Ex_5_3_FSC_part1_2

FSC-I / Ex 5.3 - 1

mathcity.org Exercise 5.3 (Solutions)

Merging man and maths Textbook of Algebra and Trigonometry for Class XI

Available online @ http://www.mathcity.org, Version: 1.0.0

Question # 1 9x − 7
(x2 +1) (x + 3)

Resolving it into partial fraction.

(x2 9x − 7 = Ax +B + C 3)
+ 1)(x + 3) (x2 + 1) (x +

Multiplying both sides by (x2 +1)(x + 3) .

9x − 7 = (Ax + B)(x + 3) + C (x2 +1) ............ (i)
Put x + 3 = 0 ⇒ x = −3 in equation (i).

9(−3) − 7 =( A(−3) + B)(0)+ C ((−3)2 +1) ⇒ − 27 − 7 = 0 + C (9 +1)

⇒ − 34 = 10C ⇒ C = − 34 ⇒ C = − 17
10 5

Now equation (i) can be written as
9x − 7 = A(x2 + 3x) + B (x + 3) + C (x2 +1)

Comparing the coefficients of x2 , x and x0 .

0 = A + C …………….…... (ii)

9 = 3A + B ……………..… (iii)

−7 = + 3B + C …………...…. (iv)

Putting value of C in equation (ii)

0 = A − 17 ⇒ A = 17
5 5

Now putting value of A in equation (iii)

9 = 3 17  + B ⇒ 9 = 51 + B ⇒ 9 − 51 = B ⇒ B = − 6
5  5 55

Hence

(x2 9x − 7 = 17 x − 6 + − 17
+1)(x + 3)
55 5

x2 +1 (x + 3)

17 x − 6 17

= 5 −5 = 17x − 6 − 17 Answer
(x + 3) 5(x2 +1) 5(x + 3)
x2 +1

Question # 2 1
(x2 +1) (x + 1)

Now Consider

(x2 1 = Ax + B + C
+1)(x + 1) x2 +1 x +1

Multiplying both sides by (x2 +1)(x +1) .

1 = (Ax + B)(x +1) + C (x2 +1)................(i)

Put x + 1 = 0 ⇒ x = −1 in equation (i)

( )1 = 0 + C (−1)2 + 1 ⇒ 1 = 2C ⇒ C=1
2
Now eq. (i) can be written as

FSC-I / Ex 5.3 - 2

1 = A(x2 + x) + B (x +1) + C (x2 +1)

Comparing the coefficients of x2 , x and x0 .

0 = A + C …………….…. (ii)

0 = A + B ………………. (iii)

1 = A + C ………………. (iv)

Putting value of C in equation (ii)

0= A+ 1 ⇒ A = −1
22

Putting value of A in equation (iii)

0=−1 +B ⇒ B=1
22

Hence (x2 1 = −1x+ 1 + 1 −x +1 + 1
+1)(x + 1)
22 2= 2 2
x +1
x2 +1 x2 +1 x +1

= −x +1 + 1 = 1− x + 1 Answer
2(x2 +1) 2(x + 1) 2(x2 +1) 2(x + 1)

Question # 3 3x + 7
(x2 + 4) (x + 3)

Resolving it into partial fraction.

(x2 3x + 7 = Ax + B + C
+ 4)(x + 3) x2 + 4 x+3

 Now do yourself , you will get 
 
 A = 2 , B = 33 and C = − 2 
13 13 13

Question # 4 x2 +15
(x2 + 2x + 5) (x −1)

Resolving it into partial fraction.

(x2 + x2 + 15 = Ax + B + C
2x + 5)(x −1) x2 + 2x + 5 x −1

⇒ x2 +15 = (Ax + B)(x −1) + C (x2 + 2x + 5).............. (i)

Put x −1 = 0 ⇒ x = 1 in equation (i)

( )(1)2 +15 = ( A(1) + B)(0) + C (1)2 + 2(1) + 5 ⇒ 1+15 = 0 + C (1+ 2 + 5)

⇒ 16 = 8C ⇒ 16 = C ⇒ C = 2
8

Now equation (i) can be written as
x2 +15 = A(x2 − x) + B (x −1) + C (x2 + 2x + 5)

Comparing the coefficients of x2 , x and x0 .
1 = A + C …………..………. (ii)
0 = − A + B + 2C ………….. (iii)
15 = − B + 5C ………….….. (iv)

Putting value of C in equation (ii).

