Category: Class 9

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 3.3 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 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

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 3.2 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 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)

Students can download Maths Chapter 3 Algebra Ex 3.1 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 3 Algebra Ex 3.1

Question 1.
Which of the following expressions are polynomials. If not give reason:
(i) \(\frac{1}{x^2}\) + 3x – 4
Solution:
(i) \(\frac{1}{x^2}\) + 3x – 4 is not a polynomial. Since the exponent of x² is not a whole number, but it is (\(\frac{1}{x^2}\) = x^-2) negative number.

(ii) x² (x – 1)
Solution:
x² (x – 1) is a polynomial.

(iii) \(\frac{1}{x}\) (x + 5)
Solution:
\(\frac{1}{x}\) (x + 5) is not a polynomial. Since the exponent of x is not a whole number, but it is (\(\frac{1}{x}\) = x^-1) negative number.

(iv) \(\frac{1}{x^{-2}}\) + \(\frac{1}{x^{-1}}\) + 7
Solution:
\(\frac{1}{x^{-2}}\) + \(\frac{1}{x^{-1}}\) + 7 is a polynomial. (\(\frac{1}{x^{-2}}\) = x² and \(\frac{1}{x^{-1}}\) = x)

(v) √5x² + √3x + √2
Solution:
√5x² + √3x + √2 is a polynomial.

(vi) m² – \(\sqrt[3]{m}\) + 7m – 10
m² –\(\sqrt[3]{m}\) + 7m – 10 is not a polynomial. Since the exponent of m is not a whole number.
(\(\sqrt[3]{m}\) = m^1/3)

Question 2.
Write the coefficient of x² and x in each of the following polynomials.
(i) 4 + \(\frac{2}{5}\) x² – 3x
Solution:
Coefficient of x² is \(\frac{2}{5}\) and coefficient of x is -3.

(ii) 6 – 2x² + 3x³ – √7x
Solution:
Coefficient of x² is -2 and coefficient of x is -√7

(iii) π x² – x + 2
Solution:
Coefficient of x² is π and coefficient of x is -1.

(iv) √3x² + √2x + 0.5
Solution:
Coefficient of x² is √3 and coefficient of x is √2

(v) x² – \(\frac{7}{2}\) x + 8
Solution:
Coefficient of x² is 1 and coefficient of x is –\(\frac{7}{2}\)

Question 3.
Find the degree of the following polynomials.
(i) 1 – √2 y² + y⁷
(ii) \(\frac{x^{3}-x^{4}+6 x^{6}}{x^{2}}\)
(iii) x³ (x² + x)
(iv) 3x⁴ + 9x² + 27x⁶
(v) 2√5p⁴ \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)
Solution:
(i) 1 – √2 y² + y⁷
The degree of the polynomial is 7.

(ii)
= x – x² + 6x⁴
The degree of the polynomial is 4.

(iii) x³ (x² + x) = x⁵ + x⁴
The degree of the polynomial is 5.

(iv) 3x⁴ + 9x² + 27x⁶
The degree of the polynomial is 6.

(v) 2√5p⁴ \(-\frac{8 p^{3}}{\sqrt{3}}+\frac{2 p^{2}}{7}\)
The degree of the polynomial is 4.

Question 4.
Rewrite the following polynomial in standard form.
(i) x – 9 + √7x³ + 6x²
Solution:
The standard form is √7x³ + 6x² – x – 9
(or) – 9 + x + 6x² + √7x³

(ii) √2x² – \(\frac{7}{2}\) x⁴ + x – 5x³
Solution:
The standard form is – \(\frac{7}{2}\) x⁴ – 5x³ + √2x² + x
(or) x + √2x² – 5x³ – \(\frac{7}{2}\) x⁴

(iii) 7x³ – \(\frac{6}{5}\) x² + 4x – 1
Solution:
The given polynomial is in standard form (or) – 1 + 4x – \(\frac{6}{5}\) x² + 7x³

(iv) y² + √5y³ – 11 – \(\frac{7}{3}\) y + 9y⁴
Solution:
The standard form is 9y⁴ + √5y³ + y² – \(\frac{7}{3}\) y – 11
(or) – 11 – \(\frac{7}{3}\) y + y² + √5y³ + 9y⁴

Question 5.
Add the following polynomials and find the degree of the resultant polynomial
(i) p(x) = 6x² – 7x + 2; q(x) = 6x³ – 7x + 15
Solution:
p(x) + q(x) = 6x² – 7x + 2 + 6x³ – 7x + 15
= 6x³ + 6x² – 7x – 7x + 2 + 15
= 6x³ + 6x² – 14x + 17
The degree of the polynomial is 3.

