# 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.2

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## Karnataka 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.2

Part – A

2nd PUC Basic Maths Partial Fractions Ex 5.2 Five Marks Questions and Answers

I. Resolve the following into partial fractions; (3 and 5 marks)

a

Question 1.
$$\frac{x}{(x+1)(x-4)}$$

Put
x = 4, 4 = A(0) +B (4 + 1) ⇒ 4 = 5B ⇒ B = $$\frac{4}{5}$$
Put x = -1, -1 = A(-1-4) +B(0) ⇒ -1 = -5A = A=$$\frac{1}{5}$$
Substituting both A & B in equation (1) we get

Question 2.
$$\frac{x+1}{(x+2)(x-3)}$$
$$\frac{x+1}{(x+2)(x-3)} =\frac{A}{x+2}+\frac{B}{x-3}$$………………….. (1)
x = 3, 3 + 1 = A(0) +B(3 + 2)
4 = 5B = B = $$\frac{4}{5}$$
Put
x = -2, -2 + 1 = A(-2 – 3) + B(0)
-1 = A(-5)
A = 1/5;
3 = -A ⇒ A = -3

Question 3.
$$\frac{7 x-1}{(1-2 x)(1-3 x)}$$
Let
$$\frac{7 x-1}{(1-2 x)(1-3 x)}=\frac{A}{1-2 x}+\frac{B}{1-3 x}$$
⇒ (7x – 1) = A(1 – 3x) + B(1 – 2x)
Put x = 0, -1 = A + B …………..1
Put x=1 6 = -2A-B …………………….2
Solving 1 & 2 (adding) -A = 5 ⇒ A = -5
B = -1-A = -1 + 5 = 4.

Question 4.
$$\frac{3-7 x^{2}}{(1-3 x)(1+2 x)(1+x)}$$
Let
$$\frac{3-7 x^{2}}{(1-3 x)(1+2 x)(1+x)}=\frac{A}{(1-3 x)}+\frac{B}{1+2 x}+\frac{C}{1+x}$$ …………… (1)
∴ 3 – 7x2 = A(1 + 2x) (1+x) +B(1 – 3x) (1 + x) +C (1 – 3x) (1 + 2x)
Put x = -1, 3 – 7 = A(0) + B(0) +C(1 +3) (1-2)
-4 = -4C ⇒ C = 1
Put x = 0, 3 = A + B + C = A + B = 2
Put x = 1, 3 – 7 = A(1 + 2) (1 + 1) + B(1 – 3) (1 + 1) +C(1 – 3) (1 + 2)
-4 = 6A – 4B – 6C ∵ C= 1
-4 + 6 = 6A – 4B = 2
3A – 2B = 1 2A + 2B = 4
⇒ A + B=2

Question 5.
$$\frac{x^{2}}{(x+1)(x+2)(x+3)}$$
Let $$\frac{x^{2}}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$$
x2 = A(x +2) (x + 3) +B(x + 1) (x + 3) + C(x + 1) (x + 2)
Put x=-1, (-1)2 = A(-1 +2) (-1 + 3) + 0 + 0
1 = 2A ⇒ A =$$\frac{1}{2}$$
Put x = -2, (-2)2 = A(0) + B(-2 + 1) (-2 + 3) + C(0)
4 = -B ⇒ b = -4
Put x = -3, (-3)2 = A(0) + B(0) +C(-3 + 1) (-3 + 2)
9 = -2C:(-1) ⇒ 9 = 2c ⇒ c = $$\frac{9}{2}$$
Equation 1 be becomes

Question 6.
$$\frac{2 x^{2}+10 x-3}{(x+1)(x-3)(x+3)}$$
Let

2x2 +10x – 3 = A[x – 3) (x + 3) + B(x + 1) (x + 3) + C(x +1) (x – 3)
Put x=-1, 2 – 10 – 3 = A(-4) (2) + B(0) + C(0)
-11 = -8A = A = $$\frac{11}{8}$$
Put x = 3, 18 + 30 – 3 = A(0) + B(4)(6) + C(0)
+45 = 24B ⇒ B = $$\frac{45}{24}=\frac{15}{8}$$
Put x = -3, 18 – 30 – 3 = C(-3 + 1) (-3 -3) = C(-2) (-6)-15 = + 12C
C = $$\frac{-15}{12}=\frac{-5}{4}$$

Question 7.
$$\frac{3 x+1}{x^{2}-6 x+8}$$

Question 8.
$$\frac{x^{2}-10 x+13}{(x+1)\left(x^{2}-5 x+6\right)}$$
Let

x2 – 10x + 13 = A(x-2)(x – 3) + B(x +1) (x – 3) + C(x + 1) (x – 2)
Put x=-1, 1 + 10 + 13 = A (-3) (-4) + 0 + 0
24 = 12A ⇒ A = $$\frac{24}{12}$$=2
Put x = 2, 4 – 20 + 13 A(0) + B(3) (-1) + C(0)
-3 = -3B ⇒ B = 1
Put x = 3, 9 – 30 + 13 = A(0) + B(0) + C(4) (1)
-8 = 47 ⇒ C= -2

Question 9.
$$\frac{3 x+20}{x^{2}+4 x}$$
Let $$\frac{3 x+20}{x^{2}+4 x}=\frac{3 x+20}{x(x+4)}=\frac{A}{x}+\frac{B}{x+4}$$ ……..(1)
3x + 20 = A(x + 4) + B(x)
Put x=0, 20 = 4A ⇒ A = 5
Put x = -4, -12 + 20 = A(0) + B(-4)
8 = -4B ⇒ B = -2
$$\frac{3 x+20}{x^{2}+4 x}=\frac{5}{x}-\frac{2}{x+4}$$

Question 10.
$$\frac{x+3}{(x-1)\left(x^{2}-4\right)}$$
Let $$\frac{x+3}{(x-1)\left(x^{2}-4\right)}=\frac{x+3}{(x-1)(x-2)(x+2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+2}$$ ……(1)
(x + 3) = A[x -2) (x + 2) +B(x – 1) (x + 2) + C(x – 1) (x – 2)
Put x= 1, 4 = A(-1)(3) ⇒ A=$$\frac{-4}{3}$$
Put x = 2, 5 = A(0) + B(1) (4) + C(O)
Put 5 = 4B ⇒ B = $$\frac{-4}{3}$$
Put x = -2, -2 + 3 = A(0) + B(0) +C(-3) (-4)
1 = 12C ⇒ C = $$\frac{1}{12}$$

Question 11.
$$\frac{x+3}{x^{3}-x}$$
Let $$\frac{x+3}{x^{3}-x}=\frac{x+3}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$$ ….. (1)
x + 3 = A(x-1) (x + 1) + B(x) (x + 1) + C(x) (x – 1)
Put x = 0 3 = -A + 0 + 0 ⇒ A=-3
Put x = 1, 4 = A(0) + B(1) (2) + 0 = 4 = 2B ⇒ B = 2
Put x = -1, 2 = A(0) + B(0) + C(-1) (-2); 2 = 2c ⇒ c = 1
∴ $$\frac{x-3}{x^{3}-x}=\frac{-3}{x}+\frac{2}{x-1}+\frac{1}{x+1}$$

Question 12.
$$\frac{1+3 x+2 x^{2}}{(1-2 x)\left(1-x^{2}\right)}$$
Let

1 + 3x + 2x2 = A(1 – x)(1 + x) +B(1 – 2x)(1 + x) + C(1 – 2x) (1 – x)
Put x = 1, 6 = A(0) + B(-1)(2) +C(0);
6 = -2B ⇒ B = -3
Put x=-1, 0) = A(0) + B(0) +C(1 + 2)(2)
6C = 0 ⇒ C = 0
Put x = 0, 1 = A + B + C = A – 3 = 1 ⇒ A = 1 + 3 = 4

Part – B

1. Resolve the following into partial fractions; (5 Marks)

Question 1.
$$\frac{4}{(x-3)(x+1)^{2}}$$
Let $$\frac{4}{(x-3)(x+1)^{2}}=\frac{A}{x-3}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}$$
∴4 = A(x + 1)2 + B(x + 1)(x – 3) + C(x – 3)
Put x = 3, 4 = A(4)2 + B(0) + C(0)
4 = 16A ⇒ A = $$\frac{1}{4}$$
Put x =-1, 4 = A(0) + B(0) + C(-1-3); Put 4 = -4C ⇒ C = -1
Put x = 0 4 = A- 3B – 3C
3B = A – 3C – 4

Question 2.
$$\frac{9}{(x+1)(x+2)^{2}}$$
Let $$\frac{9}{(x+1)(x+2)^{2}}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^{2}}$$
9 = A(x + 2)2 + B(x + 1)(x + 2) +C(x + 1)
Put x = -1, 9 = A(-1 + 2)2 + 0 + 0
A = 9
Put x= -2, 9 = A(0) + B(0) + CC – 10 = -9
Put x = 0, 9 = 4A + 2B + C
9 = 36 + 2B -9
9 = 27 + 2B $$\frac{-18}{2}$$ = B B = -9
∴ $$\frac{9}{x-1}-\frac{9}{x+2}-\frac{9}{(x+1)^{2}}$$

