Category: Class 9

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

Question 1.
Plot the following points in the coordinate system and identify the quadrants P(-7, 6), Q(7, -2), R(-6, -7), S(3, 5) and T(3, 9)
Solution:

(i) P(-7, 6) lies in the II quadrant because the x-coordinate is negative and y-coordinate is positive.
(ii) Q(7, -2) lies in the IV quadrant because the x-coordinate is positive and y-coordinate is negative.
(iii) R(-6, -7) lies in the III quadrant because the x-coordinate is negative and y-coordinate is negative.
(iv) S(3, 5) lies in the I quadrant because the x-coordinate is positive and y-coordinate is also positive.
(v) T(3, 9) lies in the I quadrant because the x-coordinate is positive and y-coordinate is also positive.

Question 2.
Write down the abscissa and ordinate of the following.
(i) P
(ii) Q
(iii) R
(iv) S

Solution:
(i) P is (-4, 4) [-4 is abscissa and 4 is ordinate]
(ii) Q is (3, 3) [3 is abscissa and 3 is ordinate]
(iii) R is (4, -2) [4 is abscissa and -2 is ordinate]
(iv) S is (-5, -3) [-5 is abscissa and -3 is ordinate]

Question 3.
Plot the following points in the coordinate plane and join them. What is your conclusion about the resulting figure?
(i) (-5, 3) (-1, 3) (0, 3) (5, 3)
Solution:

Straight line parallel to x-axis.

(ii) (0, -4) (0, -2) (0, 4) (0, 5)
Solution:

The line is on the y-axis.

Question 4.
Plot the following points in the coordinate plane. Join them in order. What type of geometrical shape is formed?
(i) (0, 0) (-4, 0) (-4, -4) (0, -4)
Solution:

The geometrical shape of the figure is square.

(ii) (-3, 3) (2, 3) (-6, -1) (5, -1)
Solution:

The shape of the geometrical figure is Trapezium.

Students can download Maths Chapter 3 Algebra Ex 3.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.7

Question 1.
Find the quotient and remainder of the following.
(i) 4x³ + 6x² – 23x + 18) ÷ (x + 3)
Solution:

∴ The quotient = 4x² – 6x – 5
The remainder = 33

(ii) (8y³ – 16y² + 16y – 15) ÷ (2y – 1)
Solution:

∴ The quotient = 4y² – 6y + 5
The remainder = -10

(iii) (8x³ – 1) ÷ (2x – 1)
Solution:

∴ The quotient = 4x² + 2x + 1
The remainder = 0

(iv) (-18z + 14z² + 24z³ + 18) ÷ (3z + 4)
Solution:

∴The quotient = 8z² – 6z + 2
The remainder = 10

Question 2.
The area of a rectangle is x² + 7x + 12. If its breadth is (x + 3) then find its length.
Solution:
Let the length of the rectangle be “l”
The breadth of the rectangle = x + 3
Area of the rectangle = length × breadth
x² + 7x + 12 = l(x + 3)

Length of the rectangle = x + 4

Question 3.
The base of a parallelogram is (5x + 4). Find its height if the area is 25x² – 16.
Solution:
Let the height of the parallelogram be “h”.
Base of the parallelogram = 5x + 4
Area of a parallelogram = 25x² – 16
∴ Base x Height = 25x² – 16
(5x + 4) × h = 25x²– 16

Height of the parallelogram = 5x – 4

Question 4.
The sum of (x + 5) observations is (x³ + 125). Find the mean of the observations.
Solution:
Sum of the observation = x³ + 125
Number of observation = x + 5

Mean = x² – 5x + 25

Question 5.
Find the quotient and remainder for the following using synthetic division:
(i) (x³ + x² – 7x – 3) ÷ (x – 3)
Solution:
p(x) = x³ + x² – 7x – 3
d(x) = x – 3 [p(x) = d(x) × q(x) + r]

x – 3 = 0
x = 3
Hence the quotient = x² + 4x + 5
Remainder = 12

(ii) (x³ + 2x² – x – 4) ÷ (x + 2)
Solution:
p(x) = x³ + 2x² -x – 4
d(x) = x + 2

x + 2 = 0
x = -2
The quotient = x² – 1
Remainder = -2

(iii) (3x³ – 2x² + 7x – 5) ÷ (x + 3)
Solution:
p(x) = 3x³ – 2x² + 7x – 5
d(x) = x + 3

x + 3 = 0
x = -3
The quotient = 3x² – 11x + 40
Remainder = -125

(iv) (8x⁴ – 2x² + 6x + 5) ÷ (4x + 1)
Solution:
p(x) = 8x⁴ – 2x² + 6x + 5
d(x) = 4x + 1

Question 6.
If the quotient obtained on dividing (8x⁴ – 2x² + 6x – 7) by (2x + 1) is (4x³ + px² -qx + 3), then find p, q and also the remainder.
Solution:
p(x) = 8x⁴ – 2x² + 6x – 7
d(x) = 2x + 1

2x + 1 = 0
2x = -1
x = –\(\frac{1}{2}\)
The quotient = \(\frac{1}{2}\) [8x³ – 4x² + 6]
= 4x³ – 2x² + 3
= 4x³ – 2x² + 0x + 3
The given quotient is = 4x³ + px² – qx + 3
(compared with the given quotient)
The value of p = -2 and q = 0
Remainder = -10

Question 7.
If the quotient obtained on dividing 3x³ + 11x² + 34x + 106 by x – 3 is 3x² + ax + b, then find a, b and also the remainder.
Solution:
p(x) = 3x³ + 11x² + 34x + 106
d(x) = x – 3

x – 3 = 0
x = 3
The quotient is = 3x² + 20x + 94
The given quotient is = 3x² + ax + b
Compared with the given quotient
The value of a = 20 and b = 94
The remainder = 388

Students can download Maths Chapter 3 Algebra Ex 3.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.6

Question 1.
Factorise the following.
(i) x² + 10x + 24
Solution:
Product = 24, sum = 10

Split the middle term as 6x and 4x
x² + 10x + 24 = x² + 6x + 4x + 24
= x(x + 6) + 4 (x + 6)
= (x + 6) (x + 4)

(ii) z² + 4z – 12
Solution:
Product = -12, sum = 4

Split the middle term as 6z and -2z
z² + 4z – 12 = z² + 6z – 2z – 12
= z(z + 6) – 2 (z + 6)
= (z + 6) (z – 2)

(iii) p² – 6p – 16
Solution:
Product = -16, sum = -6

Split the middle term as – 8p and 2p
p² – 6p – 16 = p² – 8p + 2p – 16
= p(p – 8) + 2 (p – 8)
= (p – 8) (p + 2)

(iv) t² + 72 – 17t
Solution:
Product = +72, sum = -17

Split the middle term as -9t and -8t
t² – 17t + 72 = t² – 91 – 8t + 72
= t(t – 9) – 8 (l – 9)
= (t – 9) (t – 8)

(v) y² – 16y – 80
Solution:
Product = -80, sum = -16

Split the middle term as -20y and 4y
y² – 16y – 80 = y² – 20y + Ay – 80
= y(y – 20) + 4 (y – 20)
= (y – 20) (y + 4)