(ii) h(x) = 7x³ – 6x + 1; f(x) = 7x² + 17x – 9
Solution:
h(x) + f(x) = 7x³ – 6x + 1 + 7x² + 17x – 9
= 7x³ + 7x² + 11x – 8
The degree of the polynomial is 3.

(iii) f(x) = 16x⁴ – 5x² + 9; g(x) = -6x³ + 7x – 15
Solution:
f(x) + g(x) = 16x⁴ – 5x² + 9 – 6x³ + 7x – 15
= 16x⁴ – 6x³ – 5x² + 7x + 9 – 15
= 16x⁴ – 6x³ – 5x² + 7x – 6
The degree of the polynomial is 4.

Question 6.
Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial.
(i) p(x) = 7x² + 6x – 1; q(x) = 6x – 9
Solution:
p(x) – q(x) = 7x² + 6x – 1 – (6x – 9)
= 7x² + 6x – 1 – 6x + 9
= 7x² + 6x – 6x – 1 + 9
= 7x² + 8
The degree of the polynomial is 2.

(ii) f(y) = 6y² – 7y + 2; g(y) = 7y + y³
Solution:
f(y) – g(y) = 6y² – 7y + 2 – (7y + y³)
= 6y² – 7y + 2 – 7y – y³
= -y³ + 6y² – 7y – 7y + 2
= -y³ + 6y² – 14y + 2
The degree of the polynomial is 3.

(iii) h(z) = z⁵ – 6z⁴ + z; f(z) = 6z² + 10z – 7
Solution:
h(z) – f(z) = z⁵ – 6z⁴ + z – (6z² + 10z – 7)
= z⁵ – 6z⁴ + z – 6z² – 10z + 7
= z⁵ – 6z⁴ – 6z² + z – 10z + 7
= z⁵ – 6z⁴ – 6z² – 9z + 7
The degree of the polynomial is 5.

Question 7.
What should be added to 2x³ + 6x² – 5x + 8 to get 3x³ – 2x² + 6x + 15?
Solution:
3x³ – 2x² + 6x + 15 – (2x³ + 6x² – 5x + 8)
= 3x³ – 2x² + 6x + 15 – 2x³ – 6x² + 5x – 8
= 3x³ – 2x³- 2x² – 6x² + 6x + 5x + 15 – 8
= x³ – 8x² + 11x + 7
x³ – 8x² + 11x + 7 must be added to get 3x³ – 2x² + 6x + 15.

Question 8.
What must be subtracted from 2x⁴ + 4x² – 3x + 7 to get 3x³ – x² + 2x + 1?
Solution:
2x⁴ + 4x² – 3x + 7 – (3x³ – x² + 2x + 1)
= 2x⁴ + 4x² – 3x + 7 – 3x³ + x² – 2x – 1
= 2x⁴ – 3x³ + 4x² + x² – 3x – 2x + 7 – 1
= 2x⁴ – 3x³ + 5x² – 5x + 6
2x⁴ – 3x³ + 5x² – 5x + 6 must be subtracted to get 3x³ – x² + 2x + 1.

Question 9.
Multiply the following polynomials and find the degree of the resultant polynomial:
(i) p(x) = x² – 9, q(x) = 6x² + 7x – 2
Solution:
p(x) × q(x) = (x² – 9) (6x² + 7x – 2)
= 6x⁴ + 7x³ – 2x² – 54x² – 63x + 18
= 6x⁴ + 7x³ – 56x² – 63x + 18
The degree of the polynomial is 4.

(ii) f(x) = 7x + 2, g(x) = 15x – 9
Solution:
f(x) × g(x) = (7x + 2) (15x – 9)
= 105x² – 63x + 30x – 18
= 105x² – 33x – 18
The degree of the polynomial is 2.

(iii) h(x) = 6x² – 7x + 1, f(x) = 5x – 7
Solution:
h(x) × f(x) = (6x² – 7x + 1) (5x – 7)
= 30x³ – 42x² – 35x² + 49x + 5x – 7
= 30x³ – 77x² + 54x – 7
The degree of the polynomial is 3.

Question 10.
The cost of a chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total amount paid by him in terms of x and y. If x = 10, y = 5 find the amount paid by him.
Solution:
The cost of a chocolate = (x + y)
Number of chocolates bought by Amir = x + y
Total amount paid by him = (x + y) (x + y)
= x² + xy + xy + y²
= x² + 2xy + y²
When x = 10 and y = 5
The total amount paid by him = (10)² + 2(10)(5) + (5)²
= 100 + 100 + 25 = 225

Question 11.
The length of a rectangle is (3x + 2) units and it’s breadth is (3x – 2) units. Find its area in terms of x. What will be the area if x = 20 units.
Solution:
Length of the rectangle = 3x + 2 units
Breadth of the rectangle = 3x – 2 units
Area of the rectangle = (3x + 2) (3x – 2)
= 9x² – 6x + 6x – 4
= 9x² – 4
When x = 20
Area of the rectangle = 9(20)² – 4
= 9(400) – 4
= 3600 – 4
= 3596 sq.units.

Question 12.
p(x) is a polynomial of degree 1 and q(x) is a polynomial of degree 2. What kind of the polynomial is p(x) × q(x)?
Solution:
Degree of the polynomial p(x) = 1
Degree of the polynomial q(x) = 2
Degree of p(x) × q(x) = 3
The polynomial is a cubic polynomial (or) Polynomial of degree 3.

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 2 Real Numbers Ex 2.3 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 2 Real Numbers Ex 2.3

Question 1.
Represent the following irrational numbers on the number line.
(i) \(\sqrt{3}\)
Solution:

Steps of construction:
1. Draw a line and mark a point A and B such that AB = 3 cm.
2. Mark a point C on this line such that BC = 1 cm.
3. Find the mid point of AC by drawing perpendicular bisector of AC and let it be “O”.
4. With O as centre and OC = OA as radius draw a semicircle.
5. Draw a line BD, which is perpendicular to AB at B.
6. Now BD = \(\sqrt{3}\) which can be marked in the number line as the value of BE = BD = \(\sqrt{3}\)

(ii) Represent \(\sqrt{4.7}\) on a number line.
Solution:

Steps of construction:
1. Draw a line and mark a point A and B such that AB = 4.7 cm.
2. Mark a point C on this line such that A BC = 1 cm.
3. Find the mid point of AC by drawing perpendicular bisector of AC and let it be “O”.
4. With O as centre and OC = OA as radius draw a semicircle.
5. Draw a line BD, which is perpendicular to AB at B.
6. Now BD = \(\sqrt{4.7}\), which can be marked in the number line as the value of BE = BD = \(\sqrt{4.7}\).

(iii) Represent \(\sqrt{6.5}\) on a number line.
Solution:

Steps of construction:
1. Draw a line and mark a point A and B such that AB = 6.5 cm.
2. Mark a point C on this line such that BC = 1 cm.
3. Find the mid point of AC by drawing perpendicular bisector of AC and let it be “O”.
4. With O as centre and OC = OA as radius draw a semicircle.
5. Draw a line BD, which is perpendicular to AB at B.
6. Now BD = \(\sqrt{6.5}\), which can be marked in the number line as the value of BE = BD = \(\sqrt{6.5}\)

Question 2.
Find any two irrational numbers between
(i) 0.3010011000111…. and 0.3020020002….
Solution:
Two irrational numbers between the given two rational numbers are 0.301202200222……. and 0.301303300333……..

(ii) \(\frac{6}{7}\) and \(\frac{12}{13}\)
Solution:
\(\frac{6}{7}\) = 0.\(\overline {857142}\)
\(\frac{12}{13}\) = 0.\(\overline {923076}\)
The two irrational numbers are 0.8616611666111…….. and 0.8717711777111………

(iii) \(\sqrt{2}\) and \(\sqrt{3}\)
Solution:

\(\sqrt{2}\) = 1.414
\(\sqrt{3}\) = 1.732
The two irrational numbers between \(\sqrt{2}\) and \(\sqrt{3}\) are 1.515511555……. and 1.616611666………..

Question 3.
Find any two rational numbers between 2.2360679……… and 2.236505500……….
Solution:
The two rational numbers are 2.2362 and 2.2363 (It has many answers)

Students can download Maths Chapter 2 Real Numbers Ex 2.2 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 2 Real Numbers Ex 2.2

Question 1.
Express the following rational numbers into decimal and state the kind of decimal expression.
(i) \(\frac{2}{7}\)
(ii) -5\(\frac{3}{11}\)
(iii) \(\frac{22}{3}\)
(iv) \(\frac{327}{200}\)
Solution:

(i) \(\frac{2}{7}\) = 0.2857142….
= 0.\(\overline {285714}\)
Non-terminating and recurring decimal expansion.

(ii) -5\(\frac{3}{11}\) = -5 + 0.272 = -5.272……..

= -5.\(\overline {27}\)
Non-terminating and recurring decimal expansion.

(iii) \(\frac{22}{3}\) = 7.333……..

= 7.\(\overline {3}\)
Non-terminating and recurring decimal expansion.

(iv) \(\frac{327}{200}\) = \(\frac{327}{2×100}\)

= \(\frac{3.27}{2}\)
= 1.635
Terminating decimal expansion.

Question 2.
Express \(\frac{1}{13}\) in decimal form. Find the length of the period of decimals.
Solution:

\(\frac{1}{13}\) = 0.07692307
= 0.\(\overline {076923}\)
Length of the period of decimal is 6.

Question 3.
Express the rational number \(\frac{1}{33}\) in recurring decimal form by using the recurring decimal expansion of \(\frac{1}{11}\). Hence write \(\frac{71}{33}\) in recurring decimal form.
Solution:

\(\frac{1}{11}\) = 0.0909……… = 0.\(\overline {09}\)
∴ \(\frac{1}{33}\) = \(\frac{1}{3}\) × \(\frac{1}{11}\)
= \(\frac{1}{3}\) × 0.0909 ……..
= 0.0303 …… = 0.\(\overline {03}\)
\(\frac{71}{33}\) = 2\(\frac{5}{33}\) = 2 + \(\frac{5}{33}\) = 2 + 5 × \(\frac{1}{33}\)
= 2 + 5 × 0.\(\overline {03}\)
2 + (5 × 0.030303 ……..)
2 + 0.151515 ………
2+ 0.\(\overline {15}\)
2.\(\overline {15}\)

Question 4.
Express the following decimal expression into rational numbers.
(i) 0.24
Solution:
Let x = 0.242424 ………. →(1)
100 x = 24.2424 ……… →(2)
(2) – (1) ⇒ 100 x – x = 24.2424 ……….. (-)
 0.2424 ……..
99 x = 24.0000
x = \(\frac{24}{99}\)
(or)
\(\frac{8}{33}\)

(ii) 2.327
Solution:
Let x = 2.327327327 ………. →(1)
1000 x = 2327.327327 ……… →(2)
(2) – (1) ⇒ 1000 x – x = 2327.327327 ……….. (-)
  2.327327 ……..
999 x = 2325.000
x = \(\frac{2325}{999}\)
(or)
\(\frac{775}{333}\)

(iii) – 5.132
Solution:
– 5.132 = -5 + \(\frac{1}{10}\) + \(\frac{3}{100}\) + \(\frac{2}{1000}\)
= \(\frac{-5000 + 100 +30 + 2}{1000}\) = \(\frac{-4868}{1000}\)
(or)
\(\frac{-1217}{250}\)

(iv) 3.17
Solution:
Let x = 3.1777 ………. →(1)
10 x = 31.777 ……… →(2)
100 x = 317.77 …….. →(3)
(3) – (2) ⇒ 100 x – 10 x = 317.77 ……….. (-)
 31.777 ……..
90 x = 286.000
x = \(\frac{286}{90}\)
(or)
\(\frac{143}{45}\)

