Exercise 2.1 - Chapter 2 - Algebra - 11th Business Maths Guide Samacheer Kalvi Solutions

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Chapter 2 - Algebra - 11th Business Maths Guide Samacheer Kalvi Solutions- Text Book Back Questions and Answers

Text Book Back Questions and Answers

Resolve into partial fractions for the following:

Question 1.

Solution:
Here the denominator x² – 3x + 2 is not a linear factor.
So if possible we have to factorise it then only we can split up into partial fraction.
x² – 3x + 2 = (x – 1) (x – 2)

Multiply both side by (x – 1) (x – 2)
3x + 7 = A(x – 2) + B(x – 1) …….. (2)
Put x = 2 in (2) we get
3(2) + 7 = A(2 – 2) + B(2 – 1)
6 + 7 = 0 + B(1)
∴ B = 13
Put x = 1 in (2) we get
3(1) + 7 = A(1 – 2) + B(1 – 1)
3 + 7 = A (-1) + 0
10 = A(-1)
∴ A = -10
Using A = -10 and B = 13 in (1) we get

Note: When the denominator is only two linear factors we can adopt the following method.

Question 2.

Solution:

Multiply both sides by (x – 2) (x + 1) we get
4x + 1 = A(x + 1) + B(x – 2) ……. (2)
Put x = -1 in (2) we get
4(-1) + 1 = A(-1 + 1) + B(-1 – 2)
-4 + 1 = A(0) + B(-3)
-3 = B(-3)

Put x = 2 in (2) we get
4(2) + 1 = A(2 + 1) + B(2 – 2)
8 + 1 = A(3) + B(0)
9 = 3A
A = 3
Using A = 3, B = 1 in (1) we get

Question 3.

Solution:
Here the denominator has repeated factors. So we write

Multiply both sides by (x – 1) (x + 2)² we get
1 = A(x + 2)² + B(x – 1) (x + 2) + C(x – 1) …… (2)
Put x = 1 in (2) we get
1 = A(1 + 2)² + B(1 – 1) (1 + 2) + C(1 – 1)
1 = A(3²) + 0 + 0
1 = 9A

Put x = -2 in (2) we get
1 = A(-2 + 2)² + B(-2 – 1) (-2 + 2) + C(-2 – 1)
1 = 0 + 0 + C(-3)

From (2) we have
1 = A(x + 2)² + B(x – 1) (x + 2) + C(x – 1)
0x² + 1 = A(x² + 4x + 4) + B(x² + x – 2) + C(x – 1)
Equating coefficient of x² on both sides we get
0 = A + B

Question 4.

Solution:

1 = A(x – 1) + B(x + 1) ……. (2)
Put x = 1 in (2) we get
1 = 0 + B(1 + 1)
1 = B(2)

Put x = -1 in (2) we get
1 = A(-1 – 1) + B(-1 + 1)
1 = -2A + 0

Question 5.

Solution:

x – 2 = A(x – 1)² + B(x + 2) (x – 1) + C(x + 2) ……(2)
Put x = 1 in (2) we get
1 – 2 = A(1 – 1)² + B(1 + 2) (1 – 1) + C(1 + 2)
-1 = 0 + 0 + 3C
C = −1/3
Put x = -2 in (2) we get
-2 – 2 = A(-2 – 1)² + B(-2 + 2) (-2 – 1) + C(-2 + 2)
-4 = A(-3)² + 0 + 0
-4 = 9A

From (2) we have,
0x² + x – 2 = A(x – 1)² + B(x + 2) (x – 1) + C(x + 2)
Equating coefficients of x² on both sides we get
0 = A + B

Question 6.

Solution:

2x² – 5x – 7 = A(x – 2)² + B(x – 2) + C
2x² – 5x – 7 = A(x² – 4x + 4) + B(x – 2) + C …….. (2)
Put x = 2 in (2) we get
2(2²) – 5(2) – 7 = A(0) + B(0) + C
8 – 10 – 7 = 0 + 0 + C
-9 = C
C = -9
Equating coefficient of x2 on both sides of (2) we get
2 = A
A = 2
Equating coefficient of x on both sides of (2) we get
-5 = A(-4) + B(1)
-5 = 2(-4) + B(∵ A = 2)
-5 = -8 + B
B = 8 – 5 = 3
Using A = 2, B = 3, C = -9 in (1) we get

Question 7.

Solution:
Here the denominator has three factors. So given fraction can be expressed as a sum of three simple fractions.

Multiply both sides by x² (x + 2) we get

x² – 6x + 2 = Ax(x + 2) + B(x + 2) + C(x²) ……… (2)
Put x = 0 in (2) we get
0 – 0 + 2 = 0 + B(0 + 2) + 0
2 = B(2)
B = 1
Put x = -2 in (2) we get
(-2)² – 6(-2) + 2 = 0 + 0 + C(-2)²
4 + 12 + 2 = C(4)
18 = 4C

Comparing coefficient of x² on both sides of (2) we get,
1 = A + C

Question 8.

Solution:
Here the quadratic factor x² + 1 is not factorisable.

Multiply both sides by (x + 2) (x² + 1) we get,
x² – 3 = A(x² + 1) + (Bx + C) (x + 2)
Put x = -2 we get
(-2)² – 3 = [A(-2)² + 1] + 0
4 – 3 = A(4 + 1)
1 = 5A

Equating coefficient of x² on both sides of (2) we get
1 = A + B

Equating coefficients of x on both sides of (2) we get
0 = 2B + C

Using A, B, C’s values in (1) we get

Question 9.

Solution:
Here the denominator has three factors. So given fraction can be expressed as a sum of three simple fractions.

Multiply both sides by (x – 1) (x + 3)² we get

Comparing coefficient of x² on both sides of (2) we get,
0 = A + B

Question 10.

Solution:
Here the quadratic factor x² + 4 is not factorisable.

Multiply both sides by (x + 1) (x² + 4) we get
1 = A(x² + 4) + (Bx + C) (x + 1) ……. (2)
Put x = -1 in (2) we get
1 = A((-1)² + 4) + 0
1 = A(1 + 4)

Equating coefficient of x² on both sides of (2) we get,
0 = A + B

Equating coefficient of x on both sides of (2) we get,
{∵ (Bx + C) (x + 1) = Bx² + Cx = Bx + C}
0 = B + C