(vi) a² + 10a – 600
Solution:
Product = -600, sum =10

Split the middle term as 30a and -20a
a² + 10a – 600 = a² + 30a – 20a – 600
= a(a + 30) – 20 (a + 30)
= (a + 30) (a – 20)

Question 2.
Factorise the following.
(i) 2a² + 9a + 10
Solution:
Product = 2 × 10 = 20, sum = 9

Split the middle term as 5a and 4a
2a² + 9a + 10 = 2a² + 5a + 4a + 10
= a(2a + 5) + 2 (2a + 5)
= (2a+ 5) (a+ 2)

(ii) 5x² – 29xy – 42y²
Solution:
Product = 5 × -42 = -210, sum = -29

Split the middle term as -35x and 6x
5x² – 29xy – 42y² = 5x² – 35xy + 6xy – 42y²
= 5x (x – 7y) + 6y (x – 7y)
= (x – 7y) (5x + 6y)

(iii) 9 – 18x + 8x²
Solution:
Product = 9 × 8 = 72, sum = -18

Split the middle term as -12x and -6x
9 – 18x + 8x² = 8x² – 18x + 9
= 8x² – 12x – 6x + 9
= 4x (2x – 3) – 3 (2x – 3)
= (2x – 3) (4x – 3)

(iv) 6x² + 16xy + 8y²
Solution:
Product = 6 × 8 = 48, sum = 16

Split the middle term as 4xy and 12xy
6x² + 16xy + 8y² = 6x² + 12xy + 4xy + 8y²
= 6x (x + 2y) + 4y(x + 2y)
= (x + 2y) (6x + 4y)
= 2(x + 2y) (3x + 2y)

(v) 12x² + 36x²y + 27y²x²
Solution:
3x²2 [4 + 12y + 9y²]
= 3x² [9y² + 12y + 4]
Product = 9 x 4 = 36, sum =12

Split the middle term as 6y and 6y
12x² + 36x²y + 21y²x² = 3x² [9y² + 12y + 4]
= 3x² [9y² + 6y + 6y + 4]
= 3x² [3y(3y + 2) + 2(3y + 2)]
= 3x² (3y + 2) (3y + 2)
= 3x² (3y + 2)2

(vi) (a + b)² + 9 (a + b) + 18
Solution:
Let (a + b) = x
x² + 9x + 18
Product =18, sum = 9
Split the middle term as 6x and 3x
x² + 9x + 18 = x² + 6x + 3x + 18
= x (x + 6) + 3 (x + 6)
= (x + 6) (x + 3)
But x = a + b
(a + b)² + 9(a + b) + 18 = (a + b + 6) (a + b + 3)

Question 3.
Factorise the following.
(i) (p – q)² – 6(p – q) – 16
Solution:
Let (p – q) = x
(p – q)² – 6 (p – q) – 16 = x² – 6x – 16
Product = -16, sum = -6

Split the middle term as -8x and 2x
x² – 6x – 16 = x² – 8x + 2x – 16
= x(x – 8) + 2(x – 8)
= (x – 8) (x + 2)
(But x = p – q)
= (p – q – 8) (p – q + 2)

(ii) m² + 2mn – 24n²
Solution:
Product = -24, sum = 2

Split the middle term as 6mn and -4mn
m² + 2mn – 24m² = m² + 6mn – 4mn – 24n²
= m(m + 6n) – 4n (m + 6n)
= (m + 6n) (m – 4n)

(iii) √5 a² + 2a – 3√5?
Solution:
Product = √5 × – 3√5 = -15, sum = 2

Split the middle term as 5x and -3x
√5 a² + 2a – 3√5 = √5a² + 5a – 3a – 3√5
= √5 a(a + √5) – 3(a + √5)
= (a + √5) (√5a – 3)

(iv) a⁴ – 3a² + 2
Solution:
Let a² = x
a⁴ – 3a² + 2 = (a²)² – 3a² + 2
= x² – 3x + 2
Product = 2 and sum = -3

Split the middle term as -x and -2x
x² – 3x + 2 = x² – x – 2x + 2
= x(x – 1) – 2(x – 1)
= (x – 1) (x – 2)
a⁴ – 3a² + 2 = (a² – 1)(a² – 2) [But a² = x]
= (a + 1) (a – 1) (a² – 2)

(v) 8m³ – 2m²n – 15mn²
Solution:
8m³ – 2m²n – 15mn² = m(8m² – 2mn – 15n²)
Product = 8(-15) = -120 and sum = -2

Split the middle term as -12mn and 10mn
8m³ – 2m²n – 15mn² = m[8m² – 2mn – 15n²]
= m[8m²– 12mn + 10mn- 15n²]
= m[4m (2m – 3n) + 5n(2m – 3n)]
= m(2m – 3n) (4m + 5n)

(vi) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{2}{x y}\)
Solution:

Students can download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4

Question 1.
Expand the following:
(i) (2x + 3y + 4z)²
(ii) (-p + 2q + 3r)²
(iii) (2p + 3) (2p – 4) (2p – 5)
(iv) (3a + 1) (3a – 2) (3a + 4)
Solution:
We know that (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
(i) (2x + 3y + 4z)² = (2x)² + (3y)² + (4z)² + 2(2x) (3y) + 2(3y) (4z) + 2(4z) (2x)
= 4x² + 9y² + 16z² + 12xy + 24yz + 16xz

(ii) (-p + 2q + 3r)² = (-p)² + (2q)² + (3r)² + 2(-p) (2q) + 2(2q)(3r) + 2(3r) (- p)
= p²+ 4q² + 9r² – 4pq + 12qr – 6pr

(iii) (2p + 3) (2p – 4) (2p – 5)
[Here x = 2p, a = 3, b = -4 and c = -5]
= (2p)³ + (3 – 4 – 5) (2p)² + [(3)(-4) + (-4)(-5) + (3) (-5)] 2p + (3) (-4) (-5)
= 8p³ + (-6)(4p²) + (-12 + 20 – 15) 2p + 60
= 8p³ – 24p² – 14p + 60

(iv) (3a + 1) (3a – 2) (3a + 4)
[Here x = 3a, a = 1, b = -2 and c = 4]
= (3a)³ + (1 – 2 + 4) (3a)² + [(1)(-2) + (-2) (4) + (4) (1)] (3a) + (1) (-2) (4)
= 27a³ + 3(9a²) + (-2 – 8 + 4) (3a) – 8
= 27a³ + 27a² – 18a – 8

Question 2.
Using algebraic identity, find the coefficients of x², x and constant term without actual expansion.
(i) (x + 5)(x + 6)(x + 7)
Solution:
[Here x = x, a = 5, b = 6, c = 7]
(x + a) (x + b) (x + c) = x³ + (a + b + c)x² + (ab + bc + ac)x + abc
coefficient of x² = 5 + 6 + 7
= 18
coefficient of x = 30 + 42 + 35
= 107
constant term = (5) (6) (7)
= 210

(ii) (2x + 3)(2x – 5) (2x – 6)
Solution:
[Here x = 2x, a = 3, b = -5, c = -6]
(x + a) (x + b) (x + c) = x³ + (a + b + c)x² + (ab + bc + ac)x + abc
coefficient of x² = (3 – 5 – 6)4 [(2x)² = 4x²]
= (-8) (4)
= -32
coefficient of x = [(3)(-5) + (-5)(-6) + (-6)(3)](2)
= (-15 + 30-18) (2)
= (-3) (2)
= -6
constant term = (3) (-5) (-6)
= 90

