Partial Fractions
Method 1: Substitution
Although you can substitute any two values of x, the easiest to use are x
x
1, since each makes the value of one bracket zero in the identity.
4
x
x
2
x
A(2
x)
B(1
4
1
4
2 and
x)
2
A(2
6
3B
1
A(2
3
3A
2)
B(1
B
1)
2)
2
B(1
A
1)
1
Substituting these values for A and B gives
4+x
–––––––––––
(1 + x)(2 – x)
1 + ––––
2
––––
1+x 2–x
Method 2: Equating coefficients
In this method, you write the right-hand side of
4
x
A(2
x)
B(1
x)
as a polynomial in x, and then compare the coefficients of the various terms.
4
4
x
2A
Ax
1x
(2A
B)
B
Bx
( A
B)x
Equating the constant terms:
4
2A
B
Equating the coefficients of x:
1
A
B
Solving these simultaneous equations gives A
?
●
These are simultaneous
equations in A and B.
1 and B
2 as before.
In each of these methods the identity ( ) was later replaced by equality ( ).
Why was this done?
Answers
2(2 + x)2 12–((2 1− )x)
4 − x2
x
3
x
1
2
2
2
A 1, B 0, C 1
3
A 1, B 0, C
Cha
4
–
(x – 4) (1x –to
1) 15 as a sum of partial fractions.
Write the expressions in questions
EXERCISE
5
(x − 2)(x + 3)
1
P3
x +5
(x − 1)(x + 2)
7
Answers
Using these values
5
4 Exercise 7C (Page 170)
1
2
1 – 1
(x – 2) (2x 3)
(x1 – ( 11)()3x
x
x
1
4
6
7
1)
8
2 – 1
(x – 1) (x 12)
2
1 x(x1 1)
(x 1) (2x – 1)
Investigation (Page
6 174)
3
The binomial expansion
is x
(x 1)(
1 x 3x 2.
3x 1
1
(2x(2x−2 11)(
) x(x+ 11))
2 –2
(x5– 2(viii)
) x
1 – 3
(x – 1(ix)
) (3x –81) –
1 4 –3
(2x
x2– 1) (2x – 1)2 x
83
2
5(x2 –A4)x 1,5(xB31x)0, C4 1
4)
?
●
It is the
4
4
6 is valid
The expansion
when x ! 1.
1x 2 + 22x − 42
7 (i)
Which method
(1 –isxpreferred
) (1 + 2xis
) a (2 + x)
matter of personalxpreference
for
17 22
+
(ii) 1 − 2x + x
2
(a) and (b) 9
but for (c)2 must
be (iii).
2x − x
(
1
?
●
(
Yes: Us
dy
dx
Integra
to x
2
Chapter
35 A –1, B 0, C
4
x 8+ 5174)
Exercise
7E 2
(Page
(11
2x – 1) x 2x − 1
12
2
2−8
2x2 + x −1 6
x177)
2
2 2–x 1+ 3x − 20
1 (i) ? 4 (Page
20x 18
72x
10 Investigation
4
(Page
174)
–
y=
2
(2x – 3) (x 2)
(x –11) (x 2)
6
−
x
1
2
≡
+
+
1
So
(ii)
4
10x
16x
2
2
2
It
is
the
same
as
4xexpansion
5x 2 + 13x + 10
−xx) 122x 18
2(2 + x) 5 2(216
The8 binomial
is − 3
4 − x2
9 − 25x
11 14
13
4 11x
515
(
)
(
)
(
)
(
)
x
1
2
x
–
1
13
2
x
–
5
13
x
4
(iii)
+2.1)(x − 1)(x − 3) 2 1 4x d(x.233x8x + 3)(x 2 − 4) Activ
(x 1)(x 2)(x 3)
1 (2
x x 3x
7C
Write the expressions in questions
2 1 –to2 15 as a sum of partial fractions.
19
11
2
2
6
–
(x
1
112 The expansion
Exercise 7C (Page 170)
x ! 21. (iv) – 1 – 5x – x
5
(
x
– 2) x (viii)
24(3x – 2) 6 24(3xis valid
2) when
Answer:
2
1
2
4
5
1
(
)
8
16
8
(2x 1) x 13
7 (i)
+
− ? (Page 179)
1
2
2
Which
1x(x– 1)3
3 (1 – x) is (a1 + 2x) (2 + x) 2
(1x 1method
)(x2 4is) preferred
x + 31 )
15
7
3
1(x −12)(–
13
2
3
2 (i)Yes: Using –
8 –
4 of
(3x –(ix)
– 1) Denominators
1)
(xmatter
1) of
(xpersonal
2) (x preference
3)
(x – 2) (x 3)
Type(x 2:
form
(ax
+17for
b)(cx
+
– the
2
(2x d)
(x chain
2) rule
– 1) the
(ii) 1 − 2x + x 2
(2x – 1) (2x – 1)2 x 4
15
3
x
x
+
5
2(iii).
3
2
(a)
and
(b)
but
for
(c)
must
be
du
d
y
d
y
4 2 1
5
6
4
3
2
2
1
8
(ii) 1 2x
= 4x× …
(xx−–1()(x +) 2)
x 21) + 2(x3 − x ) + (2x + 1)
5(x (–24x) − 51
xAx 1+)1,1) B 0, C 14
2()(
1 (x −
dx du dx
x 1
2
x
+
3
Exerc
a
1,
b 2, c 4, for x ! 21
5 x– 23 1A 1, B 0, as
Integrating
both sides with respect
a
sum
of
partial
fractions.
EXAMPLE
7.9
Express
1
2
1
2
2
Chapter
8
x
+
2
2
C
4
9
2
159 Exercise
−
−
–
7E
(Page
174)
7 3 (
8
2
(2x –(2x
1) −x1)(x + 4)
(2 +2xx)2 −(2x − x) (2x + 3)
to x1 x x for x ! 2
(x x –14)() 3x(x –11))
(iii)
1 (i)
x 3x 4
2 4 dy8 du
2
2
1
2 – 1
dy
1 (i) 4 20x
72x
10
–Investigation (Page 174)
4
?
(Page
177)
y7 = 13x 67x 2 dx =
du
2x + 5
(
) 2x( − 1 )
(ii)
(x –71) (x 2)
du
du dx ; 0.505%
10
112x – 32 x 2
12
SOLUTION
(iv)
2
(Page
172)
(ii)2 7D
48 10x
16x
2 4
8
It
is
the
same
as
2x 2 1+ x − 6 1
18
28x +The
3x binomial
−920 expansionExercise
x
−
is
11
(iii)
5
4 x2
3 (i)Activity
2 x x 2 8.1 (Page 181)
x 333
95 11
2
(x 21) (2x – 1)
13(2x – 5) 213(x 4) 2
2+
1 (i) 5(iii)
–
–
x
d
x
.
