Alevel FP2

8/16/2019 Alevel FP2

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F P 2

2

Algebraic techniques

Simplify 4 13

2 x x + ++

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Rewrite the fractions so that they each have denominator ( x + 1)( x + 2):

41

32

4 21 2

3 11 2 x x

x x x

x x x +

++

≡+

+ ++

+

+ +

( )( )( )

( )( )( )

Add the numerators: ≡ + + +

+ +

4 2 3 11 2

( ) ( )( )( )

x x x x

Simplify: ≡ +

+ +

7 11

1 2

x

x x ( )( )

Hence 41

32

7 111 2 x x

x x x +

++

≡ +

+ +( )( )

º means ‘is equivalent to’.º means ‘is equivalent to’.

E XA

MPL E 1

Express 5 11 1 x

x x −

+ −( )( ) as the sum of partial fractions.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Write the fraction in the required form:

5 11 1 1 1

x x x

A x

B x

−

+ − + −≡ +

( )( ) ( ) ( )

≡ − + ++ −

A x B x x x

( ) ( )( )( )

1 11 1

Compare numerators: 5 x - 1 º A( x - 1) + B( x + 1)

Substitute appropriate values of x into this identity to find A and B:

When x = 1, 4 = A(1 - 1) + B(1 + 1)i.e. 4 = 2B so B = 2

Similarly, when x = -1 A = 3

Hence, in partial fractions, 5 11 13

12

1 x

x x x x −

+ − + −≡ +

( )( ) ( ) ( )

A and B are constants. A and B are constants.

Denominators are equal so thenumerators must also be equal.Denominators are equal so thenumerators must also be equal.

You could also find A and B by comparing coefficients inthe identity

5 x - 1 º A( x - 1) + B( x + 1)

You could also find A and B by comparing coefficients inthe identity

5 x - 1 º A( x - 1) + B( x + 1)

E XA

MPL E 2

0.1

You can simplify algebraic fractions by usingcommon denominators.

See C3 and C4 for revision. This section providesbackground knowledgefor Chapters 1, 2, 3 and 6.

See C3 and C4 for revision. This section providesbackground knowledgefor Chapters 1, 2, 3 and 6.

You can split an algebraic fraction into its partial fractions.


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0 Background knowledge for unit FP2

F P2

3

You can rearrange an expression involving an algebraic fraction.

Make x the subject of the equation y x x =

+

−

2 11

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Multiply both sides of the equation by the denominator ( x

- 1): y x x

= +

−

2 11

so y ( x - 1) = 2 x + 1

Expand the bracket and collect terms in x :

yx - 2 x = y + 1

i.e. ( y - 2) x = y + 1

so x y y = +

−

12

E XA

MPL E 3

You can expand expressions of the form (1 + x )n , where n is not apositive integer, by using the binomial theorem.

a Expand ( )1 212+ x in ascending powers of x up to and

including the term in x 2.

b By substituting x = 14 into your expansion, estimate the

value of 1 5. . Give your answer to 2 decimal places.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a Apply the binomial theorem ( ) ... :( )

!1 1 1

2

2+ = + + +− x nx x n n n

( ) ( ) ( )!1 2 1 2 212 21

2

12

12

2+ = + + +( ) ( )( )−

x x x

= + − +1 122 x x

b Substitute x = 14

into the approximation ( ) :1 2 112 21

2+ ≈ + − x x x

1 2 11414

12

14

12 2+ ≈ + −( )( ) ( )

i.e. 1 5 1 21875. .≈ Hence 1 5 1 22. .≈ (to 2 decimal places)

1 2 1 51

4

12+ =( )

.1 2 1 51

4

12+ =( )

.

E XA

MPL E 4


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You can use the binomial theorem to find the expansion of analgebraic fraction.

Express x x 1 − as a series of ascending powers of x up to and

including the term in x 4.

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x x x x 1 1

1

− = − −( )

Use the binomial theorem to expand ( ) :1 1− − x

( ) ( ) ( )( )( ) ( )( )!( )( )( )

!1 11 2 31 1 22

1 2 33− = + + + +

− − − − −− − − − − x x x x

= + + + +1 2 3 x x x Multiply through by x :

x x x x x x ( ) ( )1 11 2 3− = + + + +−

= + + + + x x x x 2 3 4

Hence x x x x x x 12 3 4

− = + + + + The expansion is validfor - 1 < x < 1.

The expansion is validfor - 1 < x < 1.

E XA

MPL E 5

Exercise 0.11 Simplify these expressions.

Factorise your answers as far as possible.

a 22

41 x x + −+

b 34

11 x x − +−

c x

x x

x + ++3 1 d 2 1

11

2 1 x x x

++ −+

e x x x + +−4

11

2 Express these fractions in partial fractions.

a 4 32 3

x x x

++ −( )( )

b x x x +

+ +2

4 3( )( )

c 5 642

x x

−−

d x x x

2

27

1 2−

− +( )( )

3 Rearrange these equations to make x the subject.a y x x = −

32 1

b y x x = +2 3

4

c y x x = +

+2 1

1 d y x x =

−+

1 33 1


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4 Use the binomial theorem to express these functions as a seriesof ascending powers of x up to and including the term in x 3.a (1 + x )-1

b 11 2 3( )+ x

c ( )1 323− x

5 It is given that ( )1 112 2 31

218

116+ = + − + − x x x x

a Find the expansion of ( )1 412+ x in ascending powers of

x up to and including the term in x 3.

b By substituting the value x = −18 into your expansion inpart a estimate the value of 2.

Give your answer to 2 decimal places.

6 Express these functions as ascending powers of x up to andincluding the term in x 3.a x

x ( )1 2 2+

b x x

−+

11

c 4 31 3 1 2

x x x

++ −( )( )

Replace x with 4 x in thegiven expansion.Replace x with 4 x in thegiven expansion.


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0.2 Inequalities

You can use a graphical approach to solve an inequality. See C2 for revision. This is background knowledgefor Chapter 1.

See C2 for revision. This is background knowledgefor Chapter 1.

Use a graphical approach to solve the inequality x 2 - 5 x + 4 < 0••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Sketch the graph with equation y = x 2 - 5 x + 4:

x

y

1 4

The inequality x 2 - 5 x + 4 < 0 has solution 1 < x < 4.

x 2 - 5 x + 4 º ( x - 1)( x - 4) The graph cuts the x -axis when y = 0i.e when (x - 1)(x - 4) = 0

The graph has roots x = 1, x = 4.

x 2 - 5 x + 4 º ( x - 1)( x - 4) The graph cuts the x -axis when y = 0i.e when (x - 1)(x - 4) = 0

The graph has roots x = 1, x = 4.

x 2 - 5 x + 4 < 0 is satisfied byvalues of x for which the graph

y = x 2 - 5 x + 4 lies on or belowthe x -axis.

x 2 - 5 x + 4 < 0 is satisfied byvalues of x for which the graph

y = x 2 - 5 x + 4 lies on or belowthe x -axis.

E XA

MPL E 1

You can also solve an inequality by using algebra.

Use algebra to solve the inequality x 2 + 3 4 x ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Express the inequality in the form f( x ) 0 where f( x ) is factorised:

x 2 - 4 x + 3 0 i.e. ( x - 1)( x - 3) 0

Solve the equation f( x ) = 0 where f( x ) = ( x - 1)( x - 3):( x - 1)( x - 3) = 0 for x = 1, x = 3

Use appropriate values of x to find the sign of f( x ) in the intervals x < 1, 1 < x < 3, x > 3:

x

x > 3

3f( x) < 0f( x) > 0

1

x < 1

f( x) > 0

1 < x < 3

e.g. f(0) = 3 ( >0) f(2) = - 1 ( 0)

Hence the inequality x 2 + 3 4 x has solution x 1, x 3.

The inequality is satisfied forvalues of x for which f( x ) 0.

The inequality is satisfied forvalues of x for which f( x ) 0.

Remember to include x = 1 and

x = 3 in the solution of f( x ) 0.

Remember to include x = 1 and

x = 3 in the solution of f( x ) 0.

E XA

MPL E 2


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Exercise 0.21 Use any appropriate technique to solve these inequalities.

a x 2 - 5 x + 6 0

b 2 x 2 + 5 x - 3 < 0

c 4 x 2 - 1 > 0d x 2 - 7 x + 10 5 - x

e x 2 - 3 x + 2 2 x + 8

f 2 x 2 - 2 x - 12 > 3 - x

2 Solve the inequality x 2 - 2 x - 4 < 0 Give your answers in surd form.

3 Use a graphical approach to show that the inequality

x 2

- 6 x + 10 < 0 has no solution.

