Maths Class 9 RS Aggarwal Solutions

RS Aggarwal Class 9 Solutions Chapter 2 Polynomials Ex 2A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2A.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2A
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2B
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2C
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2D
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2E
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2F
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2G
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2H
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2I
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2J
• RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Ex 2K

Question 1.
Solution:
(i) x⁵-2x³+x+7 is a polynomial and its degree is 5
(ii) y³-√3y is a polynomial and its degree is 3

Question 2.
Solution:
(i) Degree is 1
(ii) degree is 3
(iii) degree is zero
(iv) degree is 7
(v) degree is 10
(vi) degree is 2

Question 3.
Solution:
(i) Coefficient of x³ is -5.
(ii) Coefficient of x is -2√2
(iii) Coefficient of x² is \(\frac { \pi }{ 3 } \)
(iv) Coefficient of x² is 0

Question 4.
Solution:
(i) Example of a binomial of degree 27 is 4x²⁷ – 5
(ii) Example of a monomial of degree 16 is x¹⁶
(iii) Example of trinomial of degree 3 is 2x³ + 7x + 4

Question 5.
Solution:
(i) 2x² + 4x is a quadratic polynomial. (Degree 2)
(ii) x – x³ is a cubic polynomial (Degree 3)
(iii) 2 – y – y² is a quadratic polynomial (Degree 2)
(iv) – 7 + z is a linear polynomial (Degree 1)
(v) 5t is a linear polynomial (Degree 1)
(vi) p³ is a cubic polynomial (Degree 3)

 

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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1F

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Question 1.
Solution:
We know that
a^p x a^q = a^p+q
∴ Therefore

Question 2.
Solution:
We know that
a^p ÷ a^q = a^p-q
Therefore

Question 3.
Solution:
We know that
a^p x b^p = (ab)^p
Therefore

Question 4.
Solution:
We know that
(a^p)^q =a^pq

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1E

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Rationalise the denominator of each of the followings :

Question 1.
Solution:
Here,RF of √7 is √7

Question 2.
Solution:
Here RF √3 is √3

Question 3.
Solution:
Here RF of \(\frac { 1 }{ \left( { 2+ }\sqrt { 3 } \right) }\) is \(\frac { 1 }{ \left( { 2- }\sqrt { 3 } \right) }\)

Question 4.
Solution:
Here RF is √5 + 2

Question 5.
Solution:
Here RF is 5 – 3√2

Question 6.
Solution:
Here RF is √6 + √5

Question 7.
Solution:
Here RF = √7 – √3

Question 8.
Solution:
Here RF = √3 – 1

Question 9.
Solution:
Here RF = (3-2√2)

Find the values of a and b in each of the following :

Question 10.
Solution:
\(\frac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } \), RF = √3+1
(Multiplying and dividing by √3+1)

Question 11.
Solution:
\(\frac { 3+\sqrt { 2 } }{ 3-\sqrt { 2 } } \), RF is 3+√2
(Multiplying and dividing by 3+√2)

Question 12.
Solution:
In \(\frac { 5-\sqrt { 6 } }{ 5+\sqrt { 6 } } \), RF is (5-√6)
(Multiplying and dividing by 5-√6)

Question 13.
Solution:
In \(\frac { 5+2\sqrt { 3 } }{ 7+4\sqrt { 3 } } \), RF is 7-4√3
(Multiplying and dividing by 7-4√3)

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
x = (4-√15)

Question 17.
Solution:
x = 2+√3

Question 18.
Solution:

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
(i) Draw a line segment AB = 3.2 units (cm) and extend it to C such that BC = 1 unit.

(ii) Find the mid-point O of AC.
(iii) With centre O and OA as radius draw a semicircle on AC
(iv) Draw BD ⊥ AC meeting the semicircle at D.
(v) Join BD which is √3.2 units.
(vi) With centre B and radius BD, draw an arc meeting AC when produced at E.
Then BE = BD = √3.2 units. Ans.

Question 6.
Solution:
(i) Draw a line segment AB = 7.28 units and produce is to C such that BC = 1 unit (cm)

(ii) Find the mid-point O of AC.
(iii) With centre O and radius OA, draw a semicircle on AC.
(iv) Draw a perpendicular BD at AC meeting the semicircle at D
Then BD = √7.28 units.
(v) With centre B and radius BD, draw an arc which meet AC produced at E.
Then BE = BD = √7.28 units.