4
x
25
x
3
−
−
6
x
22
x
18
x
13
x
+
10
.
1
x
3x
2
You 14
need to assume a numerator
(1of
(1 – the
x) partial fraction
–23order
x) 4(1 – 1x18) for
2 – 1with a
13
15
(iv)
21)(x– 2 2)(x 3)
2
(192x +The
1–)(expansion
x 11
− 12)(x −is 3valid
) when (x2x!+1.31)(x52x− 4x)2
6(x
(x
2)5/2 4(x 2)3/2 c
12
(
2
4– –– –2.
2x
5 – x) (1 x3)
(x – 2) x
(iv)2order
) 4, which(ii)is of
denominator
of(3xx 2+
24(3x – 2) 24
(v)
8 ) 16
(
(x 2 8(Page
2 1
179)
1)
(ii) x <
(x 2)3/2 [3(x 2) 10] c
Which
method
is preferred is a 2x – 12 ?
1 – 3
1
2
3
15
7
3
13
Bxrule
+ C is the12most general
1
–1
2 Bx
(i) 1+ CYes:
(vi)
x – Denominators
1) (3x – 1)
1) form
(2
xx +
2)(ax
) preference
3(xpersonal
2A
matter
of
(iii) for
9
(2x2 ––1() (Using
x) 2()the chain
3/2 c
Type (2:
of(xthe
+ 3b)(cx
+ d)
4 (i)
– 4)(x
≡
+
(3x
)
–
1
2
x
x
1
(
x
–
)
15
numerator
of
order
1.) 2)
2
2
(
)
(
1
–
x
3
–
x
3
2
(a)
and
(b)
but
for
(c)
must
be
(iii)
.
d
u
d
y
d
y
2 (i)
3 x++ 4
2)
x − 1 x(ii) 51+ 42x 64x–25x= … ×
(4x −+ 1)(
8
14
5(x – 4)2x 5+(x3 1)
dx du dx
(x − 1) ( 3 − x ) ( 2x + 1)
1
(iv)
1
(ii)
0,
1
2
(ii)
Exercise
8A (Page 181)
8(x a– 2)1, b8(x2, c 4) 4, for x ! 2
as a sum of partial fractions.
7.9
Express 5
2
Integrating both sides with respect
1 − 2 −
1
(x − 1–)(2x 2 + 4)
2 + 4)
9
15
Exercise
7E
(Page
174)
2
Multiplying
both
sides
by
(x
–
1)(x
gives
2
(2x – 1) x
(iii)
5 – 2x
4x 8x1(x 3 1)8 c
(2 + x) (2 − x) (2x + 3)
x 2for x ! 2
(v) (iii) 1 x to x
(iii)1 (i)
3
38
(2x 22 – 34) (8x 2)
2
SOLUTION
2
1
d
y
d
u
d
y
1
(i)
4
20x
72x
10
–
3 (i)
2 + 4) + 7(Bx
y =67
(2x – 3) (x 2)
x 2d;u0.505%
13x+
1
C)(x
–dx1)dx = 5du(i) du(ii)2 61+(x22x +1)426 c
2x4+10x
3172)16x
A(x
Can be taken
further
using
(iv)
2
Exercise
7D
(Page
(ii)
(2 − x) (1 + x )
8
You need to
with2a 4
(ii)
1
8 assume 9a numerator of order 1 for the partial2 fraction surds.
(iii) 5(x 3 2)5 c
11
2
2
3
(i)
2
x
x
5
11
33
x
x
9
3
2
Activity
8.1
(Page
181)
5
15 2 15 3
) x13(+
)
denominator
order
2.
13(2x – 5of
x 4,4which
(iii)
(ii)
1 is
(i)of
–
–
5
+
x
−
x
−
x
4 (i)
x (11– 3x) (12– x) 45(1 – x5A
2 1 4
8
8)2
(vi) 2 – 1 2–A 3 11
(iv) 6(2x 2 5)3/2 c
x 2 (2–x 2 1)
192x +–3 11
x
4
Bx
+
C
is
the
most
general
5/2
3/2
2
A
Bx
+
C
2
−
1
x
(2 – x) (51(x x)2)
12
1 5x x
3(x 2) 6 (i)c
+(ii) 2 4 (iv)
5 (i)
1+ 2
24(3x – 2) 224(3x ≡2)
of order 1.10x
– –228x – 16 –numerator
+ x)15
(2equating
(v)
(2x
(x + 1)
1coefficients.
)3/2(3x 1) c
8 B and
x − 1other
x +(2two
(x − 1)(x + 4) The
are most
found
by
– 3 2 easily
x4 – 1) unknowns,
(x 1)
(vii) C,
3/2
(ii) 2 x < 1
(x
2)
[3(x
2)
10]
c
(
)
1
2
3
3x – 1 x 15
1
2 − 9 x 31/2
13
1 2 (i)
1 2rewritten
x + 54 x2(x
(ii) (vi)
–1 3
(x 1) both
(x sides
2) (xby
3(x
) – 1)(x
2
8 9) (x 18) c
– be
(iii) 21+ may
1 – 92
Identity
as
3
(
)
(
x
2
)
Multiplying
4)
gives
x
–
2
1
3/2
2
4
(i)
Can
be
taken
further
using
(
)
(
)
(3x
4)(x
2)
c
x –1
x 2
(x – 1)
15
(
)
(
)
1– x
3–x
324 2
2 (i) 222 000
4 + 3 +
2
(ii)
surds.
16 – 2x
14
5 +(ii)
5x1) 4x …
1
2
1
4) + (Bx
C)(x
–
(x − 1) ( 32x
− x+) 3 ( 2xA(x
+ 12) +(iv)
!
1
+
(–B
+
C)x
+
(4A
–
C)
2x
+
3
(A
+
B)x
(ii)
0,
1
2
(ii)
586
(
)
8 x – 2 8(ax 1,4b) 2, c 4, for x !
