# 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions

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## Karnataka 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions

### 2nd PUC Basic Maths Partial Fractions Two Marks Questions and Answers

I. Type:-Resolve the following into Partial fractions:

Question 1.
$$\frac{3 x+5}{(x+2)(x – 3)(x+1)}$$

3x + 5=A(x – 3)(x + 1) + B(x + 2)(x + 1) +C(x + 2) (x – 3)
Put x = -, we get the value A
– 6 + 5 = A (-2 -3) (-2 + 1) + 0 + 0
-1 = A (-5) (-1) ⇒ A = $$\frac{-1}{5}$$
Put x = 3, we get the value of 13
9 + 5 = A(0) + B(3 + 2) (3 + 1) + C(0)
14 = B(5)(4) ⇒ 14 = 20 ⇒ $$B=\frac{14}{20}=\frac{7}{10}$$
Put x = – 1, we get the value.of C
(3(-1) + 5) = A(0) + B(0) + C(-1 +2) (-1-3)
– 3 + 5 = C (1) (-4)

Question 2.
$$\frac{1}{x\left(x^{2}-9\right)}$$
Let
$$\frac{1}{x(x+3)(x-3)}=\frac{A}{x}+\frac{B}{x+3}+\frac{C}{x-3}$$ ……….. (1)
1 = A(x + 3) (x – 3) + B x(x – 3) + (x (x + 3)
Put x = 1 we get A
1 = A (0 + 3) (0 – 3) + B 0(0 – 3) + C 0 (0 + 3)
1 = – 9A + 0 + 0
⇒ $$A=\frac{1}{-9}$$

Put x = 3
1 =A(3 + 3)(3 – 3) + B(3) (3 – 3 ) + C(3 (3 + 3)
1 =A(0) + B(0) + C(18)
⇒ $$C=\frac{1}{18}$$

Put x = -3
1 = A(-3 + 3) (- 3 – 3) + B (-3) (-3 -3) + C (-3) (-3 + 3)
1 = A (0) + B(- 3) (-6) + C(0)
1 = 18 B ⇒ $$B=\frac{1}{18}$$

Question 3.
$$\frac{5 x^{2}-19 x-18}{x(x+3)(x-2)}$$
Let

Put x = -3
5(-3)2 – 19 (-3) -18= A(-3 + 3) (-3 -2) + B (-3) (-3 -2) + C (-3) (-3 + 3)
5.9+ 19.3 – 18 = A(0) + B(-3)(-5) + C(0)
84 = 15B  ⇒$$B=84 / 15=\frac{28}{5}$$

Put x = 2
5 (2)2 – 19 (2) – 18 = A (2 + 3) (2 – 2) + B (2) (2 – 2) + C(2) (2 + 3)
20 – 38 – 18 = A(+0) + B(0) + C(2) (5)
-36 = 10C ⇒ $$C=\frac{-36}{10}=\frac{-18}{5}$$

Put x = 0
5(0)2 – 19(0) – 18 = A (0 + 3) (0 – 2) + B0 (0 – 2) + ((0) (0 + 3))
– 18 = A-6 + 0 + 0
– 18 = -6A
$$A=\frac{-18}{-6}$$
A = 3

Question 4.
$$\frac{x-3}{(x-1)\left(x^{2}-4\right)}=\frac{x-3}{(x-1)(x-2)(x+2)}$$
Let $$\frac{x-3}{(x-1)(x-2)(x+2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+2}$$…….(1)
(x – 3) = A(x – 2) (x + 2) +B(x -1) (x +2) +C(x -1) (x -2)
Put x = +1, we get
(x – 3) A (x – 2)(x + 2)+ B(x –  1)(x + 2)+C(x – 1)(x – 2)

Put x+1,we get
(1 – 3)=A(+1 – 2)(1 +2)+B(1- 1)(1 +2)+C(1-1)(1 – 2)
– 2 = A(-1) (3)+B(O)+C (O)
-2 = – 3A
⇒ $$A=\frac { 2 }{ 3 }$$