Question 7.
Solution:
(A) For Addition
(i) Closure property: The sum of two real numbers is always a real number.
(ii) Associative Law : (a + b) + c = a + (b + c), for all values of a, b and c.
(iii) Commutative Law : a + b = b + a for all real values of a and b.
(iv) Existance of Additive Identity : 0 is the real number such that: 0 + a = a + 0 = afor every real value of a.
(v) Existance of addtive inverse : For each real value of a, there exists a real value (-a) such that a + (-a) = (-a) + a = 0, Then (a) and (-a) are called the additive inverse of each other.
(v) Existence of Multiplicative Inverse. For each non zero real number a, there exists a real number \(\frac { 1 }{ a }\) such that a . \(\frac { 1 }{ a }\) = \(\frac { 1 }{ a }\) . a = 1
a and \(\frac { 1 }{ a }\) are called multiplicative inverse or reciprocal of each other.
(B) Multiplication
(i) Closure property: The product of two real numbers is always a real number.
(ii) Associative law : ab(c) = a(bc) for all real values of a, b and c
(iii) Commutative law : ab=ba for all real numbers a and b
(iv) Existance of Multiplicative Identity: clearly is a real number such that 1.a = a.1 = a for every value of a.

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 1 Real Numbers Ex 1C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1A
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1B
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1D
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1E
• RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1F

Question 1.
Solution:
Irrational numbers : Numbers which are not rational numbers, are called irrational numbers. Rational numbers can be expressed in terminating decimals or repeating decimals while irrational number can’t.
\(\frac { 1 }{ 2 } \) , \(\frac { 2 }{ 3 } \) , \(\frac { 7 }{ 5 } \) etc.are rational numbers and π, √2, √3, √5, √6….etc are irrational numbers

Question 2.
Solution:
(i) √4 = ±2, it is a rational number
(ii) √196 = ±14 it is a rational number
(iii) √21 It is irrational number.
(iv) √43 It is irrational number.
(v) 3 + √3 It is irrational number because sum of a rational and an irrational number is irrational
(vi) √7 – 2 It is irrational number because difference of a rational and irrational number is irrational
(vii) \(\frac { 2 }{ 3 } \)√6 . It is irrational number because product of a rational and an irrational number is an irrational number.
(viii) 0.\(\overline { 6 } \) = 0.6666…. It is rational number because it is a repeating decimal.
(ix) 1.232332333…. It is irrational number because it not repeating decimal
(x) 3.040040004…. It is irrational number because it is not repeating decimal.
(xi) 3.2576 It is rational number because it is a terminating decimal.
(xii) 2.3565656…. = 2.3 \(\overline { 56 } \) It is rational number because it is a repeating decimal.
(xiii) π It is an irrational number
(xiv) \(\frac { 22 }{ 7 } \). It is a rational number which is in form of \(\frac { p }{ q } \) Ans.

Question 3.
Solution:
(i) Let X’OX be a horizontal line, taken as the x-axis and let O be the origin. Let O represent 0.
Taken OA = 1 unit and draw AB ⊥ OA such that AB = 1 unit. Join OB, Then,

Question 4.
Solution:
Firstly we represent √5 on the real line X’OX. Then we will find √6 and √7 on that real line.
Now, draw a horizontal line X’OX, taken as x-axis

Question 5.
Solution:
(i) 4 + √5 : It is irrational number because in it, 4 is a rational number and √5 is irrational and sum of a rational and an irrational is also an irrational.
(ii) (-3 + √6) It is irrational number because in it, -3 is a rational and √6 is irrational and sum or difference of a rational and irrational is an irrational.
(iii) 5√7 : It is irrational because 5 is rational and √7 is irrational and product of a rational and an irrational is an irrational.
(iv) -3√8 : It is irrational because -3 is a rational and √8 is an irrational and product of a rational and an irrational is also an irrational.
(v) \(\frac { 2 }{ \sqrt { 5 } } \) It is irrational because 2 is a rational and √5 is an irrational and quotient of a rational and an irrational is also an irrational.
(vi) \(\frac { 4 }{ \sqrt { 3 } } \) It is irrational because 4 is a rational and √3 is an irrational number and quotient of a rational and irrational is also an irrational.

Question 6.
Solution:
(i) True.
(ii) False, as the sum of two irrational number is irrational is not always true.
(iii) True.
(iv) False, as the product of two irrational numbers is irrational is not always true.
(v) True.
(vi) True.
(vii) False as a real number can be either rational or irrational.

Hope given RS Aggarwal Solutions Class 9 Chapter 1 Real Numbers Ex 1C are helpful to complete your math homework.

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