(v) 17.215
Solution:
Let x = 17.2151515 ………. →(1)
10 x = 172.151515 ……… →(2)
100 x = 17215.1515 …….. →(3)
(3) – (2) ⇒ 1000 x – 10 x = 17215.1515 ……….. (-)
 17215.1515 ……..
990 x = 17043
x = \(\frac{17043}{990}\)
(or)
\(\frac{5681}{330}\)

(vi) -21.2137
Solution:
Let x = -21.213777 ………. →(1)
1000 x = -21213.777 ……… →(2)
100 x = -212137.77 …….. →(3)
(3) – (2) ⇒ 10000 x – 1000 x = -21213.777 ……….. (-)
-21213.777 ……..
9000 x = -190924
x = \(\frac{-190924}{9000}\)
(or)
\(\frac{-47731}{2250}\)

Question 5.
Without actual division, find which of the following rational numbers have terminating decimal expression.
(i) \(\frac{7}{128}\)
Solution:

\(\frac{7}{128}\) = \(\frac{7}{2^{7}}\)
∴ \(\frac{7}{128}\) has terminating decimal expression.

(ii) \(\frac{21}{15}\)
Solution:
\(\frac{21}{15}\) = \(\frac{7}{5}\) = \(\frac{7}{5^1}\)
\(\frac{21}{15}\) has terminating decimal expression.

(iii) 4\(\frac{9}{35}\)
Solution:
4\(\frac{9}{35}\) = \(\frac{149}{35}\)
4\(\frac{149}{5×7}\) (It is not in the form of \(\frac{P}{2^{m} × 5^{n}}\)
∴ 4\(\frac{9}{35}\) has non-terminating recurring decimal expression.

(iv) \(\frac{219}{2200}\)
Solution:

\(\frac{219}{2200}\) = \(\frac{219}{2^{3} × 5^{2} × 11}\) (It is not in the form of \(\frac{P}{2^{m} × 5^{n}}\)
∴ \(\frac{219}{2200}\) has non-terminating recurring decimal expression.

Students can download Maths Chapter 1 Set Language Ex 1.5 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 1 Set Language Ex 1.5

Question 1.
Using the adjacent Venn diagram, find the following sets:

(i) A – B
(ii) B – C
(iii) A’∪B’
(iv) A’∩B’
(v) (B∪C)’
(vi) A – (B∪C)
(vii) A – (B∩C)
Solution:
From the diagram we get
U = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8},
A= {-2,-1, 3, 4, 6}, B = {-2,-1, 5, 7, 8}
C = {-3, -2, 0, 3, 8}
A’ = U – A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 3, 4, 6}
= {-3, 0, 1, 2, 5, 7, 8}
B’ = U – B = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 5, 7, 8}
= {-3, 0, 1, 2, 3, 4, 6}
B∪C = {-2, -1, 5, 7, 8} ∪ {-3, -2, 0, 3, 8} = {-3, -2, -1, 0, 3, 5, 7, 8}
B∩C = {-2, -1, 5, 7, 8} ∩ {-3, -2, 0, 3, 8} = {-2, 8}

(i) A – B = {3, 4, 6}
(ii) B – C = {-1, 5, 7}
(iii) A’∪B’= {-3, 0, 1, 2, 5, 7, 8} ∪ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8}
(iv) A’∩B’ = {-3, 0, 1, 2, 5, 7, 8} ∩ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2}
(v) (B∪C)’ = U – (B∪C)= {-3,-2,-1,0, 1,2, 3,4, 5, 6, 7, 8} – {-3, -2, -1, 0, 3, 5, 7, 8}
= {1, 2, 4, 6}
(vi) A – (B∪C) = {-2, -1, 3, 4, 6} – {-3, -2, -1, 0, 3, 5, 7, 8} = {4, 6}
(vii) A – (B∩C) = {-2,-1, 3, 4, 6} – {-2, 8} = {-1, 3, 4, 6}

Question 2.
If K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h} then find the following:
(i) K∪(L∩M)
(ii) K∩(L∪M)
(iii) (K∪L) ∩ (K∪M)
(iv) (K∩L) ∪ (K∩M)
and verify distributive laws.
Solution:
K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h}
(i) K∪(L∩M)
(L∩M) = {b, c, d, g} ∩ [a, b, c, d, h}
= {b, c, d}
K∪(L∩M) = {a, b, d, e, f} ∪ {b, c, d}
= {a, b, c, d, e, f}

