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# NCERT Solutions for Class 11 Maths Chapter 8: Binomial Theorem - Exercise 8.1

Last updated date: 11th Aug 2024
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## NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem

NCERT Solutions for Class 11 Math Chapter 8 Exercise 8.2  have been carefully compiled and developed  on this page in the pdf format. The solutions are explained in a  stepwise method as per the latest NCERT syllabus issued by the CBSE board.. NCERT Solutions for Class 11 Math Chapter 8- Binomial Theorem Exercise 8.2 is based on the topic General and Middle Term.

 Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 8 - Binomial Theorem Exercise: Exercise - 8.2 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

### What Is The General and  Middle Term Binomial Expression?

In Binomial expression (a + b)n

The first term is nC0an, the second term is nC1an-1b, the third term is nC2an-2b2, and so on. Looking at this pattern, we can say that (r + 1)th  pattern is nCran-rbr.

The  (r + 1)th  pattern is also termed as the general term of an expression (a + b)n. It is represented by Tr + 1.

Therefore, Tr + 1.is equals to nCran-rbr.

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 8 Exercise 82 (Ex 8.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 8 Binomial Theorem Exercise 8.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

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## Access NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem

Exercise 8.2

1. Find the coefficient of ${x^5}$ in ${(x + 3)^n}$

Ans: It is known that ${(r + 1)^th}$ term, $\left( {{T_{r + 1}}} \right).$ in the binomial expansion of ${(a + b)^n}$ is given by ${{\text{T}}_{r + 1}}{ = ^ n }{{\text{C}}_{\text{r}}}{{\text{a}}^{{\text{n}} - {\text{r}}}}{{\mathbf{b}}^r}$

Assuming that ${x^2}$ occurs in the ${(r + 1)^n}$ term of the expansion ${(x + 3)^8}$, we obtain ${T_{r + 1}}{ = ^a}{C_r}{(x)^{8 - r}}{(3)^r}$

Comparing the indices of $x$ in ${x^5}$ in ${T_{r + 1}}$. We obtain $r = 3$ Thus, the coefficient of ${x^5}$ is $^8{C_3}{(3)^3} = \dfrac{{8!}}{{3!5!}} \times {3^3} = \dfrac{{8 \cdot 7 \cdot 6 \cdot 5!}}{{3 \cdot 2.5!}} * {3^3} = 1512$.

1. Find the coefficient of ${{\text{a}}^5}\;{{\text{b}}^{\text{7}}}$ in ${({\text{a}} - 2\;{\text{b}})^{12}}$

Ans: It is known that ${(r + 1)^th}$ term, $\left( {{T_{r + 2}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$

Assuming that ${a^5}{b^7}$ occurs in the ${(r + 1)^{1 = }}$ term of the expansion ${(a - 2b)^{22}}$, we obtain

${T_{r + 1}}{ = ^{12}}{C_r}{(a)^{2 - r}}{( - 2b)^r}{ = ^{12}}{C_r}{( - 2)^r}{(a)^{12 - r}}{(b)^r}$

Comparing the indices of a and $b$ in ${a^5}{b^7}$ in ${T_{r + 2}}$. We obtain $r = 7$ Thus, the coefficient of ${a^5}{b^7}$ is $^2{{\text{C}}_1}{( - 2)^7} = \dfrac{{12!}}{{7!5!}} \cdot {2^7} = \dfrac{{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7!}}{{5 \cdot 4 \cdot 3 \cdot 2 - 7!}} \cdot {( - 2)^7} = - (792)(128) = - 101376$

1. Write the general term in the expansion of ${\left( {{x^2} - y} \right)^6}$

Ans: It is known that the general term ${T_{r + 1}}\left\{ {} \right.$ which is the ${(r + 1)^n}$ term $\}$ in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}.$

Thus, the general term in the expansion of $\left( {{x^2} - {y^6}} \right)$ is ${T_{r + 1}}{ = ^6}{C_r}{\left( {{x^2}} \right)^{6 - r}}{( - y)^r} = {( - 1)^{r}}{{ ^6}C_r} - {x^{12 - 2r}} \cdot {y^r}$

1. Write the general term in the expansion of ${\left( {{x^2} - yx} \right)^{12}},x \ne 0$

Ans: It is known that the general term ${T_{r + 1}}\left\{ {} \right.$ which is the ${(r + 1)^th}$ term $\}$ in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}.$

