MS4
£4.00
GCE MARKING SCHEME
MATHEMATICS - C1-C4 & FP1-FP3
AS/Advanced
SUMMER 2009
INTRODUCTION
The marking schemes which follow were those used by WJEC for the Summer 2009
examination in GCE MATHEMATICS . They were finalised after detailed discussion at
examiners' conferences by all the examiners involved in the assessment. The conferences
were held shortly after the papers were taken so that reference could be made to the full
range of candidates' responses, with photocopied scripts forming the basis of discussion.
The aim of the conferences was to ensure that the marking schemes were interpreted and
applied in the same way by all examiners.
It is hoped that this information will be of assistance to centres but it is recognised at the
same time that, without the benefit of participation in the examiners' conferences, teachers
may have different views on certain matters of detail or interpretation.
WJEC regrets that it cannot enter into any discussion or correspondence about these
marking schemes.
PAGE
C1
C2
C3
C4
1
7
12
19
FP1
FP2
FP3
25
29
33
Mathematics C1 May 2009
Solutions and Mark Scheme
1.
(a)
Gradient of AB = increase in y
increase in x
Gradient of BC = 3/4
M1
(or equivalent)
A1
(b)
A correct method for finding C
C (3, 8)
M1
A1
(c)
Use of mAB × mL = – 1 to find gradient of L
M1
A correct method for finding the equation of L using candidate’s
coordinates for C and candidate’s gradient for L.
M1
Equation of L:
y – 8 = –4/3 (x – 3)
(or equivalent)
(f.t. candidate’s coordinates for C and candidate’s gradient for L) A1
Equation of L:
(d)
4x + 3y – 36 = 0
(convincing, c.a.o.)
A1
(i)
Substituting y = 0 in equation of L
D (9, 0)
M1
A1
(ii)
A correct method for finding the length of CD (AC)
CD = 10
(f.t. candidate’s coordinates for C and D)
M1
A1
(iii)
AC = 5
(f.t. candidate’s coordinates for C)
tan CÂD = CD = 2
(or 10/5 or equivalent)
AC
(f.t. candidate’s derived values for CD and AC)
A1
1
B1
2.
(a)
8− 7
7 −2
=
(8 − 7 )( 7 + 2)
M1
( 7 − 2)( 7 + 2)
Numerator:
8 7 + 16 − 7 − 2 7
Denominator:
7–4
8− 7 6 7 +9
=
= 2 7 +3
3
7 −2
A1
A1
(c.a.o.) A1
Special case
If M1 not gained, allow B1 for correctly simplified numerator or
denominator following multiplication of top and bottom by 7 − 2
(b)
50 = 5 2
3× 6 =3 2
14
−
= −7 2
2
50 + ( 3 × 6 ) −
3.
B1
B1
B1
14
2
= 2
(c.a.o.) B1
dy = 4x + 6
(an attempt to differentiate, at least
dx
one non-zero term correct)
M1
An attempt to substitute x = – 1 in candidate’s expression for dy
m1
dx
Gradient of tangent at P = 2
(c.a.o.) A1
y-coordinate at P = 3
Equation of tangent at P:
y – 3 = 2[x – (–1)]
(or equivalent)
(f.t. one slip provided both M1 and m1 awarded)
4.
(a)
(i)
(ii)
(b)
a = – 2·5
(or equivalent)
b = 1·75
(or equivalent)
Greatest value = – b
(or equivalent)
x2 – x – 7 = 2x + 3
An attempt to collect terms, form and solve quadratic equation
x2 – 3x – 10 = 0 ⇒ (x – 5)(x + 2) = 0 ⇒ x = 5, x = – 2
(both values, c.a.o.)
When x = 5, y = 13, when x = – 2, y = – 1
(both values f.t. one slip)
The line y = 2x + 3 intersects the curve y = x2 – x – 7 at the points
(– 2, – 1) and (5, 13)
(f.t. candidate’s points)
2
B1
A1
B1
B1
B1
M1
m1
A1
A1
E1
5.
6.
7.
(a)
y + δy = 4(x + δx) 2 – 5(x + δx) – 3
Subtracting y from above to find δy
δy = 8xδx + 4(δx) 2 – 5δx
Dividing by δx and letting δx → 0
dy = limit δy = 8x – 5
dx δx→ 0 δx
(b)
Required derivative = 7 × 3/4 × x – 1/ 4 – 2 × (– 4) × x – 5
(a)
An expression for b2 – 4ac, with at least two of a, b, c correct
b2 – 4ac = (2k) 2 – 4(k +1)(k – 1)
b2 – 4ac = 4
(c.a.o.)
candidate’s value for b2 – 4ac > 0 (⇒ two distinct real roots)
M1
A1
A1
A1
(b)
Finding critical values x = – 2, x = 3/5
– 2 ≤ x ≤ 3/5 or 3/5 ≥ x ≥ – 2 or [– 2, 3/5] or – 2 ≤ x and x ≤ 3/5
or a correctly worded statement to the effect that x lies between
– 2 and 3/5 (both inclusive)
(f.t. critical values ± 2, ± 3/5)
Note: – 2 < x < 3/5,
– 2 ≤ x, x ≤ 3/5
– 2 ≤ x x ≤ 3/5
– 2 ≤ x or x ≤ 3/5
all earn B1
B1
(a)
4
B1
M1
A1
M1
(c.a.o.) A1
2
3
B1, B1
B2
4
⎛
⎜x +
⎝
2⎞
⎛2⎞ ⎛2⎞
4
3⎛ 2 ⎞
2⎛ 2 ⎞
⎟ = x + 4 x ⎜ ⎟ + 6 x ⎜ ⎟ + 4 x⎜ ⎟ + ⎜ ⎟
x⎠
⎝ x⎠ ⎝ x⎠
⎝ x⎠
⎝ x⎠
(three terms correct) B1
(all terms correct)
B2
⎛
⎜x +
⎝
2⎞
32 16
4
2
⎟ = x + 8 x + 24 + 2 + 4
x⎠
x
x
4
(three terms correct) B1
(all terms correct)
B2
(– 1 for incorrect further ‘simplification’)
(b)
A correct equation in n, including n C 2 = 55
n = 11, – 10
(c.a.o.)
n = 11
(f.t. n = 10 from n = – 11, 10)
3
M1
A1
A1
8.