4 Solve these inequalities.a x ( x + 2)( x + 4) > 0

b ( x + 1)( x 2 - 9) 0

c x 3 + 2 x 2 + x < 0


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0.3 Modulus and exponential graphs

To sketch a graph with equation y = |f( x )|: sketch the graph y = f( x )

reflect the ‘negative’ part of your graph in the x -axis.

See C3 for revision. This is background knowledgefor Chapters 1, 4 and 5.

See C3 for revision. This is background knowledgefor Chapters 1, 4 and 5.

Sketch the graph with equation y = | x 2 - 4|••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Sketch the graph with equation y = x 2 - 4, drawing the partof the graph that lies beneath the x -axis as a broken line.

Reflect the dotted section in the x -axis:

(The reflected part has equation y = 4 - x 2)

4

–4

O x

y

y = x2 – 4

y = | x2 – 4|

y = x2 – 4

2–2

y = 4 – x2

Most graphical calculatorscan plot the graph of amodulus function.Look for the ABS key.

Most graphical calculatorscan plot the graph of amodulus function.Look for the ABS key.

The y -axis crossing point (0, 4) isthe reflection of the point (0, - 4)in the x -axis.

The y -axis crossing point (0, 4) isthe reflection of the point (0, - 4)in the x -axis.

This is the ‘negative’ part ofthe graph.

This is the ‘negative’ part ofthe graph.

Label each section of the graphwith its equation.Label each section of the graphwith its equation.

E XA

MPL E 1

You can sketch the graph with equation y = Aekx , where A and k are constants.

Sketch the graph with equation y = 3e-2 x ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

The graph with equation y = 3e- 2 x resembles that withequation y = e- x

y = 3e –2 x

y

x

3

O

When x = 0, y = 3e -2(0) = 3,i.e. the y -intercept of thegraph is 3.

When x = 0, y = 3e -2(0) = 3,i.e. the y -intercept of thegraph is 3.

The positive x -axis is anasymptote to the graph.

The positive x -axis is anasymptote to the graph.

E XA

MPL E 2


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Exercise 0.31 On separate diagrams, sketch the graphs with these equations.

Label all axis-crossing points with their values and the sectionsof the graph with their equations.a y = |2 x + 1|

b y = |2 - 3 x |

c y = | x 2 + 4 x + 3|

d y = |3 x - 2 x 2|

2 On separate diagrams, sketch the graphs of these equations.

Label all axis-crossing points with their values, giving answersin surd form where appropriate.a y = | x 3- 1|

b y = |( x + 3)2 ( x - 1)|c y = |(3 - x )( x 2 - 2)|

3 On separate diagrams, sketch the graphs with these equations.a y = 2e3 x

b y = - e- x

c y = 3e x + 1

d y = 3 - e-2 x

4 On separate diagrams, sketch the graphs with these equations.a y = |e- x - 1|

b y = |e x - 2|

5 Given that x < e x for all x 0,a show that x 2 < e2 x for all x 0

b deduce that x e-2 x is approximately zero for large positivevalues of x .


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0.4 Trigonometry

You can use a trigonometric identity to simplify an equation.You should know these identities:

cos cos sin cos sin2 2 1 1 22 2 2 2q q q q q ≡ − ≡ − ≡ −

sin sin cos2 2q q q ≡ tan tantan

2 21 2

q q

q ≡ −

tan sec2 21q q + ≡ 1 2 2+ ≡cot cosecq q

See C3 for revision. This is background knowledgefor Chapters 3 and 7.

See C3 for revision. This is background knowledgefor Chapters 3 and 7.

sin2 q + cos 2 q º 1sin2 q + cos 2 q º 1

tan sincos

q q q

≡ (for cos q ¹ 0)tan sincos

q q q

≡ (for cos q ¹ 0)

Solve the equation cos 2 q + cos q = 0 for 0 q p .Give answers in exact form.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Replace cos 2 q with 2cos 2 q - 1 in the equation:

cos 2q + cos q = 0so 2cos2 q - 1 + cos q = 0i.e. (2cos q - 1)(cos q + 1) = 0

Hence cos q = 12 or cos q = - 1

cos q = 12 when q p = 13

cos q = - 1 when q = p

Hence cos 2q + cos q = 0 has solutions q p p = 13 , for 0 q p .

This is a quadraticequation in cos q .

This is a quadraticequation in cos q .

Since 0 q p Since 0 q p

Give answers in exact form(i.e. in terms of p ).Give answers in exact form(i.e. in terms of p ).

E XA

MPL E 1

You can use identities to reduce the number of differenttrigonometric functions in an expression.

Express sin 2q cos q in terms of sin q only.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Use the double-angle identity sin 2 q º 2 sin q cos q : sin 2q cos q º (2 sin q cos q )cos q º 2 sin q cos2q º 2 sin q (1 - sin2 q )

Hence sin 2 q cos q º 2sin q (1 - sin2q )

sin 2q cos q involves twotrigonometric functions.sin 2q cos q involves twotrigonometric functions.

cos 2 q º 1 - sin2 q cos 2 q º 1 - sin2 q

RHS depends only on sin q .RHS depends only on sin q .

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MPL E 2


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Exercise 0.41 Solve these equations.

Give answers in exact form.a sin 2q = cos q for 0 q p

b sin 2q = tan q for 0 q p

c cos 2q = 3 sin q + 2 for - p q p

d 3cos 2q = cos q for −1212

p q p

2 a Express 3sin q + cos q in the form R sin(q + a ),

where R > 0 and 0 12<


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1.1

12

You need to know these standard results of differentiation:

d

d e e

x

ax ax a( ) = dd x x

ax ln( ) = 1

dd x ax a ax sin cos( ) =

dd x ax a ax cos sin

( ) = −where a is a non-zero constant.

You can use rules and standard results of differentiation to findthe first and second derivatives of a function.

See C3 and C4 for revision. This is background knowledgefor Chapters 4, 5, 6 and 7.

See C3 and C4 for revision. This is background knowledgefor Chapters 4, 5, 6 and 7.

These include the chain, productand quotient rules.

These include the chain, productand quotient rules.

0.5 Differentiation

If y = x e2 x find dd

dd

y x

y x

and 2

2 .

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Use the product rule, dd

dd x

dd

uv u v v x

u x

( ) = + , to find dd y x :

y = x e2 x so dd e y x x

x x = +( )2 12 2e ( )

Factorise: = +( )2 1 2 x x e

Use the product rule on dd y x

to find dd

2

2 x y :

dd e y x x

x = +( )2 1 2 so dd

e e2

22 22 1 2 2 y

x x x x = + +( )( ) ( )

= + = +( ) ( )2 2 2 4 12 2e e x x x x

Hence dd

dd

e e y x y

x x x x x = + = +( ) ( )2 1 4 12 2 2 2and

dd x

x x e 2e2 2( ) = using the chain rule.dd x

x x e 2e2 2( ) = using the chain rule.

E XA

MPL

E 1

Find f ¢( x ) when f( x ) = sin x (cos x + 1)••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

f( ) sin cos

sin cos sinsin sin

x x x

x x x x x

= +( )

= += +

1

12 2

Use the chain rule, d 12

12

sin2 2cos 2 cos 2 :d x

x x x ( ) = ( ) =′ = +f ( ) cos cos x x x 2

f¢( x ) is the first derivative of f( x ).f¢( x ) is the first derivative of f( x ).

Using sin 2 x º 2sin x cos x Using sin 2 x º 2sin x cos x

E XA

MPL E 2

You can use a trigonometric identity to simplify thedifferentiation process.


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13

Exercise 0.5

1 Find dd y x for these equations. Simplify each answer as far as possible.

a y = e x sin x b y = ln(1 + x 2)

c y x x = +1e d y = ln(sin 2 x )

2 Find the exact maximum value of these functions over thegiven interval. You should show each answer is a maximum.a f(q ) = sin q + cos q for 0 q p

b f(q ) = sin2 q - 1 for 0 12< q p

3 Using trigonometric identities, or otherwise, differentiate these equations.Simplify each answer as far as possible.

a y = sin xcos 2 x + cos xsin 2 x b y = 4cos2 xsin2 x

c y = (1 - 2sin2 2 x )2

4 Use implicit differentiation to find an expression for these derivatives.

a dd x y

3( ) b dd x xy 2( ) c dd

dd x y x x

2( )5 If y = ux , where u is a function of x ,

a find an expression for dd y x .

b Hence show that dd

dd

dd

2

2

2

22 y

x u x

u x

x = +

You can use the chain rule to differentiate implicitly.

