Mark Scheme (Results) Summer 2007
GCE
GCE Mathematics Further Pure Mathematics FP2 (6675)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
June 2007 6675 Further Pure Mathematics FP2 Mark Scheme Question Number
Scheme
Marks
x 2 + 4 x − 5 = ( x + 2) 2 − 9
1.
∫ √ ((
1 x + 2) − 9 2
)
dx = arcosh
x+2 3
B1 M1 A1ft
ft their completing the square, requires arcosh
x + 2⎤ 5 ⎡ ⎢⎣arcosh 3 ⎥⎦ = arcosh 3 ( −arcosh 1) 1 ⎛5 25 ⎞ ⎛5 4⎞ = ln ⎜⎜ + − 1 ⎟⎟ = ln ⎜ + ⎟ = ln 3 9 ⎝3 3⎠ ⎝3 ⎠ 3
M1 A1
(5) [5]
Alternative x 2 + 4 x − 5 = ( x + 2) 2 − 9 dx = 3sec θ tan θ dθ
Let x + 2 = 3sec θ ,
∫ √ ((
1 x + 2) − 9 2
)
dx =
∫
3sec θ tan θ dθ 2 √ ( 9sec θ − 9 )
∫
= sec θ dθ 5 arcsec ⎛5 4⎞ ⎡⎣ ln ( sec θ + tan θ ) ⎤⎦ arcsec13 = ln ⎜ + ⎟ = ln 3 ⎝3 3⎠
B1
M1 A1ft M1 A1
(5)
Question Number 2.
Scheme (a)
Marks
y 3
O
E
D
2
5
x
One ellipse centred at O Another ellipse, centred at O, touching on y-axis Intersections: At least 2, 5, and 3 shown correctly (b)
Using b 2 = a 2 (1 − e 2 ) , or equivalent, to find e or ae for D or E. 4 , 5
For S: a = 5 and b = 3 ,
e=
For T: a′ = 3 and b′ = 2 ,
5 e′ = √ , 3
ae = 4
a′e′ = √ 5
ST = √ (16 + 5) = √ 21
B1 B1 B1
(3)
M1
ignore sign with ae
A1
ignore sign with a′e′
A1 M1 A1
(5) [8]
Question Number
Scheme
3.
dy 1 ⎛ 1⎞ = ⎜ 4x − ⎟ dx 4 ⎝ x⎠
∫
1
⎛ ⎛ d y ⎞2 ⎞ ⎜ 1 + ⎜ ⎟ ⎟ dx = ⎜ dx ⎟ ⎝ ⎝ ⎠ ⎠ 2
=
∫
=
x 2 ln x + 2 4
∫
⎛ ⎛ 1 ⎞ ⎜⎜1 + ⎜ x − ⎟ 4x ⎠ ⎝ ⎝ 1
2
1 1⎞ ⎛ 2 − ⎟ dx = ⎜1 + x + 2 16 x 2 ⎠ ⎝
1
2
∫
Marks
B1
2
⎞ dx ⎟⎟ ⎠
M1 1
2
2 ⎛⎛ 1 ⎞ ⎞ ⎜⎜ ⎜ x + ⎟ ⎟⎟ dx = 4x ⎠ ⎠ ