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Class 11 NCERT Solutions- Chapter 8 Binomial Theorem – Miscellaneous Exercise on Chapter 8

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  • Last Updated : 07 Apr, 2021
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Question 1. Find a, b, and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290, and 30375, respectively.

Solutions:

As, we know that (r+1)th term of (a+b)n is denoted by,

Tr+1 = nCr an-r br

Here, it is given that first three terms of the expansion are 729, 7290 and 30375.

When, T1 = 729, T2 = 7290 and T3 = 30375

T0+1 (r=0) = nC0 an-0 b0 = an = 729  …………………(1)

T1+1 (r=0) = nC1 an-1 b1 = nan-1 b = 7290  …………………(2)

T2+1 (r=0) = nC2 an-2 b2\frac{n!}{2!(n-2)!}an-2b2 \frac{n(n-1)a^{n-2}b^2}{2}     = 30375  …………………(3)

Dividing (1) and (2), we get

\frac{na^{n-1} b}{a^n} = \frac{7290}{729}

nan-1-n b = 10

\frac{nb}{a}  = 10 ……………………….(I)

Now, dividing (3) and (2), we get

\frac{\frac{n(n-1)a^{n-2}b^2}{2}}{na^{n-1} b} = \frac{30375}{7290}

\frac{(n-1)b}{2a} = \frac{30375}{7290}

\frac{nb-b}{a} = \frac{25}{3}

\frac{nb}{a} - \frac{b}{a} = \frac{25}{3}

From (I), we can substitute

10 - \frac{b}{a} = \frac{25}{3}

\frac{b}{a}  = 10 – \frac{25}{3}

\frac{b}{a} = \frac{30-25}{3} = \frac{5}{3}  ………………….(II)

Substituting (II) in (I), we get

\frac{nb}{a}  = 10

n(\frac{5}{3})  = 10

n = \frac{10 \times3}{5}

n = 6

Substituting n = 6 in (1), we get

an = 729

a6 = 729

a = 3

Substituting a = 3 in (II), we get

\frac{b}{a} = \frac{5}{3}

b = \frac{5\times3}{3}

b = 5

Hence, a = 3, b = 5 and n = 6

Question 2. Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Solutions:

As, we know that (r+1)th term of (a+b)n is denoted by,

Tr+1 = nCr an-r br

Here, a = 3 and b = ax and n = 9.

Tr+1 = 9Cr 39-r (ax)r

Tr+1 = 9Cr (\frac{3^9}{3^r}) arxr

Tr+1 = 9Cr (\frac{3^9 \times a^r}{3^r}) x^r

So, here if you want the power of x2 and x3. Then r=2 and r=3.

r = 2, T2+1 = [^9C_2 (\frac{3^9 \times a^2}{3^2})] x^2

r = 3, T3+1[^9C_3 (\frac{3^9 \times a^3}{3^3})] x^3

Coefficient of x2 = Coefficient of x3

^9C_2 (\frac{3^9 \times a^2}{3^2}) = ^9C_3 (\frac{3^9 \times a^3}{3^3})

\frac{9!}{2! (9-2)!} \times \frac{3^9 \times a^2}{3^2} = \frac{9!}{3! (9-3)!} \times \frac{3^9 \times a^3}{3^3}

\frac{1}{2! (9-2)!} \times \frac{a^2}{3^2} = \frac{1}{3! (9-3)!} \times \frac{a^3}{3^3}

\frac{3! (9-3)!}{2! (9-2)!} = \frac{a^3 \times 3^2}{a^2 \times 3^3}

\frac{a}{3} = \frac{3! 6!}{2! 7!}

\frac{a}{3} = \frac{3}{7}

a = \frac{3\times 3}{7} = \frac{9}{7}

Hence, a = \frac{9}{7}

Question 3. Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

Solutions:

For getting the coefficient of x5, lets expand both the binomials for more clear understanding.

