + \ldots { + ^n}{C_{n - 1}}{b^{n - 1}}]$ is a natural number this shows that $(a - b)$is a factor of $({a^n} - {b^n})$,

Where n is a positive integer.

2. Evaluate $\left( {\sqrt 3 } \right. + {\left. {\sqrt 2 } \right)^6} - \left( {\sqrt 3 } \right. - {\left. {\sqrt 2 } \right)^6}$

Ans: Firstly, the expression

${\left( {a + b} \right)^6} - {\left( {a - b} \right)^6}$ is simplified by using Binomial Theorem. This can be done as

${\left( {a + b} \right)^6}$=$^6{C_0}{(a)^6}{ + ^6}{C_1}{(a)^5}b{ + ^6}{C_2}{(a)^4}{b^2}{ + ^6}{C_3}{(a)^3}{b^3}{ + ^6}{C_4}{(a)^2}{b^4}{ + ^6}{C_5}(a){b^5}{ + ^6}{C_6}{b^6}$

=${(a)^6} + 6{(a)^5}b + 15{(a)^4}{b^2} + 20{(a)^3}{b^3} + 15{(a)^2}{b^4} + 6a{b^5} + {b^6}$

Putting a=$\sqrt 3 \,and\,$b=$\sqrt 2$, we obtain

$\left( {\sqrt 3 } \right. + {\left. {\sqrt 2 } \right)^6} - \left( {\sqrt 3 } \right. - {\left. {\sqrt 2 } \right)^6}$

$= 2\left( {6 \cdot {{(\sqrt 3 )}^5}(\sqrt 2 ) + 20 \cdot {{(\sqrt 3 )}^3}{{(\sqrt 2 )}^3} + 6 \cdot (\sqrt 3 ){{(\sqrt 2 )}^5}} \right)$

=$2 \times 198\sqrt 6$

$= 396\sqrt 6$

3. Find the values of${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$

Ans: Firstly, the expression is simplified by using Binomial Theorem.

${\left( {x + y} \right)^4} + {\left( {x - y} \right)^4}$

This can be done as

${\left( {x + y} \right)^4}$=$^4{C_0}{(x)^4}{ + ^4}{C_1}{(x)^3}y{ + ^4}{C_2}{(x)^2}{y^2}{ + ^4}{C_3}x\,{y^3}{ + ^4}{C_4}{y^4}$

=${(x)^4} + 4{(x)^3}y + 6{(x)^2}{y^2} + 4x\,{y^3} + {y^4}$

${\left( {x - y} \right)^4}$=$^4{C_0}{(x)^4}{ - ^4}{C_1}{(x)^3}y{ - ^4}{C_2}{(x)^2}{y^2}{ - ^4}{C_3}x\,{y^3}{ - ^4}{C_4}{y^4}$

=${(x)^4} - 4{(x)^3}y - 6{(x)^2}{y^2} - 4x\,{y^3} - {y^4}$

Putting x=${a^2}$ and $y = \sqrt {{a^2} - 1} ,$ We obtain

${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$

$= 2\left[ {{{\left( {{a^2}} \right)}^4} + 6{{\left( {{a^2}} \right)}^2}{{\left( {\sqrt {{a^2} - 1} } \right)}^2} + {{\left( {\sqrt {{a^2} - 1} } \right)}^4}} \right]$

$= 2\left[ {\left( {{a^8}} \right) + 6\left( {{a^4}} \right)\left( {{a^2} - 1} \right) + {{\left( {{a^2} - 1} \right)}^2}} \right]$

$= 2\left[ {{a^8} + 6{a^6} - 6{a^4} + {a^4} - 2{a^2} + 1} \right]$

$= 2\left[ {{a^8} + 6{a^6} - 5{a^4} - 2{a^2} + 1} \right]$

$= 2{a^8} + 12{a^6} - 10{a^4} - 4{a^2} + 2$

4. Find an approximation of ${\left( {0.99} \right)^5}$using the first three terms of its expansion.

Ans: $0.99\, = 1 - 0.01$

${\left( {0.99} \right)^5} = {\left( {1 - 0.01} \right)^5}$

$^5{C_0}{(1)^5}{ - ^5}{C_1}{(1)^4}\left( {0.01} \right){ - ^5}{C_2}{(1)^3}{\left( {0.01} \right)^2}$

(Approximately)

