Ex 5 1 Fsc Part1

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Exercise 5.1 (Solutions)

mathcity.org

Textbook of Algebra and Trigonometry for Class XI

Merging man and maths

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

1 1 = x −1 ( x − 1)( x + 1) Resolving it into partial fraction 1 A B = + ( x − 1)( x + 1) x − 1 x + 1 Question # 1

2

Multiplying both sides by ( x − 1)( x + 1) we get 1 = A ( x + 1) + B ( x − 1) ................ (i ) Put x − 1 = 0 ⇒ x = 1 in equation (i)

1 = A (1 + 1) + B ( 0 ) ⇒ 1 = 2 A + 0



A=

1 2

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

1 = A ( 0 ) + B ( −1 − 1) ⇒ 1 = 0 − 2B ⇒

B=−

1 2

Hence 1

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

Answer

x 2 ( x 2 + 1)

x4 + x2 = ( x + 1)( x − 1) ( x 2 − 1)

Question # 2

= x2 + 2 + = x2 + 2 +

(x

x2 + 2 x2 − 1 x4 + x2 2 x4 − x 2 2

− 1) 2

A B = + ( x + 1)( x − 1) x + 1 x − 1

Multiplying both sides by ( x + 1)( x − 1)

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

⇒ 2 = − 2A + 0 ⇒

A = −1

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

2 = A ( 0 ) + B (1 + 1) So Hence

2

( x + 1)( x − 1)

(

=

)

⇒ 2 = 0 + 2B



B =1

−1 1 + x +1 x −1

x2 x2 + 1

( x + 1)( x − 1)

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

+

( x + 1)( x − 1)

2

Now consider



= x2 + 2 +

Answer



2x2 2 x2 − 2 +

2

FSc I / Ex 5.1-2

2x + 1 ( x − 1)( x + 2)( x + 3) Resolving it into partial fraction 2x + 1 A B C = + + ( x − 1)( x + 2)( x + 3) x −1 x + 2 x + 3 Multiplying both side by ( x − 1)( x + 2)( x + 3) 2 x + 1 = A ( x + 2)( x + 3) + B ( x − 1) ( x + 3) + C ( x − 1)( x + 2) ............ (i ) Put x − 1 = 0 ⇒ x = 1 in equation (i) 2(1) + 1 = A (1 + 2)(1 + 3) + B(0) + C (0) Question # 3

3 = A (3)(4) + 0 + 0

⇒ 3 = 12 A



3 = A 12



A=

1 4

Now put x + 2 = 0 ⇒ x = −2 in equation (i) 2(−2) + 1 = A (0) + B(−2 − 1)(−2 + 3) + C (0) −4 + 1 = 0 + B (−3)(1) + 0

⇒ − 3 = − 3B



B =1

Now put x + 3 = 0 ⇒ x = −3 in equation (i) 2(−3) + 1 = A (0) + B (0) + C (−3 − 1)(−3 + 2) −6 + 1 = 0 + 0 + C (−4)(−1)

⇒ − 5 = 4C



C=−

5 4

So 1 −5 2x + 1 1 4 4 = + + ( x − 1)( x + 2)( x + 3) x −1 x + 2 x + 3 1 1 5 = + − 4( x − 1) x + 2 4( x + 3) Question # 4

3x 2 − 4 x − 5 ( x − 2)( x 2 + 7 x + 10)

Answer

Q x 2 + 7 x + 10 = x 2 + 5 x + 2 x + 10 = x( x + 5) + 2( x + 5) = ( x + 5)( x + 2)

3x 2 − 4 x − 5 = ( x − 2)( x + 5)( x + 2) Now resolving into partial fraction. 3x2 − 4 x − 5 A B C = + + ( x − 2)( x + 5)( x + 2) x − 2 x + 5 x + 2  Do yourself . You will get   A = − 1 , B = 30 , C = − 5 28 7 4 

   