Thus, the general term in the expansion of ${\left( {{x^2} - yx} \right)^{12}}$ is ${T_{r + 1}}{ = ^{12}}{C_r}{\left( {{x^2}} \right)^{12 - t}}{( - yx)^r} = {( - 1)^{{r}}}{^12}{C_r} - {x^{24 - 2r}} - {y^r} = {( - 1)^{r - 2}}{C_r}{x^{24 - r}} \cdot {y^r}$

1.  Find the ${4^{{\text{th }}}}$ term in the expansion of ${(x - 2y)^{12}}$.

Ans: It is known ${(r + 1)^n}$ term, ${T_{r + 1}}$. in the binomial expansion of ${(a + b)^n}$ is given by ${{\text{T}}_{{\text{r}} + 1}}{ = ^n}{{\text{C}}_r}{{\mathbf{a}}^{n - r}}\;{{\text{b}}^r}.$

Thus, the ${4^{{\text{th }}}}$ term in the expansion of ${\left( {{x^2} - 2y} \right)^{12}}$ is ${T_4} = {T_{3 + 1}}{ = ^{12}}{C_3}{(x)^{12 - 3}}{( - 2y)^3} = {( - 1)^3} \cdot \dfrac{{12!}}{{319!}} \cdot {x^9} \cdot {(2)^3} - {y^3} = - \dfrac{{12 \cdot 11 \cdot 10}}{{3 \cdot 2}} \cdot {(2)^3}{x^9}{y^3} = - 1760{x^9}{y^3}$

1. Find the ${13^{{\text{th }}}}$ term in the expansion of ${\left( {9x - \dfrac{1}{{3\sqrt x }}} \right)^{18}},x \ne 0$

Ans: It is known ${({\text{r}} + 1)^{\text{th}}}$ term, ${T_{r + 1}}$, in the binomial expansion of ${({\text{a}} + {\text{b}})^n}$ is given by ${{\text{T}}_{\varepsilon + 1}}{ = ^n}{{\text{C}}_r}{{\text{a}}^{n - 1}}\;{{\text{b}}^r}$

Thus, the ${13^{{\text{th }}}}$ term in the expansion of ${\left( {9x - \dfrac{1}{{3\sqrt x }}} \right)^{18}}$ is ${T_{13}} = {T_{12 + 1}}{ = ^{18}}{C_{12}}{(9x)^{18 - 12}}{\left( { - \dfrac{1}{{3\sqrt x }}} \right)^{12}}$

$= {( - 1)^{12}}\dfrac{{18!}}{{12!6!}}{(9)^6}{(x)^6}{\left( {\dfrac{1}{3}} \right)^{12}}{\left( {\dfrac{1}{{\sqrt x }}} \right)^{12}}$

$= \dfrac{{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13.12!}}{{12! \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}} \cdot {x^6}\left( {\dfrac{1}{{{x^6}}}} \right) \cdot {3^{12}}\left( {\dfrac{1}{{{3^{12}}}}} \right)\quad \left[ {{9^6} = {{\left( {{3^2}} \right)}^6} = {3^{12}}} \right]$

$= 18564$

1. Find the middle terms in the expansions of ${\left( {3 - \dfrac{{{x^3}}}{6}} \right)^7}$

Ans: It is known that in the expansion of ${(a + b)^n}$, in $n$ is odd, then there are two middle terms, Namely ${\left( {\dfrac{{n + 1}}{2}} \right)^{\text{th}}}$ term and ${\left( {\dfrac{{n + 1}}{2} + 1} \right)^{\text{th}}}$ term.

Therefore, the middle terms in the expansion ${\left( {3 - \dfrac{{{x^3}}}{6}} \right)^7}\operatorname{are} {\left( {\dfrac{{7 + 1}}{2}} \right)^{\text{th}}} = {4^{{\text{th }}}}$ and ${\left( {\dfrac{{7 + 1}}{2} + 1} \right)^{{\text{th }}}} = {5^{{\text{th }}}}$ term ${T_4} = {T_{3 + 1}}{ = ^7}{C_2}{(3)^{7 - 3}}{\left( { - \dfrac{{{x^3}}}{6}} \right)^3} = {( - 1)^3}\dfrac{{7!}}{{3!4!}} - {3^4} \cdot \dfrac{{{x^2}}}{{{6^3}}}$

$= - \dfrac{{7 \cdot 6 \cdot 5.4!}}{{3 \cdot 2.4!}} \cdot {3^4} \cdot \dfrac{1}{{{2^3} \cdot {3^3}}} \cdot {x^9} = - \dfrac{{105}}{8}{x^9 }$

