Class 11 NCERT Solutions- Chapter 8 Binomial Theorem – Miscellaneous Exercise on Chapter 8
Last Updated : 05 Aug, 2024
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 = [Tex]\frac{n!}{2!(n-2)!}[/Tex]an-2b2 = [Tex]\frac{n(n-1)a^{n-2}b^2}{2} [/Tex] = 30375 …………………(3)
Dividing (1) and (2), we get
[Tex]\frac{na^{n-1} b}{a^n} = \frac{7290}{729}[/Tex]
nan-1-n b = 10
[Tex]\frac{nb}{a} [/Tex] = 10 ……………………….(I)
Now, dividing (3) and (2), we get
[Tex]\frac{\frac{n(n-1)a^{n-2}b^2}{2}}{na^{n-1} b} = \frac{30375}{7290}[/Tex]
[Tex]\frac{(n-1)b}{2a} = \frac{30375}{7290}[/Tex]
[Tex]\frac{nb-b}{a} = \frac{25}{3}[/Tex]
[Tex]\frac{nb}{a} – \frac{b}{a} = \frac{25}{3}[/Tex]
From (I), we can substitute
[Tex]10 – \frac{b}{a} = \frac{25}{3}[/Tex]
[Tex]\frac{b}{a} [/Tex] = 10 – [Tex]\frac{25}{3}[/Tex]
[Tex]\frac{b}{a} = \frac{30-25}{3} = \frac{5}{3} [/Tex] ………………….(II)
Substituting (II) in (I), we get
[Tex]\frac{nb}{a} [/Tex] = 10
[Tex]n(\frac{5}{3}) [/Tex] = 10
n = [Tex]\frac{10 \times3}{5}[/Tex]
n = 6
Substituting n = 6 in (1), we get
an = 729
a6 = 729
a = 3
Substituting a = 3 in (II), we get
[Tex]\frac{b}{a} = \frac{5}{3}[/Tex]
b = [Tex]\frac{5\times3}{3}[/Tex]
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 [Tex](\frac{3^9}{3^r})[/Tex] arxr
Tr+1 = 9Cr [Tex](\frac{3^9 \times a^r}{3^r}) x^r[/Tex]
So, here if you want the power of x2 and x3. Then r=2 and r=3.
r = 2, T2+1 = [Tex][^9C_2 (\frac{3^9 \times a^2}{3^2})] x^2[/Tex]
r = 3, T3+1 = [Tex][^9C_3 (\frac{3^9 \times a^3}{3^3})] x^3[/Tex]
Coefficient of x2 = Coefficient of x3
[Tex]^9C_2 (\frac{3^9 \times a^2}{3^2}) = ^9C_3 (\frac{3^9 \times a^3}{3^3})[/Tex]
[Tex]\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}[/Tex]
[Tex]\frac{1}{2! (9-2)!} \times \frac{a^2}{3^2} = \frac{1}{3! (9-3)!} \times \frac{a^3}{3^3}[/Tex]
[Tex]\frac{3! (9-3)!}{2! (9-2)!} = \frac{a^3 \times 3^2}{a^2 \times 3^3}[/Tex]
[Tex]\frac{a}{3} = \frac{3! 6!}{2! 7!}[/Tex]
[Tex]\frac{a}{3} = \frac{3}{7}[/Tex]
a = [Tex]\frac{3\times 3}{7} = \frac{9}{7}[/Tex]
Hence, a = [Tex]\frac{9}{7}[/Tex]
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 = 7C0 – 7C1 (x) + 7C2 (x)2 – 7C3 (x)3 + 7C4 (x)4 – 7C5 (x)5 + 7C6 (x)6 – 7C7 (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 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 6. 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 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)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 8. Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of [Tex](\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n[/Tex] 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 = [Tex](\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n [/Tex]
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 [Tex](\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n[/Tex] = Fifth term from the beginning [Tex](\frac{1}{\sqrt[4]{3}}+ \sqrt[4]{2})^n [/Tex]
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 [Tex](\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n[/Tex],
T5 = T4+1 = nC4 [Tex](\sqrt[4]{2})^{n-4} (\frac{1}{\sqrt[4]{3}})^4[/Tex]
T5 = nC4 [Tex](\sqrt[4]{2})^n(\frac{1}{\sqrt[4]{2} \times \sqrt[4]{3}})^4[/Tex]
T5 = nC4 [Tex](\sqrt[4]{2})^n (\frac{1}{2 \times 3})[/Tex]
T5 = nC4 [Tex](\frac{(\sqrt[4]{2})^n}{6}) [/Tex] …………………………….(1)
Now, Fifth term from the beginning of [Tex](\frac{1}{\sqrt[4]{3}}+ \sqrt[4]{2})^n[/Tex],
T5 = T4+1 = nC4 [Tex](\frac{1}{\sqrt[4]{3}})^{n-4} (\sqrt[4]{2})^4[/Tex]
T5 = nC4 [Tex]\times \frac{(\frac{1}{\sqrt[4]{3}})^n}{(\frac{1}{\sqrt[4]{3}})^4} \times (2)[/Tex]
T5 = nC4 [Tex]\frac{(\frac{1}{\sqrt[4]{3}})^n}{(\frac{1}{3})} (2)[/Tex]
T5 = nC4 [Tex](\frac{3}{(\sqrt[4]{3})^n}) (2)[/Tex]
T5 = nC4 [Tex](\frac{6}{(\sqrt[4]{3})^n}) [/Tex] ……………………….(2)
Now taking ratio of (1) and (2), which is equal to √6:1
[Tex]\frac{^nC_4 (\frac{(\sqrt[4]{2})^n}{6})}{^nC_4 (\frac{6}{(\sqrt[4]{3})^n})} = \sqrt{6}[/Tex]
[Tex]\frac{(\sqrt[4]{2})^n \times (\sqrt[4]{3})^n}{(6\times 6)} = \sqrt{6}[/Tex]
[Tex](\sqrt[4]{6})^n = \sqrt{6} \times 36[/Tex]
[Tex]6^{\frac{n}{4}} = 6^{\frac{5}{2}}[/Tex]
[Tex]\frac{n}{4} = \frac{5}{2}[/Tex]
n = [Tex]\frac{5\times 4}{2}[/Tex]
n = 10
Question 9. 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 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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Chapter 5 Complex Numbers And Quadratic Equations - Exercise 5.1 | Set 1For Q.11 to Q.13 find the multiplicative inverse of the given number Question 11. 4-3i Solution: Let's denote given number as a, the complement of a = [Tex]\overline{a} [/Tex] = [Tex]\overline{4-3i}[/Tex] [Tex]\overline{a} [/Tex
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Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations - Exercise 5.2
Find the modulus and the arguments of each of the complex numbers i. Exercises 1 to 2.Question 1. z = – 1 – i √3 Solution: We have, z = -1 - i√3 We know that, z = r (cosθ + i sinθ) Therefore, r cosθ = -1 ---(1) r sinθ = -√3 ----(2) On Squaring and adding (1) and (2), we obtain r2 (cos 2θ + sin 2θ) =
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Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations - Exercise 5.3
Solve each of the following equations:Question 1. x2 + 3 = 0 Solution: We have, x2 + 3 = 0 or x2 + 0 × x + 3 = 0 ---(1) Discriminant, D = b2 - 4ac from (1) , a = 1 , b = 0 and c = 3 D = (0)2 - 4*(1)*(3) D = -12 Since , x = [Tex]\frac{( -b ± \sqrt{D} }{2a}[/Tex] Therefore , x = [Tex]\frac{( -0 ± \sqr
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Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations - Miscellaneous Exercise on Chapter 5 | Set 1
Complex numbers are fundamental in higher mathematics and have a wide range of applications in engineering, physics, and applied sciences. Chapter 5 of the Class 11 NCERT textbook introduces the concept of complex numbers and their properties. This chapter also covers quadratic equations including t
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Class 11 NCERT Solutions - Chapter 5 Complex Numbers And Quadratic Equations - Miscellaneous Exercise on Chapter 5 | Set 2
Chapter 5 of Class 11 NCERT Mathematics deals with Complex Numbers and Quadratic Equations. This chapter introduces the concept of complex numbers, which are essential for solving certain types of quadratic equations that do not have real solutions. The chapter also covers the algebraic operations o
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Chapter 6: Linear Inequalities
Class 11 NCERT Solutions- Chapter 6 Linear Inequalities - Exercise 6.1 | Set 1
Question 1. Solve 24x < 100, when (i) x is a natural number. (ii) x is an integer. Solution: (i) when x is a natural number. Clearly x>0 because from definition (N =1,2,3,4,5,6.....)Now we have to divide the inequation by 24 we get x<25/6But x is a natural number that is the solution will b
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Class 11 NCERT Solutions - Chapter 6 Linear Inequalities - Exercise 6.1 | Set 2
Chapter 6 Linear Inequalities - Exercise 6.1 | Set 1Question 13. 2 (2x + 3) – 10 < 6 (x – 2) Solution: Given, 2(2x + 3) – 10 < 6 (x – 2)By multiplying we get4x + 6 – 10 < 6x – 12On simplifying we get4x – 4 < 6x – 12– 4 + 12 < 6x – 4x8 < 2x4 < xThus, the solutions of the given eq
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Class 11 NCERT Solutions - Chapter 6 Linear Inequalities - Exercise 6.2
Solve the following inequalities graphically in two-dimensional plane:Question 1: x + y < 5 Solution: Now draw a dotted line x + y = 5 in the graph (because (x + y = 5) is NOT the part of the given equation) We need at least two solutions of the equation. So, we can use the following table to dra
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Class 11 NCERT Solutions - Chapter 6 Linear Inequalities - Exercise 6.3
Solve the following system of inequalities graphically:Question 1: x ≥ 3, y ≥ 2 Solution: For equation 1: Now draw a solid line x = 3 in the graph (because (x = 3) is the part of the given equation) we need at least two solutions of the equation. So, we can use the following table to draw the graph:
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Class 11 NCERT Solutions- Chapter 6 Linear Inequalities - Miscellaneous Exercise on Chapter 6
Chapter 6 of Class 11 NCERT Mathematics delves into the topic of the Linear Inequalities. Linear inequalities are mathematical expressions that involve inequalities rather than equalities representing a range of possible values for the variable rather than a single solution. In this chapter, student
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Chapter 7: Permutations and Combinations
Class 11 NCERT Solutions - Chapter 7 Permutations And Combinations - Exercise 7.1
Chapter 7 of the Class 11 NCERT Mathematics textbook, titled "Permutations and Combinations," introduces students to fundamental concepts in counting and arranging objects. This chapter focuses on the principles of permutations and combinations, which are essential for solving various problems in pr
6 min read
Class 11 NCERT Solutions - Chapter 7 Permutations And Combinations - Exercise 7.2
Chapter 7 of the Class 11 NCERT Mathematics textbook, titled "Permutations and Combinations," delves into the principles of counting and arranging objects. This chapter covers the concepts of permutations and combinations, which are crucial for solving problems involving the arrangement and selectio
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Class 11 NCERT Solutions- Chapter 7 Permutations And Combinations - Exercise 7.3
Theorem 1: The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat is n(n-1)(n – 2). . .(n – r + 1), which is denoted by nPr = [Tex]\frac{n!}{(n-r)!}[/Tex] Theorem 2: The number of permutations of n different objects taken r at a time, wh
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Class 11 NCERT Mathematics Solutions- Chapter 7 Permutations And Combinations - Exercise 7.4
Chapter 7 of the Class 11 NCERT Mathematics textbook, "Permutations and Combinations," explores fundamental concepts of counting and arrangement. Exercise 7.4 focuses on applying these concepts to solve problems related to permutations and combinations, enhancing students' ability to handle various
6 min read
Class 11 NCERT Solutions- Chapter 7 Permutations And Combinations - Miscellaneous Exercise on Chapter 7
Chapter 7 of the Class 11 NCERT Mathematics textbook delves into the Permutations and Combinations essential topics in the probability and combinatorics. This chapter explores the fundamental principles of the counting and arrangement which are crucial for the solving various problems in the mathema
11 min read
Chapter 8: Binomial Theorem
Class 11 NCERT Solutions- Chapter 8 Binomial Theorem - Exercise 8.1
The Binomial Theorem provides a method to expand expressions that are raised to a power such as the (x + y)n. It is a crucial concept in algebra, particularly useful for expanding the polynomials and solving combinatorial problems. In this chapter, students learn to apply the Binomial Theorem to sim