Question 3.
If (x + a)(x + b)(x + c) = x³ + 14x² + 59x + 70, find the value of
(i) a + b + c
(ii) \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\)
(iii) a² + b² + c²
(iv) \(\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab}\)
Solution:
(x + a) (x + b) (x + c) = x³ + 14x² + 59x + 70
x³ + (a + b + c)x² + (ab + bc + ac)x + abc = x³ + 14x² + 59x + 70
a + b + c = 14, ab + bc + ac = 59, abc = 70
(i) a + b + c = 14

(ii) \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) = \(\frac{bc+ac+ab}{abc}\)
= \(\frac{59}{70}\)

(iii) a² + b² + c² = (a + b + c)² – 2 (ab + bc + ac)
= (14)² – 2(59)
= 196 – 118
= 78

Question 4.
Expand:
(i) (3a – 4b)³
Solution:
(a – b)³ = a³ – b³ – 3ab (a – b)
(3a – 4b)³ = (3a)³ – (4b)³ – 3(3a)(4b)(3a – 4b)
= 27a³ – 64b³ – 36ab (3a – 4b)
= 27a³ – 64b³ – 108a²b + 144ab²

(ii) [x + \(\frac{1}{y}]^{3}\)
Solution:
(a + b)³ = a³ + b³ + 3ab (a + b)

Question 5.
Evaluate the following by using identities:
(i) 98³
Solution:
98³ = (100 – 2)³ [(a – b)³ = a³ – b³ – 3ab (a – b)]
= 100³ – (2)³ – 3(100) (2) (100 – 2)
= 1000000 – 8 – 600(98)
= 1000000 – 8 – 58800
= 1000000 – 58808
= 941192

(ii) 1001³
Solution:
(1001)³ = (1000 + 1)³
[(a + b)³ = a³ + b³ + 3ab (a + b)]
= (1000)³ + 1³ + 3(1000) (1) (1000 + 1)
= 1000000000 + 1 + 3000 (1001)
= 1000000001 + 3003000
= 1003003001

Question 6.
If (x + y + z) = 9 and (xy + yz + zx) = 26, then find the value of x² + y² + z².
Solution:
x + y + z = 9; xy + yz + zx = 26
x² + y² + z² = (x + y + z)² – 2xy – 2yz – 2xz
= (x + y + z)² – 2 (xy + yz + zx)
= 9² – 2(26)
= 81 – 52
= 29

Question 7.
Find 27a³ + 64b³, If 3a + 4b = 10 and ab = 2
Solution:
3a + Ab = 10, ab = 2
27a³ + 64b³ = (3a)³ + (4b)³
[a³ + b³ = (a + b)³ – 3 ab (a + b)]
= (3a + 4b)³ – 3 × 3a × 4b (3a + 4b)
= 103 – 36ab (10)
= 1000 – 36(2)(10)
= 1000 – 720
= 280

Question 8.
Find x³ – y³, if x – y = 5 and xy = 14.
Solution:
x – y = 5, xy = 14
x³ – y³  = (x – y)³ + 3xy (x – y)
= 5³ + 3(14) (5)
= 125 + 210
= 335

Question 9.
If a + \(\frac{1}{a}\) = 6, then find the value of a³ +\(\frac{1}{a^3}\)
Solution:
a + \(\frac{1}{a}\) = 6 [a³ + b³ = (a + b)³ – 3ab (a + b)]

= 6³ – 3(6)
= 216 – 18
= 198

Question 10.
If x² + \(\frac{1}{x^2}\) = 23, then find the value of x + \(\frac{1}{x}\) and x³ + \(\frac{1}{x^3}\)
Solution:

When x = 5 [a³ + b³ = (a + b)³ – 3ab (a + b)]
= (5)³ – 3(5)
= 125 – 15
= 110
when x = -5
x³ + \(\frac{1}{x^3}\) = (-5)³ – 3(-5)
= -125 + 15
= -110
∴ x³ + \(\frac{1}{x^3}\) = ±110

Question 11.
If (y – \(\frac{1}{y})^{3}\) = 27 then find the value of y³ – \(\frac{1}{y^3}\)
Solution:

= 3³ + 3(3)
= 27 + 9
= 36

Question 12.
Simplify:
(i) (2a + 3b + 4c) (4a² + 9b² + 16c² – 6ab – 12bc – 8ca)
(ii) (x – 2y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
Solution:
x³ + y³ + z³ – 3xyz ≡ (x + y + z) (x² + y² + z² – xy – yz – zx)
(i) (2a + 3b + 4c) (4a² + 9b² + 16c² – 6ab – 12bc – 8ea)
= (2a)³ + (3b)³ + (4c)³ – 3 (2a) (3b) (4c)
= 8a³ + 27b³ + 64c³ – 72abc

(ii) (x – 2y + 3z) (x² + 4y² + 9z² + 2xy + 6yz – 3xz)
= x³ + (-2y)³ + (3z)³ – 3(x) (-2y) (3z)
= x³ – 8y³ + 27z³ + 18xyz

Question 13.
By using identity evaluate the following:
(i) 7³ – 10³ + 3³
Solution:
x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)
We know that a + b + c = 0 then a³ + b³ + c³ = 3ab
a + b + c = 7 + (-10) + 3
= 10 – 10
= 0
∴ 7³ – 10³ + 3³ = 3(7) (-10) (3)
= -630

(ii) 1 + \(\frac{1}{8}\) – \(\frac{27}{8}\)
Solution:
We know that a³ + b³ + c³ = 0 then a + b + c = 3abc

Question 14.
If 2x -3y – 4z = 0, then find 8x³ – 27y³ – 64z³.
Solution:
We know x³ +y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)
x³ + y³ + z³ = (x + y + z) (x² +y² + z² – xy – yz – zx) + 3xyz
8x³ – 27y³ – 64z³ = (2x)³ + (-3y)³ + (-4z)³
= (2x – 3y- 4z) [(2x)² + (-3y)² + (-4z)² – (2x)(-3y) – (-3y) (-4z) -(-4z)(2x)] + 3(2x)(-3y)(-4z)
= 0 (4x² + 9y² + 16z² + 6xy – 12yz + 8xz) + 72xyz
= 72xyz
8x³ – 27y³ – 64z³ = 72xyz

Students can download Maths Chapter 3 Algebra Ex 3.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.5

Question 1.
Factorise the following expressions:
(i) 2a² + 4a²b + 8a²c
(ii) ab – ac – mb + mc
Solution:
(i) 2a² + 4a²b + 8a²c = 2a²(1 + 2b + 4c)
(ii) ab – ac – mb + mc = a(b – c) – m(b – c)
= (b – c) (a – m)

Question 2.
Factorise the following expressions:
(i) x² + 4x + 4
(ii) 3a² – 24ab + 48b²
(iii) x⁵ – 16x
(iv) m² + \(\frac{1}{m^2}\) – 23
(v) 6 – 216x²
(vi) a² + \(\frac{1}{a^2}\) – 18
Solution:
(i) x² + 4x + 4 = x² + 2 × x × 2 + 2² [a² + 2ab+ b² = (a + b)²]
= (x + 2)²

