GSEB Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.1

NCERT Solutions for Class 9 Mathematics: Chapter 02 Polynomials

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Question 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) \( 4x^2 - 3x + 7 \)
(ii) \( y^2 + \sqrt{2} \)
(iii) \( 3\sqrt{t} + t\sqrt{2} \)
(iv) \( y + \frac{2}{y} \)
(v) \( x^{10} + y^3 + t^{50} \)
Answer:
(i) \( 4x^2 - 3x + 7 \) This is a polynomial with one variable, \(x\). The powers of \(x\) are non-negative integers. Therefore, this polynomial uses a single variable.
(ii) \( y^2 + \sqrt{2} \) This is a polynomial in one variable \(y\). The powers of \(y\) are whole numbers (2 for \(y^2\) and 0 for the constant \( \sqrt{2} \)). Therefore, this polynomial is in one variable.
(iii) \( 3\sqrt{t} + t\sqrt{2} \) This expression is not a polynomial because the exponent of \(t\) is \( \frac{1}{2} \), which is not a whole number. Polynomials need only whole number exponents.
(iv) \( y + \frac{2}{y} \) This expression is not a polynomial because when \( \frac{2}{y} \) is rewritten as \( 2y^{-1} \), the exponent of \(y\) is \( -1 \). Since \( -1 \) is not a whole number, this is not a polynomial in one variable.
(v) \( x^{10} + y^3 + t^{50} \) This polynomial contains three variables: \(x\), \(y\), and \(t\). All their exponents are whole numbers. Consequently, this is a polynomial involving three variables.
In simple words: Look at the variables in each expression. If there is only one type of letter (like only 'x' or only 'y') and all the powers on those letters are whole numbers (0, 1, 2, 3...), then it's a polynomial in one variable. Otherwise, it's either not a polynomial or has more than one variable.

Exam Tip: Remember, for an expression to be a polynomial, the exponents of all variables must be non-negative whole numbers (0, 1, 2, 3,...). Also, classify by the number of distinct variables present.

 

Question 2. Write the coefficients of \( x^2 \) in each of the following:
(i) \( 2 + x^2 + x \)
(ii) \( 2 - x^2 + x^3 \)
(iii) \( \frac{\pi}{2} x^2 + x \)
(iv) \( \sqrt{2}x - 1 \)
Answer:
(i) In the expression \( 2 + x^2 + x \), the coefficient of \( x^2 \) is 1.
(ii) For \( 2 - x^2 + x^3 \), the coefficient for \( x^2 \) is -1.
(iii) For \( \frac{\pi}{2} x^2 + x \), the coefficient of \( x^2 \) is \( \frac{\pi}{2} \).
(iv) In the expression \( \sqrt{2}x - 1 \), there is no \( x^2 \) term, so its coefficient is 0.
In simple words: The coefficient of \( x^2 \) is the number multiplied by \( x^2 \). If you don't see a number, it's 1 (or -1 if it's \( -x^2 \)). If \( x^2 \) isn't there at all, its coefficient is 0.

Exam Tip: Pay close attention to the sign (positive or negative) of the coefficient. If the term is absent, the coefficient is zero.

 

Question 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Answer: An example of a binomial with a degree of 35 is \( 5x^{35} + 9 \). For a monomial with a degree of 100, an example is \( 7x^{100} \).
In simple words: A binomial has two terms, and its degree is the highest power. So, for degree 35, pick any variable, give it power 35, add another term (like a number). A monomial has only one term. For degree 100, give a variable power 100, and that's it.

Exam Tip: A binomial has two terms, and a monomial has one term. The degree is determined by the highest power of the variable in the polynomial.

 

Question 4. Write the degree of each of the following polynomials:
(i) \( 5x^3 + 4x^2 + 7x \)
(ii) \( 4 - y^2 \)
(iii) \( 5t - \sqrt{7} \)
(iv) 3
Answer:
(i) In the polynomial \( 5x^3 + 4x^2 + 7x \), the highest power of \( x \) is 3 (from the term \( 5x^3 \)). Therefore, the polynomial's degree is 3.
(ii) For \( 4 - y^2 \), the highest power of \( y \) is 2. So, the degree of this polynomial is 2.
(iii) In \( 5t - \sqrt{7} \), the term \( 5t \) has the highest power of \( t \), which is 1. Thus, the polynomial's degree is 1.
(iv) The number 3 is a constant polynomial and has no variable. This means the exponent of its variable (implicitly, like \( x^0 \)) is 0. Hence, the degree of this constant polynomial is 0.
In simple words: The degree of a polynomial is the biggest exponent you find on any variable in the expression. For a number by itself, its degree is 0.

Exam Tip: Always identify the term with the highest exponent for any variable to determine the polynomial's degree. For constant terms (like just '3'), the degree is 0.

 

Question 5. Classify the following as linear, quadratic and cubic polynomials.
(i) \( x^2 + x \)
(ii) \( x - x^3 \)
(iii) \( y + y^2 + 4 \)
(iv) \( 1 + x \)
(v) \( 3t \)
(vi) \( r^2 \)
(vii) \( 7x^3 \)
Answer:
(i) The polynomial \( x^2 + x \) is quadratic because its highest degree is 2.
(ii) The polynomial \( x - x^3 \) is cubic as its highest degree is 3.
(iii) The polynomial \( y + y^2 + 4 \) is quadratic because it possesses a highest degree of 2.
(iv) The polynomial \( 1 + x \) is linear, having a highest degree of 1.
(v) The polynomial \( 3t \) is linear, as its highest degree is 1.
(vi) The polynomial \( r^2 \) is quadratic, with a highest degree of 2.
(vii) The polynomial \( 7x^3 \) is cubic because its highest degree is 3.
In simple words: Polynomials are named based on their highest power (degree). If the highest power is 1, it's linear. If it's 2, it's quadratic. If it's 3, it's cubic.

Exam Tip: Classify polynomials by their highest degree: degree 1 = linear, degree 2 = quadratic, degree 3 = cubic. Ensure you correctly identify the highest power in each expression.

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Step-by-Step Textbook Answers: Class 9 Mathematics Chapter 02 Polynomials

Official GSEB Solutions for Chapter 02 Polynomials

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