Mark Scheme (Results)
June 2011
GCE Further Pure FP3 (6669) Paper 1
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June 2011
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EDEXCEL GCE MATHEMATICS
General Instructions for Marking
1. The total number of marks for the paper is 75.
2. The Edexcel Mathematics mark schemes use the following types of marks:
•
M marks:
method marks are awarded for ‘knowing a method and attempting to apply it’, unless otherwise
indicated.
•
A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.
•
B marks are unconditional accuracy marks (independent of M marks)
•
Marks should not be subdivided.
3. Abbreviations
These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if
you are using the annotation facility on ePEN.
•
bod – benefit of doubt
•
ft – follow through
•
the symbol
•
cao – correct answer only
•
cso - correct solution only. There must be no errors in this part of the question to obtain this mark
•
isw – ignore subsequent working
•
awrt – answers which round to
•
SC: special case
•
oe – or equivalent (and appropriate)
•
dep – dependent
•
indep – independent
•
dp decimal places
•
sf significant figures
•
¿ The answer is printed on the paper
•
will be used for correct ft
The second mark is dependent on gaining the first mark
June 2011
Further Pure Mathematics FP3 6669
Mark Scheme
Question
Number
Scheme
Marks
dy
= 6 x 2 and so surface area = 2π ∫ 2 x 3 (1 + (6 x 2 ) 2 dx
dx
3 ⎤
2
⎡
(1 + 36 x 4 ) 2 ⎥
= 4π ⎢
⎣ 3 × 36 × 4
⎦
4π
Use limits 2 and 0 to give
[13860.016 − 1] = 806 (to 3 sf)
216
1.
B1
M1 A1
DM1 A1
5
B1
1M1
1A1
2DM1
2A1
Notes:
Both bits CAO but condone lack of 2π
2 ⎞
⎛
3
dy
⎛
⎞
Integrating ∫ ⎜ y 1 + ⎜ their ⎟ ⎟dx , getting k (1 + 36 x 4 ) 2 , condone lack of 2π
⎜
dx ⎠ ⎟
⎝
⎝
⎠
If they use a substitution it must be a complete method.
CAO
Correct use of 2 and 0 as limits
CAO
2.
(a) (i)
dy
x
=
+ arcsin x
dx
(1 − x 2 )
M1 A1
(2)
(ii)
At given value derivative =
1 π 2 3 +π
+ =
6
3 6
B1
(1)
(b)
dy
6e
=
dx 1 + 9e4 x
6
= −2 x
e + 9e 2 x
1M1 A1
3
5
+ e ) + 42 (e 2 x − e −2 x )
2 (e
dy
3
∴ =
dx 5cosh 2 x + 4sinh 2 x
3M1
2x
=
2x
2M1
−2 x
A1 cso
∗
(5)
8
Notes:
(a) M1
Differentiating getting an arcsinx term and a
A1 CAO
B1 CAO any correct form
GCE Further Pure Mathematics FP3 (6669) June 2011
1
1
1 ± x2
term
Question
Number
(b) 1M1
Scheme
Of correct form
Marks
ae 2 x
1 ± be4 x
1A1 CAO
2M1 Getting from expression in e4 x to e2 x and e−2 x only
3M1 Using sinh2x and cosh2x in terms of ( e2 x + e−2 x ) and ( e2 x − e−2 x )
2A1 CSO – answer given
3.