${{\text{T}}_{\text{5}}} = {{\text{T}}_{4 + 1}}{ = ^7}{{\text{C}}_4}{(3)^{7 - 4}}{\left( { - \dfrac{{{{\text{x}}^3}}}{6}} \right)^4} = {( - 1)^4}\dfrac{{71}}{{4!3!}} \cdot {3^3} \cdot \dfrac{{{{\text{x}}^{22}}}}{{{6^4}}}$

$= \dfrac{{7 \cdot 6 \cdot 5.4!}}{{4! \cdot 3 \cdot 2}} \cdot \dfrac{{{3^3}}}{{{2^4} \cdot {3^4}}} \cdot {x^{12}} = \dfrac{{35}}{{48}}{x^{12}}$

Thus, the middle terms in the expansion of ${\left( {3 - \dfrac{{{x^3}}}{6}} \right)^7}$ are $- \dfrac{{105}}{8}{x^3}$ and $\dfrac{{35}}{{48}}{{\text{x}}^{12}}$.

1.  Find the middle terms in the expansion of ${\left( {\dfrac{x}{3} + 9y} \right)^{10}}$

Ans: It is known that in the expansion of ${(a + b)^n}$, in $n$ is even, then the middle term is ${\left( {\dfrac{n}{2} + 1} \right)^n}$ term.

Therefore, the middle term in the expansion of ${\left( {\dfrac{x}{3} + 9y} \right)^{10}}$ is ${\left( {\dfrac{{10}}{2} + 1} \right)^n} = {6^{{\text{th }}}}$ ${T_4} = {T_{5 + 1}}{ = ^{30}}{C_5}{\left( {\dfrac{x}{3}} \right)^{10 - 5}}{(9y)^5} = \dfrac{{10!}}{{5!5!}} \cdot \dfrac{{{x^5}}}{{{3^5}}} \cdot {9^5} \cdot {y^5}$

$= \dfrac{{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6.5!}}{{5 \cdot 4 \cdot 3 \cdot 2.5!}} \cdot \dfrac{1}{{{3^5}}} \cdot {3^{10}} \cdot {x^2}{y^3}$

$\left[ {{9^5} = {{\left( {{3^2}} \right)}^5} = {3^{10}}} \right]$

$= 252 \times {3^5} \cdot {x^5} \cdot {y^5} = 6123{x^5}{y^5}$

Thus, the middle term in the expansion of ${\left( {\dfrac{x}{3} + 9y} \right)^{10}}$ is $61236{x^5}{y^5}$.

1.  In the expansion of ${(1 + a)^{m + n}}$, prove that coefficients of ${a^m}$ and ${a^n}$ are equal.

Ans: It is known that ${(r + 1)^{th}}$ term, $\left( {{T_{r + 1}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$

Assuming that a" occurs in the ${(r + 1)^{{\text{th }}}}$ term of the expansion ${(1 + a)^{n + m}}$, we obtain ${T_{r + 1}}{ = ^{m + n}}{C_r}{(1)^{n + m - r}}{(a)^r}{ = ^{m + n}}{C_r}{a^r}$

Comparing the indices of a in ${a^n}$ in ${T_{r + 1}}$. We obtain $r = m$ Therefore, the coefficient of ${a^n}$ is $^{m + n}{C_m} = \dfrac{{(m + n)!}}{{m!(m + n - m)!}} = \dfrac{{(m + n)!}}{{m!n!}}$

Assuming that ${a^n}$ occurs in the ${(k + 1)^{{\text{th }}}}$ term of the expansion ${(1 + a)^{m + n}}$, we obtain ${T_{k + 1}}{ = ^{m + n}}{C_k}{(1)^{m + n - k}}{(a)^k}{ = ^{m + n}}{C_k}{(a)^k}$

Comparing the indices of a in ${a^n}$ and in ${T_{k - 1}}$. We obtain ${\mathbf{k}} = {\mathbf{n}}$

Therefore, the coefficient of ${{\mathbf{a}}^{\text{n}}}$ is $^{m + n}{{\text{C}}_n} = \dfrac{{(m + n)!}}{{n!(m + n - n)!}} = \dfrac{{(m + n)!}}{{n!m!}} \ldots \ldots .(2)$

Thus, from (1) and (2), it can be observed that the coefficients of ${a^m}$ and ${a^n}$ in the expansion of ${(1 + a)^{m + n}}$ are equal.

1. The coefficients of the ${(r - 1)^{th}},{r^{th}}$ and ${(r + 1)^{th}}$ terms in the expansion of ${(x + 1)^n}$ are in the ratio $1: 3: 5$. Find $n$ and $r$.

