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

Get the free “Partial Fraction Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

Resolve the following rational expressions into partial fractions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Equating nuemarator on bothsides we get

Question 6.

Solution:

Equating numerator on both sides

(x – 2)^{2} = A(x^{2} + 1) + (Bx + c)(x)

Put x = 0

1 = A

Equating co-eff of x^{2}

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

Question 7.

Solution:

Since numerator and denominator are of same degree

we have divide the numerator by the denominator

Substituting the value in ….(1)

Question 8.

Solution:

Numerator is of greater degree than the denominator

So dividing Numerator by the denominator

⇒ 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

Question 9.

Solution:

Question 10.

Solution:

Equating Numerator on both sides we get

6x^{2} – x + 1 = A(x^{2} + 1) + (Bx + c)(x + 1)

6 + 1 + 1 = A(2) + 0 ⇒ 2A = 8 ⇒ A = 4

Equating co-eff of x^{2}

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

Question 11.

Solution:

Since Numerator and are of same degree divide Numerator by the denominator

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

Question 12.

Solution:

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

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

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

Question 4.

Solution:

Question 5.

Solution:

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