CBSE Class 11 Maths Chapter 8: Binomial Theorem NCERT Solutions

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NCERT Solutions PDF Class 11 PDF

CBSE Class 11 Mathematics, Chapter 8, Binomial Theorem, introduces students to the expansion of algebraic expressions of the form $(x+y)^n$. This section of NCERT Solutions focuses on Exercise 8.1, providing step-by-step guidance for applying the binomial theorem. The solutions demonstrate how to use the formula for binomial expansion, emphasizing the role of combinations in calculating binomial coefficients ($^nC_r$). Students will learn to expand expressions like $(x+y)^n$ and $(x-y)^n$, simplifying each term and understanding the underlying patterns. These explanations are crafted to build a solid understanding of algebraic manipulation and the structure of binomial expansions, making them an excellent resource for exam preparation and reinforcing core mathematical concepts.

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BoardCBSE
ClassClass 11
Subjectगणित-I
Session2026
LanguageEnglish
TypeNCERT Solutions
Chapter8. द्विपद प्रमेय

Chapter summary

Chapter 8, Binomial Theorem, introduces the formula for expanding binomial expressions of the form $(x+y)^n$. The NCERT Solutions for Exercise 8.1 focus on applying this theorem to expand given binomials. Students will practice using combination formulas ($^nC_r$) and understanding the alternating signs in expansions of $(x-y)^n$. The exercises cover various forms of binomials, including those with fractional or negative terms, requiring careful calculation and simplification.

Learning outcomes

  • Understand the Binomial Theorem formula for expansion.
  • Apply the binomial expansion for $(x+y)^n$ and $(x-y)^n$.
  • Calculate binomial coefficients using combinations ($^nC_r$).
  • Expand binomial expressions with positive, negative, and fractional terms.
  • Simplify expanded binomial expressions accurately.

Topics covered

Paper topics

  • Binomial Theorem
  • Expansion of binomial expressions
  • Binomial coefficients
  • Combinations ($^nC_r$)
  • General term of binomial expansion
  • Expansion of $(x+y)^n$
  • Expansion of $(x-y)^n$
  • Algebraic simplification
  • Polynomial expansion
  • Class 11 Maths
  • CBSE Mathematics
  • Exercise 8.1

Important topics

  • Binomial Theorem Formula
  • Expansion of $(x+y)^n$
  • Expansion of $(x-y)^n$
  • Calculating Binomial Coefficients ($^nC_r$)
  • General Term Application
  • Simplification of Expanded Terms
  • Handling Negative and Fractional Terms
  • Exercise 8.1 Problems

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Questions and Solutions

Question 1

Expand the expression: (1 - 2x)5
Solution:

To expand the expression $(1 - 2x)^5$, we use the Binomial Theorem formula for $(a+b)^n = {}^nC_0a^n + {}^nC_1a^{n-1}b + {}^nC_2a^{n-2}b^2 + \dots + {}^nC_nb^n$. Here, $a=1$, $b=-2x$, and $n=5$.

The expansion is:

(1-2x)^5 = {}^5C_0 (1)^5 (-2x)^0 + {}^5C_1 (1)^4 (-2x)^1 + {}^5C_2 (1)^3 (-2x)^2 + {}^5C_3 (1)^2 (-2x)^3 + {}^5C_4 (1)^1 (-2x)^4 + {}^5C_5 (1)^0 (-2x)^5

Now, we calculate the binomial coefficients and simplify each term:

  • ${}^5C_0 = 1$
  • ${}^5C_1 = 5$
  • ${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$
  • ${}^5C_3 = {}^5C_2 = 10$
  • ${}^5C_4 = {}^5C_1 = 5$
  • ${}^5C_5 = {}^5C_0 = 1$

Substituting these values and simplifying the powers of $(-2x)$:

= 1 \times 1 \times 1 + 5 \times 1 \times (-2x) + 10 \times 1 \times (4x^2) + 10 \times 1 \times (-8x^3) + 5 \times 1 \times (16x^4) + 1 \times 1 \times (-32x^5)

This simplifies to:

= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5

Thus, the expansion of $(1 - 2x)^5$ is $1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$.

Question 2

Expand the expression: $\left(\frac{2}{y} - \frac{x}{2}\right)^5$
Solution:

We use the Binomial Theorem for $(a+b)^n$, where $a = \frac{2}{y}$, $b = -\frac{x}{2}$, and $n=5$. The expansion is:

\left(\frac{2}{y} - \frac{x}{2}\right)^5 = {}^5C_0\left(\frac{2}{y}\right)^5 \left(-\frac{x}{2}\right)^0 + {}^5C_1\left(\frac{2}{y}\right)^4 \left(-\frac{x}{2}\right)^1 + {}^5C_2\left(\frac{2}{y}\right)^3 \left(-\frac{x}{2}\right)^2 + {}^5C_3\left(\frac{2}{y}\right)^2 \left(-\frac{x}{2}\right)^3 + {}^5C_4\left(\frac{2}{y}\right)^1 \left(-\frac{x}{2}\right)^4 + {}^5C_5\left(\frac{2}{y}\right)^0 \left(-\frac{x}{2}\right)^5