Question 3.
$$\frac{3 x+4}{(x+1)^{2}(x-1)}$$
Let $$\frac{3 x+4}{(x+1)^{2}(x-1)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x-1}$$
3x+4 = A(x + 1)(x – 1) + B(x – 1) + C (x + 1)2
Put x = -1, ⇒ 3 + 4 = A(0) + B(-2) + C(0)
1 = -2B ⇒ B = $$-\frac{1}{2}$$
Put x = 1, = 3 + 4 = A(0) +B(0) + C(2)2
7 = 4C ⇒ C = 5
Comparing the coefficients of x2 on both sides
0 = A +C ⇒ A = -C = $$-\frac{7}{4}$$

Question 4.
$$\frac{3 x+2}{(x-2)(x+3)^{2}}$$
Let $$\frac{3 x+2}{(x-2)(x+3)^{2}}=\frac{A}{x-2}+\frac{B}{x+3}+\frac{C}{(x+3)^{2}}$$
(3x + 2) = A(x + 3)2 + B(x – 2) (x + 3) + C(x – 2)
Put x = 2, 6 + 2 = A(2 + 3)2 + B(0)+C (0)
8 = 25A ⇒ A = $$\frac{8}{25}$$
Put X=-3, -9 + 2 = A(0) + B(0) +C(-5)
-7 = -5C ⇒ C = $$\frac{7}{5}$$
Comparing the co-efficients of x2 on both sides
0 = A + B ⇒ B = -A = $$\frac{8}{25}$$

Question 5.
$$\frac{2 x^{2}-4 x+1}{(x+2)(x-3)^{2}}$$
2x2 – 4x + 1 = A[x – 3)2 + B(x – 2) (x-3) + C (x – 2)
Put x = 2, 8-8 + 1 = A(-1)2 = A = 1
Put x = 3, 18 – 12 + 1 = A(0) + B(0) + C(3 – 2) = 7 = C
Comparing coefficient of x2 = 2 = A + B = B = 2-A = 2 -1 =1
∴ $$\frac{2 x^{2}-4 x+1}{(x-2)(x-3)^{2}}=\frac{1}{x-2}+\frac{1}{x-3}+\frac{7}{(x-3)^{2}}$$

Question 6.
$$\frac{x^{2}}{(x+1)^{2}(x-5)}$$
Let $$\frac{x^{2}}{(x+1)^{2}(x-5)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x-5}$$
x2 = A (x + 1)(x – 5) + B(x – 5) + C(x + 1)2
Put x = -1, 1 = A(0) + B(-1-5) + C(0)
1 = -6B ⇒ B = $$-\frac{1}{6}$$
Put x = 5, 25 = A(0) + B(0) + C(6)2
C = $$\frac{25}{36}$$
Compare the coefficients of x2 on both sides we get

Question 7.
$$\frac{4-7 x}{(2+3 x)(1+x)^{2}}$$
Let $$\frac{4-7 x}{(2+3 x)(1+x)^{2}}=\frac{A}{2+3 x}+\frac{B}{1+x}+\frac{C}{(1+x)^{2}}$$
⇒ 4 – 7x = A(1 + x)2 +B(1 + x) (2 + 3x) + C(2 + 3x)
Put x=-1, 11 = A(0) + B(0) + C(-1)
C = -11
Put x = 0 4 = A + 2B + 2C ⇒ 26 = A + 2B …..(1)
Compare the coefficients off of x2 on both sides
0 = A + 3B …….(2)
Equations 1 – 2 gives 26 =-B = 0 ⇒B = -26& A = 78
∴ $$\frac{4-7 x}{(2+3 x)(1+x)^{2}}=\frac{78}{2+3 x}-\frac{26}{1+x}-\frac{11}{(1+x)^{2}}$$

Question 8.
$$\frac{1+2 x}{(x+2)^{2}(x-1)}$$
Let $$\frac{1+2 x}{(x+2)^{2}(x-1)}=\frac{A}{x+2}+\frac{B}{(x+2)^{2}}+\frac{C}{x-1}$$
1 + 2x = A (x + 2) (x – 1) + B(x – 1) +C(x + 2)2
3 = A(0) + B(0) + C(9) ⇒ C = $$\frac{1}{3}$$
Put x = -2, 1 – 4 = B(-3) ⇒ B = 1
Comparing the coefficient of x2 on both sides we get