Find an expression for dd x y

2( ).••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Use the chain rule:d

ddd

dd x y y x y y

2 2( ) ( )= ×

= 2 y y x dd

dd y

y y 2 2( ) = since y 2 is beingdifferentiated with respect to y .

dd y

y y 2 2( ) = since y 2 is beingdifferentiated with respect to y .

E XA

MPL E 3


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14

Integration

You need to know these standard integrals:

1 1

ax a x x c d = +ln

e d eax ax x c

a= +1

cos sinax x ax c a( ) ( )= +d 1

sin cos( )ax x ax c ad = +− ( )

1

You can use the rules and standard results of integration tointegrate more complicated functions.

See C3 and C4 for revision. This is background knowledgefor Chapters 4 and 7.

See C3 and C4 for revision. This is background knowledgefor Chapters 4 and 7.

a is a non-zero constant.a is a non-zero constant.

These rules include integrationby substitution and by parts.

These rules include integrationby substitution and by parts.

0.6

Using the substitution u = 2 x 2, find x x x cos( )2 2 d

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Replace the term 2 x 2 with the variable u:

x x x cos( )2 2 d = xcos u d x

= x u u x cos d

4

Cancel the x terms: = 14 cosu ud

= +14 sin u c

Replace u with 2 x 2: = ( ) +14 2 2sin x c

You could also use inspection(the reverse chain rule) to findthis integral.

You could also use inspection(the reverse chain rule) to findthis integral.

u = 2 x 2 so dd

u x

x = 4 , i.e. d d x u x

=4

u = 2 x 2 so dd

u x

x = 4 , i.e. d d x u x

=4

Remember the constant ofintegration, c.

Remember the constant ofintegration, c.

E XA

MPL E 1

You can use integration to solve a differential equation.

Find the general solution of the differential equation3 2 y x x y x sec

dd =

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Separate the variables and integrate each side:

3 2 y x x y x sec d

d = so 3 y 2d y = xcos xd x

3 y 2d y = y 3 + c 1

x x x x x x c cos sin cosd = + + 2

Hence the general solution is given by y 3 = xsin x + cos x + c

i.e. y x x x c = + +sin cos3

A differential equation is anequation that involves a derivative.A differential equation is anequation that involves a derivative.

sec 1cos

x x

≡sec 1cos

x x

≡

Integrate xcos x by parts:

u x u x

= =, 1dd

and

dd

v x

x v x = =cos sin,

Integrate xcos x by parts:

x x

= =, 1dd

and

dd

v x

x v x = =cos sin,

c = c2 - c1 You only need one arbitraryconstant in the general solution ofa first order differential equation.

c = c2 - c1 You only need one arbitraryconstant in the general solution ofa first order differential equation.

E XA

MPL E 2


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15

You can express information as a differential equation to model areal-life situation.

E XA

MPL E 3

When added to a container of water, a quantity of coloureddye begins to spread out in such a way that, t seconds from

the start of the process, the dye forms a circular shape ofradius r cm and area A cm2. It is assumed that, at all times,the rate of increase of A is inversely proportional to A.After 4 seconds the area of the circle is 6 cm 2.a Formulate a differential equation for A.b Find A in terms of t . Comment on the validity of your

answer in the long term.••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a The rate at which A is increasing is dd At .

Hence dd At A∝

1

i.e. dd At

k A= for k a positive constant.

b Separate the variables and integrate each side:

dd At

k A= so A d A = k dt

hence 122 A kt c = +

i.e. A kt = +2 1c

Substitute the values t = 0, A = 0 into A kt c= +21 to find c

1:

kt c = +2 1 so 0 0 1= + c i.e. c 1 = 0

Hence A kt = 2Substitute the values t = 4, A = 6 into A kt = 2 to find k :

A kt = 2 so 6 2 4= ( )k 36 = 8k i.e. k = 4.5

Hence the equation for A in terms of t is A t t t = ( ) = =2 4 5 9 3.As t increases, the function 3 t increases without limit.Since the dye is inside a container, the equation A t = 3 is an unrealistic model for A in the long term.

P is inversely proportional toQ if P

Q∝

1 P is inversely proportional toQ if P

Q∝

1

k > 0 since the area is increasingover time.k > 0 since the area is increasingover time.

k is a constant so k dt = kt + ck is a constant so k dt = kt + c

The arbitrary constant hasbeen replaced by c1.c1 = 2c

The arbitrary constant hasbeen replaced by c1.c1 = 2c

At the start, the circle doesnot exist and so whent = 0, A = 0.

At the start, the circle doesnot exist and so whent = 0, A = 0.


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16


You can use trigonometric identities to find an integral.

Use a suitable double-angle identity to find (2cos q + 1)2 dq

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Expand the brackets:(2cos q + 1) 2 = 4cos2 q + 4cos q + 1 = 2(cos 2q + 1) + 4cos q + 1

So (2cos q + 1)2 dq = 2cos 2q + 4cos q + 3 dq

= sin 2q + 4sin q + 3q + c

Hence (2cos q + 1)2 dq = sin 2q + 4sin q + 3q + c

Using cos 2 q º 2cos 2 q - 1Using cos 2 q º 2cos 2 q - 1

2cos 2 q dq

= =( )2 2 212 sin sinq q 2cos 2 q dq

= =( )2 2 212 sin sinq q

E XA

MPL E 4

Exercise 0.61 Find these integrals. Give evaluations in exact form where appropriate.

a x sin x d x

b x (2 x 2 - 1)3 d x

c 4 x (2 x + 1)4 d x

d−1

0

x e- x d x

e

0

2p

cos q (1 + sin q )2 dq

f1

2

9 x 2 ln x d x

2 By writing tan x as sincos

x x

,

a differentiate tan x and hence find sec 2 2 x d x

b show that tan x d x = ln |sec x | + c

Use the substitution u = 2 x + 1Use the substitution u = 2 x + 1


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17

3 Find the general solution of these differential equations.Give each answer as an explicit function.

a dd y x xy = 2

b t x x

t

d

d + = 0

c secq dd e y y q =

d dd e y x x

x y = −2

4 Find the particular solution of these differential equations.In each case, make y the subject.

a x y y x 2 2d

d = for which y = 12 when x = 1

b x y y x 2 1 2− =( )

dd for which y = 1 when x = 3

c cos2 x y y x dd = for which y = 1 when x = p

5 When first noticed, a damp patch on a ceiling of a classroomhad area 25 cm 2 and was spreading at a rate of 10 cm 2 per day.t days later, the area A cm2 of the patch was increasing at a rateproportional to the square root of A.a Formulate a differential equation for A.

b Hence show that A can be modelled by the equation

A = (t + 5) 2

c Comment on the suitability of this model for A in the long term.

d Find the number of days the patch was on the ceiling beforebeing noticed. State any assumption made in arriving at your answer.

6 Use appropriate trigonometric identities to find these integrals.

a 2 sin2 q dq

b tan 2 q dq

c (sin q + cos q )2 dq


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1.1

18

Complex numbers0.7

You can use complex numbers to solve any quadratic equation. See FP1 for revision. This is background knowledgefor Chapters 3 and 5.

See FP1 for revision. This is background knowledgefor Chapters 3 and 5.

Solve the equation x 2 - 6 x + 13 = 0••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Use the quadratic formula x b b aca

= − ± −2 4

2:

x 2 - 6 x + 13 = 0 so x = − − −± −( ) ( ) ( )( )( )

6 6 4 1 132 1

2

= 6 162

± −

= 6 4

2

± i

= 3 ± 2iThe equation x 2 - 6 x + 13 = 0 has solutions x = 3 ± 2i.

− −= ×16 16 1( ) = ± 4i

− −= ×16 16 1( ) = ± 4i

E XA

MPL E 1

You can use Pythagoras’ theorem and trigonometry to find themodulus and argument of a complex number.