1 ( x − 1)(2 x − 1)(3 x − 1) Resolving it into partial fraction. 1 A B C = + + ( x − 1)(2 x − 1)(3 x − 1) x − 1 2 x − 1 3x − 1 Multiplying both side by ( x − 1)(2 x − 1)(3 x − 1) . 1 = A (2 x − 1)(3 x − 1) + B ( x − 1)(3x − 1) + C (2 x − 1)(3 x − 1) ............. (i ) Put x − 1 = 0 ⇒ x = 1 in equation (i) 1 = A (2(1) − 1)(3(1) − 1) + B (0) + C (0) ⇒ 1 = A (1)(2) + 0 + 0 Question # 5

⇒ 1 = 2A

⇒ A=

1 2

FSc I / Ex 5.1-3

1 in equation (i) 2 1  1   11 1 = A(0) + B  − 1  3   − 1  + C (0) ⇒ 1 = 0 + B  −    + 0 2   2   22 1 ⇒1= − B ⇒ B = −4 4 1 Put 3 x − 1 = 0 ⇒ 3 x = 1 ⇒ x = in equation (i) 3 1   1   2  1 1 = A(0) + B (0) + C  − 1   2   − 1 ⇒ 1 = 0 + 0 + C−  −  3   3   3  3 Put 2 x − 1 = 0 ⇒ 2 x = 1 ⇒ x =

⇒1= Hence

2 C 9



C=

9 2

1 9 −4 1 2 2 = + + ( x − 1)(2 x − 1)(3 x − 1) x − 1 2 x − 1 3x − 1 1 4 9 = − + 2( x − 1) 2 x − 1 2(3 x − 1)

Answer

x ( x − a )( x − b)( x − c) Resolving it into partial fraction. x A B C = + + ( x − a )( x − b)( x − c ) x − a x − b x − c Multiplying both sides by ( x − a )( x − b)( x − c ) . x = A ( x − b)( x − c) + B ( x − a )( x − c) + C ( x − a )( x − b) ............ (i ) Put x − a = 0 ⇒ x = a in equation (i) a = A (a − b)(a − c) + B (0) + C (0)

Question # 6

⇒ a = A (a − b)(a − c) + 0 + 0



A=

a (a − b)(a − c)



B=

b (b − a )(b − c)



B=

c (c − a )(c − b )

Now put x − b = 0 ⇒ x = b in equation (i) a = A (0) + B (b − a )(b − c ) + C (0) ⇒ a = 0 + B (b − a )(b − c ) + 0 Now put x − c = 0 ⇒ x = c in equation (i) c = A (0) + B (0) + C (c − a )(c − b) ⇒ c = 0 + 0 + C (c − a )(c − b) So x = ( x − a )( x − b )( x − c) =

a

(a − b)(a − c ) + x−a

b

(b − a )(b − c) + x−b

c

(c − a )(c − b) x−c

a b c + + (a − b)(a − c)( x − a ) (b − a )(b − c )( x − b) (c − a )(c − b)( x − c) Answer

Made By Atiq ur Rehman ( [email protected] ) URL: http://www.mathcity.org

FSc I / Ex 5.1-4

Question # 7

6 x3 + 5 x2 − 7 2x2 − x − 1 = 3x + 4 +

7x − 3 = 3x + 4 + 2 2 x − 2x + x − 1

3x + 4 2 x − x − 1 6 x3 + 5x2 − 7 6 x3 − 3 x 2 − 3 x 2

7x − 3 2x2 − x − 1



+

+

8 x + 3x − 7 8x2 − 4 x − 4 2

7x − 3 = 3x + 4 + 2 x ( x − 1) + 1( x − 1) 7x − 3 = 3x + 4 + ( x − 1)(2 x + 1)



+

+

7x − 3

Now Consider

so

7x − 3 A B = + ( x − 1)(2 x + 1) x − 1 2 x + 1  Find value of A & B yourself   13  4 You will get A = 3 and B = 3  13 4 7x − 3 4 13 3 3 = = + + x − 1 2x + 1 ( x − 1)(2 x + 1) 3( x − 1) 3(2 x + 1)

Hence 6 x3 + 5 x 2 − 7 4 13 = 3 x + 4 + + 2x2 − x − 1 3( x − 1) 3(2 x + 1)

