NCERT Exemplar Solutions Class 11 Maths Binomial Theorem — Short Answer Type Q1–11

NCERT Exemplar Problems, Class 11 Mathematics — Binomial Theorem, Short Answer Type questions 1 to 11. All 11 questions solved.

Every question here runs on one formula, the general term of $(a+b)^n$:

$$T_{r+1} = {}^nC_r\, a^{\,n-r} b^{\,r}.$$

Questions 1, 2, 4, 6 and 7 ask you to choose $r$ so that the power of $x$ comes out as required. Question 3 expands a product by taking one coefficient at a time, Question 5 asks for the middle term, Questions 8 to 10 use properties of binomial coefficients, and Question 11 splits a four-term expression into a product of two binomials. The printed key agrees with every answer on this page.

Key insight. Never expand. Write the general term, collect the power of $x$ into a single expression in $r$, set it equal to the power you want, solve for $r$, and substitute back. Check that $r$ is a whole number between $0$ and $n$; if it is not, the term does not exist.

Question 1

Find the term independent of $x$, $x \neq 0$, in the expansion of $\left(\dfrac{3x^2}{2} – \dfrac{1}{3x}\right)^{15}$.

Solution. The general term is $T_{r+1} = {}^{15}C_r \left(\dfrac{3x^2}{2}\right)^{15-r}\left(-\dfrac{1}{3x}\right)^r$. The power of $x$ is $2(15-r) – r = 30 – 3r$. For a term free of $x$ this must be zero, so $r = 10$. Substituting,

$$T_{11} = {}^{15}C_{10}\left(\frac{3}{2}\right)^{5}\left(-\frac{1}{3}\right)^{10} = 3003 \cdot \frac{3^5}{2^5}\cdot\frac{1}{3^{10}} = \frac{3003}{2^5\cdot 3^5} = \frac{3003}{7776} = \frac{1001}{2592}.$$

The sign is positive because $(-1)^{10} = 1$.

Question 2

If the term free from $x$ in the expansion of $\left(\sqrt{x} – \dfrac{k}{x^2}\right)^{10}$ is $405$, find the value of $k$.

Solution. $T_{r+1} = {}^{10}C_r (\sqrt{x})^{10-r}\left(-\dfrac{k}{x^2}\right)^r = {}^{10}C_r(-k)^r x^{\frac{10-r}{2} – 2r}$. Setting $\dfrac{10-r}{2} – 2r = 0$ gives $10 – 5r = 0$, so $r = 2$. The term is ${}^{10}C_2 k^2 = 45k^2$. Then $45k^2 = 405$, so $k^2 = 9$. Because $r$ is even, $k$ enters only as $k^2$, so both signs work.

Question 3

Find the coefficient of $x$ in the expansion of $(1 – 3x + 7x^2)(1 – x)^{16}$.

Solution. Only two products give a power of $x$ equal to one: the $1$ from the first bracket times the $x$-term of $(1-x)^{16}$, and the $-3x$ from the first bracket times the constant term of $(1-x)^{16}$. The $7x^2$ cannot contribute, because it already has degree $2$. In $(1-x)^{16}$ the constant term is $1$ and the coefficient of $x$ is $-{}^{16}C_1 = -16$. So

$$1\cdot(-16) + (-3)\cdot 1 = -19.$$

Question 4

Find the term independent of $x$ in the expansion of $\left(3x – \dfrac{2}{x^2}\right)^{15}$.

Solution. $T_{r+1} = {}^{15}C_r (3x)^{15-r}\left(-\dfrac{2}{x^2}\right)^r = {}^{15}C_r\, 3^{15-r}(-2)^r x^{15-3r}$. The power vanishes when $r = 5$:

$$T_6 = {}^{15}C_5\, 3^{10}(-2)^5 = -3003 \cdot 59049 \cdot 32 = -5674372704.$$

The key leaves this as $-{}^{15}C_5\,3^{10}\,2^5$; the number above is the same value, checked in sympy.

Question 5

Find the middle term (terms) in the expansion of (i) $\left(\dfrac{x}{a} – \dfrac{a}{x}\right)^{10}$ (ii) $\left(3x – \dfrac{x^3}{6}\right)^9$.

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 (i). With $n = 10$, which is even, there is a single middle term, $T_{\frac{n}{2}+1} = T_6$:

$$T_6 = {}^{10}C_5\left(\frac{x}{a}\right)^5\left(-\frac{a}{x}\right)^5 = -252 \cdot \frac{x^5}{a^5}\cdot\frac{a^5}{x^5} = -252.$$

Solution (ii). With $n = 9$, which is odd, there are two middle terms, $T_5$ and $T_6$:

$$T_5 = {}^9C_4 (3x)^5\left(-\frac{x^3}{6}\right)^4 = 126 \cdot 243x^5\cdot\frac{x^{12}}{1296} = \frac{189}{8}x^{17},$$

$$T_6 = {}^9C_5 (3x)^4\left(-\frac{x^3}{6}\right)^5 = -126\cdot 81x^4\cdot\frac{x^{15}}{7776} = -\frac{21}{16}x^{19}.$$

Question 6

Find the coefficient of $x^{15}$ in the expansion of $(x – x^2)^{10}$.