Find the modulus and argument of z = 3 - 3i••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

z = + −3 32 2( ) = 18

= 3 2For any complex number z = a + bi in the 4th quadrant,

arg : z ba

= − ( )tan 1

arg tan tan ( )z = =− − −( ) −1 133 1 = − 14 p

Hence z = 3 - 3i has modulus 3 2 and(principal) argument − 14

p .

The modulus of z = a + bi

is z a b= +2 2 The modulus of z = a + bi

is z a b= +2 2

Draw a diagram: z = 3 - 3i liesin the 4th quadrant.Draw a diagram: z = 3 - 3i liesin the 4th quadrant.

Give the principal argumentunless told otherwise.Give the principal argumentunless told otherwise.

Give answers in exact formunless told otherwise.Give answers in exact formunless told otherwise.

E XA

MPL E

2


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F P2

19

Exercise 0.7

1 Solve these quadratic equations.a x 2 - 2 x + 10 = 0

b 4 x 2 + 4 x + 5 = 0

c x x 2 2 2 3 0− + =2 Find the exact modulus and argument of these complex numbers.

a 5 + 5i b 2 2 3− i

c − +6 2 3i d − −6 6i

3 The diagram shows the points P and Q which representthe complex numbers z and w respectively.

OP = 4 and angle POQ = 12p .

a Given that w z − = 8 show that w = 4 3

b Given further that arg z = 16p show that arg( )w z − = 56

p

4 Let z = cos q + i sin q

Use suitable double-angle identities to show that z 2 = cos 2q + i sin 2q

O

Q

P

Im

Re4

O

Q

P

Im

Re4

You can represent complex numbers geometrically.

The diagram shows the points P and Q , which representthe complex numbers z and w respectively.

OP = 3, OQ = 4 and P QO=

13

p

Find the exact value of z w − .

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

The complex number ( z - w ) can be represented

by the vector QP .

Apply the cosine rule to find QP :

QP 2 2 23 4 2 3 4 13= + − ( )( )( )cos p = + − ×9 16 24 1

2 = 13

Hence z w − = 13

O

Q

P

4

3

Im

Re

r13

O

Q

P

4

3

Im

Re

r13

O

Q

P

4

3

Im

Re

r13

O

Q

P

4

3

Im

Re

r13

a b c bc A2 2 2 2= + − cosa b c bc A2 2 2 2= + − cos

Since ( z - w ) is represented byQP , z w QP − = .Since ( z - w ) is represented byQP , z w QP − = .

E XA

MPL E 3


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20

Answers

Exercise 0.1

1 a 6 12 1

( )( )( )

x x x

++ −

b 2 74 1 x

x x +

− +( )( )

c 2 23 4

x x x x

( )( )( )

++ +

d x x x x ( )

( ) ( )4 11 2 1

++ −

e ( )( )( )( ) x x x x

+ −+ +

2 24 1

2 a 12

33( ) ( ) x x + −+ b

24

13( ) ( ) x x + +−

c 4 21

2( ) ( ) x x + −+ d 3

11

11

2( ) ( ) ( ) x x x + − +− −

3 a x y y

= −2 3 b x y = −3

4 2

c x y y = −

−1

2 d x y y =

−+

13 1( )

4 a 1 - x + x 2 - x 3 + ¼

b 1 - 6 x + 24 x 2 - 80 x 3 + ¼

c 1 - 2 x - x 2 − 43 x 3 - ¼

5 a 1 + 2 x - 2 x 2 + 4 x 3 - ¼

b 2 ≈ 1.421875 = 1.42 (2 d. p.)

6 a x - 4 x 2 + 12 x 3 - ¼ b − + − + −1 32

78

1116

2 3 x x x .. .

c 3 + x + 17 x 2 - 11 x 3 + ¼

Exercise 0.21 a x 2, x 3 b -3 < x < 1

2 c x x < >−12

12

, d x 1, x 5

e -1 x 6 f x < −52 , x > 3

2 1 5 1 5− < < + x 4 a -4 < x < - 2, x > 0 b x - 3, - 1 x 3 c x < 0, x ¹ -1

Exercise 0.31 a

O

y = 2 x + 1 y = –(2 x + 1) y

x

1

12

–

b

O

= 2 – 3 x y = –(2 – 3 x) y

x

2

23

c

O

= x2 + 4 x + 3 y = x2 + 4 x + 3

y = –( x2 + 4 x + 3)

y

x

3

–3 –2 –1

d

O

= – x (3 – 2 x) y = – x(3 – 2 x)

y = x (3 – 2 x)

y

x32

2 a

O

y

x

1

1

b

O

y

x

9

1–3

c

O

y

x

6

3√2–√2


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Answers

3 a

O

y

x

2

b

O

y

x

–1

c

O

y

x

4

1

d

O

y

x

2

3

In 3– 12

4 a

O

y

x

1

b

O

y

x

2

In 2

Exercise 0.4

1 a q p p p = 1612

56

, , b q p p p = 0 1434

, , ,

c q p p p = − − −16 12 56, , d q p = ±16

2 a 2 16

sin q p +( ) b q p = 0 23,3 a (1 - 2sin2 q )sin q b 2(1 - cos2 q )cos q

c 12

sin 4q d cos 2q

4 b BC = k , 3k

Exercise 0.51 a e x (cos x + sin x ) b 2

1 2 x x +

c − x x e d 2cot x

2 a 2 b 03 a 3cos 3 x b 2sin 4 x c -4sin 8 x

4 a 3 2 y dy dx

b 2 2 xy y dy dx + c x x

d y dx

dy dx

22

2 2+

5 a u x dudx +

Exercise 0.61 a - x cos x + sin x + c b 1

16 2 12

4 x c − +( )

c

16

152 1 2 1

6 5

( ) ( ) x x c + − + + d -1 e 7

3 f 24ln 2 - 7

2 a ∫ sec22 x dx = 12 tan 2 x + c 3 a y A x = e 2 b x A

t = c y = - ln |c - sin q | d y = ln |e x (2 x - 2) + c |

4 a y x x = +1 b y

x x = −

+( )2 11 c y = etan x

5 a dAdt

A= 2 c The model predicts that A® ¥ as t ® ¥ .

This is unrealistic. d 5 days. This answer assumes the given modelling

assumptions held prior to the patch being noticed.6 a q − 12sin 2q + c b tan q - q + c

c q q − +12 2cos c

Exercise 0.71 a x = 1 ± 3i b x i= ±−12 c x i= ±2

2 a z z = =5 2 14, arg( ) p

b z z = = −4 13, arg( ) p

c z z = =4 3 56, ( )arg p

d z z = = −2 3 34, arg( ) p


22/70

Exam-style assessment

Further Pure FP2© Oxford University Press 2009

1. Solve the inequality x x

2 21

+

− < x

2. Find the complete set of values of x for which x x x +

+−2

312

3. Find the set of values of x for which x x x − +>

11

3

4. (a) Use algebra to solve the equation | x 2 - 5 x + 6 | = 4 x - 8

(b) On the same diagram, sketch the graphs with equations y = | x 2 - 5 x + 6 | and y = 4 x - 8

(c) Hence, or otherwise, solve the inequality | x 2 - 5 x + 6 | < 4 x - 8

5. (a) On the same diagram, sketch the graphs with equations y = | x 2 - 9 | and y = |2 x - 1 |. Label, with their coordinates, the axis-crossing points ofeach graph.

(b) Find the values of x where these two graphs intersect. Give answers insimplified surd form where appropriate.

(c) Hence solve the inequality | x 2 - 9 | |2 x - 1|

6. (a) On the same diagram, sketch the graphs with equations y = |2 x + a | and

y = |3 x + a |, where a is a positive constant. Label, with their coordinates,the axis-crossing points of each graph.

(b) Hence solve, in terms of a , the inequality |2 x + a | < |3 x + a |

7. (a) Sketch the graph with equation y = | x 2 - 2kx | where k is a positive constant.Label the stationary point with its coordinates.

(b) Hence, or otherwise, solve the inequality | x 2 - 2kx | k 2, giving your answerin terms of k .