2 Exercise 8A (Page 181)
2
1 − 2 −
1
15
2
x
1
5
5A
A
1
(iii) 18.1
5
–
2
x
2
4
x
8x 2
1 x x for2 x ! 2 (iii)
(2 + x) (2 − x) (2x + 3) (v)
1
(iii)
0 A 3+ B31 (i) 8(x 3B 1)8–1c
Equating(2coefficients
x 2 – 3) (2x 42) of
8 x :
1
3 (i) 222
The other two unknowns, B and C, are
most
easily13
found
coefficients.
2 + (ii)
2x + 461(x 2 1)6 c
67by
x 2 ; equating
x using
Can be (iv)
taken7further
5
(i)
0.505%
Exercise
7D
172)
1
(2 −–xC
) (1 + x 2) C
1 may
8
4
Identity
be (Page
rewritten
as
!
(ii) 19
Equating
constant2 terms:
3 4A
1
surds.
1
(iii) 5(x 3 2)5 c
2
3 (i) 2 x x
9 – 3 – 2
(ii) 5 + 5 x − 15 x 2 − 15 x 3
1 (i)
4 (i) A( 1, 0), x " 1
2 2
2
4
(4A – 1C)
18
– 12+– C)x3 +
(1 – 32x
1 –B)x
x(vi)
)2 + (–B
x) +(13– x)(A(+
2
(iv) 6(2x 2 5)3/2 c
This givesx x (2x(2 –1)x) – (1 x)
2 + x −1
(1 x)4
6 (i)
4 – 2x 2
c
5 (i) (a)
1
2
3/2
(ii) coefficients
:
0
A
+
B
B
–1
Equating
of
x
10
x
3
2
+
(
x
)
4
((v)
x + 115)(2x 1) (3x 1) c
– x <1
(vii)
(2x – 1) (x 2 1)
(ii)
2
(3x 2–x1)+ x3
2
1 + 1(ii)−1 xx+ 5 x 2 − 9 x 32
(b) 25
1/2
1 –terms:
1
13 4A – C
≡
(vi)
2
4
8 3(x 9) (x 18) c
Equating
C
1
(iii) constant
1
9
2
2
4)(
(i)
taken
x (1+further
4)– using
x −1 x + 4
− 1be
(x – 1) (x 2)(xCan
(x – 1)2 324
1
– x) (3 – x)
2 (i) 222 000
(ii) 3(2 2 1) 0.609
surds.
5
6
5
This gives
–
x
1
(iv)
(ii) 0, 1
(ii) 586
8(x – 2) 8(x 2 4)
2
1
1
2
2
1
2
+
≡ 2 3 +(x –2 4)7–≡(x – 1)1
2 + x 2 2− x 2(2 + x) 2(2 − x)
(2 + x)(2 − x) 10
9
●
P3
Answers
●
(
●
)
( )
●
(
!
(v)
(x
2x + 3
≡2 1 + 1 − x
5 – 2x
− 1)(2 x 2 + 4() x) − 1 x 2 + 4
(2x – 3)
x
2
Can be taken further using
surds.
(iii)
5 (i)
4x
3
8x 2
3
2 + 2x + 4
(2 − x) (1 + x 2)
(iii)
3 (i)
(ii)
18.1
2221
1
19
)
( )
!
Further algebra
E 7.10
The factor (cx + d)2 is of order 2, so it would have an order 1 numerator in the
partial fractions. However, in the case
factor there is a simpler form.
5xof2 a−repeated
3
EXAMPLE 7.11
Express 2
as a sum of partial fractions.
4x 5 x x + 1
Consider
(2x 1)2
(
This can be written as
)
P3
7
Partial fractions
3
7
Type 3: Denominators of the form (ax + b)(cx + d)2
2(2x 1) 3
SOLUTION
1)2 2
5x − 3 ≡
Let
)+
3
2
≡ 2(2x + 12x
(2x + 1) (x2x++11)2
(2x
(
)
A+ B + C
x x2 x + 1
2
3both sides by x
≡Multiplying
+
(2x + 1) (2x + 1)2
5x 2 – 3
Note
2(x
+ 1) gives
Ax(x + 1) + B(x + 1) + Cx 2
In this form, both the numerators are
x constant.
0
–3 B
x –1 px q +2 C
In a similar way, any fraction of the form
can be written as
(cx d )2
Equating coefficients of x 2: +5 A + C
A
(cx
B
d ) (cx
A
3
d )2
This gives
When expressing an algebraic fraction in partial fractions, you are aiming to find
5x 2so−you
3 would
3 +the form
2 where the
the simplest partial fractions possible,
≡ 3 −want
2
2
x +1
x (x + 1) x x
numerators are constant.
x +1
Express SE
as a sum of partial fractions.
EXERCI
(x − 1)(x − 2)2 1 Express each of the
(i)
SOLUTION
Let
following fractions as a sum of partial fractions.
4
(1 3x)(1 x)2
(ii)
(x −
Multiplying both sides by (x – 1)(x – 2)2 gives
+ 4)
x +3
x + 1 A(x(vii)
– 2)2 + B(x2– 1)(x – 2) + C(x – 1)(viii)
2
x(3x − 1)
x
x
1 (so that x – 1 0)
2 A(–1)2
2 (so that x – 22 0)
3 C
Given that
Equating coefficients of x 2:
This gives
5 − 2x
(x − 1)2(x + 2)
2x + x + 4
(2x − 3)(x + 2)
(vi)
x2 − 1
x 2(2x + 1)
2x 2 x 2
(2x 2 1)(x 1)
(ix)
4x 2 3
x(2x 1)2
Notice that you only
2
need (x – 2)2 here
(v)
and not (x – 22)3.
x +1
≡ A + B 2x++ 1C 2
(x − 1)(x − 2)2 (x −(iv)
1) (x − 2) (x − 2)
2)(x 2
(iii)
4 + 2x
(2x − 1)(x 2 + 1)
A
2
x0 2 +A 2+xB + 7 B≡ –2 A
+ Bx + C
(2x + 3)(x 2 + 4) (2x + 3) (x 2 + 4)
x +1
2 +values
3 of the constants A, B and C.
≡ 2 find
− the
(x − 1)(x − 2)2 x − 1 x − 2 (x − 2)2
3
[MEI, part]
Calculate the values of the constants A, B and C for which
x 2 − 4x + 23 ≡ A + Bx + C
(x − 5)(x 2 + 3) (x − 5) (x 2 + 3)
171
[MEI, part]
15
(x − 1) ( 3 − x ) ( 2x + 1)
1 − 2 −
1
(2 + x) (2 − x) (2x + 3)
Answer:
Exercise 7D (Page 172)
(ii)
(iii)
(iv)
(v)
5 – 2x
2
(2x 2 – 3) (x 2)
Can be taken further using
surds.
