Mark Scheme (Results) January 2010
GCE
Further Pure Mathematics FP2 (6675)
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January 2010 Publications Code UA022799 All the material in this publication is copyright Š Edexcel Ltd 2010
January 2010 6675 Further Pure Mathematics FP2 Mark Scheme
Question Number Q1
Scheme
(a) Using “ b 2 = a 2 ( e 2 − 1) ”
x2 y2 − =1 4a 2 a 2 a 2 = 4a 2 ( e 2 − 1)
e=
Leading to
M1 A1
√5
A1
2
2a a x= "x = "⇒ e e ex 5 5 a= = 2 2
(b)
ρ=
Q2
At s = 0 , ψ =
Marks
π 4
ft their e
(2) [5]
ds = tanψ dψ
B1
s = ∫ tanψ dψ = ln secψ
( + A)
0 = ln √ 2 + A ⇒ A = − ln √ 2
⎛ ⎛ secψ ⎞ ⎞ ⎜ s = ln ⎜ 2 ⎟ ⎟ ⎝ √ ⎠⎠ ⎝
GCE Further Pure Mathematics FP2 (6675) January 2010
M1 A1ft
(3)
M1 A1 M1 A1
(5) [5]
1
Question Number
Q3
Scheme
(a)
(b)
Marks
⎛ e x − e − x ⎞⎛ e x + e − x ⎞ 2sinh x cosh x = 2 ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠ e 2 x − e −2 x = 2 = sinh 2x
M1
cso
x=0 2sinh x cosh x = 6 sinh 2 x + 7 sinh x 2cosh x = 6sinh x + 7 e x + e − x = 3e x − 3e − x + 7 2 e2 x + 7 e x − 4 = 0 ( 2 e x − 1)( e x + 4 ) = 0 ex =
(2)
B1 M1 M1 A1 M1
1 2
A1
x = − ln 2
GCE Further Pure Mathematics FP2 (6675) January 2010
A1
accept ln
2
1 2
A1
(7) [9]
Question Number Q4
Scheme
Marks
I n = ∫ 1. ( ln x ) dx n
(a)
= x ( ln x ) − ⌠ ⎮ x.n ( ln x ) ⌡ n
= x ( ln x ) − n ∫ ( ln x ) n
n −1
n −1
1 × dx x
M1 A1 A1
dx
= n ( ln x ) − nI n − 1 n
cso
I 3 = x ( ln x ) − 3I 2 3
(b)
(4)
M1
(
= x ( ln x ) − 3 x ( ln x ) − 2 I1 3
A1
2
)
M1
= x ( ln x ) − 3 x ( ln x ) + 6 I1 3
2
= x ( ln x ) − 3 x ( ln x ) + 6 ( x ln x − I 0 ) 3
2
= x ( ln x ) − 3 x ( ln x ) + 6 x ln x − 6 x 3
2
M1 A1
With the limits 1 and e I 3 = e − 3e + 6e − 6e + 6 = 6 − 2e
M1 A1
(6) [10]
If I1 is integrated directly I 3 = x ( ln x ) − 3I 2 3
(b)
M1
(
= x ( ln x ) − 3 x ( ln x ) − 2 I1 3
2
)
= x ( ln x ) − 3 x ( ln x ) + 6 I1 3
2
M1
1 I1 = ∫ 1.ln x dx = x ln x − ⌠ ⎮ x × dx = x ln x − x ⌡ x
(I
= x ( ln x ) − 3x ( ln x ) + 6 x ln x − 6 x 3
3
2
With the limits 1 and e I 3 = e − 3e + 6e − 6e + 6 = 6 − 2e
GCE Further Pure Mathematics FP2 (6675) January 2010
3
Allow if quoted
M1 A1
