Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.1

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.1

9th Maths Algebra Exercise 3.1 Question 1.
Which of the following expressions are polynomials. If not give reason:
(i) \(\frac{1}{x^{2}}\) + 3x – 4
(ii) x² (x – 1)
(iii) \(\frac{1}{x}\) (x + 5)
(iv) \(\frac{1}{x^{-2}}+\frac{1}{x^{-1}}\) + 7
(v) \(\sqrt{5} x^{2}+\sqrt{3} x+\sqrt{2}\)
(vi) \(m^{2}-\sqrt[3]{m}+7 m-10\)
Solution:

9th Maths Exercise 3.1 Question 2.
Write the coefficient of x² and x in each of the following polynomials,
(i) 4 + \(\frac{2}{5} x^{2}\) – 3x
(ii) 6 – 2x² + 3x³ – \(\sqrt{7} x\)
(iii) πx² – x + 2
(iv) \(\sqrt{3} x^{2}+\sqrt{2} x\) + 0.5
(v) x² – \(\frac{7}{2} x\) + 8
Solution:

9th Maths Exercise 3.1 Solutions Question 3.
Find the degree of the following polynomials.

Solution:

9th Maths 3.1 Question 4.
Rewrite the following polynomial in standard form.

Solution:

9th Maths Algebra Exercise 3.1 Solutions Question 5.
Add the following polynomials and find the degree of the resultant polynomial.
(i) p(x) = 6x² – 7x+ 2 q(x) = 6x³ – 7x + 15
(ii) h(x) = 7x³ – 6x + 1 f(x) = 7x² + 17x – 9
(iii) f(x) = 16x⁴ – 5x² + 9 g(x) = -6x³ + 7x – 15
Solution:

9th Maths Algebra Question 6.
Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial.
(i) p(x) = 7x² + 6x – 1 q(x) = 6x – 9
(ii) f(y) = 6y² – 7y + 2 g(y) = 7y + y³
(iii) h(z) = z⁵ – 6z⁴ + z f(z) = 6z² + 10z – 7
Solution:

9th Standard Maths Exercise 3.1 Question 7.
What should be added to 2x³ + 6x² – 5x + 8 to get 3x³ – 2x² + 6x + 15?
Solution:
(2x³ + 6x² – 5x + 8) + Q(x) = 3x³ – 2x² + 6x + 15
∴ Q(x) = (3x³ – 2x² + 6x + 15) – (2x³ + 6x² – 5x + 8)

The required polynomial is x³ – 8x² + 11x + 7

9th Maths Exercise 3.1 Samacheer Kalvi Question 8.
What must be subtracted from 2x²⁴ + 4x² – 3x + 7 to get 3x³ – x² + 2x + 1 ?
Solution:
(2x⁴ + 4x² – 3x + 7) – Q(x) = 3x³ – x² + 2x + 1
Q(x) = (2x⁴ + 4x² – 3x + 7) – 3x³ – x² + 2x + 1
The required polynomial = 2x⁴ + 4x² – 3x + 7 – 3x³ + x² – 2x – 1
= 2x⁴ – 3x³ + 5x² – 5x + 6

Samacheer Kalvi 9th Maths Chapter 3 Question 9.
Multiply the following polynomials and find the degree of the resultant polynomial:
(i) p(x) = x² – 9 q(x) = 6x² + 7x – 2
(ii) f(x) = 7x + 2 g(x) = 15x – 9
(iii) h(x) = 6x² – 7x + 1 f(x) = 5x – 7
Solution:
(i) p(x) = x² – 9 q(x) = 6x² + 7x – 2
p(x) × q(x) = (x² – 9) (6x² + 7x – 2)

The required polynomial is 6x⁴ + 7x³ – 56x² – 63x + 18, degree 4.

(ii) f(x) = 7x + 2 g(x) = 15x – 9
f(x) × g(x) = (7x + 2) (15x – 9)

The required polynomial is 105x² – 33x – 18, degree 2.

(iii) h(x) = 6x² – 7x + 1 f(x) = 5x – 7
h(x) × f(x) = (6x² – 7x + 1) (5x – 7)

The required polynomial is 30x³ – 77x² + 54x – 7, degree 3.

Class 9 Maths Exercise 3.1 Solutions Question 10.
The cost of chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total amount paid by him in terms of x andy. If x = 10, y = 5 find the amount paid by him.
Solution:
Amount paid = Number of chocolates × Cost of a chocolate
= (x + y) (x + y) = (x + y)² = x² + 2xy + y²
If x = 10, y = 5
The total amount paid by him
= 10² + 2 × 10 × 5 + 5² = 100 + 100 + 25 = Rs. 225

Samacheer Kalvi Guru 9th Maths Question 11.
The length of a rectangle is (3x + 2) units and it’s breadth is (3x – 2) units. Find its area in terms of x. What will be the area if x = 20 units.
Solution:
Area of a rectangle = length × breadth
= (3x + 2) × (3x – 2) = (3x)² – 2² = [9x² – 4] Sq. units
If x = 20, Area = 9 × 20² – 4 = 9 × 400 – 4
= 3600 – 4 = 3596 Sq. units

9th Maths Exercise 3.1 In Tamil Question 12.
p(x) is a polynomial of degree 1 and q(x) is a polynomial of degree 2. What kind of the polynomial p(x) × q(x) is ?
Solution:
p(x) is a polynomial of degree 1. q(x) is a polynomial of degree 2.
Then the p(x) × q(x) will be the polynomial of degree (1 + 2) = 3 (or)
Cubic polynomial