Ans: It is known that ${(k + 1)^{th}}$ term, $\left( {{T_{k + 1}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{k + 1}}{ = ^n}{{\mathbf{C}}_k}{{\mathbf{a}}^{n - k}}{{\mathbf{b}}^k}$

Therefore, ${(r - 1)^{{\text{n }}}}$ term in the expansion of ${(x + 1)^n }$ is

$(r + 1)$ term in the expansion of ${(x + 1)^n}$ is ${T_{r - 1}}{ = ^n}{C_r}{(x)^{n - r}}{(1)^r}{ = ^n}{C_r}{x^{n - r}}$

${{\mathbf{r}}^{\text{e}}}$ term in the expansion of ${({\text{x}} + 1)^{\text{n}}}$ is ${T_r}{ = ^n}{C_{r - 1}}{(x)^{n - 1r - 1}}{(1)^{(r - 2)}}{ = ^n}{C_{r - 1}}{x^{n - r + 1}}$

Therefore, the coefficients of the ${(r - 1)^{{\text{th }}}},{r^{{\text{th }}}}$ and ${(r + 1)^{{\text{th }}}}$ terms in the expansion of ${(x + 1)^n}$ $^n{{\text{C}}_{r - 2}}{,^n}{{\text{C}}_{r - 1}}$, and $^n{{\text{C}}_r}$ are respectively. Since these coefficients are in the ratio 1: 3: 5, we obtain $\dfrac{{^n{C_{r - 2}}}}{{^n{C_{r - 1}}}} = \dfrac{1}{3}$ and $\dfrac{{^n{C_{r - 1}}}}{{^n{C_r}}} = \dfrac{3}{5}$

$\dfrac{{^n{C_{r - 2}}}}{{^n{C_{r - 1}}}} = \dfrac{{n!}}{{(r - 2)!(n - r + 2)!}} \times \dfrac{{(r - 1)!(n - r + 1)!}}{{n!}} = \dfrac{{(r - 1)(r - 2)!(n - r + 1)!}}{{(r - 2)!(n - r + 2)!(n - r + 1)!}}$

$= \dfrac{{{\mathbf{r}} - 1}}{{{\mathbf{n}} - {\mathbf{r}} + 2}}$

$\therefore \dfrac{{{\mathbf{r}} - 1}}{{{\mathbf{n}} - {\mathbf{r}} + 2}} - \dfrac{1}{3}$

$\Rightarrow 3x - 3 = n - r + 2$

$\Rightarrow n - 4r + 5 = 0 \ldots .(1)$

$\dfrac{{^n{{\text{C}}_{r - 1}}}}{{^n{{\text{C}}_r}}} = \dfrac{{{\text{n}}!}}{{({\text{r}} - 1)!({\text{n}} - {\text{r}} + 1)}} \times \dfrac{{{{\text{r}}!}({\text{n}} - {\text{r}})!}}{{{\text{n}}!}} = \dfrac{{{\text{r}}({\text{r}} - 1)!({\text{n}} - {\text{r}})!}}{{({\text{r}} - 1)!({\text{n}} - {\text{r}} + 1)({\text{n}} - {\text{r}})!}}$

$= \dfrac{{\mathbf{r}}}{{{\mathbf{n}} - {\mathbf{r}} + {\mathbf{1}}}}$

$\therefore \dfrac{r}{{n - r + 1}} = \dfrac{3}{5}$

$\Rightarrow 5r = 3n - 3r + 3$

$\Rightarrow 3n - 8r + 3 = 0$

Multiplying (1) by 3 and subtracting it from (2), we obtain $4r - 12 = 0$

$\Rightarrow {\mathbf{r}} = 3$

Putting the value of $r$ in $(1)$, we obtain $n$ $- 12 + 5 = 0$

$\Rightarrow {\text{n}} = 7$

Thus, $n = 7$ and ${\mathbf{r}} = 3$

1. Prove that the coefficient of ${x^n}$ in the expansion of ${(1 + x)^{2n}}$ is twice the coefficient of ${x^n}$ in the expansion of ${(1 + x)^{2n - 1}}$.

