NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem 2025-26

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

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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.

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