Let's calculate the coefficients and simplify the terms:

  • ${}^5C_0 = 1$
  • ${}^5C_1 = 5$
  • ${}^5C_2 = 10$
  • ${}^5C_3 = 10$
  • ${}^5C_4 = 5$
  • ${}^5C_5 = 1$

Now, substitute and simplify:

= 1 \times \frac{32}{y^5} \times 1 + 5 \times \frac{16}{y^4} \times (-\frac{x}{2}) + 10 \times \frac{8}{y^3} \times (\frac{x^2}{4}) + 10 \times \frac{4}{y^2} \times (-\frac{x^3}{8}) + 5 \times \frac{2}{y} \times (\frac{x^4}{16}) + 1 \times 1 \times (-\frac{x^5}{32})

Further simplification gives:

= \frac{32}{y^5} - \frac{5 \times 16}{2} \frac{x}{y^4} + \frac{10 \times 8}{4} \frac{x^2}{y^3} - \frac{10 \times 4}{8} \frac{x^3}{y^2} + \frac{5 \times 2}{16} \frac{x^4}{y} - \frac{x^5}{32}

= \frac{32}{y^5} - 40 \frac{x}{y^4} + 20 \frac{x^2}{y^3} - 5 \frac{x^3}{y^2} + \frac{10}{16} \frac{x^4}{y} - \frac{x^5}{32}

= \frac{32}{y^5} - \frac{40x}{y^4} + \frac{20x^2}{y^3} - \frac{5x^3}{y^2} + \frac{5x^4}{8y} - \frac{x^5}{32}

The expansion is $\frac{32}{y^5} - \frac{40x}{y^4} + \frac{20x^2}{y^3} - \frac{5x^3}{y^2} + \frac{5x^4}{8y} - \frac{x^5}{32}$.

Question 3

Expand the expression: (2x - 3)6
Solution:

We apply the Binomial Theorem for $(a+b)^n$ with $a=2x$, $b=-3$, and $n=6$. The expansion is:

(2x-3)^6 = {}^6C_0(2x)^6(-3)^0 + {}^6C_1(2x)^5(-3)^1 + {}^6C_2(2x)^4(-3)^2 + {}^6C_3(2x)^3(-3)^3 + {}^6C_4(2x)^2(-3)^4 + {}^6C_5(2x)^1(-3)^5 + {}^6C_6(2x)^0(-3)^6

First, calculate the binomial coefficients:

  • ${}^6C_0 = 1$
  • ${}^6C_1 = 6$
  • ${}^6C_2 = \frac{6 \times 5}{2 \times 1} = 15$
  • ${}^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$
  • ${}^6C_4 = {}^6C_2 = 15$
  • ${}^6C_5 = {}^6C_1 = 6$
  • ${}^6C_6 = {}^6C_0 = 1$

Now, substitute these coefficients and simplify the powers of $(2x)$ and $(-3)$:

= 1 \times (64x^6) \times 1 + 6 \times (32x^5) \times (-3) + 15 \times (16x^4) \times 9 + 20 \times (8x^3) \times (-27) + 15 \times (4x^2) \times 81 + 6 \times (2x) \times (-243) + 1 \times 1 \times 729

Performing the multiplications for each term:

= 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729

The expansion of $(2x - 3)^6$ is $64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729$.

Question 4

Expand the expression: $\left(\frac{x}{3} + \frac{1}{x}\right)^5$
Solution:

We use the Binomial Theorem for $(a+b)^n$ with $a = \frac{x}{3}$, $b = \frac{1}{x}$, and $n=5$. The expansion is:

\left(\frac{x}{3} + \frac{1}{x}\right)^5 = {}^5C_0\left(\frac{x}{3}\right)^5\left(\frac{1}{x}\right)^0 + {}^5C_1\left(\frac{x}{3}\right)^4\left(\frac{1}{x}\right)^1 + {}^5C_2\left(\frac{x}{3}\right)^3\left(\frac{1}{x}\right)^2 + {}^5C_3\left(\frac{x}{3}\right)^2\left(\frac{1}{x}\right)^3 + {}^5C_4\left(\frac{x}{3}\right)^1\left(\frac{1}{x}\right)^4 + {}^5C_5\left(\frac{x}{3}\right)^0\left(\frac{1}{x}\right)^5

The binomial coefficients are:

  • ${}^5C_0 = 1$
  • ${}^5C_1 = 5$
  • ${}^5C_2 = 10$
  • ${}^5C_3 = 10$
  • ${}^5C_4 = 5$
  • ${}^5C_5 = 1$

Now, substitute and simplify each term:

= 1 \times \frac{x^5}{243} \times 1 + 5 \times \frac{x^4}{81} \times \frac{1}{x} + 10 \times \frac{x^3}{27} \times \frac{1}{x^2} + 10 \times \frac{x^2}{9} \times \frac{1}{x^3} + 5 \times \frac{x}{3} \times \frac{1}{x^4} + 1 \times 1 \times \frac{1}{x^5}

Simplify the powers of $x$ and the coefficients:

= \frac{x^5}{243} + \frac{5x^4}{81x} + \frac{10x^3}{27x^2} + \frac{10x^2}{9x^3} + \frac{5x}{3x^4} + \frac{1}{x^5}

Further simplification leads to:

= \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}

The expansion is $\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$.