1 Inequalities


23/70

© Oxford University Press 2009

Inequalities

Exam-style mark scheme

1

Further Pure FP2

Question Solution Marks Number

1 x x

x 2 2

10

+

− − < M1

x x

+

− 1

13

0 M1

( )( )(( ) x

x x +

− + >1

1 3

2

0 M1 A1

Critical values: x = - 3, - 1 and 1 M1 A1 Considering change of sign of factors gives:

Hence x > 1 or x < - 3 A2 7

4 a ( x - 3) 2( x - 2) 2 = 16( x - 2) 2 M1oe ( x - 3) 2( x - 2) 2 - 16( x - 2) 2 = 0 M1 ( x - 2) 2[( x - 3) 2 - 16] = 0 M1 Then x = 2, 7 and - 1 (ignore x = - 1, it does not satisfy the A2 (5)

original equation).

b

O

6

2 3

y = 4 x – 8

y = |x 2 – 5 x + 6|

y

x

B2 (2)

c 2 < x < 7 A2 (2) 9


24/70

© Oxford University Press 2009 Further Pure FP2

5 a

O x

1

1–1–2–3–4 2 3 4

2

3

4

5

6

7

8

9

10

12

y

B2 (2)

b For x > 3 or - 3 < x < 0.5 then x 2 - 9 = 2 x - 1 M1 x 2 – 2 x – 8 = 0, then x = 4 or x = - 2 M1 A1 For 0.5 < x < 3 or x < - 3 then x 2 - 9 = - (2 x - 1) M1 x 2 + 2 x - 10 = 0, then x = − ±1 11 M1 Hence x = 4, - 2, ±1 11 A3 (6)

c From part b and the graph x x x − − −− +1 11 2 1 11 4, , A2 (2) 10

6 a

x

y

y = |3 x + a |

y = |2 x + a |

(0, a )

)a2–( , 0 )a3–( , 0

B3 (3)

b For − a2 < x < − a

3, then 2 x + a = - (3 x + a) M1 A1

x = − 25a A1

Hence x < - 0.4a or x > 0 A1 (4) 7

7 a

x

y

(k, k2)

2k

B3 (3)

b x 2 – 2k x - k 2 = 0, x = k 1 2±( ) M1 A1 Hence k ( )1 2− x k ( )1 2+ A2 (4) 7


25/70



1. (a) Simplify r (r + 2) - r (r - 2)

(b) Hence use the method of differences to prove that r r

n

=∑ =1 12 n (n + 1)

2. (a) Express 69 12 52r r − −

in partial fractions.

(b) Hence show that 89 12 5

15 2 13 1 3 222 r r

n n

n nr

n

− −=

+( ) −( )+( ) −( )=

∑

(c) Evaluate the series 89 12 526

15

r r r − −=∑ , giving your answer to 3 significant figures.

3. (a) Find constants A , B and C such that

84 1 2 3 2 1 2 1 2 32r r

A

r

B

r

C

r −( ) +( )≡

−( ) +

+( ) +

+( ) for allr 1

(b) Hence show that 34 1 2 3

22 1 2 321 r r

n n

n nr

n

−( ) +( )=

+( )+( ) +( )=

∑

4. (a) Show that 11

11 2

21 2r r r r r r r +( )

−+( ) +( )

=+( ) +( )

for all r 1

(b) Hence, or otherwise, find an expression for4

1 21 r r r r

n

+( ) +( )=∑

giving your answer in fully factorised form.

(c) Deduce 41 21 r r r r

n

+( ) +( )=∑ < 1 for all n > 1

5. (a) Simplify (3 r + 1) 3 - (3 r - 2) 3

(b) Hence, or otherwise, show that r r

n2

1

16=

∑ = n (n + 1)(2 n + 1)

You can use the result r r

n

=∑ =

1

12

n (n + 1) without proof.

6. (a) Express 8 2 14 1

3

2r r

r

− −−

in the form Ar + B

r

C

r 2 1 2 1−( ) +

+( ), for constants A , B

and C to be determined.

(b) Hence find an expression for 8 2 14 1

3

21

r r

r r

n − −−=

∑ giving your answer in terms ofn and in fully factorised form.

(c) Evaluate the series 8 2 14 13

27

24 r r r r − −

−=∑ , giving your answer to 3 significant figures.

2 Series


26/70


7. (a) Express 3 21 2r

r r r −

+( ) +( ) in partial fractions.

(b) Hence show that 3 21 2 1 21

r

r r r

n

n nr

n −+( ) +( )

=+( ) +( )=

∑

(c) Find the value of the positive integer N for which3 2

1 2

8

151

r

r r r r

N −

+( ) +( ) =

=

∑

8. (a) Express 312

−−( )r

r r in the form

Ar

Br

C r −

+ ++1 1 where

A , B and C

are constants.

(b) Hence, using the method of differences, show that

31

1122

−−( )

= −+( )=

∑ r r r

n

n nr

n

for n > 2


27/70

© Oxford University Press 2009

Series


2

Further Pure FP2


1 a r 2 + 2 r - r 2 + 2 r = 4 r A1 (1)

b 4 2 21 1

r r r r r r

n

r

n

= =∑ = + − −[ ]∑ ( ) ( )

4(1) = 1(3) - 1( - 1) 4(2) = 2(4) - 2(0) 4(3) = 3(5) - 3(1) 4(4) = 4(7) - 4(2) 4( n - 2) = (n - 2)( n ) - (n - 2)( n - 4) 4( n - 1) = (n - 1)( n + 1) - (n - 1)( n - 3) M1 A1 4(n ) = n (n + 2) - n (n - 2) Adding:

4(1 + 2 + 3 + … + n ) = 1 + (n - 1)( n + 1) + n (n + 2) M1 A1 = 2n 2 + 2n

Hence r n nr

n

=∑ = +

1

12

1( ) A1 (5) 6

2 a 13 5

13 1r r −

− + B1 A2 (3)

b 69 12 5 13 5 13 122 2r r r r r

n

r

n

− − = − − +( )= =∑ ∑ r = 2, 1 1

7−

r = 3, + −14

110

r = 4, + −17

113

..................

r = n - 2, + − − −1

3 111

3 5n n

r = n - 1, + − − −13 81

3 2n n M1 A1

r = n , + − − +1

3 51

3 1n n

Adding:

1 141

3 21

3 145 39 64 3 1 3 2

2− − − + =

− −+ −n n

n nn n( )( ) M1 A1

8

9 12 586

69 12 522

22r r r r r

n

r

n

− − =

− −= =∑ ∑ M1

Hence 89 12 5

15 13 23 1 3 2 15 2 13 122

2

r r n n

n nn n

nr

n

− − = − −+ − = + −+=∑ ( )( ) ( )( )( )(33 2n − ) M1 A1 (7) 10


28/70


c 89 12 526

15

r r r − −∑=

= −− − − −

∑∑==

89 2 5

89 12 52 22

5

2

15

r r r r r r

= × + −− + − +− +

( )( )( )( )

( )( )( )

15 15 2 15 145 2 45 1

75 2 415 2 15 1

= ×

× − ×

×227 1443 46

77 413 16

= 0.126 10

3 a 12 1

22 1

12 3r r r −

− + + + M1 A3 (4)

b r = 1: 1 23

15

− +

r = 2: + − +1325

17

r = 3: + − +15

27

19

….. …………..

r = n - 2: + − − − + −1

2 52

2 31

2 1n n n

r = n - 1: + − − − + +1

2 32

2 11

2 3n n n

r = n: + − − + + +1

2 12

2 11

2 3n n n M1 A1

Adding:

23

12 1

12 3

8 163 2 1 2 3

2− + + + =

++ +n n

n nn n( )( )

M1 A1

Hence 34 1 2 3

38

84 1 2 3

22 1 221

21( )( ) ( )( )

( )( )(r r r r

n n

n nr

n

r

n

− + =

− + = ++ += =

∑ ∑33) B1 A1 (6)

10


29/70


4 a LHS = ( )( )( ) ( )( )r r

r r r r r r + −

+ + = + +2

1 22

1 2 M1 A1 (2)

b 21 2

11

11 21 1r r r r r r r r

n

r

n

( )( ) ( ) ( )( )+ +∑ = + − + +∑= =

B1

r = 1: 12

16

−

r = 2: + −1

6

1

12

……… ………

r = n - 1: + − − +1

11

1( ) ( )n n n n

r = n : + + − + +1

11

1 2n n n n( ) ( )( ) M1 A1

Adding:

12

11 2

32 1 2

− + + = ++ +( )( )( )

( )( )n nn n

n n M1 A1

Hence 41 2 31 21 r r r n n

n nr

n

( )( ) ( )( )( )+ +∑ = ++ += M1 A1 (7)

c 41 23

1 23

3 2

2

21 r r r

n n

n nn n

n nr

n

( )( )( )