(ii) 3a² – 24ab + 48b² = 3[a² – 8ab + 16b²]
= 3[a² – 2 × a × 4b + (4b)²]
= 3(a- 4b)² [a² – 2ab + b² = (a – b)²]

(iii) x⁵ – 16x = x[x⁴ – 16] [a² – b² = (a + b) (a – b)]
= x[(x²)² – 4²]
= x(x² + 4) (x² – 4)
= x(x² + 4) (x² – 2²)
= x(x²+ 4) (x + 2) (x – 2)

(iv) m² + \(\frac{1}{m^2}\) – 23 = [Add + 2 and – 2 to make -23 as -25]
= m² + \(\frac{1}{m^2}\) + 2 – 2 – 23 = m² + \(\frac{1}{m^2}\) + 2 – 25
= m² + \(\frac{1}{m^2}\) + 2 × m × \(\frac{1}{m}\) – 5²

(v) 6 – 216x² = 6[1 – 36x²]
= 6[1 – (6x)²] [a² – b² = (a + b)(a – b)]
= 6(1 + 6x) (1 – 6x)

(vi) a² + \(\frac{1}{a^2}\) – 18 = a² + \(\frac{1}{a^2}\) – 2 + 2 – 18
(add -2 and +2 to make 18 as 16)
= a² + \(\frac{1}{a^2}\) – 2 × a × \(\frac{1}{a}\) – 16 [a² + b² – 2ab = (a-b)(a- b)²]

Question 3.
Factorise the following expressions:
(i) 4x² + 9y² + 25z² + 12xy + 30yz + 20xz
Solution:
[a² + b² + c² + 2ab + 2bc + 2ac = (a + b + c)²]
= (2x)² + (3y)² + (5z)² + 2(2x) (3y) + 2(3y) (5z) + 2(5z) (2x)
= (2x + 3y + 5z)²

(ii) 25x² + 4y² + 9z² – 20xy + 12yz – 30xz
Solution:
= (5x)² + (2y)² + (3z)² + 2(5x) (-2y) + 2(-2y) (- 3z) + 2(-3z) (5x)
= (5x – 2y – 3z)²

Question 4.
Factorise the following expressions:
(i) 8x³ + 125y³
(ii) 27x³ – 8y³
(iii) a⁶ – 64
Solution:
(i) 8x³ + 125y³ = (2x)³ + (5y)³ [a³ + b³ = (a + b)(a² – ab + b²)
= (2x + 5y) [(2x)² – (2x) (5y) + (5y)²]
= (2x + 5y) (4x² – 10xy + 25y²)

(ii) 27x³ – 8y³ = (3x)³ – (2y)³ [a³ – b³ = (a – b)(a² + ab + b²)]
= (3x – 2y) [(3x)² + (3x) (2y) + (2y)²]
= (3x – 2y) (9x² + 6xy + Ay²)

(iii) a⁶ – 64 = a⁶ – 2⁶
= (a²)³ – (22)³ [a² – b³ = (a- b) + (a² + ab + b²)]
= (a² – 22) [(a²)² + (a²) (2²) + (2²)²]
= (a + 2) (a – 2) (a⁴ + 4a² + 16)
= (a + 2) (a – 2) [(a²)² + 4² + 8a² – 4a²]
= (a + 2)(a- 2) [(a² + 4)² – (2a)²] {a² – b² = (a + b) (a – b)}
= (a + 2) (a – 2) [(a² + 4 + 2a) (a² + 4 – 2a)
= (a + 2) (a – 2) (a² + 2a + 4) (a² – 2a + 4)

Question 5.
Factorize the following
(i) x³ + 8y³ + 6xy – 1
(ii) l³ – 8m³ – 27n³ – 18lmn
Using the formula [a³ + b³ + c³ – 3abc] = (a + b + c) (a² + b² + c² – ab – bc – ac)
Solution:
(i) x³ + 8y² + 6xy – 1 = -(-x³ – 8y³ – 6xy + 1)
= – (-x³ – 8y³ + 1 – 6xy)
= -[(-x)³ + (-2y)³ + 1 – 3(x) (2y) (1)]
= -[-x – 2y + 1] [(-x)² + (-2y)² + 1² – (-x) (-2y) – (-2y) (1) – (1) (-x)]
= (x + 2y – 1)(x² + 4y² + 1 – 2xy + 2y + x)

(ii) l³ – 8m³ – 27n³ – 18lmn
= l³ + (-2m)³ + (-3n)² – 3(l) (-2m) (-3n)
= (l – 2m – 3n) [l² + (-2m)² + (-3n)² -1 (-2m)] – (-2m)(-3n) – (-3n)(l)
= (l – 2m – 3n) (l² + 4m² + 9n² + 2lm – 6mn + 3ln)

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1.
Check whether p(x) is a multiple of g(x) or not.
(i) p(x) = x³ – 5x² + 4x – 3; g(x) = x – 2
Solution:
p(x) = x³ – 5x² + 4x – 3
P(2) = (2)³ – 5(2)² + 4(2) – 3
= 8 – 5(4) + 8 – 3
= 8 – 20 + 8 – 3
= 16 – 23
= -7
p{2) ≠ 0
∴ p(x) is not a multiple of g(x)

Question 2.
By remainder theorem, find the remainder when p(x) is divided by g(x) where,
(i) p(x) = x³ – 2x² – 4x – 1; g(x) = x + 1
Solution:
p(x) = x³ – 2x² – 4x – 1
p(-1) = (-1)³ – 2(-1)² – 4(-1) – 1
= 1 – 2 + 4 – 1
= 4 – 4 = 0
∴ The remainder = 0

(ii) p(x) = 4x³ – 12x² + 14x – 3; g(x) = 2x – 1
Solution:
p(x) = 4x³ – 12x² + 14x – 3

= 4 × \(\frac{1}{8}\) – 12 × \(\frac{1}{4}\) + 14 × \(\frac{1}{2}\) – 3
= \(\frac{1}{2}\) – 3 + 7 – 3
= \(\frac{1}{2}\) – 6 + 7
= \(\frac{1}{2}\) + 1
= \(\frac{3}{2}\)
∴ The reminder is \(\frac{3}{2}\)

(iii) p(x) = x³ – 3x² + 4x + 50; g(x) = x – 3
Solution:
p(x) = x³ – 3x² + 4x + 50
p(3) = 3³ – 3(3)² + 4(3) + 50
= 27 – 27 + 12 + 50
= 62
The remainder is 62.

Question 3.
Find the remainder when 3x³ – 4x² + 7x – 5 is divided by (x + 3)
Solution:
p(x) = 3x³ – 4x² + 7x – 5
When it is divided by x +3,
p(-3) = 3(-3)³ – 4(-3)² + 7(-3) – 5
= 3(-27) – 4(9) – 21 – 5
= -81 – 36 – 21 – 5
= -143
The remainder is -143.

Question 4.
What is the remainder when x²⁰¹⁸ + 2018 is divided by x – 1.
Solution:
p(x) = x²⁰¹⁸ + 2018
When it is divided by x – 1,
p(1) = 1²⁰¹⁸ + 2018
= 1 + 2018
= 2019
The remainder is 2019.