(a)
Use of f (–1) = – 3
–a – 1 + 6 + 5 = – 3 ⇒ a = 2
M1
A1
(b)
Attempting to find f (r) = 0 for some value of r
M1
f (2) = 0 ⇒ x – 2 is a factor
A1
f (x) = (x – 2)(8x2 + ax + b) with one of a, b correct
M1
f (x) = (x – 2)(8x2 + 2x – 3)
A1
f (x) = (x – 2)(4x + 3)(2x – 1) (f.t. only 8x2 – 2x – 3 in above line) A1
Special case
Candidates who, after having found x – 2 as one factor, then find one
of the remaining factors by using e.g. the factor theorem, are awarded
B1 for final 3 marks
4
9.
(a)
y
O
(– 7, 0)
(1, 0)
x
(–3, – 4)
Concave up curve and y-coordinate of minimum = – 4
x-coordinate of minimum = – 3
Both points of intersection with x-axis
B1
B1
B1
y
(b)
(– 3, 0)
O
(1, 0)
x
O
(–1, – 4)
Concave up curve and y-coordinate of minimum = – 4
x-coordinate of minimum = – 1
Both points of intersection with x-axis
5
B1
B1
B1
10.
(a)
(b)
dy = 3x2 – 6x + 3
dx
Putting derived dy = 0
dx
3(x – 1)2 = 0 ⇒ x = 1
x = 1 ⇒ y = 6 ⇒ stationary point is at (1, 6)
B1
M1
A1
(c.a.o) A1
Either:
An attempt to consider value of dy at x = 1- and x = 1+
M1
dx
dy has same sign at x = 1- and x = 1+⇒ (1, 6) is a point of inflection
dx
A1
Or:
An attempt to find value of d2y at x = 1, x = 1- and x = 1+
M1
dx2
d2y = 0 at x = 1 and d2y has different signs at x = 1- and x = 1+
dx2
dx2
⇒ (1, 6) is a point of inflection
A1
Or:
An attempt to find the value of y at x = 1- and x = 1+
M1
Value of y at x = 1- < 6 and value of y at x = 1+ > 6 ⇒ (1, 6) is a point
of inflection
A1
Or:
An attempt to find values of d2y and d3y at x = 1
M1
dx2
dx3
d2y = 0 and d3y ≠ 0 at x = 1 ⇒ (1, 6) is a point of inflection
A1
dx2
dx3
6
Mathematics C2 May 2009
Solutions and Mark Scheme
1.
0
0·5
0·1
0·43173647
0·2
0·408628001
0·3
0·392507416
(3 values correct)
B1
0·4
0·379873463
(5 values correct)
B1
Correct formula with h = 0·1
M1
I ≈ 0·1 × {0·5 + 0·379873463 + 2(0·43173647 + 0·408628001
2
+ 0·392507416)}
I ≈ 0·167280861
I ≈ 0·167
(f.t. one slip)
A1
Special case for candidates who put h = 0·08
0
0·5
0·08
0·438050328
0·16
0·416666666
0·24
0·401622886
0·32
0·389759395
0·4
0·379873463
(all values correct)
B1
Correct formula with h = 0·08
M1
I ≈ 0·08 × {0·5 + 0·379873463 + 2(0·438050328 + 0·416666666
2
+ 0·401622886 + 0·389759395)}
I ≈ 0·16688288
I ≈ 0·167
(f.t. one slip)
A1
Note: Answer only with no working earns 0 marks
7
2.
3.
(a)
5 cos2θ + 2 = 3(1 – cos2θ ) – 2 cosθ.
(correct use of sin2θ = 1 – cos2θ )
M1
An attempt to collect terms, form and solve quadratic equation
in cos θ, either by using the quadratic formula or by getting the
expression into the form (a cos θ + b)(c cos θ + d),
with a × c = candidate’s coefficient of cos2θ and b × d = candidate’s constant
m1
2
8 cos θ + 2 cos θ – 1 = 0 ⇒ (4 cos θ – 1)(2 cos θ + 1) = 0
⇒ cos θ = 1, –1
A1
4 2
θ = 75·5(225)°, 284·4(775)°, 120°, 240°
(75·5°, 284·5°)
B1
(120°)
B1
(240°)
B1
Note: Subtract 1 mark for each additional root in range for each branch,
ignore roots outside range.
cos θ = +, –, f.t. for 3 marks, cos θ = –, –, f.t. for 2 marks
cos θ = +, +, f.t. for 1 mark
(b)
2x + 12° = – 32, 212°, 328°
x = 100°, 158°
(a)
(b)
(one value)
(one value)
(two values)
Note: Subtract 1 mark for each additional root in range,
ignore roots outside range.
M1
A1
A1
sin ACˆ B sin 23°
=
16
9
(substituting the correct values in the correct places in the sin rule) M1
ACˆ B = 44°, 136°
(both values) A1
(i)
Use of angle sum of a triangle = 180°
M1
BÂC = 21°
(f.t. candidate’s values for ACˆ B provided acute value < 67°) A1
(ii)
Area of triangle ABC = 1 × 16 × 9 × sin 21°
2
(substituting 16, 9 and candidate’s value for BÂC
in the correct places in the area formula)
Area of triangle ABC = 25·8 cm2.
(f.t. candidate’s acute value for BÂC)
8
M1
A1
4.
(a)
Sn = a + [a + d] + . . . + [a + (n – 1)d]
(at least 3 terms, one at each end) B1
Sn = [a + (n – 1)d] + [a + (n – 2)d] + . . . + a
Either:
2Sn = [a + a + (n – 1)d] + [a + a + (n – 1)d] + . . . + [a + a + (n – 1)d]
2Sn = [a + a + (n – 1)d] + . . .
2Sn = n[2a + (n – 1)d]
Sn = n[2a + (n – 1)d]
2
5.
(n times)
Or:
M1
(convincing)
A1
(b)
a + 7d = 46
B1
9 × [2a + 8d] = 225
B1
2
An attempt to solve the candidate’s equations simultaneously by eliminating
one unknown
M1
a = – 3, d = 7 (both values)
(c.a.o.) A1
(c)
a = 3, d = 4
Sn = n[2 × 3 + (n – 1)4]
2
Sn = n(2n + 1)
(a)
(b)
r = 108 = 3
36
t7 = 36
32
t 7= 4
(i)
(ii)
(f.t. candidate’s d)
B1
M1
(c.a.o.) A1
(c.a.o.) B1
(f.t. candidate’s value for r)
M1
(c.a.o.) A1
ar = 9
B1
a = 48
B1
1–r
An attempt to solve these equations simultaneously by eliminating a
M1
2
16r – 16r + 3 = 0
(convincing)
A1
r = 1/4, 3/4
B1
a = 36, 12
(c.a.o.) B1
9
6.