)
M1 A1
(6)
Question Number
Q5
Scheme
(a)
Marks
dy = 3sinh 3 x − 4 dx 3sinh 3x − 4 = 0 4 sinh 3 x = 3 4 3 x = arsinh 3 1 4 x = arsinh 3 3 1 ⎡ 2 = ln ⎢ 43 + √ ( 43 ) + 1 3 ⎣
(
(b)
∫ ( cosh 3x − 4 x ) dx = 1
ln 3
⎡ sinh 3 x ⎤3 A=⎢ − 2x2 ⎥ ⎣ 3 ⎦0
M1
M1
)⎤⎥⎦ = 13 ln 3
A1
sinh 3 x − 2x2 3
sinh ( ln 3) 2 ( ln 3 ) = − 3 9
=
M1 A1
M1 A1 2
M1 A1
3 − 13 eln 3 − e − ln 3 − …= − … 6 6
4 2 ( ln 3) = − 9 9 2 2 = ⎡ 2 − ( ln 3) ⎤ ⎣ ⎦ 9
(5)
M1
2
GCE Further Pure Mathematics FP2 (6675) January 2010
4
cso
A1
(6) [11]
Question Number
Q6
Scheme
x=
(a)
Marks
a dx a ⇒ =− 2 u du u
⌠ 1 ⌠ = x d ⎮ ⎮ 2 2 ⌡ x √ (a − x ) ⎮a ⎮u ⌡
⎛ a⎞ ⎜ − 2 ⎟ du ⎛ 2 a2 ⎞ ⎝ u ⎠ ⎜a − 2 ⎟ u ⎠ ⎝ 1
√
1⌠ 1 =− ⎮ du a ⌡ √ ( u 2 − 1) 1 = − arcosh u ( + A ) a 1 ⎛a⎞ = − arcosh ⎜ ⎟ ( + A ) a ⎝x⎠
(b)
B1
⌠ 1 1 ⎛5⎞ dx = − arcosh ⎜ ⎟ ⎮ 2 5 ⎝ x⎠ ⌡ x √ ( 25 − x )
M1 A1
M1 M1 A1
(6)
M1
4
⎡ 1 1⎛ ⎛ 5 ⎞⎤ ⎛5⎞ ⎛ 5 ⎞⎞ ⎢ − 5 arcosh ⎜ x ⎟ ⎥ = 5 ⎜ arcosh ⎜ 3 ⎟ − arcosh ⎜ 4 ⎟ ⎟ ⎝ ⎠⎦ 3 ⎝ ⎠ ⎝ ⎠⎠ ⎣ ⎝ 1 = [ ln 3 − ln 2] 5 1 3 = ln 5 2
M1 A1
M1 A1
(5) [11]
Alternative for part (a) x=
a dx sinh u ⇒ =− cosh u du cosh 2 u
⌠ ⌠ ⎮ 1 ⎮ dx = ⎮ ⎮ 2 2 ⎮ x √ (a − x ) ⎮ a ⎮ ⎮ cosh u ⌡ ⌡
⎛ a sinh u ⎞ ⎜− ⎟ du 2 ⎛ 2 ⎛ a ⎞2 ⎞ ⎝ cosh u ⎠ ⎜⎜ a − ⎜ ⎟ ⎟ ⎝ cosh u ⎠ ⎟⎠ ⎝ 1
√
1 1 1du = − u ( + A) ∫ a a 1 ⎛a⎞ = − arcosh ⎜ ⎟ ( + A) a ⎝ x⎠
=−
GCE Further Pure Mathematics FP2 (6675) January 2010
B1
5
M1 A1
M1 M1 A1
(6)
Question Number
Q7
Scheme
(a)
u = k arcsin 2 x ,
Marks
du 2k = dx √ (1 − 4 x 2 )
dy = cos u du dy 2k = cos u × 2 dx √ (1 − 4 x )
M1 A1
cos u = √ (1 − sin 2 u ) = √ (1 − y 2 )
dy 2k = √ (1 − y 2 ) × 2 dx √ (1 − 4 x )
√ (1 − 4 x ) dx = 2k √ (1 − y 2
dy
2
A1
)
2
⎛ ⎞ (1 − 4 x ) ⎜ ddxy ⎟ = 4k 2 (1 − y 2 ) ⎝ ⎠ 2
cso M1 A1
(5)
Using u = arcsin 2 x is marked similarly. 2
(b)
⎛ dy ⎞ dy d 2 y dy −8 x ⎜ ⎟ + (1 − 4 x 2 ) 2 = − 8k 2 y 2 dx dx dx ⎝ dx ⎠ dy d2y −4 x + (1 − 4 x 2 ) 2 = −4k 2 y dx dx
(1 − 4 x 2 )
d2y dy − 4 x + 4k 2 y = 0 2 dx dx
M1 A1 A1 M1
cso
A1
(5) [10]
Alternative for part (a) arcsin y = k arcsin 2 x 1 dy 2k = 2 dx 2 √ (1 − y ) √ (1 − 4 x )
√ (1 − 4 x ) dx = 2k √ (1 − y 2
dy
2
M1 A1 A1
)
2
⎛ ⎞ (1 − 4 x ) ⎜ ddxy ⎟ = 4k 2 (1 − y 2 ) ⎝ ⎠ 2
