HBSE 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.1

Haryana State Board HBSE 9th Class Maths Solutions Chapter 2 Polynomials Ex 2.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 2 Polynomials Exercise 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 + \(\frac {2}{y}\)
(v) x¹⁰ + y³ + t⁵⁰
Solution:
(i) We have, 4x² – 3x + 7
Since, the exponent of x in each term is a whole number.
Therefore, the given expression is a polynomial in one variable x.

(ii) We have, y² + \(\sqrt{2}\)
Since, exponent of y is a whole number.
Therefore, the given expression is a polynomial in one variable y.

(iii) We have, 3\(\sqrt{t}\) + t\(\sqrt{2}\)
Since, the exponent oft in the Ist term is \(\frac {1}{2}\) which is not a whole number.
Therefore, the given expression is not a polynomial.

(iv) We have, y + \(\frac {2}{y}\)
Since, the exponent of y in the IInd term is – 1, which is not a whole number.
Therefore, the given expression is not a polynomial.

(v) We have, x¹⁰ + y³ + t⁵⁰
Since, there are three variables x, y, t in the given expression
Therefore, it is a polynomial in three variables not in one variable.

Question 2.
Write the coefficients of x² in each of the following:
(i) 2 + x² + x
(ii) 2 – x² + x³
(iii) \(\frac {π}{2}\)x² + x
(iv) \(\sqrt{2}\)x – 1
Solution:
(i) The coefficient of x² in 2 + x² + x is 1.
(ii) The coefficient of x² in 2 – x² + x³ is \(\frac {π}{2}\).
(iii) The coefficient of x² in \(\frac {π}{2}\)x² + x is \(\frac {π}{2}\)
(iv) The coefficient of x² in \(\sqrt{2}\)x – 1 is 0 because there is no term of x² in the given expression.

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
Example of a binomial of degree 35 is 3x³⁵ – 4 and example of monomial of degree 100 is \(\sqrt{2}\)y¹⁰⁰ (you can write some more polynomials with different coefficients.)

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) We have, 5x³ + 4x² + 7x
The highest power term is 5x³ and it’s exponent is 3. So, the degree of given polynomial is 3.

(ii) We have, 4 – y²
The highest power lerm is – y² and it’s exponent is 2. So, the degree of given polynomial is 2.

(iii) We have, 5t – \(\sqrt{7}\)
The highest power term is 5t, and it’s exponent is 1. So, the degree of given polynomial is 1.

(iv) We have, 3 It is constant polynomial which degree is 0.

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) The degree of the given polynomial is 2. So, it is the quadratic polynomial.
(ii) The degree of the given polynomial is 3. So, it is the cubic polynomial.
(iii) The degree of the given polynomial is 2. So, it is the quadratic polynomial.
(iv) The degree of the given polynomial is 1. So, it is a linear polynomial.
(v) The degree of the given polynomial is 1. So, it is a linear polynomial.
(vi) The degree of the given polynomial is 2. So, it is a quadratic polynomial.
(vii) The degree of the given polynomial is 3. So, it is a cubic polynomial.