(ii) K∩(L∪M)
(L∪M) = {b, c, d, g} ∪ {a, b, c, d, h}
= {a, b, c, d, g, h}
K∩(L∪M) = {a, b, d, e, f} ∩ {a, b, c, d, g, h}
= {a, b, d }

(iii) (K∪L) ∩ (K∪M)
(K∪L) = {a, b, d, e, f} ∪ {b, c, d, g}
= {a, b, c, d, e, f, g}
(K∪M) = {a, b, d, e, f} ∪ {a, b, c, d, h}
= {a, b, c, d, e, f, h}
(K∪L) ∩ (K∪M) = {a, b, c, d, e, f, g} ∩ {a, b, c, d, e, f, h}
= {a, b, c, d, e, f}

(iv) (K∩L) ∪ (K∩M)
(K∩L) = {a, b, d, e, f) ∩ {b, c, d, g}
= {b, d}
(K∩M) = {a, b, d, e, f} ∩ {a, b, c, d, h}
= {a, b, d}
(K∩L) ∪ (K∩M) = {b, d} ∪ [a, b, d}
= {a, b, d}
From (ii) & (iv) we get, K∩(L∪M) = (K∩L) ∪ (K∩M)
From (i) & (iii) we get, K∪(L∩M) = (K∪L) ∩ (K∪M)

Question 3.
For A = {x : x ∈ Z, -2 < x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C = {-4, -1, 0, 2, 3, 4}
verify A∪(B∩C) = (A∪B) ∩ (A∪C).
Solution:
A = {-1, 0, 1, 2, 3, 4}, B = {0, 1, 2, 3, 4, 5} and C = {-4, -1, 0, 2, 3, 4}
B∩C = {0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 2, 3, 4}
= {0, 2, 3, 4}
A∪(B∩C) = {-1, 0, 1, 2, 3, 4} ∪ {0, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(1)
A∪B = {-1, 0, 1, 2, 3, 4} ∪ {0, 1, 2, 3, 4, 5}
= {-1, 0, 1, 2, 3, 4, 5}
A∪C = {-1, 0, 1, 2, 3, 4} ∪ {-4, -1, 0, 2, 3, 4}
= {-4, -1, 0, 1, 2, 3, 4}
(A∪B) ∩ (A∪C) = {-1, 0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 1, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(2)
From (1) and (2) we get A∪(B∩C) = (A∪B) ∩ (A∪C).

Question 4.
Verify A∪(B∩C) = (A∪B) ∩ (A∪C) using Venn diagrams.
Solution:

From (ii) and (v) we get A∪(B∩C) = (A∪B) ∩ (A∪C).

Question 5.
If A = {b, c, e, g, h}, B = {a, c, d, g, f}, and C = {a, d, e, g, h}, then show that A – (B∩C) = (A – B) ∪ (A – C).
Solution:
A = {b, c, e, g, h} ; B = {a, c, d, g, f}; C = {a, d, e, g, h}
B∩C = {a, c, d, g, i} ∩ {a, d, e, g, h}
= {a, d, g}
A – (B∩C) = {b, c, e, g, h} – {a, d, g}
= {b, c, e, h}…….(1)
A – B = {b, c, e, g, h} – {a, c, d, g, i}
= {b, e, h}
A – C = {b, c, e, g, h} – {a, d, e, g, h}
= {b, c}
(A – B) ∪ (A – C) = {b, e, h} ∪ {b, c}
= {b, c, e, h)……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C)

Question 6.
If A= {x : x = 6n, n∈W and n < 6}, B = {x : x = 2n, n∈N and 2 < n ≤ 9} and
C = {x : x = 3n, n∈N and 4 ≤ n < 10}, then show that A – (B∩C) = (A – B) ∪ (A – C)
Solution:
A = {0, 6, 12, 18, 24, 30}; B = {6, 8, 10, 12, 14, 16, 18}; C = {12, 15, 18, 21, 24, 27}
B∩C = {6, 8, 10, 12, 14, 16, 18} ∩ {12, 15, 18, 21, 24, 27}
= {12, 18}
A – (B∩C) = {0, 6, 12, 18, 24, 30} – {12, 18}
= {0, 6, 24, 30}………(1)
A – B = {0, 6, 12, 18, 24, 30} – {6, 8, 10, 12, 14, 16, 18}
= {0, 24, 30}
A – C = {0, 6, 12, 18, 24, 30} – {12, 15, 18, 21, 24, 27}
= {0, 6, 30}
(A – B) ∪ (A – C) = {0, 24, 30} ∪ {0, 6, 30}
= {0, 6, 24, 30}……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C).