Put x = 2,we get
2 – 3 =A(2 – 2)(2 + 2)+ 8(2 – 1)(2 + 2)+C(2 – 1)(2 – 2)
– 1 = A (O) + B( 1) (4) + C (O)
– 1 = 4B
⇒ $$B=\frac { -1 }{ 4 }$$

Put x = – 2 we get
-2 – 3 = A(-2 -2) (- 2 + 2) + B (- 2 – 1) (-2 + 2) + C(-2 – 1) (-2 – 2)
5 = A (0) + B (0) + C (-3) (- 4)
– 5 = 12C
⇒ $$C=\frac { -5 }{ 12 }$$

Put x = 1
5.1 + 3 = A(0) + B (1) (1 + 1) + C (0)
8 = B (2) ⇒ B = 4.
Put x = – 1
– 5 + 3 = A(0) + B(0) + C(-1) (-1 -1)
( -2 = C(-1)(-2) ⇒ -2 = 2C
⇒ C = -1

Question 5.
$$\frac{3-2 x}{x(x-2)(x+3)}$$

3 – 2x = A(x – 2)(x + 3)+B(x)(x + 3)+C(x)(x – 2)
Put x = 0,we get 4
3 =A(-2)(3)+B(0)+C(0)
3 = – 6A =A = $$-\frac{1}{2}$$
Put x = 2,we get B
3 – 2(2) = A(0)+B 2(2+3)+C(0)
– 1 = 10 B = $$-\frac{1}{10}$$
Put x – 3we get C
3+6=A(0)+B(0)+C(-3)(-3 – 2)
9 = 15C ⇒ C $$-\frac{9}{15}$$ $$-\frac{3}{5}$$

Question 6.
$$\frac{5 x+3}{x^{3}-x}=\frac{5 x+3}{x\left(x^{2}-1\right)}=\frac{5 x+3}{x(x-1)(x+1)}$$
Let $$\frac{5 x+3}{x^{3}-x}=\frac{5 x+3}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$$
5x+3A(x- 1)(x+ 1)+B(x)(x + 1)+C(x)(x – 1)
Put x = 0
3=A(-1)+B(0)+C(0)
A=-3

Put x= 1
5.1 +3 = A(0) + B(1)(1 + 1)+C(0)
8 = B(2) = B = 4.
Put x – 1
– 5+3 =A(0)+B(0)+C(-1)(-1 – 1)
– 2 = C(-1)(-2) -2 = 2C
= C=-1
∴ $$\frac{5 x+3}{x^{3}-x}=\frac{-3}{x}+\frac{4}{x-1}-\frac{1}{x+1}$$

Question 7.
$$\frac{3 x+20}{\left(x^{2}+4 x\right)(x+1)}=x^{2}+4 x=x(x+4)$$
3x + 20  = A(x + 4) (x + 1) + B (x) . (x + 1) + C (x) (x + 4)
Put x = 0, we get A
20 = A (4) (1) + B(0) + C (0)
20 = 4A ⇒ A = 5
Put x = – 4, we get B
3 (- 4) + 20 = A(0) + B (- 4) (- 4 + 1) + C (0)
– 12 + 20 = B(-4)(-3)

II Type:- Resolve the following into Partial fractions:

Question 8.
$$\frac{2 x^{2}+16 x+29}{(x+3)^{2}(x+4)}$$
Let
$$\frac{2 x^{2}+16 x+29}{(x+3)^{2}(x+4)}=\frac{A}{(x+3)}+\frac{B}{(x+3)^{2}}+\frac{C}{x+4}$$…………. (1)
Put x = – 4
2(-4)2 + 16 (-4) + 29 = A (- 4 + 4) (-4 + 3) + B (- 4 + 4) + C (- 4 + 3)2
2 (16)-64+ 29  = A (0) (-1) + B (0) + C(1)
32-64 + 29     = A(0) + B(0) + C(1)
3 = 1 C  ⇒ C = -3

