Ex 5 3 Fsc Part1

FSC-I / Ex 5.3 - 1

Exercise 5.3 (Solutions)

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9x − 7 ( x 2 + 1)( x + 3) Resolving it into partial fraction. 9x − 7 Ax + B C = 2 + 2 ( x + 1)( x + 3) ( x + 1) ( x + 3)

Question # 1

Multiplying both sides by ( x 2 + 1)( x + 3) .

9 x − 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) ⇒ − 34 = 10C

⇒ C =−

34 10

⇒

⇒ − 27 − 7 = 0 + C ( 9 + 1)

C =−

17 5

Now equation (i) can be written as 9 x − 7 = A ( x 2 + 3x ) + B ( x + 3) + C ( x 2 + 1) Comparing the coefficients of x 2 , x and x 0 . 0 = A + C …………….…... (ii) 9 = 3A + B ……………..… (iii) −7 = + 3B + C …………...…. (iv) Putting value of C in equation (ii) 17 17 0= A− ⇒ A = 5 5 Now putting value of A in equation (iii) 51 51  17  9 = 3  + B ⇒ 9= +B ⇒ 9 − = B 5 5  5 Hence 17

x−

6

−

⇒

B=−

6 5

17

9x − 7 5 = 52 5 + ( x + 1)( x + 3) x +1 ( x + 3) 2

17 x − 6

= Question # 2

5

x2 + 1

17

−

5

( x + 3)

=

17 x − 6 17 − Answer 5( x 2 + 1) 5( x + 3)

1 ( x + 1)( x + 1) 2

Now Consider

1 Ax + B C = 2 + ( x + 1)( x + 1) x +1 x +1 2

Multiplying both sides by ( x 2 + 1)( x + 1) . 1 = ( Ax + B)( x + 1) + C ( x 2 + 1)................ (i ) Put x + 1 = 0 ⇒ x = −1 in equation (i) 1 = 0 + C ( (−1) 2 + 1) Now eq. (i) can be written as

⇒ 1 = 2C

⇒ C =

1 2

FSC-I / Ex 5.3 - 2

1 = A( x 2 + x) + B ( x + 1) + C ( x 2 + 1) Comparing the coefficients of x 2 , x and x 0 . 0 = A + C …………….…. (ii) 0 = A + B ………………. (iii) 1 = A + C ………………. (iv) Putting value of C in equation (ii) 1 1 0= A+ ⇒ A = − 2 2 Putting value of A in equation (iii) 1 1 0=− +B ⇒ B= 2 2 Hence

1 ( x + 1)( x + 1) 2

1 1 − x+ = 22 2 +

−x + 1

1 2

=

2

1

+

2

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

3x + 7 ( x + 4)( x + 3) Resolving it into partial fraction. 3x + 7 Ax + B C = + ( x 2 + 4)( x + 3) x 2 + 4 x+3  Now do yourself , you will get  33  A = 2 13 , B = 13 and C = − 2 13 Question # 3

Answer

2

  

x 2 + 15 Question # 4 ( x 2 + 2 x + 5)( x − 1) Resolving it into partial fraction. x 2 + 15 Ax + B C = 2 + 2 ( x + 2 x + 5)( x − 1) x + 2 x + 5 x − 1

⇒ x 2 + 15 = ( Ax + B)( x − 1) + C ( x 2 + 2 x + 5).............. (i ) Put x − 1 = 0 ⇒ x = 1 in equation (i) (1)2 + 15 = ( A(1) + B ) (0) + C ( (1)2 + 2(1) + 5 )

16 =C ⇒ C = 2 8 Now equation (i) can be written as x 2 + 15 = A ( x 2 − x ) + B ( x − 1) + C ( x 2 + 2 x + 5) Comparing the coefficients of x 2 , x and x 0 . 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 ⇒ 16 = 8C

⇒

Putting value of A and C in equation (iii)

⇒ 1 + 15 = 0 + C (1 + 2 + 5)

FSC-I / Ex 5.3 - 3

0 = − (−1) + B + 2(2)

⇒ 0 =1+ B + 4 ⇒ 0 = B + 5

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

Hence

⇒

B =−5

Answer

x2 ( x 2 + 4)( x + 2) Resolving it into partial fraction. x2 Ax + B C = 2 + 2 ( x + 4)( x + 2) x +4 x+2  Now do yourself , you will get   1 , B = − 1 and C = − 1  A =  2 2  Question # 5

Question # 6

x2 + 1 x3 + 1

x2 + 1 = ( x + 1)( x 2 − x + 1)

Q x 3 + 1 = ( x + 1)( x 2 − x + 1)

Now consider x2 + 1 A Bx + C = + ( x + 1)( x 2 − x + 1) x + 1 x 2 − x + 1  Now do yourself , you will get  2 1 1  A = 3 , B = 3 and C = 3 Question # 7

  

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

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

⇒ x 2 + 2 x + 2 = ( Ax + B)( x + 1)( x − 1) + C ( x 2 + 3)( x −1) + D ( x 2 + 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 8 Now put x − 1 = 0 ⇒ x = 1 in equation (i) ⇒ 1 = − 8C

⇒

C=−

⇒ (1)2 + 2(1) + 2 = 0 + 0 + D ( (1)2 + 3) ( (1) + 1) ⇒ 5= 8D

⇒

D=

⇒ 1 + 2 + 2 = D ( 4 )( 2 )