(a)
1
1
1
=
= 2
2
x − 10 x + 34 ( x − 5) + 9 u + 9
(mark can be earned in either part (a) or (b))
1
⎡1
1
⎡1
⎛ u ⎞⎤
⎛ x − 5 ⎞⎤
I = ∫ 2 du = ⎢ arctan ⎜ ⎟ ⎥
I =∫
du = ⎢ arctan ⎜
⎟⎥
2
( x − 5) + 9
u +9
⎝ 3 ⎠⎦
⎝ 3 ⎠⎦
⎣3
⎣3
x 2 − 10 x + 34 = ( x − 5)2 + 9 so
Uses limits 3 and 0 to give
B1
2
π
Uses limits 8 and 5 to give
12
π
M1 A1
DM1 A1
12
(5)
(b) Alt 1
2
⎛ x −5+
⎛⎛ x − 5 ⎞
⎞
⎛ x−5⎞
⎜
⎜
⎟
+ ⎜
I = ln ⎜
⎟ + 1 ⎟ or I = ln ⎜
⎜ ⎝ 3 ⎟⎠
3 ⎠
⎝
⎝
⎠
⎝
( x − 5)
2
3
(
or I = ln ( x − 5) + ( x − 5) 2 + 9
)
+9 ⎞
⎟
⎟
⎠
M1 A1
Uses limits 5 and 8 to give ln(1 + 2) .
(b) Alt 2
DM1 A1
{
⎡
⎛ u ⎞⎤
du = ⎢ar sinh ⎜ ⎟ ⎥ = ln u + u 2 + 9
⎝ 3 ⎠⎦
⎣
u2 + 9
Uses limits 3 and 0 and ln expression to give ln(1 + 2) .
Uses u = x-5 to get I = ∫
1
}
(4)
9
M1 A1
DM1 A1
(4)
(b) Alt 3
Use substitution x – 5 = 3 tan θ ,
I = ∫ sec θ dθ = ln(sec θ + tan θ )
Uses limits 0 and
π
4
dx
= 3sec2 θ and so
dθ
M1 A1
DM1 A1
to get ln(1 + 2) .
(4)
Notes:
(a) B1
1M1
1A1
2DM1
2A1
CAO allow recovery in (b)
Integrating getting k arctan term
CAO
Correctly using limits.
CAO
GCE Further Pure Mathematics FP3 (6669) June 2011
2
Question
Number
(b) 1M1
1A1
2DM1
2A1
Scheme
Marks
Integrating to get a ln or hyperbolic term
CAO
Correctly using limits.
CAO
4.
(a)
⎡ x3
x 3 n(ln x) n −1
n⎤
I n = ⎢ (ln x) ⎥ − ∫ ×
dx
3
x
⎣3
⎦
M1 A1
e
e
⎡ x3
nx 2 (ln x) n −1
n⎤
= ⎢ (ln x) ⎥ − ∫
dx
3
⎣3
⎦1 1
e3 n
∴ I n = − I n −1
∗
3 3
DM1
A1cso
(4)
e
⎡ x 3 ⎤ e3 1
e3 1 ⎛ e3 1 ⎞ 2e3 1
I 0 = ∫ x dx = ⎢ ⎥ = − or I1 = − ⎜ − ⎟ =
+
3 3⎝ 3 3⎠ 9 9
⎣ 3 ⎦1 3 3
1
e3 1
e3 2
e3 3
4e3 2
I1 = − I 0 , I 2 = − I1 and I 3 = − I 2 so I 3 =
+
3 3
3 3
3 3
27 27
e
(b)
2
M1 A1
M1 A1
(4)
8
(a)1M1
1A1
2DM1
2A1
Notes:
Using integration by parts, integrating x 2 , differentiating (ln x)n
CAO
Correctly using limits 1 and e
CSO answer given
(b)1M1 Evaluating I 0 or I1 by an attempt to integrate something
1A1 CAO
2M1 Finding I 3 (also probably I1 and I 2 ) If ‘n’s left in M0
2A1 I 3 CAO
GCE Further Pure Mathematics FP3 (6669) June 2011
3
Question
Number
Scheme
Marks
5.
(a)
Graph of y =
3sinh2x
B1
Shape of −e2x
graph
B1
Asymptote: y = 13
B1
Value 10 on y axis
and value 0.7 or
1
ln ( 133 ) on x axis
2
B1
(4)
(b)
M1 A1
3 2 x −2 x
(e − e ) = 13 − 3e2 x → 9e4 x − 26e2 x − 3 = 0 to form quadratic
2
DM1 A1
1
∴ e2 x = − or 3
9
B1
1
∴ x = ln(3)
2
Use definition
(5)
9
(a) 1B1
2B1
3B1
4B1
(b) 1M1
1A1
2DM1
2A1
B1
Notes:
y = 3sinh2x first and third quadrant.