Question 7.
If A = {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6} and C = {-1, 2, 5, 6, 7}, then show that
A – (B∪C) = (A – B) ∩ (A – C).
Solution:
A= {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6}, C = {-1, 2, 5, 6, 7}
B∪C = {-1, 0, 2, 5, 6} ∪ {-1, 2, 5, 6, 7}
= {-1, 0, 2, 5, 6, 7}
A – (B∪C) = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6, 7}
= {-2, 1, 3} ………(1)
A – B = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6}
= {-2, 1, 3}
A – C = {-2, 0, 1, 3, 5}- {-1, 2, 5, 6, 7}
= {-2, 0, 1, 3}
(A- B) ∩ (A- C) = {-2, 1, 3} ∩ {-2, 0, 1, 3}
= {-2, 1, 3} ….(2)
From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).

Question 8.
IF A = {y : y = \(\frac{a + 1}{2}\), a ∈ W and a ≤ 5}, B = {y : y = \(\frac{2n – 1}{2}\), n ∈ W and n < 5} and C = {-1, \(-\frac{1}{2}\), 1, \(\frac{3}{2}\), 2} then show that A – (B∪C) = (A – B) ∩ (A – C).
Solution:

From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).

Question 9.
Verify A- (B∩C) = (A – B) ∪ (A – C) using Venn diagrams.
Solution:


From (ii) and (v) we get A- (B∩C) = (A – B) ∪ (A – C).

Question 10.
If U = {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan’s Laws for complementation.
U= {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}
(i) (A∪B)’ = A’∩B’
(ii) (A∩B)’ = A’∪B’
Solution:
(i) A∪B = {7, 8, 11, 12} ∪ {4, 8, 12, 15}
= {4, 7, 8, 11, 12, 15}
(A∪B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15}
= {10,16} ………(1)
A’ = {4, 7, 8, 10, 11, 12, 15, 16} – {7, 8, 11, 12}
= {4, 10, 15, 16}
B’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 8, 12, 15}
= {7, 10, 11, 16}
A’∩B’ = {4, 10, 15, 16} ∩ {7, 10, 11, 16}
= {10,16} ………(2)
From (1) and (2) we get (A∪B)’ = A’∩B’

(ii) A∩B = {7, 8, 11, 12} ∩ {4, 8, 12, 15}
= {8, 12}
(A∩B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {8, 12}
= {4, 7, 10, 11, 15, 16} ………(1)
A’ = {4, 10, 15, 16}
B’ = {7, 10, 11, 16}
A’∪B’ = {4, 10, 15, 16} ∪ {7, 10, 11, 16}
= {4, 7, 10, 11, 15, 16} ………(2)
From (1) and (2) we get (A∩B)’ = A’∪B’

Question 11.
Verify (A∩B)’ = A∪B’ using Venn diagrams.
Solution:

From (ii) and (i) we get (A∩B)’ = A’∪B’

Students can download Maths Chapter 2 Real Numbers Ex 2.5 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 2 Real Numbers Ex 2.5

Question 1.
Write the following in the form of \(5^n\):
(i) 625
(ii) \(\frac{1}{5}\)
(iii) \(\sqrt{5}\)
(iv) \(\sqrt{125}\)
Solution:
(i) 625 = 5⁴

(ii) \(\frac{1}{5}\) = 5^-1
(iii) \(\sqrt{5}\) = \(5^\frac{1}{2}\)
(iv) \(\sqrt{125}\) = \(\sqrt{5^3}\) = \((5^3)^\frac{1}{2} = 5^\frac{3}{2}\)

Question 2.
Write the following in the form of \(4^n\):
(i) 16
(ii) 8
(iii) 32
Solution:
(i) 16
= 4 × 4
= 4²

