GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.1

   

Gujarat Board GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x2 – 3x + 7
(ii) y2 + \(\sqrt{2}\)
(iii) 3\(\sqrt{t}\) + t\(\sqrt{2}\)
(iv) y + \(\frac { 1 }{ 2 }\)
(v) x10 + y3 + t50
Solution:
(i) 4x2 – 3x + 7
This is a polynomial in one variable x whose powers are non-negative integers. So this polynomial is in one variable.

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.1

(ii) y2 + y\(\sqrt{2}\)
This polynomial is in one variable y whose exponents are in whole numbers. So this polynomial is in one variable.

(iii) 3\(\sqrt{t}\) + t\(\sqrt{2}\)
This is not a polynomial because exponent of t is \(\frac { 1 }{ 2 }\) which is not a whole number.

(iv) y + \(\frac { 2 }{ y }\) This is not a polynomial because exponent \(\frac { 2 }{ y }\) of i.e., 2y-1 is negative which is not a whole number. So this is not a polynomial in one variable.

(v) x10 + y3 + t50
This polynomial is in three variables x, y and t. Their exponents are in whole numbers. So this is a polynomial in three variables.

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.1

Question 2.
Write the coefficients of x2 in each of the following:
(i) 2 + x2 + x
(ii) 2 – x2 + x3
(iii) \(\frac { π }{ y }\) x2 + x
(iv) \(\sqrt{t}\) – 1
Solution:
(i) 2 + x2 + x
Coefficient of x2 = 1

(ii) 2 – x2 + x3
Coefficient of x3 = – 1

(iii) \(\frac { π }{ 2 }\) x2 + x
Coefficient of x2 = 0

(iv) \(\sqrt{2x}\) – 1
Coefficient of x2 = 0

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
5x35 + 9 and 7x100.

Question 4.
Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 + 7x
(ii) 4 – y2
(iii) 5t – \(\sqrt{7}\)
(iv) 3
Solution:
(i) 5x3 + 4x2 + 7x
The term 5x3 is highest power in x whose power is 3. Therefore degree of polynomial is 3.

(ii) 4 – y2
-y2 is highest power iny which has power 2.
S0 degree of polynomial is 2.

(iii) 5t – \(\sqrt{7}\)
5t has highest power in t which has power 1. So degree of polynomial is 1.

(iv) 3 has no variable i.e., exponent of variable is 0. Therefore degree of constant polynomial is 0.

GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.1

Question 5.
Classify the following as linear, quadratic and cubic polynomials.
(i) x2 + x
(ii) x – x3
(iii) y + y2 + 4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7x3
Solution:
(i) x2 + x Quadratic (Highest degree 2)
(ii) x – x3 Cubic (Highest degree 3)
(iii) y + y2 + 4 Quadratic (Highest degree 2)
(iv) 1 + x Linear (Highest degree 1)
(v) 3t Linear (Highest degree 1)
(vi) r2 Quadratic (Highest degree 2)
(vii) 7x3 Cubic (Highest degree 3)

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