(a)
5 × x – 2 – 3 × x5/4 + c
–2
5/4
(b)
(i)
6 + 4x – x2 = x + 2
M1
An attempt to rewrite and solve quadratic equation
in x, either by using the quadratic formula or by getting the
expression into the form (x + a)(x + b),
with a × b = candidate’s constant
m1
(x – 4)(x + 1) = 0 ⇒ x = 4, y = 6 at A
(both values) A1
Note: Answer only with no working earns 0 marks
(ii)
Either:
(Deduct 1 mark if no c present)
4
4
0
0
B1,B1
Total area = ⌠(6 + 4x – x2) dx – ⌠(x + 2) dx
⌡
⌡
(use of integration)
M1
4
= [4x + (3/2)x2 – (1/3)x3 ]
0
(correct integration) B3
3
= 16 + 24 – 4 /3
(f.t. candidate’s limits in at least one integral)
Correct subtraction of integrals with correct use of 0 and
candidate’s xA as limits
= 56
(c.a.o.)
3
Or:
Area of trapezium = 16
(f.t. candidate’s coordinates for A)
m1
m1
A1
B1
4
Area under curve = ⌠( 6 + 4x – x2) dx
⌡
(use of integration)
0
2
3
M1
4
= [6x + 2x – (1/3)x ]
0
(correct integration) B2
= 24 + 32 – 64/3
(f.t. candidate’s limits)
= 104
3
Finding total area by subtracting values
Total area = 104 – 16 = 56
(c.a.o.)
3
3
10
m1
m1
A1
7.
8.
9.
(a)
Let p = logax, q = logay
Then x = ap, y = aq
(the relationship between log and power) B1
x = ap = ap– q
(the laws of indicies) B1
q
y a
logax/y = p – q
(the relationship between log and power)
logax/y = p – q = logax – logay
(convincing) B1
(b)
(5 – 2x) log 3 = log 7
(taking logs on both sides) M1
An attempt to isolate x
(no more than 1 slip)
m1
x = 1·614
(c.a.o.) A1
Note: Candidates who write down x = 1·614 without explanation are awarded
M0 m0 A0
(c)
loga (x – 3) + loga (x + 3) = loga [(x – 3) (x + 3)] (addition law)
2 loga (x – 2) = loga (x – 2)2
(power law)
(x – 3) (x + 3) = (x – 2)2
(removing logs)
x = 3·25
(c.a.o.)
B1
B1
M1
A1
(a)
A (3, – 1)
A correct method for finding the radius
Radius = 5
B1
M1
A1
(b)
Gradient AP = inc in y
inc in x
Gradient AP = 2 – (–1 ) = 3 (f.t. candidate’s coordinates for A)
7–3
4
Use of mtan × mrad = –1
Equation of tangent is:
y – 2 = – 4(x – 7)
(f.t. candidate’s gradient for AP)
3
M1
A1
M1
A1
(c)
Distance between centres of C1 and C2 = 17
B1
(f.t. candidate’s coordinates for A)
Use of the fact that distance between centres =
sum of the radii + the shortest possible length of the line QR
M1
Shortest possible length of the line QR = 4
(f.t. one slip) A1
(a)
1 × 13 × 13 × θ = 60
2
θ = 0·71
(b)
M1
A1
QR = 13ϕ, RS = 13(π – ϕ),
13ϕ = 13(π – ϕ) ± 7
ϕ = 1·84
(at least one value)
(or equivalent)
(c.a.o.)
11
B1
M1
A1
Mathematics C3 Summer 2009
Solutions and Mark Scheme
1.
h = 0·2
(h = 0·2, correct formula) M1
0⋅2
[3 + 3 ⋅ 7191397 + 4(3 ⋅ 1189742 + 3 ⋅ 4779304) + 2(3 ⋅ 2778041)]
3
(3 values) B1
(2 values) B1
≈ 2 ⋅ 6442
A1
Integral ≈
[Special case: B1 for 6 correct values, at least 5 decimal places:
3, 3·09206, 3·20936, 3·35287, 3·52292, 3·71914
]
2.
(a)
θ = 90°
cosθ + cos 3θ = 0 + 0 = 0
(choice of θ and evaluating one side) B1
2 cos 2θ cos 4θ = 2 × (−1) ×1 = −2
(other side) B1
(cosθ + cos 3θ ≡ 2 cos 2θ cos 4θ )
(b)
− 9 + cot 2θ = cos ecθ − cos ec 2θ
− 9 + cos ec 2θ − 1 = cos ecθ − cos ec 2θ
2 cos ec 2θ − cos ecθ − 10 = 0
(2 cos ecθ − 5)(cos ecθ + 2) = 0
2 1
sin θ = , −
5
2
θ = 23 ⋅ 6, 156 ⋅ 4, 210°, 330°
[Notes:
(cot 2 θ = cos ec 2θ − 1) M1
(reasonable attempt to solve) M1
A1
(23·6, 156·4) B1
(210°) B1
(330°) B1
1) Lose 1 mark for each additional value in range,
1 for each branch
2) Allow to nearest degree
3) F.T. for sinθ = + , –
3 marks
F.T. for sinθ = – , –
2 marks
F.T. for sinθ = + , +
1 mark
]
12
3.
(a)
3x 2 + 2 y
dy
dy
+ tan 2 y + 2 x sec 2 2 y
=0
dx
dx
dy
3 x 2 + tan 2 y
=−
dx
2 y + 2 x sec 2 2 y
(b)
(i)
(ii)
⎛ dy ⎞
⎜ 2 y ⎟ B1
⎝ dx ⎠
dy
⎛
⎞
2
⎜ tan 2 y + k sec 2 y ; k = 1, 2 ⎟ B1
dx
⎝
⎠
(k = 2) B1
(All correct. F.T. for last B1 if first two Bs gained) B1
⎛ (3 + 2t ) f (t ) − (1 + 4t ) g (t ) ⎞
⎜⎜
⎟⎟ M1
(3 + 2t ) 2
⎝
⎠
[ f (t ) = 4, g (t ) = 2] A1
(Give additional mark for simplification here, see last A1 of part (ii))
dy (3 + 2t )(4) − (1 + 4t )(2)
=
dt
(3 + 2t ) 2
dy
dy
= dt
dx
dx
dt
(attempt to use
dy y&
= ) M1
dx x&
(3 + 2t )(4) − (1 + 4t )(2)
1
⋅
A1
2
(3 + 2t )
(3 + 2t )
10
=
(This may have been given in (i).) A1
(3 + 2t )3
(Penalise faulty simplification on last line)
=
13
4.