Write as
(vii)
2– 1 – 3
x x 2 (2x 1)
10x – 3
a(3xsum
2 – 1) of partial
x
(a)
7x + 1
9x 2 - 1
4
5
6
fractions:
Can be taken further using
surds.
24
4x
8x 2
1)8 c
(iii)
1 3
3
8(x 3
1 2 2 6
+ c2x + 4
(ii)5 (i)
6(x (2 −1)
x) (1 + x 2)
1 (i)
x 2 + 25
(b)
( x + 7) 3
6 x + 13
2x +
10 (c)
– 1
(2x –x32) +(x5 x 2+
) 6
2
11
12
13
14
15
(d )
8
9
13(2x – 5) 13(x 4)
19
11
–
24(3x – 2) 24(3x 2)
1
(x
2
1) (x
(iv)
3
2) (x
3)
4 + 3 +
2
(x − 1) ( 3 − x ) ( 2x + 1)
1 − 2 −
1
(2 + x) (2 − x) (2x + 3)
Exercise 7D (Page 172)
1 (i)
(ii)
(iii)
(iv)
(v)
9 – 3 – 2
(1 – 3x) (1 – x) (1 – x)2
4 – 2x
(2x – 1) (x 2 1)
1 – 1
1
(x – 1)2 (x – 1) (x 2)
5
8(x – 2)
6 – 5x
8(x 2 4)
5 – 2x
2
(2x 2 – 3) (x 2)
Can be taken further using
surds.
(vi)
(vii)
324
54x +20x4 72x 2
(2 x +(ii)3)(4x 210x
+ x 16x
+ 12)
1 (i)
(iii)
3
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x x 2 (2x 1)
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Can be taken further using
surds.
2 (i)
(ii)
5 11x 33x 2
2 4
8
2
5
x
x
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– – –
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2
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4x 2
1 2x
(iv)
3 (i)
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4 (i)
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5 (i)
(ii)
6 (i)
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1 x
for x ! 2
2 4 8
7 13x 67x 2 ; 0.505%
2 4
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2 x x2
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x <1
1 – 9
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0, 121
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3
586
(iii)
18.1
2221
1
(ii)
19
(i)
A( 1, 0), x1" 1 2
7 (i)
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c 17
(a)
(ii) 41 − 2x + x 2
2
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(b) 25
(i)
2 1) 0.609
Chapter
8
1
(ii) 3(2
?
●
(Page 177)
It is the same as
4
8x 2
3
2 + 2x + 4
(2 − x) (1 + x 2)
x dx.
1
?
●
(Page 179)
Yes: Using the chain rule
dy dy du
=
×
dx du dx
Integrating both sides wit
to x
dy du
y=
dx =
du dx
(
)
(
Activity 8.1 (Page
2
(x
5
4
3(x
2)5/2
2
(x
15
2
15(3x
…
a 1, b 2, c 4, for x ! 21
(iii)
(ii)
3 (i)
2
further using
7 13Can
67xtaken
x be
; 0.505%
(iv)
2 4surds. 8
1 3
5 c
1
1
2)
Exercise
7C (Page 170)(iii) (ii)
5
5(x 5 +(viii)
(i) 2 x
x2 1
x
− (15
x 2 − 15 x 3
4
3
2
2
42x 2 18) (x 1)
(vi)
– 2–
1
2 x– x11 (2x 11)
(iv) 6(2x 2 5)3/2 c
2 + x 8− 1
6 (i)
3 5
4
(2 – x1) (x(1– 2x))– (x 3)
–
10x – 3
x) (x 2 + 1)) –
1 (2 +(ix)
(vii)
(v)
(2x
1)3/2((3x
2x – 11) (c2x – 1)2 x
15
2
x – 1)1 x
(ii) x <21(31
1
5 2 9 3
–
+ 1/2
x1, − 8Bx 0, C 1
(ii)
2 22x A
4 (x
(
)
x
(vi)
(x
9)
18) c
x
1
1
9
Can
3
(i)
– be taken further using
(1 – x3)surds.
(32– x)– 2
3
A
1,
B
0, C
4
2 (i) 222 000
(x – 4) (x – 1)
1
(ii) 0, 1
(ii) 586
2
2 – 1
4
Investigation (Page 174)
(
)
(
)
2
x
–
1
x
2
(iii) 18.1
4
x
8
x
(iii)
The binomial expansion is
1
3 53 1
1
3 (i) 222
(x2x 1+) 4 (2x – 1)
2
1 x 3x 2.
+
(i)
1
(2 − x) (12+ x 2) 2
(ii) 19
–
6
The expansion is valid when x ! 21.
(x – 2) x
(ii) 5 + 5 x − 15 x 2 − 15 x 3
4 (i) A( 1, 0), x " 1
2
41
8
Which method is preferred is a
3
7
–
2 +(x x– −
(1 x)4 of personal preference for
1)1 (3x – 1)
matter
(i)
c
5 (i) (a)
(2 + x) (x 32 + 1)
4
2
(a) and (b) but for (c) must be (iii).
8
2
1
5 ( 9 3) 5(x 1)
(b) 25
(ii) x + x 25 −x –x4
2
4
8
5 –2
9
1
7E (Page 174)
(ii) 3(2 Exercise
2 1) 0.609
(2x – 1) x
P3
9 – 3 – 2
(1 – 3x) (1 – x) (1 – x)2
4 – 2x
(2x – 1) (x 2 1)
1 – 1
1
(x – 1)2 (x – 1) (x 2) 324
5
6 – 5x
8(x – 2) 8(x 2 4)
(vi)
3
0, 1
Exercise 8A2 (Page 181)
Answers
1 (i)
(ii)
( – 2) 8(x 2 4) 1
a 1, b8 x 2,
c 4, for x ! 2
5
–
2
x
2
2
1(v) x x 2 for x ! 2
(iii)
(x 2)
(
)
–
2
x
3
4
8
2
2)3/
2)3/2 [3(x 2)
4)(x 2)3/2
Exercise 8A (Page
1 3
8(x
1 2
6(x
1)8 c
(iii)
1 3
5(x
2)5 c
(iv)
1
2
6(2x
1
15(2x
1 (i)
(ii)
(v)
(vi)
2 (i)
2
(x
3
586
(iii)
18.1
(ii)
5)3/2 c
1)3/2(3x
9)1/2(x
1
222 000
(ii)
3 (i)
1)6 c
2221
1
19
5 + 25 x − 15
x 2 − 15
x3
4
8
4 (i)
A( 1, 0), x " 1
2 + x −1
(2 + x) (x 2 + 1)
5 (i)
(a)
(1 x)4
4
(b)
225
1
(ii) 3(2
2
1
x
2
+ 54 x 2 − 89 x 3
c
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the term in x .
x dx
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Calculate
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Show that the number of organisms
M
=
R
−
0.57
,
1
(ii) Using the substitution x = !!3" tan
, find the
dx"the
x exact value
87
(i)
By
sketching
each
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the
graphs
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x
and
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−
x
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equation
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The parametric
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curve are variables.
