CBSE Class 11 Maths Important Questions - Chapter 7 Binomial Theorem

1 Mark Questions with Solutions

1. Write the general term in the expansion of $ (a + b)^n $.  

$\textbf{Solution:}$

The general term is $ T_{r+1} = \binom{n}{r} a^{n-r} b^r $.

2.  Find the coefficient of $ x^3 $ in $ (1 + x)^5 $.  

$\textbf{Solution:}$ 

The general term is $ T_{r+1} = \binom{5}{r} x^r $. For $ x^3 $, $ r = 3 $:

$\text{Coefficient} = \binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10.$

3. What is the value of $ \binom{6}{2} + \binom{6}{3} $?  

$\textbf{Solution:}$  

$\binom{6}{2} = 15, \quad \binom{6}{3} = 20.$

$\text{Sum} = 15 + 20 = 35$

4.  Expand $ (1 + x)^2 $.  

$\textbf{Solution:}$  

$(1 + x)^2 = 1 + 2x + x^2$

5. State Pascal’s Identity.  

$\textbf{Solution:}$ 

Pascal’s Identity states:

$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$

6. How many terms are there in the expansion of $ (x + y)^{10} $?  

$\textbf{Solution:}$  

The number of terms is $ n + 1 = 10 + 1 = 11 $.

7. What is the coefficient of the middle term in $ (1 + x)^8 $?  

$\textbf{Solution:}$  

For $ n = 8 $, the middle term is $ T_5 = \binom{8}{4} x^4 $:

$\text{Coefficient} = \binom{8}{4} = 70$

8. Write the expansion of $ (2a - b)^0 $.  

$\textbf{Solution:}$  

Any expression raised to $ 0 $ is $ 1 $.

9. State the formula for the sum of all binomial coefficients in $ (x + y)^n $.  

$\textbf{Solution:}$  

The sum is:

$\sum_{r=0}^n \binom{n}{r} = 2^n$

10. Find the value of $ \binom{8}{3} $.  

$\textbf{Solution:}$  

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

2 Marks Questions with Solutions

1. Expand $ (1 - x)^4 $ using the Binomial Theorem.  

Solution:  

$(1 - x)^4 = 1 - 4x + 6x^2 - 4x^3 + x^4$

2.  Find the coefficient of $ x^5 $ in $ (2 + x)^6 $.  

Solution:  

The general term is $ T_{r+1} = \binom{6}{r} (2)^{6-r} x^r $. For $ x^5 $, $ r = 5 $:

$\text{Coefficient} = \binom{6}{5} (2)^1 = 6 \times 2 = 12$

3. Write the 5th term in the expansion of $ (x - 2y)^8 $.  

Solution:  

The general term is $ T_{r+1} = \binom{8}{r} x^{8-r} (-2y)^r $. For $ r = 4 $:

$T_5 = \binom{8}{4} x^4 (-2y)^4 = 1120x^4y^4$

4.  If $ (a + b)^6 = \sum_{r=0}^6 T_{r+1} $, write the expression for $ T_4 $.  

Solution: 

The general term is $ T_{r+1} = \binom{6}{r} a^{6-r} b^r $. For $ r = 3 $:

$T_4 = \binom{6}{3} a^{3} b^3 = 20a^3b^3$

5. Prove that $ \binom{n}{r} = \binom{n}{n-r} $.  

Solution:  

Using the formula $ \binom{n}{r} = \frac{n!}{r!(n-r)!} $, we see:

$\binom{n}{r} = \binom{n}{n-r}$

3 Marks Questions with Solutions

1. Expand $ (2x - 3)^5 $ completely.  

Solution:  

Using the Binomial Theorem:

$(2x - 3)^5 = \sum_{r=0}^5 \binom{5}{r} (2x)^{5-r} (-3)^r$

Expanding:

$= 32x^5 - 240x^4 + 720x^3 - 1080x^2 + 810x - 243$

2. Find the term containing $ x^4 $ in the expansion of $ \left(1 + \frac{1}{x}\right)^7 $.  

Solution:

The general term is $ T_{r+1} = \binom{7}{r} (1)^{7-r} \left(\frac{1}{x}\right)^r $. For $ x^4 $, $ r = 3 $:

$T_4 = \binom{7}{3} \cdot 1 \cdot \frac{1}{x^3} = 35 \cdot \frac{1}{x^3}$

3. Expand the expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$.

Ans. By using Binomial Theorem, the expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$ can be expanded as

\[\begin{gathered} {\left( {\frac{2}{x} - \frac{x}{2}} \right)^5} = {}^5{C_0}{\left( {\frac{2}{x}} \right)^5} - {}^5{C_1}{\left( {\frac{2}{x}} \right)^4}\left( {\frac{x}{2}} \right) + {}^5{C_2}{\left( {\frac{2}{x}} \right)^3}{\left( {\frac{x}{2}} \right)^2} - {}^5{C_3}{\left( {\frac{2}{x}} \right)^2}{\left( {\frac{x}{2}} \right)^3} + {}^5{C_4}{\left( {\frac{2}{x}} \right)^1}{\left( {\frac{x}{2}} \right)^4} \\ - {}^5{C_5}{\left( {\frac{x}{2}} \right)^5} \\ = \frac{{32}}{{{x^5}}} - 5\left( {\frac{{16}}{{{x^4}}}} \right)\left( {\frac{x}{2}} \right) + 10\left( {\frac{8}{{{x^3}}}} \right)\left( {\frac{{{x^2}}}{4}} \right) - 10\left( {\frac{4}{{{x^2}}}} \right)\left( {\frac{{{x^3}}}{8}} \right) + 5\left( {\frac{2}{x}} \right)\left( {\frac{{{x^4}}}{{16}}} \right) - \frac{{{x^5}}}{{32}} \\ = \frac{{32}}{{{x^5}}} - \frac{{40}}{{{x^3}}} + \frac{{20}}{x} - 5x + \frac{5}{8}{x^3} - \frac{{{x^5}}}{{32}} \\ \end{gathered}\]

4. Find ${\left( {a + b} \right)^4} - {\left( {a - b} \right)^4}$. Hence, evaluate ${\left( {\sqrt 3  + \sqrt 2 } \right)^4} - {\left( {\sqrt 3  - \sqrt 2 } \right)^4}$.

Ans. Using Binomial Theorem, the expressions, ${\left( {a + b} \right)^4}$ and ${\left( {a - b} \right)^4}$ , can be expanded as 

\[\begin{gathered} {\left( {a + b} \right)^4} = {}^4{C_0}{a^4} + {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} + {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} \\ {\left( {a - b} \right)^4} = {}^4{C_0}{a^4} - {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} - {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} \\ \end{gathered} \]

Therefore,

\[\begin{gathered} {\left( {a + b} \right)^4} - {\left( {a - b} \right)^4} = {}^4{C_0}{a^4} + {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} + {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} -  \\ \left[ {{}^4{C_0}{a^4} - {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} - {}^4{C_3}a{b^3} + {}^4{C_4}{b^4}} \right] \\ = 2\left( {{}^4{C_1}{a^3}b + {}^4{C_3}a{b^3}} \right) \\ = 2\left( {4{a^3}b + 4a{b^3}} \right) \\ = 8ab\left( {{a^2} + {b^2}} \right) \\ \end{gathered} \]

By putting $a = \sqrt 3 $ and $b = \sqrt 2 $, we obtain

\[\begin{gathered} {\left( {\sqrt 3  + \sqrt 2 } \right)^4} - {\left( {\sqrt 3  - \sqrt 2 } \right)^4} = 8\left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right)\left[ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} \right] \\ = 8\sqrt 6 \left( {3 + 2} \right) \\ = 40\sqrt 6  \\ \end{gathered} \]

5 Marks Questions

1. 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

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$

\[ = 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}\]

\[ = 27{x^6} - 54a{x^5} + 117{a^2}{x^4} - 116{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}\].

2. 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$.

Practice Questions of Class 11 Maths Chapter 7 Binomial Theorem Class 11 

1. Find the number of terms and the middle term of the following expression:
   (x/3 + 9y)10.

2. Show that 6n - 5n always leaves remainder 1 when divided by 25, using the binomial theorem.

3. Find (x + y)4 - (x - y)4 and then evaluate (√5 + √6)4 - (√5 - √6)4.

4. Find the coefficient of m5 in (m + 3)8.

5. Expand: (x + 1⁄x)6

6. Find the middle terms in the expansions of (3 - a3⁄6)7

7. Find the middle term of (2x - x2⁄4)9.

8. Expand: (x⁄3 + 1⁄x)5

9. In the expansion of (1 + p)q+r, prove that coefficients of pq and pr are equal.

10. Find the term which is independent of x in the expansion of (3⁄2 x2 - 1⁄3x)6.

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Conclusion

The Binomial Theorem is a key topic in Class 11 Maths, providing a systematic approach to expanding expressions and solving complex problems efficiently. By practising the important questions provided here, students can strengthen their understanding of binomial expansions, coefficients, and properties. These questions cover a range of difficulty levels, ensuring thorough preparation for CBSE exams. Mastering this chapter not only boosts exam performance but also lays the foundation for advanced mathematical concepts in higher studies. Keep practising and revising regularly for the best results!