NCERT Solutions Class 11 – Chapter 7 Binomial Theorem – Miscellaneous Exercise

Question 1. If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.

[Hint write an = (a – b + b)ⁿ and expand]

Solutions:

To prove that (a – b) is a factor of (aⁿ – bⁿ), 

aⁿ – bⁿ = k (a – b) where k is some natural number or constant.

a can be written as = a – b + b

aⁿ = (a – b + b)ⁿ = [(a – b) + b]ⁿ

[(a – b) + b]ⁿ = ⁿC₀ (a – b)ⁿ + ⁿC₁ (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1 + ⁿC_n bⁿ

aⁿ = (a – b)ⁿ + n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1 + bⁿ

Now, aⁿ – bⁿ will be

aⁿ – bⁿ = [(a – b)ⁿ + n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1 + bⁿ] – bⁿ

aⁿ – bⁿ = (a – b)ⁿ + n (a – b)^n-1 b + ……….ⁿC_n-1(a – b)b^n-1

Taking (a-b) common, we have

aⁿ – bⁿ = (a – b) [(a –b)^n-1 + n (a – b)^n-2 b + …… + ⁿC_n-1 b^n-1]

aⁿ – bⁿ = (a – b) k

Where k = [(a –b)^n-1 + n (a – b)^n-2 b + …… + ⁿC_n-1 b^n-1] is a natural number

Hence, it is proved a – b is a factor of aⁿ – bⁿ, where n is a positive integer

Question 2. Evaluate (√3​+√2​)⁶−(√3​−√2​)⁶.

Solutions:

Using binomial theorem the expression (a + b)⁶ and (a – b)⁶, can be expanded as follows:

(a + b)⁶ = ⁶C0 a⁶ + ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² + ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ + ⁶C₅ a b⁵ + ⁶C₆ b⁶

(a – b)⁶ = ⁶C₀ a⁶ – ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² – ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ – ⁶C₅ a b⁵ + ⁶C₆ b⁶

Now adding them,

(a + b)⁶ – (a – b)⁶ = ⁶C₀ a⁶ + ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² + ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ + ⁶C₅ a b⁵ + ⁶C₆ b⁶ – [⁶C₀ a⁶ – ⁶C₁ a⁵ b + ⁶C₂ a⁴ b² – ⁶C₃ a³ b³ + ⁶C₄ a² b⁴ – ⁶C₅ a b⁵ + ⁶C₆ b⁶]

(a + b)⁶ – (a – b)⁶ = 2[⁶C₁ a⁵ b + ⁶C₃ a³ b³ + ⁶C₅ a b⁵]

Substituting a = √3 and b = √2, we get

(√3 + √2)⁶ – (√3 – √2)⁶ = 2 [6 (√3)⁵ (√2) + 20 (√3)³ (√2)³ + 6 (√3) (√2)⁵]

= 2 [54(√6) + 120 (√6) + 24 √6]

= 2 (√6) (198)

= 396 √6

Question 3. Find the value of [Tex](a^2 + \sqrt{a^2-1})^4 + (a^2 – \sqrt{a^2-1})^4[/Tex].

Solutions:

Using binomial theorem the expression (x+y)⁴ and (x – y)⁴, can be expanded as follows:

(x + y)⁴ = ⁴C₀ x⁴ + ⁴C₁ x³ y + ⁴C₂ x² y² + ⁴C₃ x y³ + ⁴C₄ y⁴

(x – y)⁴ = ⁴C₀ x⁴ – ⁴C₁ x³ y + ⁴C₂ x² y² – ⁴C₃ x y³ + ⁴C₄  y⁴

Now adding them,

(x + y)⁴ + (x – y)⁴ = ⁴C₀ x⁴ + ⁴C₁ x³ y + ⁴C₂ x² y² + ⁴C₃ x y³ + ⁴C₄ y⁴ + [⁴C₀ x⁴ – ⁴C₁ x³ y + ⁴C₂ x² y² – ⁴C₃ x y³ + ⁴C₄ y⁴]

(x + y)⁴ + (x – y)⁴ = 2[⁴C₀ x⁴ + ⁴C₂ x² y² + ⁴C₄ y⁴]