3
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2 line" l. 4 ! where
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y toofbea continuous
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Show
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1
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,
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x
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10 allThe
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A andfind
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square
of
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x vectors
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e−t cos
t!3
,u.+Give
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xexpression
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coordinates
of
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.
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differential
equation
and
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1
=
i
+
j
+
2k
+
&
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+
j
−
k#.
has
equation
r
ib, interval
where the
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yourexactly
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the form
has
two
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0 <real
x <numbers
!.
[3]
11
xin
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7
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.
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expressing
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Show
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does not
intersect
passing
through
A
(ii) On an Argand diagram,
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complex
numbers
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Show
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!
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x
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.
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7
(a) The
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number
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[4]
relation
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! sinlocus.
x
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Express
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+ 4i
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Find
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the
plane
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the
line l and
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first
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express
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i
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9709/33/M/J/15
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x
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x
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by
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2
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3
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3
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#
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4
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1
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.
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The
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6
(ii) Calculate the maximum value of arg "(ii)
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lyingafter
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shadedthe
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Express
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1
d
R
8
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to
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4
decimal
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(iii)
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the curve
and the axes.
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,
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1
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A
Bx +variables.
C (ii) When
where(b)
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x
are
taken
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be
continuous
x
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B answer
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. (i)
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and
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giving
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to
3 significant
figures.
Express
in
the
form
+
+
1
2
9
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Solve the differential
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89
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and including
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93
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Give
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9709/31/M/J/14
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the
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In theThe
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A isnumber
a point
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of
a circle wit
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6
.
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Given
also
that
x
$3
2x − 7x − 1
10
e
10
The
line
l
has
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=
4i
−
9j
+
9k
+
%
!−2i
+
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−
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centre
A
meets
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circumference
at
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and
C
.
The
7
Let f!inx"the
= form sin 23y# = 3..
conjugate of w. Find w, giving your answer in
[1]the form x
(ii) Using
the
substitution
value
of
!x −
2"!x + 3"4x = !!3" tan ", find theisexact
bounded by the circumference of the circle and the arc with
(iii) Show that the number of organisms never the
reaches
48,that
however
largeof
t becomes.
[2]
(i) shaded
Show
theof
the
perpendicular
A to l is
region
islength
equal
to
halfdiagram,
the area shade
of thefrom
circle.
3
(b)
On a sketch
an Argand
the
region
wh1
(iii)
equation
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(i) Hence
Expresssolve
f!x" the
in partial
[5]
1
# 3#
which satisfy both the inequalities $ ' − 2i$ ≤ 2 and 0 ≤
d='x,0,
1 $3
2 The
x −!3
x+
(ii)
line l lies in the 2plane
sin 2$with
− # equation ax + by − 3& + 1
+
x
"
6
−Show
−→
−in
−→including
value
of
'of
$ xfor
points
this region,
givingin
your
(i) the
that$of
cos
2and
$ to
= band
.
values
a
.
(ii) points
Hence A
obtain
the
expansion
ofvectors
f!x"1 in given
ascending
powers
,
up
x2 . answer
OA
=
2i
−
j
+
3k
and
OB
=
5k.term
The line
10 The
and
B
have
position
by
4$ i + j + the
3
giving your answers correct to 3 significant Tfigures.
[4]l
[5]
single
[4]
= i +answer
j + 2k +as&a"3i
+ j −logarithm.
k#.
hasexpressing
equation ryour
(ii) Use the xiterative formula
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2014points A and B have position vectors 9709/32/O/N/14
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through
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and B.
!
x − 8O
x+9
2
equation
x permitted.
+ −y =
5. sin x in the form R1 cos!−1x +2&sin
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Let
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= expanding
.
Throughout
this question
the
use
of
a
calculator
is
not
7 98 (i)
By
cos!
x
+
45$",
express
cos!
x
+
45$"
!!2"
",
$ = 2 cos
!1>−0xand
"!2 −0$x"<2 & < 90$. Give the value of R correct to 4 significant figures
4$n
the value
T4 the point A. Given+1
(ii)where
FindRthe
equation
of the plane
containing
the line 1l and
yourand
answer
in the
2 sin
x
position
vector ofpoint
the point
of th
The
shows
the curve
yand
= ev satisfy
cos the
x forequations
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x ≤ 2the
%, and
its maximum
M . of intersection
(a) ofdiagram
The
complex
numbers
u
&
correct
to
2
decimal
places.
[5]
+xby
c$ = d.fractions.
[6]
form axf!
(i) Express
" in+ partial
[5]
with initial value $1 = 1, to determine $ correct to 2 deci
T2
u
+
2
v
=
2i
and
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v =the
3.
=
sin
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,
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exact
value
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shaded region bounded by
(i)
Using
the
substitution
u
iteration
to area
4 decimal
(ii) Hence solve the equation
(ii) Hence
obtain
of f!x"1in ascending
powers
ofplane
x, upqtohas
andanincluding
in [5]
x2x. + by + c
(ii)
A
second
equationthe
of term
the form
the curve
and the
the expansion
axes.
−2x
The diagram
the curve
ycos!
= 10e
sinboth
for
xThe
≥x 0.
The
labelled
T2 , angle be
Solve theshows
equations
for u and
vx, giving
answers
in
theqstationary
form
x + ithe
ypoints
, where
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y are
1 , [5]
contains
lineare
AB
, and
theTreal.
acute
+ 45$"
−4x!!2"
sin
=plane
2,
T(ii)
,
…
as
shown.
[5]
3 Find the x-coordinate of M , giving your
answer
correct
to 3 decimal
places.
[6]
© UCLES
2013the
9709/31/O/N/13
equation
of q.
for
0$
<
x
<
360$.