$= 1 - 0.05 + 0.001$

$= 1.001 - 0.05$

=$= 0.951$

Thus, the value of ${\left( {0.99} \right)^5}$is approximately 0.951

5. Expand using Binomial Theorem ${\left( {1 + \dfrac{x}{2} - \dfrac{2}{x}} \right)^4},\,x \ne 0$

Ans: ${\left( {1 + \dfrac{x}{2} - \dfrac{2}{x}} \right)^4}$

${ = ^n}{C_0}\left( {1 + {{\dfrac{x}{2}}^4}} \right){ - ^n}{C_1}{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{2}{x}} \right) - {\,^n}{C_2}{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^2}{\left( {\dfrac{2}{x}} \right)^2}{ - ^n}{C_3}\left( {1 + {{\dfrac{x}{2}}^4}} \right){\left( {\dfrac{2}{x}} \right)^3}{ - ^n}{C_4}{\left( {\dfrac{2}{x}} \right)^4}$

$= \left( {1 + {{\dfrac{x}{2}}^4}} \right) - 4{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{2}{x}} \right) + \,6\left( {1 + x + {{\dfrac{x}{4}}^2}} \right)\left( {\dfrac{4}{{{x^2}}}} \right) - 4\left( {1 + \dfrac{x}{2}} \right)\left( {\dfrac{8}{{{x^3}}}} \right) + \left( {\dfrac{{16}}{{{x^4}}}} \right)$

$= \left( {1 + {{\dfrac{x}{2}}^4}} \right) - {\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{8}{x}} \right) + \,\left( {\dfrac{8}{{{x^2}}}} \right) + \dfrac{{24}}{x} + 6 - \left( {\dfrac{{32}}{{{x^3}}}} \right) + \left( {\dfrac{{16}}{{{x^4}}}} \right)$…..(1)

Again, by using the Binomial Theorem, we obtain

${\left( {1 + \dfrac{x}{2}} \right)^4}{ = ^4}{C_0}{(1)^4}{ + ^4}{C_1}{(1)^3}\left( {\dfrac{x}{2}} \right){ + ^4}{C_2}{(1)^2}{\left( {\dfrac{x}{2}} \right)^2}{ + ^4}{C_3}\,{\left( {\dfrac{x}{2}} \right)^3}{ + ^4}{C_4}{\left( {\dfrac{x}{2}} \right)^4}$

$= 1 + 4 \times \dfrac{x}{2} + 6 \times \dfrac{{{x^4}}}{4} + 4 \times \dfrac{{{x^3}}}{8} + \dfrac{{{x^3}}}{{16}}$

$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}}$…..(2)

${\left( {1 + \dfrac{x}{2}} \right)^3}{ = ^3}{C_0}{(1)^3}{ + ^3}{C_1}{(1)^2}\left( {\dfrac{x}{2}} \right){ + ^3}{C_2}(1){\left( {\dfrac{x}{2}} \right)^2}{ + ^3}{C_3}\,{\left( {\dfrac{x}{2}} \right)^3}$

$= \,1 + \dfrac{{3x}}{2} + \dfrac{{3{x^2}}}{4} + \dfrac{{{x^3}}}{8} + \dfrac{{{x^3}}}{8}$…… (3)

From (1), (2), and (3) we obtain

${\left( {\left( {1 + \dfrac{x}{2}} \right) - \dfrac{2}{x}} \right)^4}$

$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - \left( {\dfrac{8}{x}} \right)\left( {1 + \dfrac{{3x}}{2} + \dfrac{{3{x^2}}}{4} + \dfrac{{{x^3}}}{8}} \right) + \dfrac{8}{{{x^2}}} + \dfrac{{24}}{x} + 6 - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}}$

$= 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - \dfrac{8}{x} - 12 - 6x - {x^2} - \dfrac{8}{{{x^2}}} + \dfrac{{24}}{x} + 6 - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}}$

$= \dfrac{{16}}{x} + \dfrac{8}{{{x^2}}} - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}} - 4x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - 5$.

6. Find the expansion of ${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$using binomial theorem.

Ans: Using the Binomial Theorem, the given expression

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$Can be expanded as

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$

${ = ^3}{C_0}{\left( {3{x^2} - 2ax} \right)^3}{ - ^3}{C_1}{\left( {3{x^2} - 2ax} \right)^2}\left( {3{a^2}} \right){ + ^3}{C_2}\left( {3{x^2} - 2ax} \right){\left( {3{a^2}} \right)^2}{ - ^3}{C_3}{\left( {3{a^2}} \right)^3}$

$= {\left( {3{x^2} - 2ax} \right)^3} + 3\left( {9{x^4} - 12a{x^3} + 4{a^2}{x^2}} \right)\left( {3{a^2}} \right) + 3\left( {3{x^2} - 2ax} \right)\left( {9{a^4}} \right) + \left( {2{a^6}} \right)$

$= {\left( {3{x^2} - 2ax} \right)^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 36{a^4}{x^2} + 81{a^4}{x^2} - 54{a^5}x + 27{a^6}$

$= {\left( {3{x^2} - 2ax} \right)^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$…. (1)