(a)
( f ′( x) = 0)
2(2 x − 3)e 2 x + 2e 2 x − 4
(Attempt to find f ′( x), – 4 present)
( (2 x − 3) f ( x) + e 2 x g ( x) )
( f ( x) = ke 2 x , k = 1, 2; g ( x) = 2 )
(k = 2)
(4 x − 6 + 2)e 2 x − 4
(4 x − 4)e 2 x − 4 = 0
( x − 1)e 2 x − 1 = 0
(b)
x
f ′( x)
1
2
–1
53·6
M1
M1
A1
A1
(equate f ′( x) to zero) M1
(result – convincing) A1
(Attempt to find values or signs) M1
(correct values) A1
Change of sign indicates presence of root between 1 and 2.
x0 = 1 ⋅1, x1 = 1⋅1108..., x2 = 1 ⋅1084..., x3 = 1 ⋅1089...
( x1 ) B1
x3 ≈ 1⋅1089
( x3 correct to 4 decimal places) B1
Check 1·10895, 1·10885
x
f ′(x)
1·10885
1·10895
– 0·00008
+ 0·001
(Attempt to find values) M1
(Correct signs or values) A1
Change of sign indicates presence of root
so root is 1·1089 correct to 4 decimal places.
Note: (b) must involve ‘change of sign’ (o.e.) at least once.
14
A1
5.
⎛ f ( x)
⎞
, f ( x) ≠ 1⎟ M1
⎜
2
⎝ 3 + 2x
⎠
( f ( x) = 4 x ) A1
(a)
4x
3 + 2x2
(b)
x2
+ 2 x tan −1 x
2
1+ x
( x 2 f ( x) + g ( x) tan −1 x) M1
1
⎛
⎞
, g ( x) = 2 x ⎟ A1
⎜ f ( x) =
2
1+ x
⎝
⎠
(c)
10(5 + 7 x 2 )9 .14 x
(10(5 + 7 x 2 )9 . f ( x)) M1
( f ( x) = 14 x) A1
(Simplified answer) A1
(F.T. for f ( x) = 7 x, i.e. 70x)
= 140 x(5 + 7 x 2 )9
6.
(a)
9 x − 7 ≤ 3,
x≤
10
9
B1
and
9 x − 7 ≥ −3
4
x≥
9
4
10
≤x≤
9
9
M1
(answer must involve the word ‘and’ o.e.) A1
(Note: lose A1 for the omission of ‘ =’)
Alternative Scheme: (9 x − 7) 2 ≤ 9
81x 2 − 126 x + 40 ≤ 0
(9 x − 10)(9 x − 4)
Any method
Correct answer
(b)
B1
M1
A1
5 x +1 = 9
5x =8
(a x = b; a = 5, b = 8) B1
8
5
(both answers, F.T. a, b) B1
x=±
15
7.
(a)
Penalise the absence of constant one mark once only in part (a)
(i)
1
− cos 5 x (+C )
5
(ii)
−
⎛
k
3
3⎞
⎜⎜
; k = − , ± ⎟⎟ M1
2
2
4⎠
⎝ ( 2 x + 7)
3⎞
⎛
⎜ k = − ⎟ A1
4⎠
⎝
3
( +C )
4(2 x + 7) 2
3
⎛
⎜ k ln 5 x + 3 ;
⎝
⎡2
⎤
⎢⎣ 5 ln 5 x + 3 ⎥⎦
0
(b)
1
(k cos 5 x; k = ± , − 5) M1
5
1⎞
⎛
⎜ k = − ⎟ A1
5⎠
⎝
may be omitted;
k = 2,
2⎞
⎟ M1
5⎠
2⎞
⎛
⎜ k = ⎟ A1
5⎠
⎝
2
[ln18 − ln 3]
5
≈ 0 ⋅ 717...
=
8.
(k (ln18 − ln 3); F.T. for any k except k = 1) M1
A1
y
(4, 9)
(shape and coordinates of
one point) B1
(another point) B1
(another point) B1
(6, 0)
O
(1, 0)
Note: 3 correct points, no
graph: B2
x
2 or less correct points,
no graph: B0
Special Case: All correct with left x translation: B1
16
9.
(a)
Domain of fg = (0, ∞)
Range = (0, ∞)
(b)
3e 2 ln 4 x = 12
B1
B1
( fg (x), correct order and equating) M1
ln 16 x 2
3e
= 12
2
48 x = 12
1
x=±
2
1
x = (since domain (0, ∞) )
2
[ Alternatively:
3e 2 ln 4 x = 12
e 2 ln 4 x = 4
2 ln 4 x = ln 4
ln 4 x = ln 2
1
x=
2
(correct attempt to use laws of logs) m1
A1
(F.T. 12 x 2 = 12) A1
A1
(correct order and equating) M1
A1
(taking logs) m1
(o.e.) A1
A1 ]
17
10.
(a)
⎛ f ( x)
⎞
⎜⎜
; f ( x) ≠ ±1, ± 2 ⎟⎟ M1
2
2
⎝ (3x + 2)
⎠
− 2 × −1
× 6x
(3x 2 + 2) 2
12 x
=
(3 x 2 + 2) 2
f ′( x) =
Since 12 x > 0,
( f ( x) = 12 x) A1
1
> 0, f ′( x) > 0
(3x + 2) 2
2
( 12 x > 0 ) B1
⎛
⎞
1
⎜⎜
> 0 ⎟⎟
2
2
⎝ (3 x + 2)
⎠
(b)
Range is (0, 1)
(c)
Let y = 1 −
y −1 =
B1
⎛
−2 ⎞
⎟ M1
⎜⎜ y − 1 =
( ) ⎟⎠
⎝
2
3x + 2
2
−2
3x 2 + 2
2
= 3x 2 + 2
1− y
2
− 2 = 3x 2
1− y
x=±
x=
B1
A1
⎞
2⎛ 1
⎜⎜
− 1⎟⎟
3 ⎝1− y ⎠
2⎛ y ⎞
⎜
⎟
3 ⎜⎝ 1 − y ⎟⎠
f −1 ( x) =
o.e. (must have ±
(simplified form not required)
) A1
o.e. (domain x > 0) B1
2⎛ x ⎞
⎟ (simplified form not required)
⎜
3 ⎝1− x ⎠
(F.T. candiate’s expression) B1
Domain of f −1 is (0, 1)
o.e. (F.T. from (a) )
) B1
)
Range of f −1 is (0, ∞)
18
Mathematics C4 Summer 2009
Solutions and Mark Scheme
1.
