Bihar Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.1 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions 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.

- 4x² – 3x + 7
- y² + \(\sqrt{2}\)
- 3\(\sqrt{t}\) + t\(\sqrt{2}\)
- y + \(\frac { 2 }{ y }\)
- x
^{10}+ y³ + t^{50}

Solution:

- In 4x² – 3x + 7, all the indices of x are whole numbers so it is a polynomial in one variable x.
- In y² + \(\sqrt{2}\) the index of y is a whole number so it is a polynomial in one variable y.
- 3\(\sqrt{t}\) + t\(\sqrt{2}\) = 3t
^{\(\frac { 2 }{ y }\)}+ \(\sqrt{2}\)t, here the exponent of the first term is \(\frac { 1 }{ 2 }\), which is not a whole number. Therefore, it is not a polynomial. - y + \(\frac { 2 }{ y }\) = y + 2y
^{-1}here the exponent of the second term is – 1, which is not a whole number and so it is not a polynomial. - x
^{10}+ y³ + t^{50}is not a polynomial in one variable as three variables x, y, t occur in it.

Question 2.

Write the coefficients. of x² in each of the following:

- 2 + x² + x
- 2 – x³ + x³
- \(\frac { π }{ 2 }\) x² + x
- \(\sqrt{2}\)x – 1

Solution:

Coefficient of x² :

- in 2 + x² + x is 1
- in 2 – x² + x³ is – 1
- in \(\frac { π }{ 2 }\) x² + x \(\frac { π }{ 2 }\)
- in \(\sqrt{2x}\) – 1 is 0.

Question 3.

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Solution:

- Binomial of degree 35 may be taken as x
^{35}+ 4x - Monomial of degree 100 may be taken as 5x
^{100}.

Question 4.

Write the degree of each of the following polynomials :

Solution:

- The highest power term is 5x³ and the exponent is 3. So, the degree is 3.
- The highest power term is – y² and the exponent is 2. So, the degree is 2.
- The highest power term is 51 and the exponent is 1. So, the degree is 1.
- The only term here is 3 which can be written as 3x° and so the exponent is 0. Therefore, the degree is 0.

Question 5.

Classify the following as linear, quadratic and cubic polynomials :

- x² + x
- x – x³
- y + y² + 4
- 1 + x
- 3t
- r²
- 7x³

Solution:

- The highest degree of x² + x is 2, so it is a quadratic.
- The highest degree of x – x³ is 3, so it is a cubic.
- The highest degree of y + y² + 4 is 2, so it is a quadratic.
- The highest degree of x in 1 + x in 1. So it is a linear polynomial.
- The highest degree of t in 3t is 1. So it is a linear polynomial.
- The highest degree of r in r² is 2. So, it is a quadratic polynomial.
- The highest degree of x in 7x³ is 3. So, it is cubic polynomial.