Shape of y = −e2x correct intersects on positive axes.
Equation of asymptote, y = 13, given. Penlise ‘extra’asymptotes here
Intercepts correct both
Getting a three terms quadratic in e2 x
Correct three term quadratic
Solving for e2x
CAO for e2 x condone omission of negative value.
CAO one answer only
GCE Further Pure Mathematics FP3 (6669) June 2011
4
Question
Number
6.
(a)
Scheme
Marks
n = (2 j- k ) × (3i + 2 j + 2k ) = 6i – 3j – 6k o.a.e. (e.g. 2i – j – 2k )
M1 A1
(2)
(b)
Line l has direction 2i –2j – k
Angle between line l and normal is given by (cos β or sinα ) =
α = 90 − β = 63 degrees to nearest degree.
4+2+2 8
=
9
9 9
A1 awrt
(4)
(c) Alt 1 Plane P has equation r.(2i − j − 2k ) = 1
1 − (−7) 8
Perpendicular distance is
=
3
9
M1 A1
M1 A1
(c) Alt 2 Parallel plane through A has equation r.
Plane P has equation r.
2i − j − 2k 1
=
3
3
B1
M1 A1ft
(4)
10
2i − j − 2k −7
=
3
3
M1 A1
M1
So O lies between the two and perpendicular distance is
1 7 8
+ =
3 3 3
A1
(4)
(c) Alt 3
22 + 22 + 12 = 3
8 8
Perpendicular distance is ‘3’ sin α = 3 × =
9 3
Distance A to (3,1,2) =
M1A1
M1A1
(4)
(c) Alt 4 Finding Cartesian equation of plane P: 2x – y – 2z – 1 = 0
n α + n2 β + n3γ + d
2(1) − 1(3) − 2(3) − 1 8
d= 1
=
=
2
2
2
3
n1 + n2 + n3
22 + 12 + 22
M1 A1
M1A1
(4)
Notes:
(a) M1
A1
(b) B1
M1
1A1ft
2A1
(c) 1M1
1A1
2M1
2A1
Cross product of the correct vectors
CAO o.e.
CAO
Angle between ‘ 2i – j – 2k ’ and 2i –2j – k, formula of correct form
8/9ft
CAO awrt
Eqn of plane using 2i – j – 2k or dist of A from O or finding length of AP
Correct equation (must have = ) or A to (3,1,2) = 3
Using correct method to find perpendicular distance
CAO
GCE Further Pure Mathematics FP3 (6669) June 2011
5
Question
Number
7.
(a)
Scheme
Marks
Det M = k ( 0 − 2) + 1(1 + 3) + 1( −2 – 0 ) = −2k + 4 − 2 = 2 – 2k
M1 A1
(2)
⎛k 1 3⎞
⎜
⎟
M = ⎜ -1 0 -2 ⎟ so cofactors =
⎜ 1 -1 1 ⎟
⎝
⎠
( -1 A mark for each term wrong)
-1
1 ⎞
⎛ -2
1 ⎜
⎟
−1
- 4 k - 3 k + 1⎟
M =
2 − 2k ⎜⎜
⎟
⎝ - 2 2k - 3 1 ⎠
T
(b)
⎛ -2
⎜
⎜ -4
⎜ -2
⎝
⎞
⎟
k - 3 k + 1⎟
2k - 3 1 ⎟⎠
-1
1
M1
M1 A3
(5)
(c)
⎛ x′ ⎞ ⎛ 4 ⎞
⎛ 4⎞
⎜
⎟
⎜
⎟
Let ( x, y , z ) be on l1 . Equation of l2 can be written as ⎜ y′ ⎟ = ⎜ 1 ⎟ + λ ⎜⎜ 1 ⎟⎟ .