(a)
3x
A
B
C
≡
+
+
2
2
2+ x
(1 + x) (2 + x) 1 + x (1 + x)
3x ≡ A(1 + x)(2 + x) + B(2 + x) + C (1 + x) 2
x = −1
x = −2
x2
(b)
∫
(correct form) M1
(correct attempt to clear fractions
and substitute for x) M1
− 3 = B(1)
B = −3
− 6 = C (−1) 2
C = −6
0 = A+C
A=6
(2 constants) A1
(3rd constant) A1
(F.T. one slip)
⎛ 6
3
6 ⎞
⎜⎜
⎟ dx
−
−
2
0 1+ x
(1 + x) 2 + x ⎟⎠
⎝
1
1
3
⎡
⎤
= ⎢6 ln(1 + x) +
− 6 ln(2 + x)⎥
1+ x
⎣
⎦0
(F.T. candidate’s equivalent work)
= 6 ln 2 +
≈ 0 ⋅ 226
⎛ 3 ⎞
⎜
⎟ B1
⎝1+ x ⎠
(logs) B1, B1
3
− 6 ln 3 − 6 ln1 − 3 + 6 ln 2
2
(must be at least 3 decimal places)
19
C.A.O. B1
2.
(Use of sin 2θ = 2 sin θ cosθ ) M1
3 × 2 sin θ cosθ = 2 sin θ
sin θ = 0
or
A1
3 cosθ = 1
1
cosθ =
3
A1
θ = 0°, 180°, 360°
) (F.T. one slip)
)
No workings shown – no marks
70 ⋅ 5°, 289 ⋅ 5°
3.
(a)
(b)
4.
R=2
tan α = 3 , α = 60°
(any method)
B1
B1
B1
M1
A1
2 cos(θ − 60°) = 1
1
(F.T. R and α ) M1
cos(θ − 60°) =
2
θ − 60° = −60°, 60°, 300°
(one value) A1
θ = 0°, 120°, 360°
(A2 for 3 answers, A1 for 2 answers
A2
A0 for 1 answer, lose 1 for more than 3 answers)
Volume = π ∫
π
cos 2 2 x dx
8
0
= (π ) ∫
π
8
0
(must contain limits) B1
1 + cos 4 x
dx
2
(cos 2 2 x = a + b cos 4 x; a, b ≠ 0) M1
1
1⎞
⎛
⎜ a = , b = ⎟ A1
2
2⎠
⎝
π
⎡ x sin 4 x ⎤ 8
= (π ) ⎢ +
8 ⎥⎦ 0
⎣2
A1
⎛π 1
⎞
= (π )⎜ + − 0 − 0 ⎟
⎝ 16 8
⎠
=
(correct use of limits) m1
π ⎛π
1⎞
⎜ + ⎟ or 1 ⋅ 0095
2 ⎝ 8 4⎠
(C.A.O.) A1
[ If substitution used, marks are gained after
1
cos 2 u = a + b cos 2u
2
20
M1 ]
5.
(a)
dy 3t 2
=
dx 2t
⎛ dy y& ⎞
⎜ = ⎟ M1
⎝ dx x& ⎠
3t
2
(simplified form) A1
=
Equation of tangent is
3
y − p 3 = p( x − p 2 )
2
3
2 y − 2 p = 3 px − 3 p 3
3 px − 2 y = p 3
(b)
(use of any method) M1
(convincing) A1
Substitute x = q 2 , y = q 3
(substitution of x = q 2 , y = q 3 and p = 2 )
3 pq 2 − 2q 3 = p 3
When p = 2 ,
6q 2 − 2q 3 = 8
q 3 − 3q 2 + 4 = 0
(convincing)
2
(q + 1)(q − 4q + 4) = 0
(attempt to solve)
q = −1 or q = 2
Disregard q = 2 (as this relates to point P)
[Alternatively:
y − q3
=3
x − q2
q 3 − 3q 2 + 4 = 0
6.
(a)
2x
∫ ( x + 3) e dx = ( x + 3)
M1
A1
M1
A1
A1
(must have gradient 3) M1
(convincing) A1 ]
e 2x
− ∫ 1. e 2 x dx
2
( ( x + 3) f ( x) − ∫ Af ( x) dx; f ( x) ≠ k , A = 1, 3) M1
( f ( x) = k e 2 x ) A1
1
⎛
⎞
⎜ k = , A = 1⎟ A1
2
⎝
⎠
e 2x e 2x
= ( x + 3)
−
+C
2
4
C.A.O. (must contain C) A1
21
(b)
∫
2
3
−
1
1
2
2u
[ ]
1
= −u2
⎛ k ⎞
⎜ 1 ⎟ M1
⎝u2 ⎠
1⎞
⎛
⎜ k = − ⎟ A1
2⎠
⎝
du
2
(integration, any k, no limits) A1
3
(correct use of limits) m1
C.A.O. (either answer) A1
Answer only gains 0 marks
= 2 + 3 ≈ 0 ⋅ 318
7.
(a)
dP
= − kP 3
dt
(b)
∫p
(allow ± k) B1
dP
−
3
= − ∫ k dt
(separation of variables & attempt to integrate
1
= − kt + C
2 p2
(C may be omitted, n ≠ 1 ) A1
t = 0, P = 20
(attempt to find C) M1
1
=C
800
1
1
∴ − 2 = −kt −
2p
800
1
1
∴ 2 = 2kt +
p
400
1
1
= At +
( A = 2k )
2
p
400
∴ −
(c)
1
, any n) M1
pn
t = 1, P = 10
1
1
= A+
100
400
3
∴ A=
400
(F.T. similar work) A1
(convincing) A1
(attempt to find A) M1
A1
1
3
1
=
+
25 400 400
15
3
=
t
400 400
t =5
(substitute p = 5) m1
(F.T. one slip) A1
22
8.
(a)
(i)
AB = i − 2 j + 3k
B1
Equation of AB is
r = 3i + 4 j + 7k + λ (i − 2 j + 3k )
[ Alternative:
(1 − λ )a + λb
r = ............
(ii)
( r = a + λAB, o.e.) M1
A1
(a, b substituted) M1
A1
(all correct) A1 ]
Assume AB and L intersect. Equate coefficients of i, j (o.e.).
(3 + λ ) = 5 + 3μ
4 − 2λ = 6 − 2μ
(F.T. candidate’s values) M1
A1
Solve for λ , μ ,
5
3
λ=− , μ=−
2
2
(attempt to solve for λ , μ ) m1
(one value; F.T. one slip) A1
Check k coefficient (o.e.)
L.H.S. = 7 + 3λ = −
1
2
(attempt to check) m1
1
2
(Terms check so lines intersect)
1
Point of intersection is i + 9 j − k .