Substituting x = a² and y = [Tex]\sqrt{a^2-1} [/Tex], we get

[Tex](a^2 + \sqrt{a^2-1})^4 + (a^2 – \sqrt{a^2-1})^4 = 2[(a^2)^4 +6 (a^2)^2 (\sqrt{a^2-1})^2 + (\sqrt{a^2-1})^4][/Tex]

= 2[a⁸ + 6a⁴ (a²-1) + (a²-1)²]

= 2[a⁸ + 6a⁶ – 6a⁴ + (a⁴ + 1 – 2(a²)(1))]

= 2[a⁸ + 6a⁶ – 6a⁴ + a⁴ + 1 – 2a²]

= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2

Question 4. Find an approximation of (0.99)⁵ using the first three terms of its expansion.

Solutions:

To make 0.99 in binomial form,

0.99 = 1 – 0.01

Now by applying binomial theorem, we get

(0. 99)⁵ = (1 – 0.01)⁵

Taking first three terms of its expansion, we have

= ⁵C₀ (1)⁵ – ⁵C₁ (1)⁴ (0.01) + ⁵C₂ (1)³ (0.01)²

= 1 – 5 (0.01) + 10 (0.01)²

= 1 – 0.05 + 0.001

= 0.951

Approximation of (0.99)⁵ = 0.951.

Question 5. Expand using Binomial Theorem [Tex](1+\frac{x}{2} – \frac{2}{x})^4[/Tex], x≠0.

Solutions:

Grouping [Tex](1+\frac{x}{2} – \frac{2}{x})^4  [/Tex] in binomial form, we have

[Tex][(1+\frac{x}{2}) – \frac{2}{x}]^4[/Tex]

Comparing it with (a+b)ⁿ,

a = [Tex](1+\frac{x}{2})  [/Tex], b = [Tex]\frac{2}{x}  [/Tex] and n = 4

[Tex][(1+\frac{x}{2}) – \frac{2}{x}]^4  [/Tex] = ⁴C₀ [Tex](1+\frac{x}{2})^4  [/Tex] –  ⁴C₁ [Tex](1+\frac{x}{2})^3 (\frac{2}{x})  [/Tex] + ⁴C₂ [Tex](1+\frac{x}{2})^2 (\frac{2}{x})^2  [/Tex] – ⁴C₃ [Tex](1+\frac{x}{2}) (\frac{2}{x})^3  [/Tex] + ⁴C₄ [Tex](\frac{2}{x})^4[/Tex]

= [Tex](1+\frac{x}{2})^4 – 4 (1+\frac{x}{2})^3 (\frac{2}{x}) + 6 (1+(\frac{x}{2})^2 (\frac{4}{x^2}) – 4 (1+\frac{x}{2}) (\frac{8}{x^3}) + (\frac{16}{x^4})[/Tex]

= [Tex](1+\frac{x}{2})^4 –  (\frac{8}{x}) (1+\frac{x}{2})^3 + (\frac{24}{x^2}) (1+\frac{x}{2})^2  – 4 (\frac{8}{x^3} + (\frac{x}{2})(\frac{8}{x^3})) + (\frac{16}{x^4})[/Tex]

= [Tex](1+\frac{x}{2})^4 –  (\frac{8}{x}) (1+\frac{x}{2})^3 + (\frac{24}{x^2}) (1+\frac{x}{2})^2  – 4 (\frac{8}{x^3} + \frac{4}{x^2}) + (\frac{16}{x^4})[/Tex]

Now, lets get the value of [Tex](1+\frac{x}{2})^4, (1+\frac{x}{2})^3  [/Tex] and [Tex](1+\frac{x}{2})^2[/Tex]

[Tex](1+\frac{x}{2})^2 = 12 + (\frac{x}{2})^2 + 2(1)(\frac{x}{2})[/Tex]

[Tex](1+\frac{x}{2})^2 = 1 + \frac{x^2}{4} + x[/Tex]