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Class 11 NCERT Solutions - Chapter 8 Binomial Theorem - Exercise 8.2
Question 1. Find the coefficient of x5 in (x+3)8 Solution: The (r+1)th term of (x+3)8 is given by Tr+1 = 8Cr(x)8-r(3)r (eq1). Therefore for x5 we need to get 8-r =5 (Because we need to find x5. Therefore, power ox must be equal to 5) So we get r=3. Now, put r=3 in eq1. We get, Coefficient of x5 = 8C
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Class 11 NCERT Solutions- Chapter 8 Binomial Theorem - Miscellaneous Exercise on Chapter 8
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
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Chapter 9: Sequences and Series
Class 11 NCERT Solutions- Chapter 9 Sequences And Series - Exercise 9.1
Chapter 9 of the NCERT Class 11 Mathematics textbook covers Sequences and Series which are fundamental concepts in algebra and mathematical analysis. Exercise 9.1 specifically deals with the problems that help students understand and solve various types of sequences and series. Mastery of these prob
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Class 11 NCERT Solutions- Chapter 9 Sequences And Series - Exercise 9.2
Chapter 9 of the NCERT Class 11 Mathematics textbook focuses on the Sequences and Series a fundamental topic in algebra. Exercise 9.2 deals with the various problems related to arithmetic and geometric progressions in which are key concepts in understanding the behavior and relationships of the sequ
9 min read
Class 11 NCERT Solutions- Chapter 9 Sequences And Series - Exercise 9.3 | Set 1
Question 1. Find the 20th and nth terms of the G.P 5/2, 5/4, 5/8, ... Solution: According to the question G.P: 5/2, 5/4, 5/8, ... So, first term(a) = 5/2 So, the common ratio(r) = [Tex]\frac{a_1}{a}=\frac{\frac54}{\frac52}= \frac12[/Tex] Find: 20th and nth terms of the given G.P So, the nth term of
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Class 11 NCERT Solutions- Chapter 9 Sequences And Series - Exercise 9.3 | Set 2
Question 17. If the 4th, 10th and 16th terms of a G.P. are x, y, and z, respectively. Prove that x, y, z are in G.P. Solution: Let the first term of G.P. be a and common ratio be r. According to the question a4 = ar3 = x ......(1) a10 = ar9 = y ......(2) a16 = ar15 = z ......(3) Now divide eq(2) by
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Class 11 NCERT Solutions- Chapter 9 Sequences And Series - Exercise 9.4
Find the sum to n terms of each of the series in Exercises 1 to 7.Question 1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … Solution: Given: Series = 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … To find nth term, we have nth term, an = n ( n + 1) So, the sum of n terms of the series: [Tex]S_n=\displaystyle\sum^n_{k=1}a_k=
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Class 11 NCERT Solutions- Chapter 9 Sequences And Series - Miscellaneous Exercise On Chapter 9 | Set 1
Question 1. Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term. Solution: Let the first term and common difference of the A.P. be a and d respectively. (m+n)th term of the A.P. = a+(m+n−1)d (m−n)th term of the A.P. = a+(m−n−1)d Thus, L.H.S = a+(m+n−1)d + a+(
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Class 11 NCERT Solutions- Chapter 9 Sequences And Series - Miscellaneous Exercise On Chapter 9 | Set 2
Question 17. If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P. Solution: We are given, a, b, c and d are in G.P. Therefore, we have b2 = ac … (1) c2 = bd … (2) ad = bc … (3) We need to prove (an + bn), (bn + cn), (cn + dn) are in G.P. i.e., => (bn+ cn)2 = (an + bn)
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