Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

   

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Resolve the following rational expressions into partial fractions

11th Maths Exercise 2.9 Answers Question 1.
11th Maths Exercise 2.9 Answers Samacheer Kalvi
Solution:
11th Maths Exercise 2.9 Samacheer Kalvi

11th Maths Exercise 2.9 Question 2.
11th Std Maths Exercise 2.9 Solutions Samacheer Kalvi
Solution:
Ex 2.9 Class 11 Maths Samacheer Kalvi

11th Std Maths Exercise 2.9 Solutions Question 3.
11th Maths Exercise 2.9 Solutions Samacheer Kalvi
Solution:
Exercise 2.9 Class 11 Samacheer Kalvi
Samacheer Kalvi 11th Maths Solution

Ex 2.9 Class 11 Maths Question 4.
Samacheer Kalvi 11 Maths Solutions
Solution:
Samacheer Kalvi 11th Maths

11th Maths Exercise 2.9 Solutions Question 5.
Samacheer Kalvi 11th Maths Example Sums
Solution:
Samacheer Kalvi Guru 11th Maths
Equating nuemarator on bothsides we get
Samacheer Kalvi 11th Maths Solutions

Exercise 2.9 Class 11 Question 6.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 146
Solution:
11 Maths Samacheer Kalvi
Equating numerator on both sides
(x – 2)2 = A(x2 + 1) + (Bx + c)(x)
Put x = 0
1 = A
Equating co-eff of x2
1 = A + B
(i.e.,) 1 + B = 1 ⇒ B = 0
put x = 1
A(2) + B + C = 0 (i.e.,) 2A + B + C = 0
2 + 0 + C = 0 ⇒ C = -2
11 Maths Solutions Samacheer

Samacheer Kalvi 11th Maths Solution Question 7.
11th Maths Chapter 2 Exercise 2.9 Samacheer Kalvi
Solution:
Since numerator and denominator are of same degree
we have divide the numerator by the denominator
11th Maths Book Volume 1 Solutions Samacheer Kalvi
Substituting the value in ….(1)
Samacheer 11th Maths Solution

Samacheer Kalvi 11 Maths Solutions Question 8.
11th Samacheer Maths Solutions
Solution:
Numerator is of greater degree than the denominator
So dividing Numerator by the denominator
Samacheer Kalvi Class 11 Maths Solutions
⇒ 21x + 31 = A(x + 3) + B(x + 2)
Put x = -3
-63 + 31 = B(-1)
B = 32
Put x = -2
-42 + 31 = A(1) + B(0)
A = -11
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 17

Samacheer Kalvi 11th Maths Question 9.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 18
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 19
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 20

Samacheer Kalvi 11th Maths Example Sums Question 10.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 21
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 22
Equating Numerator on both sides we get
6x2 – x + 1 = A(x2 + 1) + (Bx + c)(x + 1)
6 + 1 + 1 = A(2) + 0 ⇒ 2A = 8 ⇒ A = 4
Equating co-eff of x2
6 = A + B
(i.e.,) 4 + B = 6 ⇒ B = 6 – 4 = 2
put x = 0
1 = A+ C
4 + C = 1 ⇒ C = 1 – 4 = -3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 23

Samacheer Kalvi Guru 11th Maths Question 11.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 24
Solution:
Since Numerator and are of same degree divide Numerator by the denominator
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 25
equating Numerator on both sides we get
x – 5 = A(x + 3) + B(x – 1)
Put x = -3
-3 -5 = A(0) + B(-4)
-4B = -8 ⇒ B = 2
Put x = 1
1 – 5 = A(4) + B(0)
4A = -4 ⇒ A = -1
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 26

Samacheer Kalvi 11th Maths Solutions Question 12.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 27
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 28

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 Additional Questions

11 Maths Samacheer Kalvi Question 1.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 29
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 30

11 Maths Solutions Samacheer Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 31
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 32

11th Maths Chapter 2 Exercise 2.9 Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 33
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 34
Equating nemerator on b/s
9 = A(x+2)2 + B(x – 1)(x + 2) + C(x – 1)
Put x = -2
9 = A(0) + B(0) + C(-3)
-3C = 9 ⇒ C = -3
Put x = 1
9 = A (1 + 2)2 + B (0) + C(0)
9A = 9
A = 1
Put x = 0
9 = 4A – 2B – C
9 = 4(1) – 2B + 3
9 – 7 = -2B
2 = -2B
B = -1
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 35

11th Maths Book Volume 1 Solutions Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 36
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 133

Samacheer 11th Maths Solution Question 5.
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 38
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 39
0 = 0 + B(1 + 2)
3B = 0 ⇒ B = 0
Put x = -2
(-2)3 – 1 = A(-2 – 1) + B(0)
-8 – 1 = -3A
-9 = -3A
A = 9/3 ⇒ A = 3
Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 40