( )( )+ + = ++ + =

++ +

∑=

Now n 2 + 3n < n 2 + 3n + 2

As n > 1, n 2 + 3n + 2 > 0 so thatn n

n n

2

23

3 21

++ +

<

Therefore 41 211r r r

r

n

( )( )+ +∑=<

9

5 a 27 r 3 + 27 r 2 + 9 r + 1 - (27 r 3 - 54 r 2 + 36 r - 8) = 9(9 r 2 - 3 r + 1) M1 A1 (2)

b 9 9 3 1 3 1 3 221

3 3

1( ) ( ) ( )r r r r

r

n

r

n

− +∑ = + − − ∑= =

B1

r = 1: 4 3 - 13r = 2: + 73 - 43

….. ……….r = n - 1: + (3 n - 2) 3 - (3 n - 5) 3 M1 A1

r = n : + ( 3 n + 1) - (3 n - 2) 3 M1 A1 Adding:

(3n + 1) 3 - 1 = 9n (3 n 2 + 3n + 1) B1

Hence ( ) ( )9 3 1 3 3 121

3r r n n nr

n

− +∑ = + +=

r n n n r nr

n

r

n2

1

2

1

19

3 3 1 3= =∑ = + + + ∑ −( ) B1

= 19

[n (3n 2 + 3n + 1) + 32 n (n + 1) - n ]

= n18

[6n 2 + 6n + 2 + 3n + 3 - 2]

= n

6[2n 2 + 3n + 1] M1 A1 (9)

= 16n (n + 1)(2 n + 1)

11


30/70


6 a 2 12 2 1

12 2 1

r r r

− − + +( ) ( ) M1 A3 (4)

b 8 2 14 121

r r

r r

n 3 − −−

∑=

= + + − −∑=2 1

2 2 11

2 2 11r

r r r

n

( ) ( )

= ++

−−

∑∑==

2 12

12 1

12 111

r r r r

n

r

n

( ) ( )

= n (n + 1) + −1213 1

+ −15

13

+ .....

+ −− −1

2 11

2 3n n

+ −+ −1

2 11

2 1n n

= + +

+ −n n

n( )112

12 1 1

= + − +n n n

n( )1

2 1

= + + −[ ]

+n n n

n

( )( )1 2 1 12 1

= ++ = +

+n n n

n

n n

n( ) ( )2 3

2 12 3

2 1

2 2

c = − =∑ − ×∑∑===

24 5149

36 1513

2

1

6

1

24

7

24 ( )r r r

» 558 (3sf)

7 a − + + − +1 5

14

2r r r M1 A3 (4)

b 3 21 21 5

14

211r

r r r r r r r

n

r

n −+ + = + + − +∑∑

−== ( )( )

= + −−1 5243

− + −12

53 1

− + −13

54

45

+ .....

− − ++ −1

15 4

1n n n

− + −+ +1 5

14

2n n n

= + − + −− + +152

12

11

42n n

= + + − −+ +12 4 4

1 2n n

n n( )( )

= + + − −+ +n n n

n n

2

3 2 3 21 2( )( )

= + +n

n n

2

1 2( )( ) as required.


31/70


c 15 N 2 = 8( N + 1)( N + 2) B1 7 N 2 - 24 N - 16 = 0 (7 N + 4)( N - 4) = 0, Hence N = 4 M1 A1 (3) 11

8 a 1 13 2

1r r r − − + + B1 A1 (2)

b 31

11

3 2122 1

−−∑ = − − + +∑= =r

r r r r r r

n

r

n

( ) B1

r = 2: 1 32

23

− +

r = 3: + − +12

1 24

r = 4: + − +13

34

25

…… ………..

r = n - 2: +−

−−

+−

1

3

3

2

2

1n n n

r = n - 1: + − − − +1

23

12

n n n

r = n : + − − + +1

13 2

1n n n M1 A1

Adding:

1 3212

2 3 21

1 21

− + + − + + = + +−n n n n n M1 A1

Hence 31

1122

−−

∑ = −+=r

r r

nn nr

n

( ) ( ) A1 (6)

8


32/70



1. (a) Using de Moivre’s theorem, or otherwise, prove that

cos 4q

º 8cos4 q

- 8cos2 q

+ 1 (b) Solve the equation cos 4 q + 4cos 4 q = 0 for 0 q p When not exact, give each answer correct to 2 decimal places.

2. The transformation T from the z-plane to the w-plane is given by

w = z z + +1 2i , z ¹ - 1 - 2i

(a) Show that T maps the line y = 2 x to part of the real axis in the w-plane.

(b) Find the locus of points in the z-plane which are mapped to theline u = 0 in the w-plane.

3. (a) Shade on an Argand diagram the region R given by| z - 1 | | z - i |

The transformation T from the z-plane to the w-plane is given by

w = z z+

−3i , z ¹ i

(b) Show that T maps | z - 1 | = | z - i | to a circle in the w-plane.Give the cartesian equation of this circle.

(c) On a separate Argand diagram shade the region in the w-plane

which is the image of R under T . 4. Point P represents the complex number z where | z - 3 | = 2 | z - 3i |

(a) Use algebra to show that the locus of P is a circle, giving the centreC and radius of this circle.

Point Q represents the complex number z where arg( z + 1 - 4i) = 34

p

(b) On the same Argand diagram sketch the locus of P and the locus of Q ,marking clearly the point A where the two loci intersect.

(c) Find the complex number a which represents A and express the equation

of the tangent to this circle at A in the form| z + 1 - 4i | = | z - b| for b a complex number to be stated.

5. (a) Solve the equation z3 = 4 - 4 3i giving your answers in the form r eiq where r > 0 and exact q , - p < q p

(b) Illustrate your values from a on an Argand diagram.

(c) Hence, or otherwise, show that, if p and q are two distinct cuberoots of 4 - 4 3i then | p + q | = 2

3 Further complex numbers


33/70


6. (a) Solve the equation z4 = 2 3 + 2i giving your answers in exact modulus-argument form.

Give arguments as principal values.

(b) Illustrate your values from ( a ) on an Argand diagram.

(c) Prove that the points representing these values form the vertices of a square.

(d) State the two possible values of | p - q| where p and q are two distinctfourth roots of 2 3 + 2i

7. (a) Given that z = cos q + isin q , use de Moivre’s theorem to show that

zn - 1 zn

= 2isin nq where n is a positive integer.

(b) Express 4sin 3 q in the form Asin q + Bsin 3 q for integers A and B tobe stated.

O

y

R

ir

The diagram shows the curve with equation y = 2 3sin q for 0 q p . R is theregion bounded by this curve, the q -axis and the lines q = 0 and q = p

(c) Find the exact volume formed when region R is rotated once around the q -axis.

8. Point P represents the complex number z, where | z + 2i | = k | z - 3 - i |,where k is a constant. Point A, which represents the complex number 6 + 4i,lies in the locus of P .

(a) Show that k = 2

(b) Use algebra to show that the locus of P is a circle C .Give the centre and radius of this circle.

Point Q represents the complex number z, where arg( z - 4 - 2i) = 14

p

(c) Shade, on a single Argand diagram, the region of points on or inside

C and which satisfy 14

p arg( z - 4 - 2i) p and find the exact areaof this region.

The transformation T from the z-plane to the w-plane is given by

w = z z

+

− −

23

ii, z ¹ 3 + i

(d) Show that T maps the locus of P to a circle in the w-plane.


34/70


9. Points A and B represent the complex numbers 0 - 3i and 4 + 0i respectively.

Point P represents the complex number z, where arg z z

+

−( ) =34 23i p (a) Sketch on an Argand diagram the locus of P . Indicate on your sketch the

position of points A and B.

(b) Given that point Q in this locus is such that AQ = BQ

( i) show that AQ = 53 3 , ( ii) find the exact area of triangle AQB .

10. Point P represents the complex number z, where | z - 2 | = 2 | z - 4i |

Point Q represents the complex number z where arg z z

−

−( ) =24 14i p (a) Use algebra to show that the locus of P is a circle. Give the centre and

exact radius of this circle.

(b) Given that p = z z

−

−

24i

, where z is the complex number which

belongs to both of these loci ( i) show that p = 1 + i, ( ii) find z.

