NCERT Exemplar Solutions Class 11 Maths Binomial Theorem — Long Answer and Objective Type Q12–24

NCERT Exemplar Problems, Class 11 Mathematics — Binomial Theorem, Long Answer and Objective Type questions 12 to 24. All 13 questions solved.

Questions 12 to 17 are the longer ones: a middle term (12, 13), a ratio of terms from the two ends (14), two identities from splitting $(x+a)^n$ into odd and even terms (15), a coefficient formula (16), and a product with a binomial (17). Questions 18 to 24 are multiple choice and lean on three facts: the number of terms, the symmetry ${}^nC_r = {}^nC_{n-r}$, and the ratio ${}^nC_r : {}^nC_{r+1} = (r+1) : (n-r)$. The printed key agrees with every answer here.

Key insight. Most of these questions turn on one move: write the general term $T_{r+1} = {}^nC_r a^{n-r}b^r$ and compare it with whatever is asked, whether that is “equals $1120$”, “same coefficient”, or “power equals $p$”. The rest is solving a small equation in $r$ or $n$.

Question 12

If $p$ is a real number and if the middle term in the expansion of $\left(\dfrac{p}{2} + 2\right)^8$ is $1120$, find $p$.

Finding the middle term is no longer part of the current NCERT Class 11 syllabus, but it stays in the Exemplar, so it is solved here.

Solution. $n = 8$ is even, so the middle term is $T_5$:

$$T_5 = {}^8C_4\left(\frac p2\right)^4 2^4 = 70\cdot\frac{p^4}{16}\cdot 16 = 70p^4.$$

Setting $70p^4 = 1120$ gives $p^4 = 16$. For real $p$, $p^2 = 4$ (as $p^2 = -4$ is impossible), so $p = \pm 2$.

Question 13

Show that the middle term in the expansion of $\left(x – \dfrac1x\right)^{2n}$ is $\dfrac{1\times 3\times 5\times\cdots\times(2n-1)}{n!}\times(-2)^n$.

Solution. There are $2n + 1$ terms, so the middle one is $T_{n+1}$:

$$T_{n+1} = {}^{2n}C_n\, x^{n}\left(-\frac1x\right)^n = (-1)^n\,\frac{(2n)!}{n!\,n!}.$$

The powers of $x$ cancel, so the middle term is free of $x$. Now split $(2n)!$ into its odd and even factors:

$$(2n)! = \big[1\cdot3\cdot5\cdots(2n-1)\big]\big[2\cdot4\cdot6\cdots 2n\big] = \big[1\cdot3\cdots(2n-1)\big]\cdot 2^n\, n!.$$

Substituting,

$$T_{n+1} = (-1)^n\frac{1\cdot3\cdots(2n-1)\cdot 2^n\, n!}{n!\,n!} = \frac{1\cdot3\cdot5\cdots(2n-1)}{n!}\,(-2)^n.$$

Question 14

Find $n$ in the binomial $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^n$ if the ratio of the $7^{\text{th}}$ term from the beginning to the $7^{\text{th}}$ term from the end is $\dfrac16$.

Solution. The $7^{\text{th}}$ term from the beginning is $T_7 = {}^nC_6\,(2^{1/3})^{n-6}(3^{-1/3})^6$. The $7^{\text{th}}$ term from the end is the $7^{\text{th}}$ term of the expansion taken in the reverse order, that is of $\left(3^{-1/3} + 2^{1/3}\right)^n$, so it is ${}^nC_6\,(3^{-1/3})^{n-6}(2^{1/3})^6$. The binomial coefficients cancel in the ratio:

$$\frac{(2^{1/3})^{n-12}}{(3^{-1/3})^{n-12}} = \left(2^{1/3}\cdot 3^{1/3}\right)^{n-12} = 6^{\frac{n-12}{3}}.$$

Setting this equal to $\dfrac16 = 6^{-1}$ gives $\dfrac{n-12}{3} = -1$, so $n = 9$.

Question 15

In the expansion of $(x+a)^n$, if the sum of odd terms is denoted by $O$ and the sum of even terms by $E$, prove that (i) $O^2 – E^2 = (x^2 – a^2)^n$ (ii) $4OE = (x+a)^{2n} – (x-a)^{2n}$.