[Tex](1+\frac{x}{2})^3  [/Tex] = ³C₀ (1)³ + ³C₁ (1)² [Tex](\frac{x}{2})  [/Tex] + ³C₂ (1) [Tex](\frac{x}{2})^2  [/Tex] + ³C₃ [Tex](\frac{x}{2})^3[/Tex]

[Tex](1+\frac{x}{2})^3 = 1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}[/Tex]

[Tex](1+\frac{x}{2})^4  [/Tex] = ⁴C₀ (1)⁴ + ⁴C₁ (1)³ [Tex](\frac{x}{2})  [/Tex] + ⁴C₂ (1)² [Tex](\frac{x}{2})^2  [/Tex] + ⁴C₃ (1) [Tex](\frac{x}{2})^3  [/Tex] + ⁴C₄ [Tex](\frac{x}{2})^4[/Tex]

[Tex](1+\frac{x}{2})^4 = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16}[/Tex]

Now, substituting these values in the main equation, we get

= [Tex]1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} –  (\frac{8}{x}) (1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}) + (\frac{24}{x^2}) (1 + \frac{x^2}{4} + x)  – 4 (\frac{8}{x^3} + \frac{4}{x^2}) + (\frac{16}{x^4})[/Tex]

= [Tex]1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} –  (\frac{8}{x} + (\frac{8}{x})(\frac{3x}{2}) + (\frac{8}{x})(\frac{3x^2}{4}) + (\frac{8}{x})(\frac{x^3}{8})) + (\frac{24}{x^2} + (\frac{24}{x^2})(\frac{x^2}{4}) + (\frac{24}{x^2}) (x))  – (\frac{32}{x^3} + \frac{16}{x^2}) + (\frac{16}{x^4})[/Tex]

= [Tex]1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} –  \frac{8}{x} – 12 – 6x – x^2 + \frac{24}{x^2} + 6+ \frac{24}{x}  – \frac{32}{x^3} – \frac{16}{x^2}) + (\frac{16}{x^4})[/Tex]

= [Tex]\frac{16}{x} + \frac{8}{x^2} – \frac{32}{x^3} + \frac{16}{x^4} – 4x + \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} –  5[/Tex]

Question 6. Find the expansion of (3x²– 2ax + 3a²)³ using binomial theorem.

Solutions:

Grouping (3x²– 2ax + 3a²)³ in binomial form, we have

[3x² + (- 2ax + 3a²)]³

Comparing it with (a+b)ⁿ,

a = 3x², b = -a (2x-3a) and n = 3

[3x² + (-a (2x-3a))]³

= ³C₀ (3x²)³ +  ³C₁ (3x²)² (-a (2x-3a)) +  ³C₂ (3x²) (-a (2x-3a))² +  ³C₃ (-a (2x-3a))³

= 27x⁶ +  3 (9x⁴) (-a) (2x-3a) +  3 (3x²) (-a)² (2x-3a)² + (-a)³ (2x-3a)³

= 27x⁶ + (-54ax⁵ + 81a²x⁴) +  9a²x² (2x-3a)² – a³ (2x-3a)³

Now, lets get the value of (2x-3a)² and (2x-3a)³.

(2x-3a)² = (2x)² + (3a)² – 2(2x)(3a)

(2x-3a)² = 4x² + 9a² -12xa

(2x-3a)³ = (2x)³ – (3a)³ – 3(2x)(3a)(2x-3a)

(2x-3a)³ = 8x³ – 27a³ – 36x²a +54xa²

Now, substituting these values in the main equation, we get

= 27x⁶ – 54ax⁵ + 81a²x⁴ +  9a²x² (4x² + 9a² -12xa) – a³ (8x³ – 27a³ – 36x²a + 54xa²)

= 27x⁶ – 54ax⁵ + 81a²x⁴ + 36a²x⁴ + 81a⁴x² -108x³a³ – (8a³x³ – 27a⁶ – 36x²a⁴ + 54xa⁵⁾

= 27x⁶ – 54ax⁵ + 117a²x⁴ + 81a⁴x² -108x³a³ – 8a³x³ + 27a⁶ + 36x²a⁴ – 54xa⁵

= 27x⁶ – 54ax⁵+ 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