Question 5

Expand the expression: $\left(x + \frac{1}{x}\right)^6$
Solution:

We use the Binomial Theorem for $(a+b)^n$ with $a=x$, $b=\frac{1}{x}$, and $n=6$. The expansion is:

\left(x + \frac{1}{x}\right)^6 = {}^6C_0 x^6 \left(\frac{1}{x}\right)^0 + {}^6C_1 x^5 \left(\frac{1}{x}\right)^1 + {}^6C_2 x^4 \left(\frac{1}{x}\right)^2 + {}^6C_3 x^3 \left(\frac{1}{x}\right)^3 + {}^6C_4 x^2 \left(\frac{1}{x}\right)^4 + {}^6C_5 x^1 \left(\frac{1}{x}\right)^5 + {}^6C_6 x^0 \left(\frac{1}{x}\right)^6

Calculate the binomial coefficients:

  • ${}^6C_0 = 1$
  • ${}^6C_1 = 6$
  • ${}^6C_2 = 15$
  • ${}^6C_3 = 20$
  • ${}^6C_4 = 15$
  • ${}^6C_5 = 6$
  • ${}^6C_6 = 1$

Substitute these values and simplify each term:

= 1 \times x^6 \times 1 + 6 \times x^5 \times \frac{1}{x} + 15 \times x^4 \times \frac{1}{x^2} + 20 \times x^3 \times \frac{1}{x^3} + 15 \times x^2 \times \frac{1}{x^4} + 6 \times x \times \frac{1}{x^5} + 1 \times 1 \times \frac{1}{x^6}

Simplify the powers of $x$ in each term:

= x^6 + 6x^{5-1} + 15x^{4-2} + 20x^{3-3} + 15x^{2-4} + 6x^{1-5} + x^{-6}

This results in:

= x^6 + 6x^4 + 15x^2 + 20x^0 + 15x^{-2} + 6x^{-4} + x^{-6}

Writing with positive exponents:

= x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}

The expansion of $\left(x + \frac{1}{x}\right)^6$ is $x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$.

Common mistakes

  • Errors in calculating binomial coefficients ($^nC_r$).
  • Incorrectly applying the alternating signs in $(x-y)^n$ expansions.
  • Mistakes in simplifying powers of terms, especially with fractions or negative bases.
  • Algebraic errors during the simplification of the final expanded terms.

Revision tips

  • Memorize the binomial expansion formulas for $(x+y)^n$ and $(x-y)^n$.
  • Practice calculating $^nC_r$ values quickly and accurately.
  • Pay close attention to the signs and powers of each term in the expansion.
  • Work through each example and exercise step-by-step to ensure understanding of the process.

Practice MCQs

Q1. What is the general term in the expansion of $(x+y)^n$ according to the Binomial Theorem?

Q2. In the expansion of $(x-y)^n$, what is the sign of the term containing $y^r$?

Q3. Which formula is used to calculate the binomial coefficients?

Q4. For the expansion of $(1-2x)^5$, what is the coefficient of the $x^2$ term?

Q5. What is the value of $^5C_2$?

Frequently asked questions

What is the main purpose of the Binomial Theorem?

The Binomial Theorem provides a systematic way to expand algebraic expressions of the form $(x+y)^n$, where $n$ is a non-negative integer, without having to multiply the binomial by itself $n$ times.

How do I calculate the binomial coefficients like $^nC_r$?

Binomial coefficients are calculated using the combination formula: $^nC_r = \frac{n!}{r!(n-r)!}$. For example, $^5C_2 = \frac{5!}{2!3!} = 10$.

What is the difference between expanding $(x+y)^n$ and $(x-y)^n$?

The expansion of $(x+y)^n$ has all positive terms. The expansion of $(x-y)^n$ has alternating signs, starting with positive, then negative, positive, and so on, because $(x-y)^n = (x+(-y))^n$.

Are these NCERT Solutions for Class 11 Maths Chapter 8 suitable for exam preparation?

Yes, these solutions provide clear, step-by-step explanations for expanding binomial expressions, which is essential for mastering the concepts in Chapter 8 and performing well in exams.

What if the binomial expression has fractions or negative numbers?

The same Binomial Theorem applies. You need to carefully substitute the fractional or negative terms into the formula and pay close attention to the powers and signs during calculation and simplification.

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