Put x = -3
2 (-3)2 + 16 (-3) + 29 = A (-3 + 4) (- 3 + 3) + B (- 3 + 4) + C (- 3 + 3)2
2 (9) – 48 + 29    = A(1)(0) + B(1) + C(0)
18 – 48 + 29       = A(0) + B(1) + C(0)
– 1 = B ⇒ B = -1

Comparing the coeff of x2
2 = A + C ⇒ 2 =A + (-3) ⇒ A = 2 + 3 = 5
∴ A = 5

Question 9.
$$\frac{3 x+5}{(x+2)^{2}(x-3)}$$
Let
$$\frac{3 x+5}{(x+2)^{2}(x-3)}=\frac{A}{(x+2)}+\frac{B}{(x+2)^{2}}+\frac{C}{(x-3)}$$ ………………….. (1)
3x + 5 = A (x + 2) (x – 3) + B (x – 3) + C (x + 2)2
Put x = – 2
3(-2) +5 = A (- 2 + 2) (- 2 – 3) + B (- 2 – 3) + C (- 2 + 2)2
-6 + 5 = A(0) (-5) + B(-5) + C(0)
-1 = A (0) + B (- 5) + C (0)
-1 = -5B
⇒ $$B=\frac { -1 }{ -5 }$$
$$B=\frac { 1 }{ 5 }$$

Put x = 3
3 (3) + 5 = A (3 + 2) (3 – 3) + B (3 – 3) + C (3 + 2)2
9 + 4 = A (5) (0) + B (0) + C (25)
14 = A (0) + B (0) + C (25)
14 = 25C
⇒ $$C=\frac { 14 }{ 25 }$$
∴ Comparing coeffcient of x2 . we get 0 = A + C

Question 10.
$$\frac{1+2 x}{(x+2)^{2}(x-1)}$$

1 + 2x = A (x + 2) (x – 1) + B (x – 1) + C (x + 2)2
Put x = -2
– 2 + 2 (- 2) = A (- 2 + 2) (- 2 – 1) + B (- 2 – 1) + C (- 2 + 2)2
-2 – 4 = A (0) + B(- 3) + C (0)2
-6 = B (-3)
$$B=\frac { -6 }{ -3 }$$
B = 2

Put x = 1
1+2(1) = A(1+2)(1 – 1) + B(1- 1) + C( 1 + 2)2
1 + 2 = A (0) + B (0) + C (3)2
3 = 9C
$$C=\frac {3}{9}$$ $$\frac {1}{3}$$

Put x = 0
0 + 2(0) = A (0 + 2) (0 – 1) + B (0 – 1) + C (0 + 2)2
0 = A(2)(-1) + B(-1) + C(2)2
0 = -2A – B + 4C

Question 11.
$$\frac{3 x+5}{(x+2)^{2}(x-3)}$$

3x + 5 = A (x + 2) (x – 3) + B (x – 3) + C (x + 2)2
Put x = – 2
3(- 2) + 5 = A (- 2 + 2) (- 2 – 3) + B (- 2 – 3) + C (- 2 + 2)2
– 6 + 5  = A (0) + B (-5) + C (0)
– 1 = B (-5)
– 1 = – 5B
⇒ $$B=\frac { -1 }{ -5 }$$
$$B=\frac { 1 }{ 5 }$$

Put x = 3
3 (3) + 5 = A (3 + 2) (3 – 3) + B (3 – 3) + C (3 + 2)2
9 + 5 = A (0) + B (0) + C ( 5)2
14 = C (25)
$$C=\frac { 14 }{ 25 }$$

Put x = 0
3 (0) + 5 = A(0 + 2)(0 – 3) + B(0 – 3) + C(0 + 2)2
5 = A(2)(-3) + B(-3) + C(4)
5 = -6A – 3B + 4C

Question 12.
$$\frac{3 x+5}{(x+2)^{2}(x+3)}$$

3x + 5 = A (x + 2) (x + 3) + B (x + 3) + C (x + 2)2
Put x = -2
3 (- 2) + 5 = A (- 2 + 2) (- 2 – 3) + B (- 2 + 3) + B (- 2 + 3) + C (- 2 + 2)2
– 6 + 5 = A (0) + B (+ 1) + C (0)
-1 = B (1)
B = -1