[4]
© UCLES 2015
9709/32/M/J/15
(i) line
Find
the
x-coordinates
T1 +and
T+2%locus
, !−2i
giving
x-coordinate
to 3"decimal
[6]
10 The
lan
has
equation
r = 4iof
−sketch
9j
9k
+
jeach
− 2k".
Thecomplex
point Acorrect
has
position
vector places.
3i #+" 8j
+=
5k.
(b)
On
Argand
diagram,
the
representing
numbers
satisfying
+ i#
© UCLES 2014
9709/33/M/J/14
[Turn
over1
and the locus representing complex numbers w satisfying arg!w − 2" = 34 #. Find the least value
(i) Show
the
length
of the perpendicular
from
to l25.
is 15.
1 points
Aof TBn is greater
C A
(ii)
It is
that
the x-coordinate
than
Find the least possible value of n. [5]
[4]
of
# " given
−that
w # for
[5]
8
(i) Express
in on
thethese
formloci.
+ +
.
[4]
2
2
x 2x + 1
x !2x + 1"
x
(ii) The line l lies in the plane with equation ax + by − 3& + 1 = 0, where a and b are constants. Find
© UCLES 2013
9709/32/M/J/13
thevariables
values ofxaand
andybsatisfy
.
[5]
the differential equation
9 (ii) The
C
dy
y = x29709/32/O/N/14
!2x + 1" ,
© UCLES 2014
dx
a
Past Paper Questions
1.
2.
3.
4.
5.
6.
7.
9709/31/M/J/14
y = 1 when x = 1. Solve the differential
equation and find the exact value of y when x = 2.
Give your value of y in a form not involving logarithms.
[7]
© UCLES 2014
and
for y in terms
.+ ax arg"
$in−denoted
2i#
= point
%by pand
−given
3$that
= $ $that
− 3t
3 9 The
x + of
3x tof
3 is
(x1,
). 2
It$show
By
first
expressing
partial
6 fractions,
(ii)
State
the gradient
the +curve
at
the
(−
)$isand
sketch
2 polynomial
73
2x2 + 5x + 2
8
Two lines have equations
9
The
percentage
mass
of B remaining at time t is
intersect
at the
. Express
(i)(ii)Find
the
value
of point
aof
. ythePinitial
4 the complex number repres
2
1
Find the quotient and remainder when 2x2 is divided
by
x
+
2.
[3]
4
x
x+3
approached
by
p
as
t
becomes
large.
3
the exact
value of+r5correct
5 dxto=38significant
5
1
p value of2& and the #
− lnthe
9. co
7
(a) The complex number u is defined2by
, where
2 u=
xroots
+ 5xofa+the
2 2iequation
4 "this
5 ", find the real
r = ! 1 " + s!−1 " (ii)
and When
r = !a has
+ t!value,
p(x) =
+
0
6
The equation of a curve is 3x2−4
− 4xy + y23= 45.
−2
4 form x + iy, where x and y are real.
1
+
3
x
(i) Express
u in−the
8. 2 Expand !
given
tan43x =
x, where
is a constant
in ascending R
powers 6
of x It
upisto
and that
including
thek tan
term
in x2 , ksimplifying
theand tan x ≠ 0
8 intersect.
(a) Show that ' 4x ln x dx = 56 ln 2 − 12. *
"1
+
2
x
#
where
p
is
a
constant.
It
is
given
that
the
lines
3
(i) Find the gradient of the curve at the 4pointThe
(2, variables
−3(ii)
). Find
[4]
the2value
a forby
which
arg(u ) = 4equation
π[4]
, where u* den
x and
θ are of
related
the
differential
2
coefficients.
10
(i)
given
that 2 tantan
2x(√
+2
5
tan
x
=
0.
Denoting
tan x by t, for
(i) It
Byisfirst
expanding
x
+
x
)
,
show
that
x
3
2
(i)
Find
the
value
of
p
and
determine
the
coordinates
of
the
point
of
intersection.
[5]
d
x
1%
O
that
either
t
=
0
or
t
=
(
t
+
0.8
)
.
p at which the gradient is 1.
(ii) Show that there are no points on the curve
24
sin 2(θ3k − =
(tan
x +21x[3]
) =cos
2θ3.,
1
)
k
−
4x to find
exact
value
of ' po
(b) On
Use
the substitution
u = sin
dshade
θ thethe
(b)
a sketch
of an Argand
diagram,
region
whose
2
7xthe
− 3equation
x+2
0
(ii)
Find
of
the
plane
containing
the
two
lines,
giving
your
answer
in
the
form
√
(ii)
It
is
given
that
there
is
exactly
one
real
value
of
t
satisfying
which
satisfy
both
the
inequalities
|"|
<
2
and
|"|
<
|"
−
2equa
− 2i
3 The
Express
in of
partial
fractions.
[5]
1 π . When
1 π, x =
diagram
shows
part
the
curve
y
=
cos
(
x
)
for
x
≥
0,
where
x
is
in
radians.
The
shaded
region
2
where
0
<
θ
<
θ
=
0.
Solve
the
differential
9.
2
x"x+x cand
+"1#
ax + by
=ydare
, where
a, bby
, cthe
anddifferential
d are2integers.
[5]
by
calculation
thatequation
this 12
valuetan
lies3xbetween
1.2when
and 1.3.
7
The variables
related
equation
(ii)
Hence
the
k = 4, g
terms
of θsolve
, and◦simplifying
as
farx
as possible.
between the curve, the axes and the line x =
px,in
where
by R.your
The answer
area =ofk9709/31/M/J/13
Rtan
is
equal
© UCLES
2013
◦ p > 0, is denoted
0
<
x
<
180
.
√
to 1.
dy (iii)
6x e3xUse the iterative
tn+1 = 3 (tn + 0.8) to find the value of
5x − formula
x2
=
.
2
8
(i)
Express
in
partial fractions.
(i) Solve the equation
[3]
2 the result of each iteration
9 − 7x + 8$4
xx − 1$ = $ x − 3$.pd2x
2 tanto
5=decimal
places.
10. 49
y
(iii)
Show
that
the
equation
33xgraphs,
tanshow
no [5]
root
in the inte
(
1
+
x
)(
2
+
x
)
(i) Express
in
partial
fractions.
√
5
(i) By sketching a suitable
pair of
the equation
− 2kcos
px has that
2
2
O
M
(i) Use the substitution
(
x
)
d
x
.
Hence
show
that
sin
p
=
.
