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NCERT Solutions

Class 11

Maths

Chapter 7 Binomial Theorem Miscellaneous Exercise

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

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

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 Chapter 7 NCERT Solutions, students may confidently solve any related problem and improve their overall mathematical skills. Get the latest CBSE Class 11 Maths Syllabus here.

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

CBSE Class 11 Maths Chapter 7 Other Study Materials

S. No	Important Links for Chapter 7 Binomial Theorem
1	Class 11 Binomial Theorem Revision Notes
2	Class 11 Binomial Theorem Important Questions
3	Class 11 Binomial Theorem Formula
4	Class 11 Binomial Theorem NCERT Exemplar Solution
5	Class 11 Binomial Theorem RD Sharma Solutions
6	Class 11 Binomial Theorem RS Aggarwal 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.

S. No	NCERT Solutions Class 11 Maths All Chapters
1	Chapter 1 - Sets Solutions
2	Chapter 2 - Relations and Functions Solutions
3	Chapter 3 - Trigonometric Functions Solutions
4	Chapter 4 - Complex Numbers and Quadratic Equations Solutions
5	Chapter 5 - Linear Inequalities Solutions
6	Chapter 6 - Permutations and Combinations Solutions
7	Chapter 7 - Binomial Theorem Solutions
8	Chapter 8 - Sequences and Series Solutions
9	Chapter 9 - Straight Lines Solutions
10	Chapter 10 - Conic Sections Solutions
11	Chapter 11 - Introduction to Three Dimensional Geometry Solutions
12	Chapter 12 - Limits and Derivatives Solutions
13	Chapter 13 - Statistics Solutions
14	Chapter 14 - Probability Solutions

Important Related Links for CBSE Class 11 Maths

S.No.	Important Study Material for Maths Class 11
1	CBSE Class 11 Maths NCERT Books
2	CBSE Class 11 Maths Revision Notes
3	CBSE Class 11 Maths Important Questions
4	CBSE Class 11 Maths NCERT Solutions
5	CBSE Class 11 Maths NCERT Exemplar Solution
6	CBSE Class 11 Maths RS Aggarwal Solutions
7	CBSE Class 11 Maths RD Sharma Solutions
8	CBSE Class 11 Maths Formulas
9	CBSE Class 11 Maths Syllabus
10	CBSE Class 11 Maths Sample Papers

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

1. What is the correct method to find a specific term, like the 4th term, in an expansion of (a + b)ⁿ using the NCERT solution approach?

To find any specific term in a binomial expansion, the correct method is to use the formula for the general term, which is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. For the 4th term (T₄), you must set r+1 = 4, which means r = 3. Then, substitute n, a, b, and r=3 into the formula to calculate the exact term and its coefficient.

2. How do you determine the middle term(s) in the expansion of (x + y)ⁿ as per the Class 11 Maths NCERT syllabus?

The method to find the middle term depends on the value of the exponent 'n':

If 'n' is even: There is only one middle term, which is the (n/2 + 1)th term.
If 'n' is odd: There are two middle terms, which are the ((n+1)/2)th and the ((n+1)/2 + 1)th terms.

You find the position first and then use the general term formula to calculate its value.

3. Why is the general term, Tᵣ₊₁, considered the most important tool for solving questions in NCERT Solutions for Binomial Theorem?

The general term Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ is crucial because it acts as a universal formula for the entire expansion. Instead of writing out all the terms, you can use it to directly:

Find any specific term (e.g., the 5th term).
Calculate the middle term(s).
Determine the coefficient of a specific power of x.
Find the term independent of x.

It turns most problems into a simple exercise of finding the correct value of 'r'.

4. What is a common mistake when finding the term independent of x in an expansion, and what is the correct NCERT method?

A common mistake is incorrectly combining the powers of x from the terms in the binomial. The correct method is to first write the general term Tᵣ₊₁. Then, gather all powers of x and simplify them into a single expression involving 'r'. To find the term independent of x, you must set this final exponent of x equal to zero and solve for 'r'. If 'r' is a non-negative integer, a term independent of x exists.

5. How do Vedantu's NCERT Solutions for Class 11 Maths Chapter 7 help master the chapter for the 2025-26 session?

Vedantu's NCERT Solutions for Class 11 Maths Chapter 7 provide detailed, step-by-step answers that align with the latest CBSE 2025-26 pattern. They clarify the correct methodology for complex problems, such as finding coefficients or proving divisibility, helping students avoid common errors and build a strong conceptual foundation for exams.

6. How is the Binomial Theorem applied to solve problems involving divisibility as seen in the NCERT miscellaneous exercise?

To solve divisibility problems, you strategically express a number as a binomial. For instance, to check if 6ⁿ - 5n is divisible by 25, you write 6ⁿ as (1 + 5)ⁿ. You then expand (1 + 5)ⁿ using the theorem. After the expansion, you can isolate terms and show that the entire expression simplifies to a form like 25k, proving it is divisible by 25.

7. What are the key properties of Binomial Coefficients (ⁿCᵣ) needed to solve Chapter 7 exercises efficiently?

Understanding the properties of binomial coefficients is essential for solving problems faster. The key properties include:

Symmetry: ⁿCᵣ = ⁿCₙ₋ᵣ, which means the coefficient of the r-th term from the beginning is equal to the r-th term from the end.
Pascal's Rule: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.
Sum of Coefficients: The sum of all coefficients in the expansion of (a+b)ⁿ can be found by setting a=1 and b=1.

8. How does the problem-solving approach change when finding a coefficient in (axᵖ + b/xq)ⁿ compared to a simple (a+b)ⁿ expansion?

The fundamental approach of using the general term remains the same, but the algebraic complexity increases. For (axᵖ + b/x^q)ⁿ, after writing the general term, you must carefully apply the laws of exponents to combine the powers of x from both parts (x^p(n-r) from the first part and x^-qr from the second). This requires meticulous calculation to find the final exponent of x before equating it to the required power.