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

Last Updated : 19 Apr, 2024
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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 + ……….nCn-1(a – b)bn-1 + nCn bn

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

Now, an – bn will be

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

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

Taking (a-b) common, we have

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

an – bn = (a – b) k

Where k = [(a –b)n-1 + n (a – b)n-2 b + …… + nCn-1 bn-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 a66C1 a5 b + 6C2 a4 b26C3 a3 b3 + 6C4 a2 b46C5 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 a66C1 a5 b + 6C2 a4 b26C3 a3 b3 + 6C4 a2 b46C5 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 x44C1 x3 y + 4C2 x2 y24C3 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 x44C1 x3 y + 4C2 x2 y24C3 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)55C1 (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



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