[6]
(3 − x)(1 + x =) u % toy+1find #% cos
y
It(ii)
is given
that
y = the
2 when
x = 0.4 Solve
differential
equation
andthird-party
hence
findmaterial
theprevious
value
when
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Hence
solve
equation
− 1 the
$4
−(iv)
3$ tocorrect
to
3values
significant
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included
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reproduce
items
where
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0= Permission
2is
1of
Using
the
of t found
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question,
5holders,
x x− =
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publisher
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copyright
but
items
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2[8]
x = 0.5, giving your answer correct to 2 decimal
places.
(ii)
in
ascending
po
2 Hence obtain the expansion of
2
9 − 7xbe
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82xcos
to make
amends at the earliest possible opportunity.
2 (1 + x)(2 + x )
2
pn ascending
The
diagram
shows
the
curve
y
=
x
ln
x
and
its
minimum
point
M
.
(ii)
Hence
obtain
the
expansion
of
in
powers
of
x
,
up
to
and
including
the
−1 3 −
2
tan
2
x
+
5
tan
x
=
0
1
3ininitial
xwith
is
radians,
hasp1a =root
in
theCambridge
interval
0 of
< x <Group.
π . Camb
(ii) Use the iterative formula pn+1 = (sin
",in
value
the value
University
Cambridge
Examinations
is 1,
partto
of find
the
Assessment
3 − x!)(
1 +2xp2of)where
term
xInternational
. where
2
5
For each of the3 following curves, find the gradient
at
the
point
the
curve
crosses
the
y
-axis:
n Local Examinations Syndicate (UCLES), which is itself a department of the University of Camb
in x to. 2 decimal places. Give theCambridge
[5]
−
π iteration
≤exact
x ≤ πvalues
. to 4 decimal
2
(i) Verify
Find
the
of1the
coordinates
of M . [3]
pterm
correct
result
offor
each
places.
(ii)
by
calculation
that
this
root
lies
between
1
and 1.4.
©
UCLES
2012
9709/33/O/N/12
62 line l has equation r = ! 3 " + λ ! 1 ".
68
8
The point P
the
1 +has
x2 coordinates (−1, 4, 11) and
9
Two planes have equations−4x + 2y − 23" = 7 and 2x + [3]
y + 3" = 5.
(i) y =
;
2x
(ii) Show
Find2 the
exact
of satisfies
the area the
of the
shaded region bounded
2value
1
+
e
(iii)
that
this
root
also
equation
2
10 (a) Without
solve the equation
iw x==(e.
2 − 2i) .
[3]
4xusing
− 7xa−calculator,
1
3
Calculate
the acute angle between the planes. [4]
(x) =the perpendicular
. distance from P to3 l.(i) line
11. 8 Let
(i) fFind
32x − 3)
2 [4]
++
1)(
(ii) 2x3 + (5xxy
yan
= 8.
xrepresenting
= cos−1 !
−−→
(b)
(i)curve
Sketch
Argand
diagram atshowing
the
region
R consisting
of. At
points
the".
6
A
certain
is
such
that
its
gradient
a
point
(
x
,
y
)
is
proportional
to
xy
the
point
(
1,
2
)
the
6−
7
With respect to the origin O, the position vectors (ii)
of two
points
A and
B are given
byline
OAof
= iintersection
+ 2j +
2kx2 of the planes
Find
a vector
equation
for the
complex
" where
(ii)
Find
the
of fractions.
the
plane which
contains
P and l, working,
giving
your
answer
in the roots
form
(i)
Express
f (equation
x) innumbers
partial
[5]of the compl
−−→
−−→
−−
→ the
gradient
is 4.
10 through
(a)
Showing
find
two square
and
OB
i ++4jc."The
P alies
the dline
A and Byour
, and AP = λ AB
.
ax=+3by
= d ,point
where
, b,on
c and
are
integers.
[5]
©
UCLES
2012
9709/33/M/J/12
|" −
4(iv)
− 4iUse
|to≤ the
2.anorigin
[2]in part (iii) t
answers
in theOform
x +byiybased
, whereonx and
are 2exact.
6
The points P and6Q have position vectors,
relative
, given
iterative
formula
the yequation
x −1
49
(i)
By
setting
up
and
solving
a
differential
equation,
show
that
the
equation
of
the
curve
is
y
=
2e
. to 4 decimal
−
−
→
10
places. Give the
# =f ((x1)+dx2λ
(ii)Show
Showthat
thatOP
=−→
[5]
y result of each iteration
−
−.−→
(i)
)8i +− (ln
2 #+32$λ. )j + (2 − 2λ 2)kdecimal
[2]
[7]
(ii) For the2 complex
numbers
byOn
points
the
is given that
(b)
a sketch
of
shade the region whose poin
OP
= 7i + represented
7j − 5k 69
and
OQ
= in
−5i
+ region
jan+ Argand
k. R, itdiagram,
4x2 + 5x + 3
% show
whichthat
satisfy the inequality |% − 3i | ≤ 2.
Find the greatest valu
12. 9 (ii)
© UCLES 2012
9709/31/M/J/12
By first
expressing
in
partial
fractions,
equating
cos
in
,curve.
find
the
value
of λ for
which
2 the
|"|
qandand
α) ≤and
argterms
" ≤ β of
. the
TheBy
mid-point
ofexpressions
PQ
is
A≤at
. AOP
The
plane
#
isBOP
perpendicular
to3λthe
line
PQ
and passes
through
A.
(ii)
State
the gradient
curve
the≤point
(−
1,
2
sketch
[2]
2xof
+the
5xpoint
+for
2 pcos
6
(i)
Express
cos
x
+
sin
x
in
the
form
R
cos
(
x
)−, M
where
R > 0 an
bisects
angleequations
AOB. r = 3i − 2j +2k + λ (−i + 2j + k) and r = 4i + 4j + 2k + µ (ai +−bαj [5]
9
TheOP
lines
l andthe
m have
k
)
4
valueanswers
of R andcorrect
the value
ofsignificant
α correct figures.
to 2 decimal
Find
the values
,q
, α and
βanswer
,5giving
[6]places.
respectively,
a and
constants.
4x2 +
x + 3 inyour
(i) Find 16
thewhere
equation
of bof
#are
,pgiving
your
the form ax +
by + cto
$ =3 d
.
[4]
P
#
2
d
x
=
8
−
ln
9.
[
10
5to
When λ has2 this
value, verify
that2AP
=2
OAand
: OB
.
