- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.1
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.2
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.3
- NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.4
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1
Ex
2.1 Class 9 Maths 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 +
√2 (iii) 3√t + t√2
(iv) y + 2/y (v)
x10 + y3 + t50
Solution:
(i) 4x2 – 3x + 7
It
is a polynomial in one variable, i.e., x, because each exponent of x is a whole
number.
(ii)
y2 + √2
It is a polynomial in one variable, i.e., y, because each exponent of y is a
whole number.
(iii)
3√t + t√2 = 3(t)1/2 + √2.t
It is not a polynomial, because one of the exponents of t is 1/2, which is not a whole number.
(iv) y + 2/y = y + 2.y-1
It is not a polynomial, because one of the exponents of y is -1, which is not a
whole number.
(v) x10 + y3 + t50
Here, exponent of every variable is a whole
number, but x10 + y3 + t50 is a
polynomial in three variables, i.e., in x, y and t.
So, it is not a polynomial in one variable.
Ex 2.1 Class 9 Maths Question 2.
Write the coefficients of x2 in
each of the following:
(i)
2 + x2 + x (ii) 2 – x2 +
x3
(iii) π/2 . x2 +
x (iv) √2 x
– 1
Solution:
(i) The given polynomial is 2 + x2 +
x.
The coefficient of x2 is 1.
(ii)
The given polynomial is 2 – x2 + x3.
The coefficient of x2 is -1.
(iii)
The given polynomial is π/2 . x2 + x.
The coefficient of x2 is π/2.
(iv)
The given polynomial is √2 x – 1. It can be written as 0.x2 + √2
x – 1.
The coefficient of x2 is 0.
Ex 2.1 Class 9 Maths Question 3.
Give one example each of a binomial
of degree 35, and of a monomial of degree 100.
Solution:
(i) A binomial of degree 35 can be written as 2x35 +
5.
(ii)
A monomial of degree 100 can be written as 4y100 – 3.
Ex 2.1 Class 9 Maths Question 4.
Write the degree of each of the
following polynomials.
(i)
5x3 + 4x2 + 7x (ii) 4 – y2
(iii)
5t – √7 (iv)
3
Solution:
(i) The given polynomial is 5x3 +
4x2 + 7x.
The highest power of the variable x is 3.
So, the degree of the polynomial is 3.
(ii)
The given polynomial is 4 – y2.
The
highest power of the variable y is 2.
So, the degree of the polynomial is 2.
(iii)
The given polynomial is 5t – √7.
The
highest power of the variable t is 1.
So,
the degree of the polynomial is 1.
(iv)
The given polynomial is 3.
Since,
3 can be written as 3x0. That is 3 = 3x0 [∵ x0 = 1]
So, the degree of the polynomial is 0.
Ex 2.1 Class 9 Maths 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 highest degree of x2 + x
is 2. So, it is a quadratic polynomial.
(ii)
The highest degree of x – x3 is 3. So, it is a cubic
polynomial.
(iii) The highest degree of y + y2 + 4 is 2. So, it
is a quadratic polynomial.
(iv) The highest degree of 1 + x is 1. So, it is a linear
polynomial.
(v)
The highest degree of 3t is 1. So, it is a linear polynomial.
(vi)
The highest degree of r2 is 2. So, it is a quadratic
polynomial.
(vii)
The highest degree of 7x3 is 3. So, it is a cubic polynomial.
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