2
R.H.S. = 1 + μ = −
C.A.O. A1
(dependent on M1, m1 earlier)
(b)
(3i − 2 j + 2k ) . (2i + j − 2k ) = 0
6−2−4 = 0
(therefore vectors are perpendicular)
23
(correct method of finding scalar product) M1
A1
9.
1
1
1⎛ 1⎞1
(1 + 4 x) 2 = 1 + (4 x) + ⎜ − ⎟ (4 x) 2 + ...
2
2⎝ 2⎠2
2
= 1 + 2 x − 2 x + ....
Valid for x <
(first line with possibly 4 x 2 ) M1
(1 + 2 x) A1
(−2 x 2 ) A1
1
4
B1
(1 + 4k + 16k 2 ) = 1 + 2(k + 4k 2 ) − 2(k + 4k 2 ) + ...
(correct substitution for x
and attempt to evaluate) M1
= 1 + 2k + 8k 2 − 2k 2 + ....
= 1 + 2k + 6k 2 + ....
(F.T. quadratic in x) A1
[ Alternative:
First principles with three terms
Answer
10.
9k 2 = 3b 2
b 2 = 3k 2
( b 2 has a factor 3)
b has a factor 3
a and b have a common factor – contradiction
( 3 is irrational)
24
M1
A1 ]
B1
B1
B1
(must mention contradiction) B1
A/AS Level Maths – FP1 – June 2009 – Markscheme
n
n
n
r =1
r =1
r =1
S n = ∑ r 3 + 2∑ r 2 + ∑ r
1.
n 2 (n + 1) 2 2n(n + 1)(2n + 1) n(n + 1)
+
+
=
4
6
2
2
n(n + 1)(3n + 3n + 8n + 4 + 6)
=
12
n(n + 1)(n + 2)(3n + 5)
=
12
α + β = −3
αβ = 4
2.
(a)
detA = 1(6 – 5) + 2( 3 – 4) + 3(10 – 9) = 2
⎡ 1 −1 1 ⎤
Cofactor matrix = ⎢⎢ 11 − 7 1 ⎥⎥
⎢⎣− 7 5 − 1⎥⎦
⎡ 1 11 − 7 ⎤
Adjugate matrix = ⎢⎢− 1 − 7 5 ⎥⎥
⎢⎣ 1
1 − 1 ⎥⎦
[A1 for 1,2 or 3 errors]
⎡ 1 11 − 7 ⎤
1⎢
Inverse matrix = ⎢− 1 − 7 5 ⎥⎥
2
⎢⎣ 1
1 − 1 ⎥⎦
(b)
A1A1A1
m1
A1
B1
Consider
α + β + αβ = 1
cao
2
2
αβ + α β + αβ = αβ + αβ (α + β )
=–8
cao
2 2
α β = 16
The required equation is
x 3 − x 2 − 8 x − 16 = 0
[FT from above]
3.
M1
[Award 0 marks for answer only : FT from their adjugate]
⎡ x⎤
⎡ 1 11 − 7 ⎤ ⎡13 ⎤
⎢ y ⎥ = 1 ⎢− 1 − 7 5 ⎥ ⎢13 ⎥
⎢ ⎥ 2⎢
⎥⎢ ⎥
⎢⎣ z ⎥⎦
⎢⎣ 1
1 − 1 ⎥⎦ ⎢⎣22⎥⎦
⎡1 ⎤
= ⎢⎢3⎥⎥
⎢⎣2⎥⎦
[FT from their inverse]
M1A1
M1
A1
B1
M1A1
M1A1
M1
A2
A1
M1
A1
25
4.
(a)
(b)
5.
(9 + 7i)(3 + i)
(3 − i)(3 + i)
27 − 7 + 21i + 9i
=
9 +1
= 2 + 3i
mod(z) = 13 , arg(z) = 0.983 (56.3°)
[FT on their z]
M1
z=
A1A1
A1
B1B1
The statement is true for n = 1 since the formula gives1/2 which is correct.
B1
Let the statement be true for n = k, ie
k
1
k
=
∑
k +1
r =1 r ( r + 1)
Consider
k +1
k
1
1
1
=
+
∑
∑
(k + 1)(k + 2)
r =1 r ( r + 1)
r =1 r ( r + 1)
k
1
=
+
k + 1 (k + 1)(k + 2)
k (k + 2) + 1
=
(k + 1)(k + 2)
(k + 1) 2
(k + 1)(k + 2)
k +1
=
k+2
True for n = k ⇒ true for n = k + 1, hence proved by induction.
=
6.
M1
M1
A1
A1
A1
A1
A1
det(A) = λ (−4 − λ2 ) + 3λ − 8 + 2(2λ + 3) = −(λ3 − 3λ + 2)
M1A1
This is zero when λ = 1 (so the matrix is singular)
B1
3
2
2
λ − 3λ + 2 = (λ − 1)(λ + λ − 2) = (λ − 1) (λ + 2)
M1
[For candidates using long division award M1 for 2 terms]
Therefore λ = 1 is the only positive value giving a zero determinant.
A1
(b)(i)
x + y + 2z = 2
M1
–3y – 3z = – 6
A1
– 2y – 2z = – 4
A1
[Award the M1 for 1 correct row]
The equations are consistent because the 3rd equation is a multiple of
A1
the 2nd
(ii)
Put z = α
M1
y = 2 – α, x = – α
cao
A1
(a)
26
7.
Putting z = x + iy,
[Award for attempting to use]
( x − 1) 2 + y 2 = 2 ( x + 2) 2 + y 2
A1
x − 2 x + 1 + y = 4( x + 4 x + 4 + y )
x 2 + y 2 + 6 x + 5 = 0 [Accept any multiple]
( x + 3) 2 + y 2 = 4
Centre (– 3, 0), radius = 2
[FT from their circle]
2
8.
(a)
2
2
⎡0
Reflection matrix = ⎢⎢− 1
⎢⎣ 0
⎡1
Translation matrix = ⎢⎢0
⎢⎣0
2
− 1 0⎤
0 0⎥⎥
0 1⎥⎦
0 h⎤
1 k ⎥⎥
0 1 ⎥⎦
⎡1 0 h ⎤ ⎡ 0 − 1 0 ⎤
T = ⎢⎢0 1 k ⎥⎥ ⎢⎢− 1 0 0⎥⎥
⎢⎣0 0 1 ⎥⎦ ⎢⎣ 0
0 1⎥⎦
⎡ 0 − 1 h⎤
= ⎢⎢− 1 0 k ⎥⎥
⎢⎣ 0
0 1 ⎥⎦
(b)(i) We are given that
⎡ 0 − 1 h ⎤ ⎡1 ⎤ ⎡ 2 ⎤
⎢ − 1 0 k ⎥ ⎢ 2 ⎥ = ⎢1 ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢⎣ 0
0 1 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣1 ⎥⎦
giving
h=4
k=2
(ii) EITHER
A general point on the line is (λ, 3λ + 2).