Expand
in ascending
powers
of
,x up
including
the term in x , simplifying [1]
the] ◦
13.71 (iii)
2 :xPB
x
+
5
+
(a) The (complex
number u is defined
by
u
=
,
where
the
constant
a
is
real.
(ii)
Hence
solve
the
equation
cos
2
θ
+
3
sin
2
θ
=
2,
for 0 < θ < 90◦
2
+
x
)
0
(i) Given
that l and
mthrough
intersect,
show
that
a +x2i
(ii)
The
straight
line
P
parallel
to
the
-axis
meets
#
at
the
point
B
.
Find
the
distance
AB
coefficients.
[4] ,
correct
to
3
significant
figures.
(i) Express u in the
form x + iy, where
are real.
[2]
© UCLES
2011
9709/32/O/N/11
2ax −and
b
=y4.
[[5]
4]
2
12
+
8
x
−
x
14.8 Let f (x(ii)
)=
. for which
3
Ocomplex
u*) = 34 π , tan
where
u.
2 of
10
(i) It isFind
given
that
+ 5 tan22xx =arg
0. (Denoting
x byu*t, denotes
form anthe
equation
in tconjugate
and henceofshow
(2 − the
x)(
4value
+2xtan
) 2xa√
e
3
(ii)equation
Given
also
l calculator,
and
findthe
thegradient
values ofofa the
andcurve
b. at the point for which
[4]
[3]
2 7 The
ofusing
y =(are
Show
that
that
either
tathat
=curve
0 aor
tis= m
t + perpendicular,
0.8
). . the
[4]
(a)
Without
solve
equation
1 + e2x
2
−
x
A
Bx + C The diagram shows the curve y = x √
e .
9
x(i)
= lnOn
3 is
.f (and
[4]
x)that
inban
the
form
+one
.value
[4]
3
(b)
a sketch
of
Argand
diagram,
shade
the
region
whoseof
points
represent
complex
numbers
".
*=
(iii)
have
these
values,
find
the
position
vector
the
point
of intersection
of)l. and
m
50 a
3
w
+
2i
w
17
+
8i,
2
(ii)Express
ItWhen
is
given
there
is exactly
real
of
t
satisfying
the
equation
t
=
(t + 0.8
Verify
2−x
4
+
x
which
satisfy
both
the
inequalities
|"|
<
2
and
|"|
<
|"
−
2
−
2i
|
.
[4]
[2]
by calculation
that this value lies between
1.3.that the area of the shaded region
[2] by the curv
(i) and
Show
bounded
© UCLES1.2
2011
where w* denotes
the complex conjugate
of
w. Give your answer
in the form a + b9709/32/O/N/11
i.
[4]
1
17
◦
◦
√
25
equal
to
2
−
.
3
(i)
Express
8
cos
θ
+
15
sin
θ
in
the
form
R
cos
(
θ
−
α
)
,
where
R
>
0
and
0
<
α
<
90
.
Give
the
value
# f (x) formula
(ii)
thatiterative
dx = ln! t2 ". = 3 (t + 0.8) to find the value of
[5]
(iii)Show
Use the
e3t correct to 3 decimal places. Give
n+1loci
n
05
x2
(b) ofThe
Inα an
Argand
diagram,
10 (a)
complex
uthe
and
w satisfy
the equations
correct
tox numbers
2−
decimal
places.
[3]
(i) Express
in partial
fractions.
[5]
the result of each iteration
to 5 decimal
places.
[3]
15.8
(ii)
Find
the
x
-coordinate
of
the
maximum
point
M
on the curve.
(1 + x)(2 + x2 )
1
− 15
w==sin
4i and
and
= all
5.
$ −θu2i#
$ $uw
− 3$
= $solutions
$ − 3i$ in the interval 0◦ < θ < 360◦ .
(ii) Hence solve the equationarg"
8 cos
+
12, giving
6% θ =
(iv) Using the values of t found in previous parts
of the question, solve the equation
9
[4]
Find the
x-coordinate
the point
P at
which
the tangent to th
5x − x2all(iii)
y for u and w, giving
Solve
the at
equations
answers
in the
form x + iby
yof
, where
x and
y rare
intersect
the
P. Express
the complex
represented
form
ei& ,real.
giving
(ii) Hence
obtain
the point
expansion
of
in2number
ascending
powers of x,Pupintothe
and
including
the
2
[5]
0
(12+tan
xof)(2r2xcorrect
++x5 )tantox 3=significant
the exact
value of & and the value
figures.
[5]
3
term in x .
[5]
≤ xa ≤sketch
π . the
4
During
an−π
experiment,
of organisms
at time
t days
is denoted
N , where
N [3]
is
(b) for(i)
On
of number
an Argand
diagram,
shade the
region
whose
points by
represent
complex
© UCLESpresent
2011
9709/32/M/J/11
1
treated as anumbers
continuous
variable.
It
is
given
that
4 satisfying the inequalities |% − 2 + 2i | ≤ 2, arg % ≤ − 4 π and Re % ≥ 1, where Re %
9 8 Two
have
xpart
y56−%ln
2. "2=−712.
and 2x + y + 3" = 5.
[5]
(a) planes
Showdenotes
that equations
' the
4x real
ln x d
x+=2of
[5]
N
by copyright
included has been sought and cleared where possible. Every reasonable
Permission to reproduce items where third-party owned materialdprotected
−0.02t is 0.5
2
=
1.2e
N
.
effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will
dthe
t value
(ii) amends
Calculate
theangle
greatest
possible
of Re % for points lying in the shaded region.
[1]
(i) toCalculate
the
acute
planes.
[4]
be pleased
make
at the
earliest
possiblebetween
opportunity.
1%
24
When
tUse
= 0,the
the
number
of u
organisms
is 100.
University
Cambridge
International
Examinations
is part of present
the Cambridge
Assessment Group. Cambridge Assessment
= sin
4isx itself
to
find
the
exact
value
' cos3 4x dx. is the brand name of University
[5]of
(b)ofLocal
substitution
Examinations
Syndicate
(UCLES),
whichline
a department
of the University
of Cambridge.
(ii)
Find
a vector
equation
for
the
of
intersection
of the of
planes.
[6]
©Cambridge
UCLES
2012
9709/33/M/J/12
0
(i)2012
Find an expression for N in terms of t. 9709/33/O/N/12
[6]
© UCLES
7
y
© UCLES 2013
10
9709/31/M/J/13
(ii) State what happens to the
y number of organisms present after a long time.
e
x
[1]