(c) On a single Argand diagram, sketch the locus of P and the locus of Q.

11. (a) Solve the equation z5 = 1 giving your answers in the form e iq for - p < q p

(b) Given that w is any complex 5th root of 1

(i) state the value of 1 + w + w2

+ w3

+ w4

(ii) hence find the value of1 1 2

4

+( ) +( )w ww


35/70


Further complex numbers


3


1 a (cos sin ) cos sinq q q q + = +i i4 4 4 M1 (cos sin ) cos cos sin cos sinq q q q q q q + = + −i i4 4 3 2 24 6 M1 - 4i cos q sin 3 q + sin 4 q Considering the real parts:

cos 4 q = cos 4 q - 6cos 2 q sin 2 q + sin 4 q = − − + −cos cos ( cos ) ( cos )4 2 2 2 26 1 1q q q q M1 A1 (4) = − +8 8 14 2cos cosq q

b cos 4 q + 4cos 4 q = 0 Þ 8c4 - 8c2 + 1 + 4c4 = 0 where c = cos q

Þ 12 c4

- 8c2

+ 1 = 0 Þ c2 8 4

24= ±

Hence cos ,2 1216

q =

Þ cos ,q = ± ±12

16

cos ,q q p p q p = ± ⇒ =12

14

34

0for

cos . , .q q q p = ± ⇒ =16

1 150 1 991 0 for

Hence q p p = 14 34 1 15 1 99 2, , . , . ( )c c decimalplaces

2 a Let P be any point on y = 2 x , then P = (k , 2k ), where k ∈ M1 w k ik

k ik i= ++ + +

22 1 2

= ++ + + = ++ +

k ik i k

k ik i

( )( )

( )( )( )

1 21 2 1

1 21 1 2

B1

= + ∈ ≠ −k

k k

1 1, A1 (3)

b w iv x iy x iy i= + = ++ + +0 1 2 B1

= ++ + + × + − +

+ − + x iy

x i y x i y x i y( ) ( )( ) ( )( ) ( )1 2

1 21 2

M1

= + + + + −+ + +

x x y y i y x x y

2 2

2 22 2

1 2( )

( ) ( ) B1

Considering the real part of w = 0: B1 x 2 + x + y2+ 2 y = 0 ( x + 0.5) 2 + ( y + 1) 2 = 5

4 , M1

Hence the locus is a circle centre ( - 0.5, - 1), radius 52 (excluding the point ( - 1, - 2)) A1 (6) 9


36/70


3 a

O

1

1

–2

2

11–2 2

y

x

B2 (2)

b z iwwi u iv

u iv= +

− = + +

+ −3

13

1( ) M1

= − +− + × − −

− −( )( )

( )( )

31

11

v iuu iv

u ivu iv M1

= + − + − − +− +

( ) ( )( )

3 3 31

2 2

2 2u v i u u v v

u v B1

Since | z - 1 | = | z - i|, then the real part of z = the imaginary part of z: A1 (4) Then 3 u + v - 3 = u2 - u - 3v + v2

u2 - 4u + v2 - 4v + 3 = 0 (u - 2) 2 + (v - 2) 2 = 5

c

O

4

5

y

x

B2 (2)

10

4 a ( x - 3) 2 + y2 = 4[ x 2 + ( y - 3) 2] M1 x 2 + 2 x + y2 - 8 y + 9 = 0 B1 ( x + 1) 2 + ( y - 4) 2 = 8 C (- 1, 4), radius = 2 2 A2 (4)

b

O

4

6

8

2

–5

A

C

y

x

3 r

4

B3 (3)

c From Argand diagram: a = - 3 + 6i A1 | z + 1 - 4 i| = | z + 5 - 8 i| so b = - 5 + 8 i A1 (2) 9


37/70


5 a 4 4 3 8 123

2− = −( )i i B1oe q

p = =− 3 8, r A1

z ei n3 3 28= − +( )

p p M1

z ei n= − +( )2 3 3 2

p p A1

If n = 0, then z e i0 92= − p

If n = 1, then z ei

1

592=

p

and z ei

3

792= −

p

A2 (6)

b z 1

z 0

z 2

2

–

–2

y

x

2 r

3

r

9

B2 (2)

c | z0| = | z1| = | z2| = 2 and the angle between any two distinctroots p and q = 2

3p , B1oe

p q+ = =( )2 2 23cos p M1 A1 (3) 11 6 a z i4 4 3

212= +

B1oe

= + + +( ) ( )4 2 26 6cos sinp p p p n i n B1 z in n= + + +( ) ( )2 24 2 24 2cos sinp p p p M1 If n = 0 then z i0 2 24 24= +( )cos sinp p

If n = 1 then z i1 213

24

13

24= +

( )cos sinp p and A2

z i2 21124

1124= −( )cos sinp p A1 (7)

If n = 2 then z i3 22324

2324= −( )cos sinp p B2 (2)


38/70


b z 1

z 0

z 2

z 3O

y

x

c | z0| = | z1| = | z2| = | z3| = 2 and

arg( z1) - arg( z0) = p

2 = arg( z3) - arg( z1) = arg( z2) - arg( z3) B2oe (2)

Then z0, z1, z2, z3 are the vertices of a square.

d | p - q| = 2 or 2 2 11

7 a zn = cos nq + isin nq M1 z- n = cos nq - isin nq M1 zn - z- n = 2i sin nq A1 (3)

b (2 i sin q )3 = −( ) z z13

= z3 - 3 z + −3 13 z z = − − −( ) z z z z3 31 13 = 2 i sin 3 q - 6 i sin q Hence 4 sin 3 q = - sin 3 q + 3sin q so A = 3, B = - 1

c V = p p

0

4sin 3 q d q = p p

0

(3sin q - sin 3 q )d q M1

= +−

=

p q q p

p

3 313163

0cos cos M2

A1 (4) 12


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8 a |6 + 6 i| = k |3 + 3 i| Þ k = 2 M1 A1 (2)

b x 2 + ( y + 2) 2 = 4[( x - 3)2 + ( y - 1)2] M1 x 2 + y2 + 4 y + 4 = 4( x 2 - 6 x + 9 + y2 - 2 y + 1) B1 3 x 2 - 24 x + 3 y2 - 12 y + 36 = 0 ( x - 4) 2 + ( y - 2)2 = 8 M1 Centre (4, 2), radius = 2 2 A2 (5)

c

O

4

2C

5

y

x

r

4

B3

Area = 1212

34

2 8r q p = × × M1

= 3p

A1 (5)

d w z i

z i= =+− −

23

2 (using part a above) M1oe

|w| = 2 is a circle centre (0, 0) radius = 2 A1 (2) 14

9 a

O

–2

–4

Q

A

B

arg( z + 3 i )

y

x

arg( z + 3 i ) – arg( z – 4) =2 r3

arg( z – 4)

B3 (3)

b i Angle ABQ = p

6 , AB = 4 32 2+ = 5 M1

AQ = 521

6

52

23

53

3× = × =cos

p M1 A1

ii Area = 12

53

23

25 312

32

( ) =sin p M1 A1 (5) 8


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10 a ( x - 2) 2 + y2 = 2[ x 2 + ( y - 4) 2] M1 x 2 - 4 x + 4 + y2 = 2 x 2 + 2 y2 - 16 y + 32 x 2 + 4 x + y2 - 16 y + 28 = 0 B1 ( x + 2) 2 ( y - 8) 2 = 40 M1 Centre ( - 2, 8) and radius 2 10 A2 (5)

b i Since | z - 2 | = 2 | z - 4 i| and arg z z i−−( ) =24 14 p , then B1

| p | = 2 and arg( ) p = 14 p p i i= = ++( )2 14 4cos sinp p M1 A1 ii z ip p=

−−

4 21 , substitute p = 1 + i: M1 A1 (6)

iii= = +

− −4 4 2 4 6

c

O

4

6

8

10

12

14

2

(2, 0)

(0, 4)

(–2, 8)

–5 5

P

y

x

B3 (3)

14

11 a z i n n5 2 21 1= = =+e e i(0 p p ) B1

z ei n=25

p M1

If n = 0, then z0 = ei0 = 1 A1

If n = ±1, then z e z ei i

1

25

2

25= = −

p p

, A2

If n = ±2, then z e z ei i

3

45

4

45= =

−p p , A2 (7)

b i (w5 – 1) = (w – 1)( w4 + w3 + w2 + w + 1) = 0 A1 Hence w4 + w3 + w2 + w + 1 = 0 M1 ii w3 + w2 + w + 1 = - w4 A1 (3)

( )( )1 12

4 1+ + = −w w

w

10


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1. (a) Find the general solution of the differential equation

tan( ) ,2 2 014 x y x

y x

dd = < −dd

where

(a) Find the general solution of this differential equation, giving youranswer in the form y = f( x )

(b) Given that the curve passes through the point P (0, 5) find the equation of C .