Consider
⎡ 0 − 1 4⎤ ⎡ λ ⎤ ⎡2 − 3λ ⎤
⎢− 1 0 2⎥ ⎢3λ + 2⎥ = ⎢ 2 − λ ⎥
⎢
⎥⎢
⎥ ⎢
⎥
⎢⎣ 0
0 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦
x = 2 − 3λ ; y = 2 − λ
Eliminating λ, 3y – x = 4
cao
27
M1
A1
A1
M1
A1A1
B1
B1
B1
M1
A1
A1
M1
M1
A1
M1A1
OR
9.
(a)(i)
(ii)
(b)
(c)
Let ( x, y ) → ( x′, y′)
⎡ 0 − 1 4⎤ ⎡ x ⎤ ⎡ x′ ⎤
⎢ − 1 0 2 ⎥ ⎢ y ⎥ = ⎢ y ′⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢⎣ 0
0 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦
− y + 4 = x′ or y = 4 − x′
− x + 2 = y′ or x = 2 − y′
y = 3x + 2 → 4 − x′ = 6 − 3 y′ + 2
equation of image is 3 y − x = 4
M1
M1
A1
cao
ln f ( x) = ln( x x ) + ln(e −2 x ) = x ln x − 2 x
f ′( x)
x
= ln x + − 2 = ln x − 1
f ( x)
x
f ′( x) = f ( x)(ln x − 1)
(so a = 1, b = – 1)
f ( x)
f ′′( x) = f ′( x)(ln x − 1) +
x
At a stationary point,
ln x = 1
x=e
(2.72)
−e
y=e
(0.066)
From (b), f ′′(e) > 0
so it is a minimum.
[Award 0 marks for answer only]
28
M1
A1
B1
B1B1
B1
B1
B1
B1
B1
B1
B1
A/AS Maths – FP2 – June 2009 – Markscheme
1.
2.
(a) h is not continuous
Any valid reason, eg f is not defined for x = 0, f(x) jumps from large
negative to large positive going through 0.
(b(i) Attempting to compare g ( x) with g (− x).
g is even.
(ii) Attempting to compare h( x) with h(− x).
h is odd.
u = tan x ⇒ du = sec 2 xdx
and [0,π/6] → [0, 1 / 3 ]
1/
I=
∫
3
du
3 − (1 + u 2 )
[Must be correct here]
B1
B1
M1
A1
M1
A1
B1
B1
M1
0
1/ 3
=
∫
0
du
A1
2 − u2
1/ 3
u ⎤
⎡
= ⎢sin −1 ( )⎥
A1
2 ⎦0
⎣
= 0.421
A1
Any valid reason, eg the denominator would be the square root of a
negative number towards the upper limit.
B1
3.
4.
Modulus = 16
Argument = 2π/3
− 8 + 8 3i = 16[cos(2π / 3) + i sin( 2π / 3)]
B1
B1
B1
(−8 + 8 3i)1/4 = 161 / 4 [cos(2π / 3 × 1 / 4) + isin(2π / 3 × 1 / 4)]
= 2(cos(π/6) + isin(π/6))
Second root = 2(cos(2π/3) + isin(2π/3))
Third root = 2(cos(7π/6) + isin(7π/6))
Fourth root = 2(cos(5π/3) + isin(5π/3))
[FT from their modulus and argument]
M1
A1
A1
A1
A1
The equation can be rewritten
sin2θ + 2sin2θcosθ = 0
sin2θ(1 + 2cosθ) = 0
nπ
sin2θ = 0 gives θ =
2
1
cosθ = − gives θ = 2nπ ± 2π/3
2
29
M1A1
A1
M1A1
M1A1
5.
6.
1
A
B
C
≡
+
+
( x + 1)( x + 2)( x + 3) x + 1 x + 2 x + 3
A( x + 2)( x + 3) + B ( x + 1)( x + 3) + C ( x + 1)( x + 2)
=
M1
( x + 1)( x + 2)( x + 3)
x = -1 gives A = 1/2 ; x = -2 gives B = -1 ; x = -3 gives C = 1/2
A1A1A1
5
(b)
I = [1 / 2 ln( x + 1) − ln( x + 2) + 1 / 2 ln( x + 3)]0
B1B1
[B1 for 1 error]
1
M1
= (ln 6 − ln 49 + ln 8 + ln 4 − ln 3)
2
1 ⎛ 6×8× 4 ⎞
= ln⎜
A1
⎟
2 ⎝ 49 × 3 ⎠
⎛8⎞
A1
= ln⎜ ⎟
⎝7⎠
(a)
(a)
(b)
Let
dy dy / dθ
b cos θ
=
=−
a sin θ
dx dx / dθ
Equation of tangent is
b cos θ
y − b sin θ = −
( x − a cos θ )
a sin θ
bxcosθ + aysinθ = ab(cos 2 θ + sin 2 θ )
whence the printed result.
⎛ a
⎞
Putting y = 0, P is ⎜
,0 ⎟
⎝ cos θ ⎠
b ⎞
⎛
Putting x = 0, Q is ⎜ 0,
⎟
⎝ sin θ ⎠
b ⎞
⎛ a
,
Therefore R is ⎜
⎟
⎝ 2 cos θ 2 sin θ ⎠
M1A1
M1A1
A1
M1A1
A1
A1
Eliminating θ,
cos θ =
a
b
; sin θ =
2x
2y
a2
b2
+
= cos 2 θ + sin 2 θ = 1
2
2
4x
4y
30
M1
M1A1
7.
(a)
(b)
8.
z − n = cos(− nθ) + isin( −nθ)
M1
A1
= cos nθ − i sin nθ
n
−n
z + z = cos nθ + isinnθ + (cos nθ − i sin nθ )
M1
= 2cosnθ
AG
Using the result in (a),
2cos2θ - 4cosθ +3 = 0
M1
2
M1
2(2 cos θ − 1) − 4 cos θ + 3 = 0
2
4 cos θ − 4 cos θ + 1 = 0
A1
2
M1
(2 cos θ − 1) = 0
1
cos θ =
A1
2
3
A1
sin θ = ±
2
1
3
±
i
A1
The roots are
2 2
[The ± must be there for the last 2 marks; accept cos(π/3) ± isin(π/3]
(a) By long division,
x+4
x − 1 x 2 + 3x
x2 − x
(Must be 2x or 4x for M1)
4x
M1A1
4x − 4
4
Therefore,
4
x −1
[FT 2 from above]
4
f ′( x) = 1 −
(b)
( x − 1) 2
[FT their expression from (a)]
Stationary points occur when
4
1=
( x − 1) 2
x = −1, 3
y = 1, 9
f ( x) = x + 4 +
31
A1
B1
M1
A1
A1
(c)
The asymptotes are x = 1 and y = x + 4.