Question 5.
For what value of k is the polynomial
p(x) = 2x³ – kx² + 3x + 10 exactly divisible by x – 2
Solution:
p(x) = 2x³ – kx² + 3x + 10
When it is exactly divided by x – 2,
P(2) = 0
2(2)³ – k(2)² + 3(2) + 10 = 0
2(8) – k(4) + 6 + 10 = 0
16 – k(4) + 6 + 10 = 0
16 – 4k + 6 + 10 = 0
32 – 4k = 0
32 = 4k
∴ k = \(\frac{32}{4}\)
= 8
The value of k = 8

Question 6.
If two polynomials 2x³ + ax² + 4x – 12 and x³ + x² – 2x + a leave the same remainder when divided by (x – 3), find the value of a and also find the remainder.
Solution:
p(x₁) = 2x³ + ax² + 4x – 12
When it is divided by x – 3,
p(3) = 2(3)³ + a(3)² + 4(3) – 12
= 54 + 9a + 12 – 12
= 54 + 9a ……….(R₁)
p(x₂) = x³ + x² – 2x + a
When it is divided by x – 3,
p(3) = 3³ + 3² – 2(3) + a
= 27 + 9 – 6 + a
= 30 + a ………(R₂)
The given remainders are same (R₁ = R₂)
∴ 54 + 9a = 30 + a
9a – a = 30 – 54
8a = -24
∴ a = -24/8
= -3
Consider R₂,
Remainder = 30 – 3
= 27

Question 7.
Determine whether (x – 1) is a factor of the following polynomials:
(i) x³ + 5x² – 10x + 4
Solution:
p(x) = x³ + 5x² – 10x + 4
p(1) = 1³ + 5(1) – 10(1) + 4
= 1 + 5 – 10 + 4
= 10 – 10
= 0
∴ x – 1 is a factor of p(x)

(ii) x⁴ + 5x² – 5x + 1
Solution:
p(1) = 1⁴ + 5(1)² – 5(1) + 1
= 1 + 5 – 5 + 1
= 7 – 5
= 2
= 0
∴ x – 1 is not a factor of p(x)

Question 8.
Using factor theorem, show that (x – 5) is a factor of the polynomial
2x³ – 5x² – 28x + 15
Solution:
p(x) = 2x³ – 5x² – 28x + 15
x – 5 is a factor
p(5) = 2(5)³ – 5(5)² – 28(5) + 15
= 250 – 125 – 140 + 15
= 265 – 265
= 0
∴ x – 5 is a factor of p(x)

Question 9.
Determine the value of m, if (x + 3) is a factor of x³ – 3x² – mx + 24.
Solution:
p(x) = x³ – 3x² – mx + 24
when x + 3 is a factor
P(-3) = 0
(-3)³ – 3(-3)² – m(-3) + 24 = 0
-27 – 27 + 3m + 24 = 0
-54 + 24 + 3m = 0
-30 + 3m = 0
3m = 30
m = \(\frac{30}{3}\)
= 10
The value of m = 10

Question 10.
If both (x-2) and (x – \(\frac{1}{2}\)) are the factors of ax² + 5x + b, then show that a = b.
Solution:
p(x) = ax² + 5x + b
when (x-2) is a factor
P(2) = 0
a(2)² + 5(2) + b = 0
4a + 10 + b = 0
4a + b = -10 …….(1)
when (x – \(\frac{1}{2}\)) is a factor
p(\(\frac{1}{2}\)) = 0
a\((\frac{1}{2})^2\) + 5(\(\frac{1}{2}\)) + b = 0
Multiply by 4
a + 10 + 4b = 0
a + 46 = -10 …….(2)
From (1) and (2) we get
4a + b = a + 4b
4a – a = 4b – b
3a = 3b
a = b
Hence it is proved.

Question 11.
If (x – 1) divides the polynomial kx³ – 2x² + 25x – 26 without remainder, then find the value of k.
Solution:
p(x) = kx³ – 2x² + 25x – 26
When it is divided by x – 1
P(1) = 0
k(1)³ – 2(1)² + 25(1) – 26 = 0
k – 2 + 25 – 26 = 0
k + 25 – 28 = 0
k – 3 = 0
k = 3
The value of k = 3

Question 12.
Check if (x + 2) and (x – 4) are the sides of a rectangle whose area is x² – 2x – 8 by using factor theorem.
Solution:
Let the area of a rectangle be p(x)
p(x) = x² – 2x – 8
When x + 2 is the side of the rectangle
p(-2) = (-2)² – 2(-2) – 8
= 4 + 4 – 8
= 8 – 8
= 0
When x – 4 is the side of the rectangle.
P(4) = (4)² – 2(4) – 8
= 16 – 8 – 8
= 16 – 16
= 0
(x + 2) and (x – 4) are the sides of a rectangle

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 1.
Find the value of the polynomial f(y) = 6y – 3y² + 3 at
(i) y = 1
(ii) y = -1
(iii) y = 0
Solution:
(i) When y = 1
f(y) = 6y – 3y² + 3
f(1) = 6(1) – 3(1)² + 3
= 6 – 3 + 3 = 6

(ii) When y = – 1
f(y) = 6y – 3y² + 3
f(-1) = 6(-1) – 3(-1)² + 3
= – 6 – 3 + 3
= – 6

(iii) When y = 0
f(y) = 6y – 3y² + 3
f(0) = 6(0) – 3(0)² + 3
= 0 – 0 + 3
= 3

Question 2.
If p(x) = x² – 2√2x + 1, find p(2√2).
Solution:
p(x) = x² – 2√2x + 1
p(2√2) = (2√2)² – 2√2 (2√2) + 1
= 8 – 8 + 1
= 0 + 1
= 1

Question 3.
Find the zeros of the polynomial in each of the following.
(i) P(x) = x – 3
Solution:
p( 3) = 3 – 3
= 0
p(3) is the zero of p(x)

(ii) p(x) = 2x + 5
Solution:

= -5 + 5
= 2(0)
= 0
Hence –\(\frac{5}{2}\) is the zero of p(x).

(iii) q(y) = 2y – 3
Solution:

= 2 × 0
= 0
Hence \(\frac{3}{2}\) is the zero of q(y).

(iv) f(z) = 8z
Solution:
f(0) = 8 × 0
= 0
Hence 0 is the zero of f(z)

(v) p(x) = ax when a ≠ 0
Solution:
p(0) = a(0)
= 0
Hence, 0 is the zero of p(x)

(vi) h(x) = ax + b, a ≠ 0, a, b∈R
Solution:

Hence –\(\frac{b}{a}\) is the zero of h(x).

Question 4.
Find the roots of the polynomial equations.
(i) 5x – 6 = 0
Solution:
5x = 6
x = \(\frac{6}{5}\)
\(\frac{6}{5}\) is the root of the polynomial.

(ii) x + 3 = 0
Solution:
x = -3
-3 is the root of the polynomial.

(iii) 10x + 9 = 0
Solution:
10x = -9
x = –\(\frac{9}{10}\)
–\(\frac{9}{10}\) is the root of the polynomial.

(iv) 9x – 4 = 0
Solution:
9x = 4
x = \(\frac{4}{9}\)
\(\frac{4}{9}\) is the root of the polynomial.