Put x = – 3
3 (-3) + 5 = A (3 + 2) (- 3 + 3) + B (- 3 + 3) + C (- 3 + 2)2 -9 + 5
= A (0) + B (0) + C (- 1 )2
– 4 = C (1)
$$C=\frac { 1 }{ 4 }$$
= – 4 = C (1)
⇒ $$C=\frac {-1 }{ 4 }$$

Put x = 0
3 (0) + 5 = A(0 + 2)(0 + 3) + B(0 + 3) + C(0 + 2)2
5 = A (2) (+ 3) + B (+ 3) + C (4)
5 = 6A + 3B + 4C
5 = 6 A -3 + 14
8 – 14 = 6A
6 A = – 6
A = – 1

Question 13.
$$\frac{x-3}{(x-1)\left(x^{2}-4\right)}$$
Consider (x – 1) (x2 – 4)
(x – 1) (x2 – 22)
(x- 1)(x + 2) (x-2)

x – 3 = A (x + 2) (x – 2) + B (x – 1) (x – 2) + C (x – 1) (x + 2)
Put x = 1
1 – 3 = A (1 + 2) (1 -2) + B (1 – 1) (1 -2) + C (1 – 1)(1 + 2)
-2 = A (3) (- 1) + B (0) + C (0)
-2 = -3A + B (0) + C (0)
-2 = -3 A
$$A=\frac {2}{3}$$

Put x = 2
2 – 3 = A(2 + 2) (2 – 2) + B (2 – 1) (2 – 2) + C (2 – 1)(2 + 2)
– 1 = A (0) + B (0) + C (1) (4)
-1 = C (4)
$$C=\frac {-1}{4}$$

Put x = – 2
-2 – 3 = A (-2+ 2) (-2 – 2) +B (-2 – 1) (-2 – 2) +C (-2 – 1)(-2 + 2)
-5 = A (0) + B (- 3) (- 4) + C (0)
– 5 = 12B
⇒ $$B=\frac {-5}{12}$$

Question 14.
Resolve $$\frac{2 x^{2}+16 x+29}{(x+3)^{2}(x+4)}$$  into partial fraction

2x2+ 16x + 29 = A(x + 3)(x + 4>+B(x + 4) + C(x + 3)2
Put x = – 3
2 (- 3)2 + 16 (- 3) + 29 = A (- 3 + 3) (- 3 + 4) + B (- 3 + 4) + C (- 3 + 3)2
18 – 48 + 29 = A(0) + B(1) + C(0)
-1       = B
⇒ B = – 1

Put x = – 4
2 (-4)2 + 16 (-4) + 29 = A (-4 + 3) (-4 + 4) + B (-4 + 4) + C (-4 + 3)2
32 – 64 + 29 = A (0) + B (0) + C (- 1)2
-3 = C
⇒ C = -3

Put x = 0
2 (0)2 + 16 (0) + 29 = A (0 + 3) (0 + 4) + B (0 + 4) + C (0 + 3)2
29 = A (3) (4) + B (4) + C (3)2 29 = 12A + 4B + 9C
29 =12A + 4(-1) + 9(-3)
29 = 12A – 4 – 27
12A-29 + 4 + 27
12A = 60
A = $$\frac{60}{12}$$ ⇒ A = 5

Question 15.
Improper fraction:
Resolve $$\frac{x^{2}-2}{x^{2}+x-12}$$ into Partial Fraction
To convert Improper to proper.

Multiply both sides by (x + 4) (x – 3)
Then we get (10-x) = A(x – 3) + B (x + 4)
Put x = – 4
10 + 4 = A (-4 – 3)
14 = A (- 7)
⇒ $$A=\frac{14}{-7}$$
⇒ A = -2

Put B = 3
10 – 3 = B(3 + 4)
7 = 7B
⇒ $$B=\frac{7}{7}$$
⇒ B = 1
Putting these value of A and B in (2) we get

Question 16.
$$\frac{2 x^{2}+3 x+2}{x^{2}-x-2}$$