GCE Further Pure Mathematics FP2 (6675) January 2010
6
cso M1 A1
(5)
Question Number Q8
Scheme
Marks
(a) Eliminating y between y = mx + c and y 2 = 4ax
( mx + c ) = 4ax m2 x 2 + 2 ( mc − 2a ) x + c 2 = 0 2
For a tangent
“ b 2 − 4ac = 0 ” 2 4 ( mc − 2a ) − 4m 2 c 2 = 0 c=
Leading to
(b)
M1 A1 M1
a m
cso A1
a passes through ( 4a, 5a ) m a 5a = 4ma + m 2 4 m − 5m + 1 = 0 ( 4m − 1)( m − 1) = 0
(4)
y = mx +
M1 A1 M1
1 m= , 1 4 y=
1 x + 4a, 4
A1
y = x+a
both
(c) Obtaining the points of contact between the tangents and the curve 1 e.g. Substituting m = , c = 4a into the quadratic in part (a) 4 2 1 2 ⎛x ⎞ x − 2ax + 16a 2 = ⎜ − 4a ⎟ = 0 ⇒ x = 16a 16 ⎝4 ⎠ Point has coordinates (16a, 8a )
A1
(5)
M1 A1
Substituting m = 1, c = a into the quadratic in part (a) x 2 − 2ax + a 2 = ( x − a ) = 0 ⇒ x = a 2
Point has coordinates ( a, 2a )
A1
With all methods, M1 is for finding the x- or y-coordinate of one point of contact. A1 is for any two coordinates correct. The second A1 is for all four coordinates correct. RQ 2 = (16a − a ) + ( 8a − 2a ) = 261a 2 2
RQ = a √ 261
2
( = 3a √ 29)
M1 A1
(5) [14]
GCE Further Pure Mathematics FP2 (6675) January 2010
7
Question Number Q8
Scheme
Marks
Alternatives for part (a) 1 Let the point of contact of l and C be ( 2at , at 2 ) d y y& 2a 1 = = = dx x& 2at t 1 l has the form y = x+c t 1 2at = at 2 + c ⇒ c = at ( 2at , at 2 ) lies on l t a c = at = m a 2 Eliminate y between y = mx + and y 2 = 4ax m 2 a⎞ ⎛ ⎜ mx + ⎟ = 4ax m⎠ ⎝ m=
M1 A1 M1
cso A1
a2 = 4ax m2 a2 m 2 x 2 − 2ax + 2 = 0 m 2 a⎞ ⎛ ⎜ mx − ⎟ = 0 m⎠ ⎝
(4)
m 2 x 2 + 2ax +
M1 A1 M1
a cso m Note: Alternative 2 allows the coordinates of R and Q in part (c) to be a written down using x = 2 . m
As this is a complete square, l is a tangent to C when c =
A1
(4)
Alternative for part (b) y = mx +
a x = + at passes through ( 4a, 5a ) m t
4a + at t t 2 − 5t + 4 = 0 ( t − 1)( t − 4 ) = 0 t = 1, 4 1 y = x + 4a, y = x + a 4 5a =
GCE Further Pure Mathematics FP2 (6675) January 2010
8
M1 A1 M1 A1
both
A1
(5)
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