Question 5.
Verify whether the following are zeros of the polynomial, indicated against them,or not.
(i) p(x) = 2x – 1, x = \(\frac{1}{2}\)
Solution:
p (\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\)) – 1
= 1 – 1
= 0
∴ \(\frac{1}{2}\) is the zero of the polynomial.

(ii) p(x) = x³ – 1, x = 1
Solution:
p(1) = 1³ – 1
= 1 – 1
= 0
∴ 1 is the zero of the polynomial

(iii) p(x) = ax + b, x = \(\frac{-b}{a}\)
Solution:
p(\(\frac{-b}{a}\)) = a(\(\frac{-b}{a}\)) + b
= -b + b
= 0
∴ \(\frac{-b}{a}\) is the zero of the polynomial. a

(iv) p(x) = (x + 3) (x – 4); x = -3, x = 4
Solution:
P(-3) = (-3 + 3) (-3 – 4)
= (0) (-7)
= 0
P( 4) = (4 + 3) (4 – 4)
= (7) (0)
= 0
∴ -3 and 4 are the zeros of the polynomial.

Question 6.
Find the number of zeros of the following polynomials represented by their graphs.

Solution:
(i) Number of zeros = 2 (The curve is intersecting the x-axis at 2 points)
(ii) Number of zeros = 3 (The curve is intersecting the x-axis at 3 points)
(iii) Number of zeros = 0 (The curve is not intersecting the x-axis)
(iv) Number of zeros = 1 (The curve is intersecting at the origin)
(v) Number of zeros = 1 (The curve is intersecting the x-axis at one point)

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.4

Question 1.
If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find
(i) (P∪Q)∪R
(ii) (P∩Q)∩S
(m) (Q∩S)∩R
Solution:
P = {1, 2, 5, 7, 9}; Q = {2, 3, 5, 9, 11}; R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}
(i) P∪Q = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11}
= {1, 2, 3, 5, 7, 9, 11}
(P∪Q)∪R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9}
= {1, 2, 3, 4, 5, 7, 9, 11}

(ii) P∩Q = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11}
= {2, 5, 9}
(P∩Q)∩S = {2, 5, 9} ∩ {2, 3, 4, 5, 8}
= (2, 5}

(iii) Q∩S = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8}
= {2, 3, 5}
(Q∩S)∩R = {2, 3, 5} ∩ {3, 4, 5, 7, 9}
= {3, 5}

Question 2.
Test for the commutative property of union and intersection of the sets
P = {x : x is a real number between 2 and 7} and
Q = {x : x is an irrational number between 2 and 7}
Solution:
P is a real number set
Q is a set of irrational number
∴ Q⊂P
P∪Q= Q∪P = P
∴ Union of sets is commutative.
P∩Q = Q∩P = Q
∴ Intersection of sets is commutative.

Question 3.
If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets.
Solution:
When union of sets is associative
A∪(B∪C) = (A∪B)∪C
(B∪C) = {m, n, q, s, t) ∪ {m, n, p, q, s}
= {m, n, p, q, s, t}
A∪(B∪C) = {p, q, r, s} ∪ {m, n, p, q, s, t}
= {m, n, p, q, r, s, t} ……..(1)
(A∪B) = {p, q, r, s} ∪ {m, n, q, s, t}
= {m, n, p, q, r, s, t}
(A∪B)∪C = {m, n, p, q, r, s, t} ∪ {m, n, p, q, s}
= {m, n, p, q, r, s, t} ……….(2)
From (1) and (2) it is verified that A∪(B∪C) = (A∪B)∪C

Question 4.
Verify the associative property of intersection of sets for A = {-11, √2, √5, 7},
B = {√3, √5, 6, 13} and C = {√2, √3, √5, 9}.
Solution:
When intersection of sets is associative
A∩(B∩C) = (A∩B)∩C
(B∩C) = {√3, √5, 6, 13} ∩ {√2, √3, √5, 9}
= {√3, √5}
A∩(B∩C) = {-11, √2, √5, 7} ∩ {√3, √5}
{√5} ………(i)
(A∩B) = {-11, √2, √5, 7} ∩ {√3, √5 ,6, 13}
= {√5}
(A∩B)∩C = {√5} n {√2, √3, √5, 9}
= {√5}……..(2)
From (1) and (2) it is verified that A∩(B∩C) = (A∩B)∩C

Question 5.
If A={ x : x = 2ⁿ, n ∈ W and n < 4}, B = {x : x = 2n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets.
Solution:
A = {x : x = 2ⁿ, n ∈ W and n < 4}
A = {1, 2, 4, 8}
B = {x : x = 2n, n ∈ N and n ≤ 4}
B = {2, 4, 6, 8}
C ={0, 1, 2, 5, 6}
When intersection of sets is associative
A∩(B∩C) = (A∩B)∩C
(B∩C) = {2, 4, 6, 8} ∩ {0, 1, 2, 5, 6}
= {2, 6}
A∩(B∩C)= {1 ,2, 4, 8} ∩ {2, 6}
= {2}……….(1)
(A∩B) = {1, 2, 4, 8} ∩ {2, 4, 6, 8}
= {2, 4, 8}
(A∩B)∩c= {2, 4, 8} n {0, 1, 2, 5, 6}
= {2}………..(2)
From (1) and (2) we get A∩(B∩C) = (A∩B)∩C

Students can download Maths Chapter 1 Set Language Ex 1.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.3

Question 1.
Using the given venn diagram, write the elements of

(i) A
(ii) B
(iii) A∪B
(iv) A∩B
(v) A – B
(vi) B – A
(vii) A’
(viii) B’
(ix) U
Solution:
(i) A = {2, 4, 7, 8, 10}
(ii) B = {3, 4, 6, 7, 9, 11}
(iii) A∪B = {2, 3, 4, 6, 7, 8, 9, 10, 11}
(iv) A∩B = {4, 7}
(v) A – B = {2, 8, 10}
(vi) B – A = {3, 6, 9, 11}
(vii) A’ = {1, 3, 6, 9, 11, 12}
(viii) B’ = {1,2, 8, 10, 12}
(ix) U = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12}

Question 2.
Find A∪B, A∩B, A – B and B – A for the following sets
(i) A = {2, 6, 10, 14} and B = {2, 5, 14, 16}
Solution:
A∪B = {2, 6, 10, 14} ∪ {2, 5, 14, 16}
= {2, 5, 6, 10, 14, 16}
A∩B = {2, 6, 10, 14} ∩ {2, 5, 14, 16}
= {2, 14}
A – B = {2, 6, 10, 14} – {2, 5, 14, 16}
= {6, 10}
B – A = {2, 5, 14, 16} – {2, 6, 10, 14}
= {5, 16}

(ii) A = {a, b, c, e, u} and B = {a, e, i, o, u}
Solution:
A∪B = {a, b, c, e, u} ∪ {a, e, i, o, u}
= {a, b, c, e, i, o, u}
A∩B = {a, b, c, e, u} ∩ {a, e, i, o, u}
= {a, e, u}
A – B = {a, b, c, e, u} – {a, e, i, o, u}
= {b, c}
B – A = {a, e, i, o, u} – {a, b, c, e, u}
= {i, o}