1 = A + 2 ⇒ 1− 2 = A ⇒ A = −1

Putting value of A and C in equation (iii)

FSC-I / Ex 5.3 - 3

0 = − (−1) + B + 2(2) ⇒ 0 = 1 + B + 4 ⇒ 0 = B + 5 ⇒ B =−5

Hence (x2 + x2 + 15 = (−1)x − 5 + 2
2x + 5)(x −1) x2 + 2x + 5 x −1

= −x−5 + 2 Answer
x2 + 2x + 5 x −1

Question # 5 x2
(x2 + 4) (x + 2)

Resolving it into partial fraction.

(x2 x2 = Ax + B + C
+ 4)(x + 2) x2 +4 x+2

 Now do yourself , you will get 
 
 A = 12 , B = −1 and C = − 1 
2

Question # 6 x2 +1 = (x + x2 +1 x + 1) Q x3 +1 =(x +1)(x2 − x +1)
x3 +1 1) (x2 −

Now consider

x2 +1 + 1) = A + Bx + C
(x +1) (x2 − x x +1 x2 − x +1

 Now do yourself , you will get 
 
 A = 23 ,B= 1 and C = 1 
3 3

Question # 7 x2 + 2x + 2
(x2 + 3) (x +1) (x −1)

Consider

(x2 x2 + 2x + 2 = Ax + B + C + D
+ 3)(x + 1)(x −1) x2 + 3 x +1 x −1

⇒ x2 + 2x + 2 = ( Ax + B)(x +1)(x −1) + C (x2 + 3)(x −1) + D(x2 + 3)(x +1).............(i)

Put x + 1 = 0 ⇒ x = −1 in equation (i)

( )(−1)2 + 2(−1) + 2 = 0 + C (−1)2 + 3 ((−1) −1) + 0 ⇒ 1− 2 + 2 = C (4)(−2)

⇒ 1 = −8C ⇒ C=−1
8

Now put x −1 = 0 ⇒ x = 1 in equation (i)

( )⇒ (1)2 + 2(1) + 2 = 0 + 0 + D (1)2 + 3 ((1) + 1) ⇒ 1+ 2 + 2 = D(4)(2)

⇒ 5= 8D ⇒ D=5
8

Equation (i) can be written as

x2 + 2x + 2 = ( Ax + B)(x2 −1) + C (x3 − x2 + 3x − 3) + D (x3 + x2 + 3x + 3)

⇒ x2 + 2x + 2 = A(x3 − x) + B (x2 −1)+ C (x3 − x2 + 3x − 3) + D(x3 + x2 + 3x + 3)

Comparing the coefficients of x3 , x2 , x and x0 .

0 = A + C + D ……………...…. (ii)

1= B − C + D ……………….…. (iii)

2 = − A + 3C + 3D ……………. (iv)

FSC-I / Ex 5.3 - 4

2 = − B − 3C + 3D ………….…. (v)

Putting values of C and D in (ii)

0=A− 1 + 5 ⇒ 0=A+ 1 ⇒ A=− 1
88 2 2

Putting values of C and D in (iii)

1 = B −  − 1  + 5 ⇒ 1= B+ 1 + 5 ⇒ 1= B+ 3
 8  8 88 4

⇒ 1− 3 = B ⇒ B=1
4 4

Hence

(x2 x2 + 2x + 2 = −1x+ 1 + −1 + 5
+ 3)(x +1)(x −1) 8
24 8
x +1
x2 + 3 x −1

= −2x + 1 + −1 + 5 = −2x + 1 + −1 + 5
8 4(x2 + 3) 8(x +1) 8(x −1)
4 8
x +1
x2 + 3 x −1

= 1− 2x − 1 + 5 Answer

4( x 2 + 3) 8(x + 1) 8(x −1)

Question # 8 1
(x −1)2 (x2 + 2)

Resolving it into partial fraction.

1 + 2) = A+ B + Cx+D
(x −1)2 (x2 x −1 (x −1)2 x2 + 2

⇒ 1 = A(x −1)(x2 + 2) + B (x2 + 2) + (C x + D)(x −1)2 .............(i)

Put x −1 = 0 ⇒ x = 1 in equation (i)

( )1 = 0 + B (1)2 + 2 + 0 ⇒ 1 = 3B ⇒ B = 1
3

Now equation (i) can be written as
1 = A(x3 − x2 + 2x − 2) + B (x2 + 2) + (C x + D)(x2 − 2x +1)

⇒ 1 = A(x3 − x2 + 2x − 2) + B (x2 + 2) + C( x3 − 2x2 + x) + D (x2 − 2x + 1)

Comparing the coefficients of x3 , x2 , x and x0 .