Question 9.
$$\frac{9 x-27}{(2+1)(x-2)^{2}}$$
Let $$\frac{9 x-27}{(2+1)(x-2)^{2}}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{(x-2)^{2}}$$
9x – 27 = A(x – 2)2 + B(x + 1)(x – 2) + C(x + 1)
Put x = -1, -36 = A(-3)2 + B(0) + C(0)
-36 = 9A ⇒ A=-4
Put x = 2, 18 – 27 = A(0) + B(0) +C(2+1)
-9 = 3C ⇒ C =-3
Comparing the coefficients of x? on both sides
0 = A + B = B = -A ⇒ B =-(-4) = 4

Question 10.
$$\frac{2 x+5}{(x+2)(x-1)^{2}}$$
Let $$\frac{2 x+5}{(x+2)(x-1)^{2}}=\frac{A}{x+2}+\frac{B}{x-1}+\frac{C}{(x-1)^{2}}$$
2x + 5 = A(x – 1)2 + B(x – 1)(x + 2) +C(x + 2)
Put x = -2, -4 + 5 = A(-3)2 + 0 + 0
1 = 9A ⇒ A =$$\frac{1}{9}$$
Put x= 1, 2 + 5 = A(0) + B(0) + C(1 + 2)
7 = 3C ⇒ C = $$\frac{7}{3}$$
Comparing the coeff of x2 on both sides
0 = A+B ⇒ B = A = $$-\frac{1}{9}$$

Question 11.
$$\frac{x}{(1+2 x)^{2}(1-3 x)}$$
Let $$\frac{x}{(1+2 x)^{2}(1-3 x)}=\frac{A}{1+2 x}+\frac{B}{(1+2 x)^{2}}+\frac{C}{1-3 x}$$
X = A(1 + 2x) (1 – 3x) + B(1 – 3x) + C(1 + 2x)2
Put x = 0 we get A + B + C = 0
Comparing co efficient of x2 both sides
0 = -6A + 40 ⇒ 6A = 4C ⇒3A = 20
We have A+B+C = 0
⇒ 3A + 3B + 3C = 0
2C + 3B + 3C = 0
5C = -3B
: 3A = 2C
Comparing the coefficient of x both sides

Part – C

III. Resolve into partial fractions: (5 Marks)

Question 1.
$$\frac{2 x^{2}-7 x+1}{x^{2}-3 x-4}$$
$$\frac{2 x^{2}-7 x+1}{x^{2}-3 x-4}$$
This is an improper fraction
∴ Convert into proper fraction by actual division.

9- x = A(x + 1) + B(x – 4)
Put x = 4, 9-4 = A(4 + 1) + B(0)
5 = 5A ⇒ A = 1
Put x=-1, 9-(-1) = A(0) +B(-1-4)
10 = -5B ⇒ B = -2

Question 2.
$$\frac{4 x^{2}-3 x+5}{(2-x)(1+x)}$$

x + 13 = A(1 + x) + B(2 – x)
Put x = 2, 15 = A[1 + 2) + B(0)
15 = 3A ⇒ A = $$\frac{15}{3}$$ = 5
Put x=-1, -1 + 13 = A(0) +B(2-(-1))
12 = 3B ⇒ B = 4

Question 3.
$$\frac{x^{3}+7 x^{2}+17 x+11}{x^{2}+5 x+6}$$

(x – 1) = A(x + 3) + B(x + 2)
Put x = -2, -3 = A(-2 + 3) + B(0)
-3 = A ⇒ A=-3
Put X=-3, -4 = A(0) + B(-3 + 2)
-4 = -B ⇒ B = 4

Question 4.
$$\frac{2 x^{2}+3 x+2}{x^{2}-x-2}$$

∴ 5x + 6 = A (x + 1) +B(x – 2)
Put x = 2, 16 = A(3) ⇒ A = $$\frac{16}{3}$$
Put x = -1, 1 = A(0) + B(-3)=B = $$\frac{1}{-3}$$
1 2x+3x+2-21 16 1 x – x – 2 =2+3(x-2) 3(x+1)

Question 5.
$$\frac{2 x^{3}+x^{2}-x-3}{x(x-1)(2 x+3)}$$
-1 = 5B ⇒ B = $$-\frac{1}{5}$$