Solution. The odd-numbered terms $T_1, T_3, T_5, \dots$ carry even powers of $a$, and the even-numbered terms carry odd powers of $a$. Replacing $a$ by $-a$ therefore leaves $O$ unchanged and flips the sign of $E$. So

$$(x+a)^n = O + E, \qquad (x-a)^n = O – E.$$

(i) Multiply the two: $(O+E)(O-E) = O^2 – E^2 = (x+a)^n(x-a)^n = (x^2-a^2)^n$.

(ii) Square and subtract: $(O+E)^2 – (O-E)^2 = 4OE$, which equals $(x+a)^{2n} – (x-a)^{2n}$.

Question 16

If $x^p$ occurs in the expansion of $\left(x^2 + \dfrac1x\right)^{2n}$, prove that its coefficient is $\dfrac{(2n)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}$.

Solution. The general term is

$$T_{r+1} = {}^{2n}C_r\,(x^2)^{2n-r}\left(\frac1x\right)^r = {}^{2n}C_r\,x^{4n-2r-r} = {}^{2n}C_r\,x^{4n-3r}.$$

If $x^p$ occurs, then $4n – 3r = p$, so $r = \dfrac{4n-p}{3}$ and $2n – r = \dfrac{6n – 4n + p}{3} = \dfrac{2n+p}{3}$. The coefficient is

$${}^{2n}C_r = \frac{(2n)!}{r!\,(2n-r)!} = \frac{(2n)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}.$$

Question 17

Find the term independent of $x$ in the expansion of $(1 + x + 2x^3)\left(\dfrac32x^2 – \dfrac{1}{3x}\right)^9$.

Solution. The general term of the second factor is

$$T_{r+1} = {}^9C_r\left(\frac32\right)^{9-r}\left(-\frac13\right)^r x^{18-3r}.$$

A constant term in the product comes from three pairings, each needing the power of $x$ from the second factor to cancel the power in the first:

• $1 \times$ the $x^0$ term: $18 – 3r = 0$, so $r = 6$.
• $x \times$ the $x^{-1}$ term: $18 – 3r = -1$ has no integer solution, so this gives nothing.
• $2x^3 \times$ the $x^{-3}$ term: $18 – 3r = -3$, so $r = 7$.

For $r = 6$: ${}^9C_6\left(\frac32\right)^3\left(\frac13\right)^6 = 84\cdot\frac{27}{8}\cdot\frac{1}{729} = \frac{7}{18}$.

For $r = 7$: $2\cdot{}^9C_7\left(\frac32\right)^2\left(-\frac13\right)^7 = 2\cdot36\cdot\frac94\cdot\left(-\frac{1}{2187}\right) = -\frac{2}{27}$.

Adding, $\dfrac{7}{18} – \dfrac{2}{27} = \dfrac{21 – 4}{54} = \dfrac{17}{54}$.

Choose the correct answer from the given options in Questions 18 to 24 (M.C.Q.).

Question 18

The total number of terms in the expansion of $(x+a)^{100} + (x-a)^{100}$ after simplification is

Solution. In $(x+a)^{100}$ every term is ${}^{100}C_r x^{100-r}a^r$, and in $(x-a)^{100}$ the same term carries $(-1)^r$. Adding, the terms with odd $r$ cancel and those with even $r$ double. The surviving terms have $r = 0, 2, 4, \dots, 100$, which is $51$ terms.

Question 19

Given the integers $r > 1$, $n > 2$, and coefficients of the $(3r)^{\text{th}}$ and $(r+2)^{\text{nd}}$ terms in the binomial expansion of $(1+x)^{2n}$ are equal, then

Solution. The coefficients are ${}^{2n}C_{3r-1}$ and ${}^{2n}C_{r+1}$. They are equal when the lower indices are equal or add up to $2n$.

• $3r – 1 = r + 1$ gives $r = 1$, which the condition $r > 1$ rules out.
• $(3r-1) + (r+1) = 2n$ gives $4r = 2n$, so $n = 2r$.

Question 20

The two successive terms in the expansion of $(1+x)^{24}$ whose coefficients are in the ratio $1:4$ are

Solution. Let the two terms have coefficients ${}^{24}C_r$ and ${}^{24}C_{r+1}$. Using ${}^nC_r : {}^nC_{r+1} = (r+1):(n-r)$,

$$\frac{{}^{24}C_r}{{}^{24}C_{r+1}} = \frac{r+1}{24-r} = \frac14 \;\Rightarrow\; 4r + 4 = 24 – r \;\Rightarrow\; r = 4.$$

The coefficients ${}^{24}C_4$ and ${}^{24}C_5$ belong to $T_5$ and $T_6$. Check: ${}^{24}C_4 = 10626$ and ${}^{24}C_5 = 42504 = 4 \times 10626$.