(ii) 8
= 4 × 2
= 4 × \(\left(2^{2}\right)^{\frac{1}{2}} \)
= 4 \(\times 4^{\frac{1}{2}} \)
= 4\(^{1+\frac{1}{2}} \)
= 4\(^{\frac{2+1}{2}} \)
= 4\(^{3 / 2}\)

(iii) 32
= 4 × 4 × 2
= 4² × \(\left(2^{2}\right)^{\frac{1}{2}} \)
= 4\(^{2} \times 4^{\frac{1}{2}} \)
= 4\(^{2+\frac{1}{2}} \)
= 4\(^{\frac{4+1}{2}} \)
= 4\(^{\frac{5}{2}} \)

Question 3.
Find the value of
(i) (49)\(^\frac{1}{2}\)
(ii) (243)\(^\frac{2}{5}\)
(iii) (9)\(^\frac{-3}{2}\)
(iv) \((\frac{64}{125})^\frac{-2}{3}\)
Solution:
(i) 49\(^\frac{1}{2}\) = \((7^2)^\frac{1}{2}\) = 7\(^{2 × \frac{1}{2}}\) = 7
(ii) (243)\(^\frac{2}{5}\) = \((3^5)^\frac{2}{5}\) = 3\(^{5 × \frac{2}{5}}\) = 3² = 9

(iii) \(9^{\frac{-3}{2}}=\left(3^{2}\right)^{\frac{-3}{2}}=3^{2 \times \frac{-3}{2}}=3^{-3}=\frac{1}{3^{3}}=\frac{1}{27}\)
(iv) \(\left(\frac{64}{125}\right)^{\frac{-2}{3}}=\left(\frac{4^{3}}{5^{3}}\right)^{\frac{-2}{3}}=\left[\left(\frac{4}{5}\right)^{3}\right]^{\frac{-2}{3}}=\left(\frac{4}{5}\right)^{3 \times \frac{-2}{3}}=\left(\frac{4}{5}\right)^{-2}=\frac{4^{-2}}{5^{-2}}=\frac{5^{2}}{4^{2}}=\frac{25}{16} \)

Question 4.
Use a fractional index to write:
(i) \(\sqrt{5}\)
(ii) \(\sqrt[2]{7}\)
(iii) (\(\sqrt[3]{49})^{5}\)
(iv) \((\frac{1}{\sqrt[3]{100}})^{7}\)
Solution:
(i) \(\sqrt{5}\) = (5)\(^\frac{1}{2}\)
(ii) \(\sqrt[2]{7}\) = 7\(^\frac{1}{2}\)
(iii) \((\sqrt[3]{49})^{5}=\left[(49)^{\frac{1}{3}}\right]^{5}=\left[\left(7^{2}\right)^{\frac{1}{3}}\right]^{5}=\left(7^{\frac{2}{3}}\right)^{5}=7^{\frac{2}{3} \times 5}=7^{\frac{10}{3}}\)
(iv) \(\left(\frac{1}{\sqrt[3]{100}}\right)^{7}=\left[\frac{1}{\sqrt[3]{10^{2}}}\right]^{7}=\left[\frac{1}{\left(10^{2}\right)^{1 / 3}}\right]^{7}=\left[\frac{1}{10^{2 / 3}}\right]^{7}=\left(10^{\frac{-2}{3}}\right)^{7}=10^{\frac{-2}{3} \times 7}=10^{\frac{-14}{3}}\)

Question 5.
Find the 5th root of:
(i) 32
(ii) 243
(iii) 100000
(iv) \(\frac{1024}{3125}\)
Solution:
(i) \(\sqrt[5]{32}=(32)^{\frac{1}{5}}=\left(2^{5}\right)^{\frac{1}{5}}=2^{5 \times \frac{1}{5}} \) = 2

(ii) \(\sqrt[5]{243}=(243)^{\frac{1}{5}}=\left(3^{5}\right)^{\frac{1}{5}}=3^{5 \times \frac{1}{5}}\) = 3

(iii) \(\sqrt[5]{100000}=(100000)^{\frac{1}{5}}=\left(10^{5}\right)^{\frac{1}{5}}\)
= \(10^{5}\times{\frac{1}{5}}\)