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8/16/2019 Alevel FP2
1/70
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8/16/2019 Alevel FP2
2/70
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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8/16/2019 Alevel FP2
3/70
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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8/16/2019 Alevel FP2
4/70
F P 2
4
0 Background knowledge for unit FP2
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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8/16/2019 Alevel FP2
5/70
F P2
0 Background knowledge for unit FP2
5
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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8/16/2019 Alevel FP2
6/70
F P 2
6
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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8/16/2019 Alevel FP2
7/70
0 Background knowledge for unit FP2
F P2
7
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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8/16/2019 Alevel FP2
8/70
F P 2
8
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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8/16/2019 Alevel FP2
9/70
F P2
0 Background knowledge for unit FP2
9
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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8/16/2019 Alevel FP2
10/70
F P 2
10
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 .
E XA
MPL E 2
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8/16/2019 Alevel FP2
11/70
F P2
0 Background knowledge for unit FP2
11
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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8/16/2019 Alevel FP2
12/70
F P 2
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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8/16/2019 Alevel FP2
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F P2
0 Background knowledge for unit FP2
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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8/16/2019 Alevel FP2
14/70
F P 2
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
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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 =
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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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8/16/2019 Alevel FP2
15/70
F P2
0 Background knowledge for unit FP2
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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8/16/2019 Alevel FP2
16/70
F P 2
16
0 Background knowledge for unit FP2
You can use trigonometric identities to find an integral.
Use a suitable double-angle identity to find (2cos q + 1)2 dq
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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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17/70
F P2
0 Background knowledge for unit FP2
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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F P 2
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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8/16/2019 Alevel FP2
19/70
0 Background knowledge for unit FP2
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 − .
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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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8/16/2019 Alevel FP2
20/70
F P 2
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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F P2
21
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
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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
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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
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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
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Exam-style assessment
Further Pure FP2© Oxford University Press 2009
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
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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
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Series
Exam-style mark scheme
2
Further Pure FP2
Question Solution Marks Number
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
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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
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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
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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.
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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
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Exam-style assessment
Further Pure FP2© Oxford University Press 2009
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
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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.
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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
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Further complex numbers
Exam-style mark scheme
3
Question Solution Marks Number
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
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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
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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)
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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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Exam-style assessment
Further Pure FP2© Oxford University Press 2009
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
Exam-style mark scheme
4
Question Solution Marks Number
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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Exam-style assessment
Further Pure FP2© Oxford University Press 2009
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
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