Question 21

The coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ and $(1+x)^{2n-1}$ are in the ratio

Solution. The coefficients are ${}^{2n}C_n$ and ${}^{2n-1}C_n$. Then

$$\frac{{}^{2n}C_n}{{}^{2n-1}C_n} = \frac{(2n)!}{n!\,n!}\cdot\frac{n!\,(n-1)!}{(2n-1)!} = \frac{2n}{n} = 2.$$

Question 22

If the coefficients of the $2^{\text{nd}}$, $3^{\text{rd}}$ and the $4^{\text{th}}$ terms in the expansion of $(1+x)^n$ are in A.P., then the value of $n$ is

Solution. The A.P. condition is $2\,{}^nC_2 = {}^nC_1 + {}^nC_3$, that is

$$n(n-1) = n + \frac{n(n-1)(n-2)}{6}.$$

Divide by $n$ and multiply by $6$: $6(n-1) = 6 + (n-1)(n-2)$, which simplifies to $n^2 – 9n + 14 = 0$, so $n = 2$ or $n = 7$. The value $n = 2$ must be rejected: $(1+x)^2$ has only three terms, so there is no fourth term whose coefficient could be in the progression. (It satisfies the equation only because ${}^2C_3 = 0$.) That leaves $n = 7$: the coefficients $7, 21, 35$ are in A.P. with common difference $14$.

Question 23

If $A$ and $B$ are coefficients of $x^n$ in the expansions of $(1+x)^{2n}$ and $(1+x)^{2n-1}$ respectively, then $\dfrac AB$ equals

Solution. This is Question 21 asked as a single number: $A = {}^{2n}C_n$, $B = {}^{2n-1}C_n$, and $\dfrac AB = \dfrac{2n}{n} = 2$.

Question 24

If the middle term of $\left(\dfrac1x + x\sin x\right)^{10}$ is equal to $7\dfrac78$, then the value of $x$ is

Finding the middle term is no longer part of the current NCERT Class 11 syllabus, but it stays in the Exemplar, so it is solved here.

Solution. The middle term of a power $10$ is $T_6$:

$$T_6 = {}^{10}C_5\left(\frac1x\right)^5(x\sin x)^5 = 252\sin^5 x.$$

The powers of $x$ cancel. Also $7\frac78 = \frac{63}{8}$, so $252\sin^5x = \frac{63}{8}$ gives $\sin^5 x = \frac{63}{8\cdot252} = \frac1{32}$, and $\sin x = \frac12$. The general solution of $\sin x = \sin\frac\pi6$ is $x = n\pi + (-1)^n\dfrac\pi6$. Option (A) would miss the solutions $\dfrac{5\pi}{6} + 2n\pi$, and (B) is wrong for odd $n$.

Common mistakes

• Question 12: stopping at $p^4 = 16$ and writing $p = 2$ only. The question says $p$ is real, so both $2$ and $-2$ satisfy $p^4 = 16$.
• Question 14: reading the $7^{\text{th}}$ term from the end as $T_{n-7}$. The $k^{\text{th}}$ term from the end is $T_{n-k+2}$, so the $7^{\text{th}}$ from the end is $T_{n-5}$. Reversing the binomial, as done above, avoids the index slip altogether.
• Question 15: forgetting that $(x-a)^n = O – E$ only because odd-numbered terms carry even powers of $a$. Students sometimes assign the sign change to $O$ instead of $E$ and the algebra then fails to close.
• Question 17: counting only the $r = 6$ term and missing the $2x^3$ pairing at $r = 7$. When a product has several terms, test each one against the target power.
• Question 18: answering $101$, thinking both expansions have $101$ terms and adding the counts. The odd-$r$ terms cancel and the even-$r$ terms merge, leaving $51$.
• Question 22: accepting $n = 2$ because it satisfies the equation. The fourth term must exist, so $n \geq 3$.
• Question 24: choosing (A) or (B) after getting $\sin x = \frac12$. The general solution has the factor $(-1)^n$ because $\sin\frac{5\pi}6 = \sin\frac\pi6$.

Practise next

• The rest of this unit: Short Answer Type Q1–11 and Fill in the Blanks and True or False Q25–40.
• NCERT Class 11 Maths, Binomial Theorem, Exercise 7.1 for general-term practice.
• NCERT Class 11 Maths, Binomial Theorem, Miscellaneous Exercise for coefficient comparison problems.