3. (a) Express 212 y −

in partial fractions.

(b) Hence show that the particular solution of the equation 2 1 2 x y y x

dd

+ = ,

where y > 1, x > 0, and for which y = 2 when x = 13 is given by y = 1

1+

−

x x

4. (a) Show that x 2 1+ is an integrating factor for the differential equation

112

1 x

y x

y x

dd

++

=

(b) Hence find the particular solution of this differential equation for which y = 13 when x = 0. Sketch the graph of this solution.

5. (a) Find the general solution of the differential equation dd y x

+ 2 ytan x = 1

(b) Hence show that the particular solution of the given equation for which y = 12

when x = 1

4p is given by y = 1

2sin 2 x .

6. As part of a training exercise, an athlete undertakes a 20 km run. After t hoursshe has run x km. During the run she varies her speed in such a way that therate of increase of x is directly proportional to x multiplied by the distanceshe has left to run.

(a) Formulate a differential equation for x

(b) Given that after 1 hour she has run 10 km and that after 90 minutes she has run15 km, solve this differential equation to show that

x

x t

209 1

−=

−

(c) Calculate the total distance run after 2 hours.

4 First order differential equations


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7. (a) Use integration by parts to show that x e x d x = ( x - 1)e x + c

(b) Hence find the general solution of the differential equation

( x + 1) dd y x

+ ( x + 2) y = x

8. The equation of a curve C satisfies the differential equation x dd y x

- y = 2 y2 x

(a) Show that the substitution y = ux transforms the given differentialequation into the equation dd

u x = 2u

2 x

(b) Find the general solution for u and hence find y in terms of x .

(c) Given that C passes through the point with coordinates ( - 1, 1),find the equation of the curve C and sketch its graph.

9. The equation of a curve C satisfies the differential equation

x 2 dd y x

- xy = 2 y( y + x ) (1)

(a) Show that the substitution y = ux transforms equation (1) into the differentialequation d

du

x u u

x =

+( )2 1 (2)

(b) Hence show that the general solution of equation (1) can be expressed as

y = Ax Ax

3

21 −,

for A an arbitrary constant.

The curve C passes through the point (1, - 2).

(c) Find the equation of C and the coordinates of all the points where C intersectsthe line y = x .


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First order differential equations


4


1 a dy y

x x

dx = 2 22cossin

B1

ln | y| = ln |sin 2 x | + ln c M3 y = k sin 2 x A1 (4)

b - 2 = k sin p 6

, then k = - 4 M1 A1

y = - 4sin 2 x A1 ft (3)

O

–4

y

xr4 B1

8

2 a dy y

x x

dx − = +44

2 1 B1

ln | y - 4 | = 2 22 1

− +( ) x dx M2 ln | y - 4| = 2 x - ln |2 x + 1| + ln c M1

y ce x

x − =

+4

2

2 1 B1

y ce x

x = + +4

2

2 1 A1 (6)

b 5 = 4 + c A1 c = 1

y e x x

= + +42

2 1 A1 ft (2) 8

3 a 11

11 y y− − +

M1 A1 (2)

b 212

dy y

dx x − =

B1

11

11 y y

dx x

dy− − +

= M1

ln ln ln y y

x c−+ = +11 M2oe

y y kx −+ =

11 A1

k = 1 since when y x = =( )2 13 A1 y - 1 = yx + x M1 y x

x = +

−11

A1 (8) 10


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4 a dydx

x x

y x ++

=2 1 B1

e e x

x x

dx x 22

112

1 2 1+ +∫ = = +ln( )

M1 A1 (3)

b y x ( )2121+ = x x dx c( )2

121+ + M1

y x x c( ) ( )212 2

321 11

3+ + += M1 A1

c = 0 A1

y = 13( x 2 + 1) A1ft

O x

2

B2 (7)

10

5 a The integrating factor is e e x x dx

x 2

2 2(tan )

lncos sec∫ = =− M1 ysec 2 x = (sec 2 x ) dx = tan x + c M1 A1 (3)

b Substitute y = 12

and x = p 4

, c = 0 A1

Particular solution:

y x x x x

x = = =tan

secsin cos sin2 2

2 M1 A1 (3) 6

6 a dx

dt kx x = −( )20 A1 (1)

b dx x x ( )20 − = kdt M1

120

1 120 x x dx kt c+ −( ) = + B1 M1 A1 1

20 20ln x

x Kt C − = + M1 A1

Substitute x = 10 and t = 1: K + C = 0 (1)

Substitute x = 15 and t = 1.5: 3 2 3110

K C + = ln (2) M1

From (1) and (2): K = 110

3ln and C = − 110 3ln A2

Hence 120 20

110

1 3ln ln x x

t − ( )= − M1 A1 (11)

i.e. ln ln ln ln x x

t t t 20

1 2 3 1 9 9 1− ( ) ( )= − = − = −

Therefore x x

t

209 1− = − as required

c When t = 2, x x 20

9− = Therefore x = 180 - 9 x , so x = 18 km 12


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7 a Let u = x , then du = dx B1 Let dv = e x dx , v = e x

xe x dx = xe x - e x dx = ( x - 1) e x + c M1 A1 (3)

b dydx

x x

x x

y+ =++ +21 1

M1

The integrating factor is e e x e

x x

dx x

dx x

++

+ +

∫

= ∫

= +

21

11

1

1( ) M1 A1 ( x + 1) e x y =

x x +1( x + 1) e

x dx = xe x dx M1 A1

(x + 1)e x y = ( x - 1) e x + c so y x e c x e

x

x = −( ) +

+( )1

1 M1 A1

10

8 a dydx

dudx

u x = + M1

x u x ux u x dudx + − =( ) 2 2 3 M1

dudx

u x = 2 2 A1 (3)

b u- 2du = 2 x dx B1

− = +1 2u x c M1 A1

− = + x y x c2 B1

y x x c

= −+2

A1 (5)

c c = 0 A1

y x

= −1 A1oe

O

–2

y

x

2

B1 (3)

11


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9 a Substitute y = ux anddydx

dudx

u x = + in (I): M1

x u x x ux ux ux x dudx

2 2+ − = +( ) ( ) ( ) B1 du

dx u u

x = +2 1( ) A1 (3)

b du

u u

dx

x ( )+ =

1

2

1 11u u

du− +( ) = 2dx x B1 ln u - ln(u + 1) = 2ln x + ln A M1

uu

Ax + =12 M1

u - Ax 2u = Ax 2 M1

u Ax Ax

=−

2

21

Substitute u y x = : Bloe

y Ax Ax

=−

3

21 A1 (6)

c − = −2 1 A

A, then A = 2 A1

y x x

=−2

1 2

3

2 A1

Substitute y = x :

x x x = −21 23

2

x (1 - 2 x 2) = 2 x 3 M1

x (4 x 2 - 1) = 0 B1 (0, 0), (0.5, 0.5) and ( - 0.5, - 0.5) A2 (6) 15


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1. (a) Find the general solution of the differential equation dd

dd

2

2 0 y

x y x

k + = , giving y in terms of x and the positive constant k .

(b) Describe the behaviour of y as x increases, where x > 0

2. An object is oscillating about a fixed point P . After t seconds the object is x metresfrom P where x is modelled by the differential equation

dd

2

2 x t

+ 4 x = 0

(a) Given that x = 3 when t = 0 and when t = 14

p , solve this differential equation to

show that x = 3cos 2 t + 3sin 2 t

(b) Express x in the form R sin (2 t + a ), where R > 0 and 0 < a < 1

2p are both exact.

(c) Hence find the maximum possible distance of the object from point P .Give your answer in simplified surd form.

3. (a) Find the general solution of the differential equation

4 42

2dd

dd

y x

y x + + 5 y = 2cos x - 8sin x

A curve C has equation which satisfies this differential equation.

The curve crosses the x -axis when x = 0 and whe

Alevel FP2

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