5 8

Equation (i) can be written as x 2 + 2 x + 2 = ( Ax + B)( x 2 − 1) + C ( x 3 − x 2 + 3x − 3) + D ( x 3 + x 2 + 3x + 3) ⇒ x 2 + 2 x + 2 = A( x 3 − x) + B ( x 2 − 1) + C ( x3 − x 2 + 3x − 3) + D ( x 3 + x 2 + 3x + 3) Comparing the coefficients of x 3 , x 2 , x and x 0 . 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) 1 5 1 1 0= A− + ⇒ 0= A+ ⇒ A=− 8 8 2 2 Putting values of C and D in (iii) 1 5 3  1 5 1= B −−  + ⇒ 1= B + + ⇒ 1= B + 8 8 4  8 8 ⇒ 1−

3 =B 4

⇒

B=

1 4

Hence 1

− x+

1

−

1

5

x + 2x + 2 4 + 8 + 8 = 22 ( x + 3)( x + 1)( x − 1) x +3 x +1 x −1 2

2

−2 x + 1

= =

4

x2 + 3

+

1 − 2x

4( x + 3) 2

−

1

5

8 +

8

x +1 −

x −1

1

8( x + 1)

+

=

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

8( x − 1)

Answer

1 ( x − 1)2 ( x 2 + 2) Resolving it into partial fraction. 1 A B Cx+D = + + 2 2 2 2 ( x − 1) ( x + 2) x − 1 ( x − 1) x +2 Question # 8

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

⇒ 1 = A ( x3 − x 2 + 2 x − 2) + B ( x 2 + 2) + C ( x3 − 2 x 2 + x ) + D ( x 2 − 2 x + 1) Comparing the coefficients of x 3 , x 2 , x and x 0 . 0 = A + C ……………………. (ii) 0 = − A + B − 2C + D …..…….. (iii) 0 = 2 A + C − 2 D …………..… (iv) 1 = −2 A + 2 B + D ……………. (v) Multiplying eq. (iii) by 2 and adding in (iv) 0 = −2 A + 2 B − 4C + 2 D 0 = 2A + C − 2D 0= 2 B − 3C Putting value of B in above 2 2 2 1 0 = 2   − 3C ⇒ 0 = − 3C ⇒ 3C = ⇒ C= 3 3 9  3 Putting value of C in eq. (ii) 2 2 0= A+ ⇒ A=− 9 9

FSC-I / Ex 5.3 - 5

Putting value of A and B in eq. (v) 4 2 4 2  2 1 1 = −2  −  + 2   + D ⇒ 1 = + + D ⇒ 1 − − = D 9 3 9 3  9 3 Hence 2 x+ − 1 1 −2 1 9 9 3 = + + 9 2 2 2 2 ( x − 1) ( x + 2) x − 1 ( x − 1) x +2

( ) (

=

−2

9+

1

x − 1 ( x − 1)

2

+

9

x +2 2

D=−

1 9

)

2x − 1 3

⇒

=

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

x4 Question # 9 1 − x4 1 1 1 = −1 + = − + 1 − x4 (1 − x 2 )(1 + x 2 ) 1 = −1 + (1 − x)(1 + x)(1 + x 2 )

−1 1 − x 4 x4 x 4 −1 −

+

1

Now consider

1 A B Cx + D = + + 2 (1 − x)(1 + x )(1 + x ) 1 − x 1 + x 1 + x 2  Now find values of A, B, C and D yourself .  1 1 1  You will get A = 4 , B = 4 , C = 0 and D = 2

  

So 1 1 (0) x + 12 1 4 4 = + + (1 − x)(1 + x )(1 + x 2 ) 1 − x 1 + x 1 + x2 1 1 1 = + + 4(1 − x) 4(1 + x) 2(1 + x 2 )

Hence x4 1 1 1 = −1 + + + 4 1− x 4(1 − x) 4(1 + x ) 2(1 + x 2 )

Answer

x2 − 2 x + 3 Question # 10 Q x 4 + x 2 + 1 = x4 + 2 x 2 + 1 − x 2 x 4 + x2 + 1 = ( x 2 + 1) 2 − x 2 2 x − 2x + 3 = ( x 2 + 1 + x )( x 2 + 1 − x) = 2 2 ( x + x + 1)( x − x + 1) = ( x 2 + x + 1)( x 2 − x + 1) Now Consider x2 − 2 x + 3 Ax + B Cx+D = + ( x 2 + x + 1)( x 2 − x + 1) x 2 + x + 1 x 2 − x + 1

⇒ x 2 − 2 x + 3 = ( Ax + B)( x 2 − x + 1) + (Cx + D)( x 2 + x + 1).............. (i ) ⇒ x 2 − 2 x + 3 = A( x 3 − x 2 + x ) + B ( x 2 − x + 1) + C ( x3 + x 2 + x) + D( x 2 + x + 1) Comparing the coefficients of x 3 , x 2 , x and x 0 . 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 +

−

+

−

−

B −D 2= ⇒ 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) 5 5 1 3= + D ⇒3− = D ⇒ D= 2 2 2 Putting value of B and D in (iii) 5 1 5 1 1= − A + + C + ⇒ 1− − =− A + C 2 2 2 2 ⇒ − 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 (1) x +

5

(−1) x +

1

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

2

2x + 5

= = =

2

x + x +1 2

−2 x + 1

+

2x + 5

2( x 2 + x + 1) 2x + 5

2( x + x + 1) 2

2

x − x +1 2

+ +

−2 x + 1

2( x 2 − x + 1) 1 − 2x

2( x 2 − x + 1)

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