0 = A + C ……………………. (ii)
0 = −A + B − 2C + D …..…….. (iii)
0 = 2A + C − 2D …………..… (iv)
1 = −2 A + 2B + D ……………. (v)
Multiplying eq. (iii) by 2 and adding in (iv)

0 = −2 A + 2B − 4C + 2D
0 = 2A + C − 2D
0 = 2B − 3C

Putting value of B in above

0 = 2 1  − 3C ⇒ 0 = 2 − 3C ⇒ 3C = 2 ⇒ C=2
3  3 3 9

Putting value of C in eq. (ii)

0= A+ 2 ⇒ A=−2
99

FSC-I / Ex 5.3 - 5

Putting value of A and B in eq. (v)

1 = −2  − 2  + 2 1  + D ⇒ 1= 4 + 2 + D ⇒ 1− 4 − 2 = D ⇒ D = −1
 9  3  93 93 9

Hence

( ) ( )1 = − 2 + 1 + 2 x+ − 1
9 3 9 9
(x −1)2 (x2
+ 2) x −1 (x −1)2 x2 + 2

− 2 1 2x −1
9 3
= + + 9 = −2 + 1 + 2x −1
x −1 (x −1)2 9(x −1)
x2 + 2 3(x −1)2 9(x2 + 2)

Question # 9 x4 −1

1− x4 1− x4 x4

= − 1 + 1 1 = −1 + 1 x4 −1
− x4 (1 − x2 ) (1 + x2 )
−+
= −1 + 1
x)(1 + x)(1 + 1

(1 − x2)

Now consider

(1 − 1 x2) = A + B + Cx + D
x)(1 + x)(1 + 1− x 1+ x 1+ x2

 Now find values of A, B,C and D yourself . 
 
 You will get A= 1 , B= 1 ,C =0 and D = 1 
4 4 2

So

1 = 1 + 1 + (0)x + 1
x)(1 + x)(1 + 4 4 2

(1 − x2) 1− x 1+ x 1+ x2

= 1 x) + 1 x) + 1 x2)
4(1 − 4(1 + 2(1 +

Hence

x4 =−1 + 1 x) + 1 x) + 1 x2) Answer
1− x4 4(1 − 4(1 + 2(1 +

Question # 10 x2 − 2x + 3 Q x4 + x2 +1= x4 + 2x2 +1− x2
x4 + x2 +1 = (x2 + 1)2 − x2
=(x2 +1 + x)(x2 +1− x)
= (x2 + x2 − 2x + 3 x + 1) = (x2 + x +1) (x2 − x + 1)
x + 1) (x2 −

Now Consider

(x2 + x2 − 2x + 3 x + 1) = Ax+ B + Cx+D
x + 1)(x2 − x2 + x +1 x2 − x +1

⇒ x2 − 2x + 3= ( Ax + B)(x2 − x +1) +(Cx + D)(x2 + x +1)..............(i)

⇒ x2 − 2x + 3= A(x3 − x2 + x) + B (x2 − x + 1)+ C(x3 + x2 + x) + D(x2 + x + 1)

Comparing the coefficients of x3 , x2 , x and x0 .

0 = A+ C ……………………..… (ii)

1 = − A + B + C + D ……………. (iii)

−2 = A − B + C + D …………...… (iv)

FSC-I / Ex 5.3 - 6

3 = B + D …………………….... (v)

Subtracting (ii) and (iv)
0= A +C
−2= A−B +C+D

+ −+ − −

2= B − D

⇒ 2 = B − D …………… (vi)

Adding (v) and (vi)
3=B+ D
2=B−D
5 = 2B

⇒ B=5
2

Putting value of B in (v)

3= 5 + D ⇒3− 5= D ⇒ D= 1
22 2

Putting value of B and D in (iii)

1= − A + 5 + C + 1 ⇒ 1− 5 − 1 =− A + C
22 22

⇒ − 2 = − A + C ……………. (vii)

Adding (ii) and (vii)
0= A+C
−2 = − A + C

−2 = 2C ⇒ C = −1

Putting value of C in equation (ii)
0 = A−1 ⇒ A =1

Hence

(x2 + x 2 − 2x + 3 x + 1) = (1) x + 5 + (−1)x + 1
x + 1)(x2 −
2 2

x2 + x +1 x2 − x +1

2x + 5 −2x + 1

= x2 2 + x2 2

+ x +1 − x +1

= 2x + 5 + −2 x +1

2(x2 + x + 1) 2(x2 − x +1)

= 2x + 5 + 1− 2x Answer

2(x2 + x + 1) 2(x2 − x +1)

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