⎜ z′ ⎟ ⎜ 7 ⎟
⎜3⎟
⎝ ⎠ ⎝ ⎠
⎝ ⎠
⎛x⎞
⎛ x′ ⎞
⎛x⎞
⎛ −2 − 1 1 ⎞ ⎛ 4 + 4 λ ⎞
1 ⎜
⎜
⎟
⎟
⎜
⎟
⎟⎜
⎟
−1 ⎜
Use ⎜ y ⎟ = M ⎜ y′ ⎟ with k = 2. i.e. ⎜ y ⎟ =
−4 − 1 3 ⎟ ⎜ 1 + λ ⎟
⎜
⎜ z ⎟ −2 ⎜ −2 1 1 ⎟ ⎜ 7 + 3λ ⎟
⎜z⎟
⎜ z′ ⎟
⎝ ⎠
⎝
⎠⎝
⎠
⎝ ⎠
⎝ ⎠
⎛ x ⎞ ⎛ 3λ + 1 ⎞
⎜ ⎟ ⎜
⎟
∴ ⎜ y ⎟ = ⎜ 4λ − 2 ⎟
⎜ z ⎟ ⎜ 2λ ⎟
⎝ ⎠ ⎝
⎠
and so (r - a)×b = 0 where a = i - 2j and b = 3i +4 j + 2k or equivalent
or r = a + λb where a = i - 2j and b = 3i +4 j + 2k or equivalent
B1
M1
M1 A1
B1ft
(5)
12
Notes:
(a) M1 Finding determinant at least one component correct.
A1 CAO
(b) 1M1
2M1
1A1
2A1
3A1
(c) 1B1
1M1
2M1
A1
2B1
Finding matrix of cofactors or its transpose
Finding inverse matrix, 1/(det) cofactors + transpose
At least seven terms correct (so at most 2 incorrect) condone missing det
At least eight terms correct (so at most 1 incorrect) condone missing det
All nine terms correct, condone missing det
Equation of l2
Using inverse transformation matrix correctly
Finding general point in terms of λ .
CAO for general point in terms of one parameter
ft for vector equation of their l1
GCE Further Pure Mathematics FP3 (6669) June 2011
6
Question
Number
Scheme
Marks
8.
(a)
2 x 2 yy′
xb 2 b cosh θ
b cosh θ
− 2 = 0 → y′ = 2 =
=
or
a2
b
ya
a sinh θ
a sinh θ
b cosh θ
( x − a cosh θ )
So y − b sinh θ =
a sinh θ
2
2
2
2
∴ ab(cosh θ − sinh θ ) = xb cosh θ − ya sinh θ and as (cosh θ − sinh θ ) =1
xb cosh θ − ya sinh θ = ab ∗
dy
Uses
=
dx
dy
dθ
dx
dθ
M1 A1
M1
A1cso
(4)
(b)
P is the point (
M1 A1
a
, 0)
cosh θ
(2)
(c)
l2 has equation x = a and meets l1 at Q (a,
b(cosh θ − 1)
)
sinh θ
M1 A1
(2)
a(cosh θ + 1)
b(cosh θ − 1)
, Y=
2 cosh θ
2sinh θ
2
2
⎛ cosh θ + 1 − 2 cosh θ + sinh θ ⎞
4Y 2 + b 2 = b 2 ⎜
⎟
sinh 2 θ
⎝
⎠
2
⎛ 2 cosh θ − 2 cosh θ ⎞
= b2 ⎜
⎟
sinh 2 θ
⎝
⎠
⎛ (cosh θ + 1)(cosh θ − 1)2 cosh θ ⎞
X (4Y 2 + b 2 ) = ab 2 ⎜
⎟
2 cosh θ sinh 2 θ
⎝
⎠
2
2
Simplify fraction by using cosh θ − sinh θ = 1 to give x(4 y 2 + b2 ) = ab2 ∗
(d) Alt 1 The mid point of PQ is given by X =
1M1 A1ft
2M1
3M1
4M1
A1cso
(6)
(d) Alt 2 First line of solution as before
4Y 2 + b 2 = b 2 ( coth 2 θ + cosech 2θ − 2 coth θ cosechθ + 1)
1M1A1ft
2M1
= b 2 ( 2 coth 2 θ − 2 coth θ cosechθ )
3M1
X (4Y 2 + b 2 ) = ab 2 ( coth θ (coth θ − cosechθ ) (1 + sechθ ))
4M1
Simplify expansion by using coth 2 θ − cosech 2θ = 1 to give x(4 y 2 + b2 ) = ab2 ∗ A1cso
(6)
14
GCE Further Pure Mathematics FP3 (6669) June 2011
7
Question
Number
8.