Again, by using the Binomial Theorem, we obtain

${\left( {3{x^2} - 2ax} \right)^3}$

${ = ^3}{C_0}{\left( {3{x^2}} \right)^3}{ - ^3}{C_1}{\left( {3{x^2}} \right)^2}\left( {2ax} \right){ + ^3}{C_2}\left( {3{x^2}} \right){\left( {2ax} \right)^2}{ - ^3}{C_3}{\left( {2ax} \right)^3}$

$= \left( {27{x^6}} \right) - 3\left( {9{x^4}} \right)\left( {2ax} \right) + 3\left( {3{x^2}} \right)\left( {4{a^2}{x^2}} \right) - 8{a^3}{x^3}$

$= 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3}$……… (2)

From (1) and (2), we obtain

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$

$= 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$

$= 27{x^6} - 54a{x^5} + 117{a^2}{x^4} - 116{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$.

## Conclusion

NCERT Solutions for Class 11 Maths Miscellaneous Exercise in Chapter 7 on the Binomial Theorem are essential for scoring this key concept. These solutions offer clear explanations, helping students solve complex problems easily. By regularly practising these exercises, students can improve their understanding and use of the binomial theorem. This practice improves their confidence and prepares them well for exams and future math challenges.

## Class 11 Maths Chapter 7: Exercises Breakdown

 Exercise Number of Questions Exercise 7.1 14 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 7 - Binomial Theorem Miscellaneous Exercise

1. What is the Binomial Theorem in NCERT solutions of Miscellaneous Exercise Class 11 Chapter 7?

The Binomial Theorem of Miscellaneous Exercise Class 11 Chapter 7 provides a formula for expanding expressions raised to any positive integer power. It is essential for understanding polynomial expansions. This theorem is expressed as a sum involving terms of the form. Each term represents a specific combination of the variables.

2. Why is the Binomial Theorem important in NCERT solutions of Miscellaneous Exercise Class 11 Chapter 7?

In NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 7 the binomial theorem simplifies the expansion of binomials. It is widely used in various fields of mathematics such as algebra, calculus, and probability theory. This makes it a critical tool for solving complex mathematical problems. Understanding this theorem helps in efficiently handling polynomial expansions and related calculations.

3. What are binomial coefficients in NCERT solutions of Binomial Theorem Class 11 Miscellaneous Exercise?

In the Binomial Theorem Class 11 Miscellaneous Exercise, binomial coefficients are the numerical factors in the binomial expansion. They are calculated using combinations. These coefficients determine the weight of each term in the expansion. They play a crucial role in the precise calculation of terms in the binomial expression.

4. What is Pascal's Triangle in NCERT solutions of Binomial Theorem Class 11 Miscellaneous Exercise?

Pascal's Triangle of Binomial Theorem Class 11 Miscellaneous Exercise is a triangular array of numbers where each number is the sum of the two directly above it. This pattern is used to find binomial coefficients. Pascal's Triangle simplifies the process of identifying coefficients in binomial expansions.

5. What is the importance of the NCERT solutions of class 11 maths chapter 7 miscellaneous exercise?

The NCERT solutions of class 11 maths chapter 7 miscellaneous exercise include a variety of problems. It helps reinforce the understanding and application of the Binomial Theorem. By practising diverse questions, students can grasp the concepts more deeply. This exercise ensures comprehensive preparation for exams by covering all aspects of the chapter.

6. What is the main topic of the Miscellaneous Exercise on Chapter 7 Class 11?

Miscellaneous Exercise on Chapter 7 Class 11 focuses on Permutations and Combinations, which are essential concepts in counting and probability. These concepts help in determining the number of ways to arrange or select items.

7. How do NCERT Solutions help with the Miscellaneous Exercise of Chapter 7?

Miscellaneous Exercise on Chapter 7 Class 11 provides step-by-step explanations and detailed solutions for the Miscellaneous Exercise. They help students understand complex problems and learn the correct methods to solve them, ensuring thorough preparation.

8. What types of problems are included in the Binomial Theorem Miscellaneous Exercise?

The Binomial Theorem Miscellaneous Exercise includes problems on permutations, combinations, factorials, and the application of these concepts in various situations. These problems range from simple to complex, providing comprehensive practice.

9. How can students effectively solve the problems in the Binomial Theorem Miscellaneous Exercise?

Students should carefully read each problem, understand the concepts involved, and apply the correct formulas. Practising regularly and referring to NCERT Solutions for guidance can help in solving problems accurately.

10. What are the benefits of using NCERT Solutions for Binomial Theorem Miscellaneous Exercise?

NCERT Solutions Binomial Theorem Miscellaneous Exercise helps students understand difficult concepts with ease. They provide clear explanations and solve problems step-by-step, which improves comprehension and problem-solving abilities. Using these solutions ensures better preparation for exams.