Answer

2 x3 + x 2 − 5 x + 3 Question # 8 1 2 x3 + x2 − 3x 3 2 3 2 2 x + x − 3 x 2 x + x − 5x + 3 −2 x + 3 3 2 =1 + 3 2 x + x − 3x 2 x + x 2 − 3x − − + − 2x + 3 −2 x + 3 −2 x + 3 =1 + = 1 + x(2 x 2 + x − 3) x(2 x 2 + 3 x − 2 x − 3) −2 x + 3 −2 x + 3 =1 + =1 + x ( x(2 x + 3) − 1(2 x + 3) ) x(2 x + 3)( x − 1) Now consider 3 − 2x A B C = + + x(2 x + 3)( x − 1) x 2 x + 3 x − 1 ⇒ 3 − 2 x = A(2 x + 3)( x − 1) + Bx ( x − 1) + C x (2 x + 3) ............... (i ) Put x = 0 in equation (i) 3 − 2(0) = A ( 2(0) + 3)( (0) − 1) + B (0) + C (0) ⇒ 3 − 0 = A ( 0 + 3)( −1) + 0 + 0 ⇒ 3 = − 3A



Now put 2 x + 3 = 0 ⇒ 2 x = −3 ⇒ x = −

A = −1 3 2

in equation (i)

 3  3 3   3 5 3 − 2  −  = A(0) + B  −   − − 1  + C (0) ⇒ 3 + 3 =0+ B −   − + 0  2  2 2   2 2 15 8  4 ⇒ 6 = B ⇒ B = ( 6)  ⇒ B = 4 5  15  Now put x − 1 = 0 ⇒ x = 1 in equation (i)

3 − 2(1) = A ( 0 ) + B ( 0 ) + C (1)( 2(1) + 3) ⇒ 1 = 0 + 0 + 5C

⇒ C=

1 5

FSc I / Ex 5.1-5

So

8 1 3 − 2x −1 1 8 1 5 = + + 5 =− + + x(2 x + 3)( x − 1) x 2 x + 3 x − 1 x 5(2 x + 3) 5( x − 1) 2 x3 + x 2 − 5 x + 3 1 8 1 =1− + + 3 2 2 x + x − 3x x 5(2 x + 3) 5( x − 1)

Hence

Answer

( x − 1)( x − 3)( x − 5) ( x − 2)( x − 4)( x − 6)

Question # 9

( x − 1)( x 2 − 3 x − 5 x + 15) = ( x − 2)( x 2 − 4 x − 6 x + 24) =

1

( x − 1)( x − 8 x + 15) ( x − 2)( x 2 − 10 x + 24) 2

x 3 − 12 x 2 + 44 x − 48 x 3 − 9 x 2 + 23 x − 15

x 3 − 8 x 2 + 15 x − x 2 + 8 x − 15 x3 − 10 x 2 + 24 x − 2 x 2 + 20 x − 48 x 3 − 9 x 2 + 23 x − 15 = 3 x − 12 x 2 + 44 x − 48 3 x 2 − 21x + 33 3 x 2 − 21x + 33 = 1+ 3 = 1 + x − 12 x 2 + 44 x − 48 ( x − 2)( x − 4)( x − 6) Now Suppose 3 x 2 − 21x + 33 A B C = + + ( x − 2)( x − 4)( x − 6) x − 2 x − 4 x − 6  Find value of A, B and C yourself     You will get A = 3 8 , B = 3 4 , C = 15 8  =

3

So

Hence

3



x3 −12 x 2 + 44 x − 48 +



15

3 x 2 − 21x + 33 = 8 + 4 + 8 ( x − 2)( x − 4)( x − 6) x − 2 x − 4 x − 6 3 3 15 = + + 8( x − 2) 4( x − 4) 8( x − 6) ( x − 1)( x − 3)( x − 5) 3 3 15 = 1+ + + Answer ( x − 2)( x − 4)( x − 6) 8( x − 2) 4( x − 4) 8( x − 6)

Question # 10

+

3 x − 21x + 33 2

1 (1 − ax)(1 − bx )(1 − cx)