(1 + 2x)6 = 6C0 + 6C1 (2x) + 6C2 (2x)2 + 6C3 (2x)3 + 6C4 (2x)4 + 6C5 (2x)5 + 6C6 (2x)6

= 1 + 6 (2x) + 15 (2x)2 + 20 (2x)3 + 15 (2x)4 + 6 (2x)5 + (2x)6

= 1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6

(1 – x)7 = 7C07C1 (x) + 7C2 (x)27C3 (x)3 + 7C4 (x)47C5 (x)5 + 7C6 (x)67C7 (x)7

= 1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7

Now, the product will be seen as follows

(1 + 2x)6 (1 – x)7 = (1 + 12 x + 60x2 + 160 x3 + 240 x4 + 192 x5 + 64x6) (1 – 7x + 21x2 – 35x3 + 35x4 – 21x5 + 7x6 – x7)

Here, what we can see that x5 will be obtained when two terms will be multiplied of having sum of power of x as 5. Those two terms will be as follows:

First binomial Second binomial
1st term 6th term
2nd term  5th term
3rd term 4th term
4th term 3rd term
5th term 2nd term 
6th term 1st term

So, 

Coefficient of x5 = (1)(-21) + (12)(35) + (60)(-35) + (160)(21) + (240)(-7) + (192)(1)

Coefficient of x5 = -21 + 420 – 2100 + 3360 – 1680 + 192

Coefficient of x5 = -21 + 420 – 2100 + 3360 – 1680 + 192

Coefficient of x5 = 171

Hence, the coefficient of x5 in the expression (1+2x)6 (1-x)7 is 171.

Question 4. 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 5. 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 6. Find the value of (a^2 + \sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4.

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 = \sqrt{a^2-1} , we get

(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]

= 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 7. 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 8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n is √6:1.

Solutions:

As, here it is said we have to calculate

(Fifth term from the beginning : Fifth term from the end) of the Binomial = (\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n

Instead of taking fifth term from the end, lets reverse the binomial term and take it from beginning.

Fifth term from the end of binomial (\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n = Fifth term from the beginning (\frac{1}{\sqrt[4]{3}}+ \sqrt[4]{2})^n

Lets move further with this

As, we know that (r+1)th term of (a+b)n is denoted by,

Tr+1 = nCr an-r br

Fifth term from the beginning of (\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n

T5 = T4+1 = nC4 (\sqrt[4]{2})^{n-4} (\frac{1}{\sqrt[4]{3}})^4

T5 = nC4 (\sqrt[4]{2})^n(\frac{1}{\sqrt[4]{2} \times \sqrt[4]{3}})^4

T5 = nC4 (\sqrt[4]{2})^n (\frac{1}{2 \times 3})

T5 = nC4 (\frac{(\sqrt[4]{2})^n}{6})  …………………………….(1)

Now, Fifth term from the beginning of (\frac{1}{\sqrt[4]{3}}+ \sqrt[4]{2})^n,

T5 = T4+1 = nC4 (\frac{1}{\sqrt[4]{3}})^{n-4} (\sqrt[4]{2})^4

T5 = nC4 \times \frac{(\frac{1}{\sqrt[4]{3}})^n}{(\frac{1}{\sqrt[4]{3}})^4} \times (2)

T5 = nC4 \frac{(\frac{1}{\sqrt[4]{3}})^n}{(\frac{1}{3})} (2)

T5 = nC4 (\frac{3}{(\sqrt[4]{3})^n}) (2)

T5 = nC4 (\frac{6}{(\sqrt[4]{3})^n})   ……………………….(2)

Now taking ratio of (1) and (2), which is equal to √6:1

\frac{^nC_4 (\frac{(\sqrt[4]{2})^n}{6})}{^nC_4 (\frac{6}{(\sqrt[4]{3})^n})} = \sqrt{6}

\frac{(\sqrt[4]{2})^n \times (\sqrt[4]{3})^n}{(6\times 6)} = \sqrt{6}

(\sqrt[4]{6})^n = \sqrt{6} \times 36

6^{\frac{n}{4}} = 6^{\frac{5}{2}}

\frac{n}{4} = \frac{5}{2}

n = \frac{5\times 4}{2}

n = 10

Question 9. Expand using Binomial Theorem (1+\frac{x}{2} - \frac{2}{x})^4, x≠0.

Solutions:

Grouping (1+\frac{x}{2} - \frac{2}{x})^4   in binomial form, we have

[(1+\frac{x}{2}) - \frac{2}{x}]^4

Comparing it with (a+b)n,

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

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

(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})

(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})

(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})

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

(1+\frac{x}{2})^2 = 12 + (\frac{x}{2})^2 + 2(1)(\frac{x}{2})

(1+\frac{x}{2})^2 = 1 + \frac{x^2}{4} + x

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

(1+\frac{x}{2})^3 = 1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}

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

(1+\frac{x}{2})^4 = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16}

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

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})

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})

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})

\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

Question 10. 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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