B1B1
y
x
G2
(d)
Since f(-3) = f(0) = 0, part of the answer is [-3,0].
Consider
x( x + 3)
= 10
( x − 1)
x 2 − 7 x + 10 = 0
x = 2, 5
−1
f ( A) = [−3,0] ∪ [ 2,5]
32
B1
M1
A1
A1
A1
A/AS Level Mathematics – FP3 – June 2009 – Mark Scheme
1.
2.
The equation can be rewritten
1 + 2 sinh 2 θ = 6 sinh θ − 3
sinh 2 θ − 3 sinh θ + 2 = 0
(sinh θ − 1)(sinh θ − 2) = 0
sinhθ = 1,2
θ = ln(1 + 2 ), ln(2 + 5 )
f(0) = 0
ex
f ′( x) = −
; f ′(0) = −1
(2 − e x )
f ′′( x) = −
2e x
; f ′′(0) = −2
(2 − e x ) 2
2e x (2 − e x ) 2 + 2e x 2(2 − e x )e x
; f ′′′(0) = −6
(2 − e x ) 4
The Maclaurin series is
2x 2 6x3
0−x−
−
+ ... = − x − x 2 − x 3 + ...
2
6
[FT on their derivatives]
f ′′′( x) = −
3.
dx = 2coshudu ; [0,2] → [ 0, sinh −1 1]
-1
sinh 1
∫
I=
0
2 cosh udu
(4 sinh 2 u + 4) 3 / 2
sinh −1 1
=
∫
0
1
=
4
2 cosh udu
8 cosh 3 u
sinh −1 1
∫ sec h
2
M1A1
A1
M1
A1
B1B1
B1
B1B1
B1B1
B1B1
M1A1
B1B1
M1
A1
u dθ
A1
0
1
sinh −1 1
= [tanh u ]0
4
1
= tanh sinh −1 (1)
4
= 0.18
[Award 0 marks for answer only]
33
A1
A1
A1
4.
EITHER
2y
dy
dy 2a
= 4a ⇒
=
dx
dx
y
a
CSA = 2π ∫ y 1 +
0
B1
4a 2
dx
y2
a
= 2π ∫ 4ax 1 +
0
M1A1
a
dx
x
A1
a
= 4 aπ ∫ x + a dx
A1
0
= 4 aπ .
[
2
( x + a) 3 / 2
3
]
a
[
8
aπ 2 3 / 2 a 3 / 2 − a 3 / 2
3
= Answer given
=
A1
0
]
A1
OR
dx
dy
= 2at , = 2a
dt
dt
B1B1
2
2
⎛ dx ⎞ ⎛ dy ⎞
CSA = 2π ∫ y ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
M1
1
= 2π ∫ 2at 4a 2t 2 + 4a 2 dt
A1
0
1
= 8πa ∫ t t 2 + 1 dt
2
A1
0
1
⎡ (t 2 + 1)3 / 2 ⎤
= 8πa ⎢
⎥
3
⎣
⎦0
= Answer given
2
34
M1A1
5.
(a)
•
O
initial line
G1
(b)
1
Area =
2
π /2
1
2
π /2
1
2
π /2
=
=
∫ (2 + cosθ ) dθ
2
M1
0
∫ (4 + 4 cosθ + cos
2
θ ) dθ
A1
0
1
1
∫ (4 + 4 cosθ + 2 + 2 cos 2θ )dθ
M1A1
0
π /2
(c)
1 ⎡ 9θ
1
⎤
= ⎢ + 4 sin θ + sin 2θ ⎥
2⎣ 2
4
⎦0
[Limits not required until here]
9
= π + 2 (5.53)
8
Consider
y = rsinθ = (2 + cosθ)sinθ
dy
= − sin 2 θ + cos θ (2 + cos θ )
dθ
dy
= 0 giving
At a stationary point,
dθ
2 cos 2 θ + 2 cos θ − 1 = 0
3 −1
cos θ =
2
θ = 1.20, r = 2.37
35
A1
A1
M1
A1
M1
A1
A1
A1A1
π /4
6.
(a)
In =
∫ tan
n−2
x tan 2 xdx
n−2
x(sec 2 x − 1)dx
M1
0
π /4
=
∫ tan
m1A1
0
[
]
π /4
1
tan n −1 x 0 − I n − 2
n −1
1
− I n−2
=
n −1
=
π /4
(b)
I0 =
∫ d x = [x ]
π /4
0
0
=
π
4
1
I4 = − I2
3
1
= − (1 − I 0 )
3
π 2
(0.119)
= −
4 3
7.
A1A1
M1A1
M1
A1
A1
f ′( x) = sinh x − x cosh x , f ′(0) = 0
M1A1
f ′′( x) = − x sinh x, f ′′(0) = 0
A1
(ii) Any valid method, eg looking at the behaviour of f ′( x) or f ′′( x)
around x = 0 or noting that f is an even function.
M1
Concluding that it is not a stationary point of inflection
A1
b)(i)
2coshα - αsinhα = 0
M1
α sinh α
=2
A1
cosh α
whence
αtanhα = 2
(a)(i)
(ii) Let f(α) = αtanhα - 2
f(2) = -0.0719…
M1
f(2.1) = 0.0379…
The change of sign shows that the value of α lies between 2 and 2.1.
A1
[Accept consideration of f(α) = 2coshα – αsinhα]
(iii) Let
2
M1
y=
tanh x
dy
2
A1
=−
× sec h 2 x
2
dx
tanh x
= 0.137 < 1 in modulus when x = 2.05
A1
therefore convergent.
36
(iv)
Taking the initial value as 2.05, successive values are
2.05
2.067407830
2.065063848
2.065374578
2.065333300
2.065338782
Thus α = 2.0653 correct to 4dps.
B1
B1
2.0653
(c)
Area =
∫ (2 cosh x − x sinh x)dx
M1
0
= [2 sinh x ]0
2.0653
− [x cosh x ]0
2.0653
= [3 sinh x − x cosh x ]
= 3.365
2.0653
0
GCE Maths C1-C4 & FP1-FP3 MS (Summer 2009)
30 July 2009
37
2.0653
+
∫ cosh xdx
A1A1A1
0
A1
A1
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