Resolving it into partial fraction. 1 A B C = + + (1 − ax)(1 − bx)(1 − cx ) 1 − ax 1 − bx 1 − cx Multiplying both sides by (1 − ax)(1 − bx)(1 − cx) . 1= A (1 − bx )(1 − cx) + B (1 − ax)(1 − cx) + C (1 − ax)(1 − bx) ............ (i ) 1 in equation (i). Put 1 − ax = 0 ⇒ ax = 1 ⇒ x = a 1  1 b  c    1= A  1 − b ⋅ 1 − c ⋅  + B (0) + C (0) ⇒ 1= A  1 − 1 −  + 0 + 0 a  a a  a    ( a − b )( a − c ) ⇒ A = a2  a − b  a − c  ⇒ 1= A  ⇒ 1 = A   a2 ( a − b )( a − c )  a  a 

FSc I / Ex 5.1-6

Find value of B & C yourself as A.     2 2 b c You will get B =  ,C=  (b − a ) (b − c ) (c − a )(c − b)  a

Hence

2

b

2

c

2

1 (a − b ) ( a − c ) (b − a ) (b − c ) (c − a )(c − b ) = + + (1 − ax)(1 − bx)(1 − cx ) 1 − ax 1 − bx 1 − cx a2 b2 c2 = + + (a − b)(a − c)(1 − ax) (b − a )(b − c )(1 − bx) (c − a )(c − b)(1 − cx) Answer x2 + a 2 ( x 2 + b 2 )( x 2 + c 2 )( x 2 + d 2 )

Question # 11 Put y = x 2 in above.

y + a2 ( y + b 2 )( y + c 2 )( y + d 2 ) Now consider y + a2 A B C = + + 2 2 2 2 2 ( y + b )( y + c )( y + d ) y + b y+c y + d2

⇒ y + a 2 = A ( y + c 2 )( y + d 2 ) + B ( y + b2 )( y + d 2 ) + C ( y + b2 )( y + c 2 ) ........... (i ) Put y + b2 = 0 ⇒ y = −b 2 in equation (i) −b 2 + a 2 = A (−b2 + c 2 )(−b2 + d 2 ) + B (0) + C (0) ⇒ a − b = A (c − b )(d − b ) + 0 + 0 2

2

2

2

2

a 2 − b2 A= 2 (c − b 2 )(d 2 − b 2 )



2

Now put y + c 2 = 0 ⇒ y = −c 2 in equation (i)

−c 2 + a 2 = A (0) + B (−c 2 + b 2 )(−b 2 + d 2 ) + C (0) ⇒ a 2 − c 2 = 0 + B (b2 − c 2 )(d 2 − c 2 ) + 0



B=

a 2 − c2 (b 2 − c 2 )(d 2 − c 2 )

Now put y + d 2 = 0 ⇒ y = −d 2 in equation (i)

−d 2 + a 2 = A (0) + B (0) + C (−d 2 + b 2 )(−d 2 + c 2 ) ⇒ a − d = 0 + 0 + C (b − d )(c − d ) 2

2

2

2

2

2



a2 − d 2 C= 2 (b − d 2 )(c 2 − d 2 )

Hence a −b 2

2

a −c 2

2

a −d 2

2

2 2 2 2 2 2 2 2 2 2 2 2 y + a2 (c − b ) (d − b ) (b − c ) ( d − c ) (b − d ) ( c − d ) = + + ( y + b 2 )( y + c 2 )( y + d 2 ) y + b2 y + c2 y + d2

a 2 − b2 a 2 − c2 a2 − d 2 = 2 + + (c − b 2 )(d 2 − b 2 )( y + b 2 ) (b 2 − c 2 )(d 2 − c 2 )( y + c 2 ) (b 2 − d 2 )(c 2 − d 2 )( y + d 2 ) Since y = x 2 a 2 − b2 a 2 − c2 a2 − d 2 = 2 + + (c − b 2 )(d 2 − b 2 )( x 2 + b 2 ) (b 2 − c 2 )(d 2 − c 2 )( x 2 + c 2 ) (b 2 − d 2 )(c 2 − d 2 )( x 2 + d 2 ) Answer Made By Atiq ur Rehman ( [email protected] ) URL: http://www.mathcity.org

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