Ans: It is known that ${(r + 1)^{{\text{th }}}}$ term, $\left( {{T_{r + 1}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{{\text{C}}_r}{{\text{a}}^{n - r}}\;{{\text{b}}^r}$

Assuming that ${x^n}$ occurs in the ${(r + 1)^n}$ term of the expansion of ${(1 + x)^{2n}}$, we obtain ${T_{r + 1}}{ = ^{2n}}{{\text{C}}_r}{(1)^{2n - r}}{({\text{x}})^r}{ = ^{2n}}{{\text{C}}_{\text{r}}}{({\text{x}})^r}$

Comparing the indices of $x$ in ${x^n}$ and in ${T_{r + 1}}$, we obtain $r = n$ Therefore, the coefficient of ${x^n}$ in the expansion of ${(1 + x)^{2n}}$ is $^{2n}{{\text{C}}_n} = \dfrac{{(2n)!}}{{n!(2n - n)!}} = \dfrac{{(2n)!}}{{n!n!}} = \dfrac{{(2n)!}}{{{{(n!)}^2}}} \ldots \ldots .(1)$

Assuming that ${x^n}$ occurs in the ${(k + 1)^{{\text{in }}}}$ term of the expansion of ${(1 + x)^{2n - 2}}$, we obtain

Comparing the indices of $x$ in ${x^n}$ and in ${T_{k + 1}}$, we obtain $k = n$ Therefore, the coefficient of ${x^n}$ in the expansion of ${(1 + x)^{2n - 1}}$ is $^{2n - 1}{{\text{C}}_n} = \dfrac{{(2n - 1)!}}{{n!(2n - 1 - n)!}} = \dfrac{{(2n - 1)!}}{{n!(n - 1)!}}$

$= \dfrac{{2n \cdot (2n - 1)!}}{{2n \cdot n!(n - 1)!}} = \dfrac{{(2n)!}}{{2n!n!}} = \dfrac{1}{2}\left[ {\dfrac{{(2n)!}}{{{{(n!)}^2}}}} \right] \ldots \ldots .(2)$

From (1) and (2), it is observed that $\dfrac{1}{2}\left( {^{2n}{C_n}} \right){ = ^{2n - 1}}{C_n}$

${ \Rightarrow ^{2n}}{{\text{C}}_n} = 2\left( {^{2n - 1}{{\text{C}}_n}} \right)$

Therefore, the coefficient of ${x^n}$ expansion of ${(1 + x)^{2n}}$ is twice the coefficient of ${x^n}$ in the expansion of ${(1 + x)^{2n - 1}}$. Hence proved.

1.  Find a positive value of $m$ for which the coefficient of ${x^2}$ in the expansion ${(1 + x)^m}$ is 6 .

Ans: It is known that ${(r + 1)^{{\text{th }}}}$ term, $\left( {{T_{r + 1}}} \right)$, in the binomial expansion of ${(a + b)^n}$ is given by ${T_{r + 1}}{ = ^n}{C_r}{a^{n - r}}{b^r}$

Assuming that ${{\text{x}}^2}$ occurs in the ${({\text{r}} + 1)^{\text{n}}}$ term of the expansion of ${(1 + {\text{x}})^{\text{n}}}$, we obtain ${T_{r + 1}}{ = ^n}{C_r}{(1)^{n - r}}{(x)^r}{ = ^m}{C_n}{(x)^r}$

Comparing the indices of $x$ in ${x^2}$ and in ${T_{r + 1}}$, we obtain $r = 2$ Therefore, the coefficient of ${x^2}$ is $^m{{\text{C}}_2}$ It is given that the coefficient of ${x^2}$ in the expansion ${(1 + x)^m}$ is 6 . ${\therefore ^m}{{\text{C}}_2} = 6$

$\Rightarrow \dfrac{{m!}}{{2!(m - 2)!}} = 6$

$\Rightarrow \dfrac{{m(m - 1)(m - 2)!}}{{2 \times (m - 2)!}} = 6$

$\Rightarrow m(m - 1) = 12$

$\Rightarrow {m^2} - m - 12 = 0$

$\Rightarrow {m^2} - 4m + 3m - 12 = 0$

$\Rightarrow m(m - 4) + 3(m - 4) = 0$

$\Rightarrow (m - 4)(m + 3) = 0$

$\Rightarrow (m - 4) = 0$ or $(m + 3) = 0$

$\Rightarrow m = 4$ or $m = - 3$

Thus, the positive value of $m$, for which the coefficient of ${x^2}$ in the expansion ${(1 + x)^n}$ is 6 , is $4 .$

## NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Exercise 8.2

Opting for the NCERT solutions for Ex 8.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 8.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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## FAQs on NCERT Solutions for Class 11 Maths Chapter 8: Binomial Theorem - Exercise 8.1

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