(iii) A = {x : x ∈ N, x ≤ 10} and B = {x : x ∈ W, x < 6}
Solution:
A = {1, 2, 3, 4, 5, 6,7, 8, 9, 10} and B = {0, 1, 2, 3, 4, 5}
A∪B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∪ {0, 1, 2, 3, 4, 5}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A∩B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 3, 4, 5}
= {1, 2, 3, 4, 5}
A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} – {0, 1, 2, 3, 4, 5}
= {6, 7, 8, 9, 10}
B – A = {0, 1, 2, 3, 4, 5} – {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {0}

(iv) A = Set of all letters in the word “mathematics” and B = Set of all letters in the word “geometry”
Solution:
A = {m, a, t, h, e, i, c, s}
B = {g, e, o, m, t, r, y}
A∪B = {m, a, t, h, e, i, c, s} ∪ {g, e, o, m, t, r, y}
= {a, c, e, g, h, i, m, o, r, s, t, y}
A∩B= {m, a, t, h, e, i, c, s} ∩ {g, e, o, m, t, r, y}
= {e, m, t}
A – B = {m, a, t, h, e, i, c, 5} – {g, e, o, m, t, r, y}
= {a, c, h, i, s}
B -A = {g, e, o, m, t, r, y} – {m, a, t, h, e, i, c, s}
= {g, o, r, y}

Question 3.
If U = {a, b, c, d, e, f, g, h), A = {b, d, f, h} and B = {a, d, e, h}, find the following sets.
(i) A’
Solution:
A’ = U – A
= {a, b, c, d, e, f, g, h} – {b, d, f, h)
= {a, c, e, g}

(ii) B’
Solution:
B’ = U – B
= {a, b, c, d, e, f, g, h} – {a, d, e, h}
= {b, c, f, g}

(iii) A’∪B’
Solution:
A’∪B’ = {a, c, e, g} ∪ {b, c,f, g}
= {a, b, c, e, f, g}

(iv) A’∩B’
Solution:
A’∩B’ = {a, c, e, g} ∩ {b, c, f, g}
= {c, g}

(v) (A∪B)’
Solution:
A∪B = {b, d, f, h} ∪ {a, d, e, h}
= {a, b, d, e, f, h}
(A∪B)’ = U – (A∪B)
= {a, b, c, d, e, f, g, h} – {a, b, d, e, f, h}
= {c, g}

(vi) (A∩B)’
Solution:
(A∩B) = {b, d, f, h) ∩ {a, d, e, h)
= {d, h}
(A∩B)’ = U – (A∩B)
= {a, b, c, d, e, f, g, h} – {d, h}
= {a, b, c, e, f, g}

(vii) (A’)’
Solution:
A’ = {a, c, e, g}
(A’)’ = U – A’
= {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, h}

(viii) (B’)’
Solution:
B’ = {b, c, f, g}
(B’)’ = U – B’
= {a, b, c, d, e, f, g, h) – {b, c, f, g)
= {a, d, e, h}

Question 4.
Let U = {0, 1, 2, 3, 4, 5, 6, 7} A = {1, 3, 5, 7} and B = {0, 2, 3, 5, 7}, find the following sets.
(i) A’
Solution:
A’ = U – A
= {0, 1, 2, 3, 4, 5, 6, 7} – {1, 3, 5, 7}
= {0, 2, 4, 6}

(ii) B’ = U – B
Solution:
= {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 3, 5, 7}
= {1, 4, 6}

(iii) A’∪B’
Solution:
A’∪B’ = {0, 2, 4, 6,}∪{1, 4, 6}
{0, 1, 2, 4, 6}

(iv) A’∩B’
Solution:
A’∩B’ = {0, 2, 4, 6,}∩{1, 4, 6}
{4, 6}

(v) (A∪B)’
Solution:
A∪B = {1, 3, 5, 7}∪{0, 2, 3, 5, 7}
= {0, 1, 2, 3, 5, 7}
(A∪B)’ = U – (A∪B)
{0, 1, 2, 3, 4, 5, 6, 7} – {0, 1, 2, 3, 5, 7}
{4, 6}

(vi) (A∩B)’
Solution:
(A∩B)= {1, 3, 5, 7}∩{0, 2, 3, 5, 7}
= {3, 5, 7}
(A∩B)’ = U – (A∩B)
= {0, 1, 2, 3, 4, 5, 6, 7} – {3, 5, 7}
= {0, 1, 2, 4, 6}

(vii) (A’)’
A’ = {0, 2, 4, 6}
(A’)’ = U – A’
= {0, 1, 2, 3, 4, 5, 6, 7} – {0, 2, 4, 6}
= {1, 3, 5, 7}

(viii) (B’)’
B’ = {1, 4, 6}
(B’)’ = {0, 1, 2, 3, 4, 5, 6, 7} – {1, 4, 6}
= {0, 2, 3, 5, 7}

Question 5.
Find the symmetric difference between the following sets.
(i) P = {2, 3, 5, 7, 11} and Q = {1, 3, 5, 11}
Solution:
Use anyone of the formula to find A & B
AΔB = (A – B)∪(B – A) or AΔB = (A∪B) – (A∩B)
P∪Q = {2, 3, 5, 7, 11} ∪ {1,3,5, 11}
= {1, 2, 3, 5, 7, 11}
P∩Q = {2, 3, 5, 7, 11}∩{1, 3, 5,11}
= {3, 5, 11}
PΔQ = (P∪Q) – (P∩Q)
= {1, 2, 3, 5, 7, 11} – {3, 5, 11}
= {1, 2, 7}
(OR)
P – Q = {2, 3, 5, 7, 11} – {1, 3, 5, 11}
= {2,7}
Q – P = {1, 3, 5, 11} – {2, 3, 5, 7, 11}
= {1}
PΔQ = (P – Q)∪(Q – P)
= {2, 7} ∪ {1}
= {1, 2, 7}

(ii) R = {l, m, n, o, p} and S = {j, l, n, q}
Solution:
R- S = {l, m, n, o, p} – {j, l, n, q}
= {m, o, p}
S – R = {j, l, n, q} – {l, m, n, o, p}
= {j, q}
RΔS = (R – S)∪(S – R)
= {m, o, p} – {j, q} = {j, m, o, p, q)
(OR)
R∪S = {l, m, n, o, p} ∪ {j,l,n,q}
= {l, m, n, o, p, j, q}
R∩S = {l, m, n, o,p} ∩ {j, l, n, q}
= {l, n}
RΔS = (R∪S) – (R∩S)
= {l, m, n, o, p, j, q} – { l, n}
= {m, o, p, j, q}

(iii) X = {5, 6, 7} and Y = {5, 7, 9, 10}
Solution:
X∪Y = {5, 6, 7} ∪ {5, 7, 9, 10}
= {5 ,6, 7, 9, 10}
X∩Y = {5, 6, 7} ∩ {5, 7, 9, 10}
= {5, 7}
XΔY = (X∪Y) – (X∩Y)
= {5, 6, 7, 9, 10} – {5, 7}
= {6, 9, 10}
OR
X – Y = {5, 6, 7} – {5, 7, 9, 10} = {6}
Y – X = {5, 7, 9, 10} – {5, 6, 7} = {9, 10}
XΔY = (X – Y) ∪ (Y – X)
= {6}∪{9, 10}
= {6, 9, 10}

Question 6.
Using the set symbols, write down the expressions for the shaded region in the following

Solution:

Question 7.
Let A and B be two overlapping sets and the universal set U. Draw appropriate Venn diagram for each of the following,
(i) A∪B
(ii) A∩B
(iii) (A∩B)’
(iv) (B – A)’
(v) A’∪B’
(Vi) A’∩B’
(vii) What do you observe from the Venn diagram (iii) and (v)?
Solution:
(i) A∪B

(ii) A∩B

(iii) (A∩B)’

(iv) (B – A)’

(v) A’∪B’

(Vi) A’∩B’

(vii) What do you observe from the diagram (iii) and (v)?
From the diagram (iii) and (v) we get (A∩B)’ = A’∪B’

Students can download Maths Chapter 1 Set Language Ex 1.2 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.2

Question 1.
Find the cardinal number of the following sets.
(i) M = {p, q, r, s, t, u}
(ii) P = {x : x = 3n + 2, n ∈ W and x < 15}
(iii) Q = {y : y = \(\frac{4}{3n}\), n ∈ N and 2 < n ≤ 5}
(iv) R = {x : x is an integer, x ∈ Z and – 5 ≤ x < 5}
(v) S = The set of all leap years between 1882 and 1906.
Solution:
(i) n (M) = 6
(ii) n (P) = 5 [n = {0, 1, 2, 3 . . . . 14}]
(iii) Since n = {3, 4, 5} ; n (Q) = 3
(iv) X = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4} ∴ n (R) = 10
(v) S = {1884, 1888, 1892, 1896, 1904}; n (S) = 5

Question 2.
Identify the following sets as finite or infinite.
(i) X = The set of all districts in Tamilnadu.
(ii) Y = The set of all straight lines passing through a point.
(iii) A = {x : x ∈ Z and x < 5}
(iv) B = {x : x² – 5x + 6 = 0, x ∈ N}
Solution:
(i) Finite
(ii) Infinite set (many lines can be drawn from a point)
(iii) Infinite set {A = ……. -2, -1, 0, 1, 2, 3, 4}
(iv) Finite set [x² – 5x + 6 = 0 ⇒ (x – 3) (x – 2) = 0; x = 3 and 2]

Question 3.
Which of the following sets are equivalent or unequal or equal sets?
(i) A = The set of vowels in the English alphabets.
B = The set of all letters in the word “VOWEL”
(ii) C = {2, 3, 4, 5}
D = {x : x ∈ W, 1 < x < 5}
(iii) X = {x : x is a letter in the word “LIFE”}
Y = {F, I, L, E}
(iv) G = {x : x is a prime number and 3 < x < 23}
H = {x : x is a divisor of 18}
Solution:
(i) Equivalent set [n(A) = n(B) = 5] ∴ A ≈ B
(ii) Unequal sets [C = {2, 3, 4, 5}; D = {2, 3, 4}]
(iii) Equal sets [X = {L, I, F, E}; Y = {F, I, L, E} [n(X) = 4 = n(Y)] ∴ X ≈ Y
(iv) Equivalent sets [G = {5, 7, 11, 13, 17, 19}; H = {1, 2, 3, 6, 9, 18}]
[n(G) = n(H) = 6 ∴ G ≈ H)]

Question 4.
Identify the following sets as null set or singleton set.
(i) A = {x : x ∈ N, 1 < x < 2}
(ii) B = The set of all even natural numbers which are not divisible by 2
(iii) C = {0}.
(iv) D = The set of all triangles having four sides.
Solution:
(i) Null set [No natural numbers is in between 1 and 2]
(ii) Null set [All the even natural numbers are not divisible by 2]
(Hi) Singleton set [n (C) = 1]
(iv) Null set [All the triangles has 3 sides]

Question 5.
State which pairs of sets are disjoint or overlapping?
(i) A = {f, i, a, s} and B = {a, n, f, h, s}
A = {f, i, a, s} and B = {a, n, f, h, s}
A and B are overlapping sets

(ii) C = {x : x is a prime number, x > 2} and D = {x : x is an even prime number}
C= {3, 5, 7…….}
D = {2}
C and D are disjoint sets

(iii) E = {x : x is a factor of 24} and F = {x : x is a multiple of 3, x < 30}
E = {1, 2, 3, 4, 6, 8, 12, 24}
F = {3, 6, 9, 12, 15, 18, 21, 24, 27} [Hint: E ∩ F = {3, 6, 24, …….}]
E and F are overlapping sets

Question 6.
If S = {square, rectangle, circle, rhombus, triangle}. List the elements of the following subset of S.
(i) The set of shapes which have 4 equal sides.
(ii) The set of shapes which have radius.
(iii) The set of shapes in which the sum of all interior angles is 180°.
(iv) The set of shapes which have 5 sides.
Solution:
(i) Subset of S = {square, rhombus}
(ii) Subset of S = {circle}
(iii) Subset of S = {triangle}
(iv) Subset of S = { }

Question 7.
If A = {a,{a, b}}, write all the subsets of A.
Solution:
A = {a, {a, b}}
Subset of A are Ø, {a}, {a, b}, {a, {a, b}} (or) { }, {a}, {a,b, {a,{a,b}}
P(A) = {Ø, {a}, {a, b}, {a {a, b}} (or) {{ }, {a}, {a,b, {a,{a,b}}

Question 8.
Write down the power set of the following sets.
(i) A = {a, b}
(ii) B = {1, 2, 3}
(iii) D = {p, q, r, s}
(iv) E = Ø
Solution:
(i) A = {a, b)
P(A) = {{},{a},{b}, {a, b}}

(ii) B = {1, 2, 3}
P(B) = {{}, {1}, {2}, {3}, {1,2}, {2, 3}, {1,3}, {1,2,3}}

(iii) D = {p, q, r, s}
P(D) = {{},{p},{q},{r},{s},{p, q} {p, r} {p, s}
{q, r}, {q, s}, {r, s}, {p, q, r} {q, r, s}
{p, r, s} {p, q, s} {p, q, r, s}}

(iv) E = Ø
P(E) = {{}}
Note: (empty set is the subset of all the sets)

Question 9.
Find the number of subsets and the number of proper subsets of the following sets.
(i) W = {red, blue, yellow}
(ii) X = {x² : x ∈ N, x² ≤ 100}
Solution:
(i) W = {red, blue, yellow}
n (W) = 3
The number of subsets of W = n [P(W)] = 2^m
= 2³ = 8
Number of proper subsets of W = n[P(W)] – 1
= 8 – 1
= 7 (or)
Number of proper subsets of W = 2^m – 1
= 2³ – 1 = 8 – 1 = 7

(ii) X = {x² : x ∈ N, x² ≤ 100}.
X= {1,2, 3, 4, …. 10}
n(X) = 10
The number of subsets of X = n[P(X)]
= 2^m
= 2¹⁰ = 1024
Number of proper subsets of X = 2^m – 1
= 1024 – 1
= 1023

Question 10.
(i) If n(A) = 4, find n[P(A)]
(ii) If n(A) = 0, find n[P(A)]
(Hi) If n[P(A)] = 256, find n(A)
Solution:
(i) n (A) = 4
n [P(A)] = 2^m = 2⁴
= 16

(ii) n (A) = 0
n [P(A)] = 2^m = 2° = 1

(iii) n [P(A)] = 256

2^m = 2⁸
∴ n (A) = 8