(a)1M1
1A1
2M1
2A1
Scheme
Finding gradient in terms of θ
CAO
Finding equation of tangent
CSO (answer given) look for ±(cosh 2θ − sinh 2θ )
(b)M1 Putting y = 0 into their tangent
A1ft P found, ft for their tangent o.e.
(c) M1 Putting x = a into their tangent.
A1 CAO Q found o.e.
(d)
1M1
1A1
2M1
3M1
4M1
2A1
For Alt 1 and 2
(d)
1M1
1A1
2M1
3M1
4M1
2A1
For Alts 3, 4 and 5
Finding expressions, in terms of sinh θ and cosh θ but must be adding
Ft on their P and Q,
2
2
Finding 4 y + b
Simplified, factorised, maximum of 2 terms per bracket
2
2
Finding x(4 y + b ) , completely factorised, maximum of 2 terms per bracket
CSO
Finding expressions, in terms of sinh θ and cosh θ but must be adding
Ft on their P and Q
Getting cosh θ in terms of x
y or y 2 in terms of cosh θ or sinh θ in terms of x and y
Getting equation in terms of x and y only. No square roots.
CSO
GCE Further Pure Mathematics FP3 (6669) June 2011
8
Marks
Question
Number
8(d)
Alt 3
Scheme
a(cosh θ + 1)
b(cosh θ − 1)
, Y=
2 cosh θ
2sinh θ
a
coshθ =
2x − a
b(coshθ -1) b(a − x)
sinhθ =
=
2y
(2 x − a) y
Marks
As main scheme
X=
1M1 A1ft
cosh θ in terms of x
2M1
sinh θ in terms of x and y
3M1
2
2
Using cosh 2θ - sinh 2θ = 1
⎛ a ⎞ ⎛ b(a − x) ⎞
⎟ =1
⎜
⎟ −⎜
⎝ 2 x − a ⎠ ⎝ (2 x − a ) y ⎠
Simplifies to give required equation
⎡⎣ y 2 4 x ( a − x ) = b 2 (a − x) 2 , x(4 y 2 + b 2 ) = ab 2 ⎤⎦
4M1
A1cso
(6)
Alt 4
a(cosh θ + 1)
b(cosh θ − 1)
, Y=
2 cosh θ
2sinh θ
a
coshθ =
2x − a
2
b (coshθ - 1) 2 b 2 (coshθ - 1)
=
y2 =
4(cosh 2θ - 1) 4(coshθ + 1)
X=
⎛ 2a − 2 x ⎞
b ⎜
⎟
2x − a ⎠
o.e
y2 = ⎝
⎛ 2x ⎞
4⎜
⎟
⎝ 2x − a ⎠
Simplifies to give required equation
2
2
As main scheme
1M1 A1ft
cosh θ in terms of x
2M1
y 2 in terms of cosh θ only
3M1
Forms equation in x and y only
4M1
A1 cso
(6)
Alt 5
a(cosh θ + 1)
b(cosh θ − 1)
, Y=
2 cosh θ
2sinh θ
a
coshθ =
2x − a
⎛ b(coshθ - 1) ⎞ ⎛ b(coshθ - 1) ⎞
y=⎜
⎟
⎟=⎜
⎝ 2sinhθ ⎠ ⎝ 2 cosh 2θ - 1 ⎠
X=
Eliminate
and forms equation in x and y
Simplifies to give required equation
GCE Further Pure Mathematics FP3 (6669) June 2011
9
As main scheme
1M1 A1ft
cosh θ in terms of x
2M1
y in terms of cosh θ only
3M1
4M1
A1cso
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