Solution. $T_{r+1} = {}^{10}C_r\, x^{10-r}(-x^2)^r = {}^{10}C_r(-1)^r x^{10+r}$. For $x^{15}$ we need $r = 5$, and the coefficient is ${}^{10}C_5(-1)^5 = -252$.

Question 7

Find the coefficient of $\dfrac{1}{x^{17}}$ in the expansion of $\left(x^4 – \dfrac{1}{x^3}\right)^{15}$.

Solution. $T_{r+1} = {}^{15}C_r\, x^{4(15-r)}\left(-x^{-3}\right)^r = {}^{15}C_r(-1)^r x^{60-7r}$. We need $60 – 7r = -17$, so $7r = 77$ and $r = 11$. The coefficient is ${}^{15}C_{11}(-1)^{11} = -1365$.

Question 8

Find the sixth term of the expansion $\left(y^{1/2} + x^{1/3}\right)^n$, if the binomial coefficient of the third term from the end is $45$.

Solution. The third term from the end has the same binomial coefficient as the third term from the beginning, ${}^nC_2$, because ${}^nC_r = {}^nC_{n-r}$. So $\dfrac{n(n-1)}{2} = 45$, giving $n^2 – n – 90 = 0$, so $(n-10)(n+9) = 0$. Since $n$ is a positive integer, $n = 10$. Then

$$T_6 = {}^{10}C_5\left(y^{1/2}\right)^5\left(x^{1/3}\right)^5 = 252\, y^{5/2}x^{5/3}.$$

Question 9

Find the value of $r$, if the coefficients of the $(2r+4)^{\text{th}}$ and $(r-2)^{\text{th}}$ terms in the expansion of $(1+x)^{18}$ are equal.

Solution. The coefficient of the $m^{\text{th}}$ term is ${}^{18}C_{m-1}$. So we need ${}^{18}C_{2r+3} = {}^{18}C_{r-3}$. Two binomial coefficients ${}^nC_a$ and ${}^nC_b$ are equal when $a = b$ or $a + b = n$.

• $2r + 3 = r – 3$ gives $r = -6$, which is not allowed.
• $(2r+3) + (r-3) = 18$ gives $3r = 18$, so $r = 6$.

Check: the $16^{\text{th}}$ and $4^{\text{th}}$ terms have coefficients ${}^{18}C_{15} = {}^{18}C_3$. They are equal.

Question 10

If the coefficients of the second, third and fourth terms in the expansion of $(1+x)^{2n}$ are in A.P., show that $2n^2 – 9n + 7 = 0$.

Solution. The three coefficients are ${}^{2n}C_1 = 2n$, ${}^{2n}C_2 = n(2n-1)$ and ${}^{2n}C_3 = \dfrac{2n(2n-1)(2n-2)}{6} = \dfrac{2n(2n-1)(n-1)}{3}$. Three numbers are in A.P. when twice the middle one equals the sum of the other two:

$$2n(2n-1) = 2n + \frac{2n(2n-1)(n-1)}{3}.$$

Divide by $2n$ (valid since $n \neq 0$): $(2n-1) = 1 + \dfrac{(2n-1)(n-1)}{3}$. Multiply by $3$:

$$6n – 3 = 3 + (2n^2 – 3n + 1) \;\Rightarrow\; 2n^2 – 9n + 7 = 0.$$

The roots are $n = 1$ and $n = \tfrac72$; the question asks only for the equation.

Question 11

Find the coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3)^{11}$.

Solution. Factorise the base: $1 + x + x^2 + x^3 = (1+x)(1+x^2)$. So the expression is $(1+x)^{11}(1+x^2)^{11}$. To reach $x^4$, the power from $(1+x)^{11}$ (any power) and the power from $(1+x^2)^{11}$ (an even power) must add to $4$:

The total is $330 + 605 + 55 = 990$.

Common mistakes

• Questions 1 and 4: writing the exponent of $x$ as $15 – r + 2r$ instead of $2(15-r) – r$. Students forget that the $2$ in $x^2$ multiplies the whole exponent $15 – r$, or that the $2$ in $\frac1{x^2}$ multiplies $r$.
• Question 2: taking only $k = 3$. Since $r = 2$ is even, $(-k)^2 = k^2$ and the sign of $k$ is invisible, so $k = -3$ is just as valid.
• Question 3: forgetting the cross term $-3x \times 1$ and counting only $1 \times (-16x)$, which gives $-16$ instead of $-19$.
• Question 5(ii): giving one middle term for $n = 9$. An odd $n$ gives $n+1 = 10$ terms, an even count, so the middle falls between two terms and both $T_5$ and $T_6$ count.
• Question 8: treating “third term from the end” as ${}^nC_3$. The third from the end is $T_{n-1}$, with coefficient ${}^nC_{n-2} = {}^nC_2$.
• Question 9: using ${}^{18}C_{2r+4}$ for the $(2r+4)^{\text{th}}$ term. The $m^{\text{th}}$ term has lower index $m – 1$, not $m$.
• Question 11: expanding the four-term base directly. Factorising to $(1+x)(1+x^2)$ is the whole trick.

Practise next

• The rest of this unit: Long Answer and Objective Type Q12–24 and Fill in the Blanks and True or False Q25–40.
• NCERT Class 11 Maths, Binomial Theorem, Exercise 7.1 for the standard expansion questions.
• NCERT Class 11 Maths, Binomial Theorem, Miscellaneous Exercise for coefficient and term-comparison problems.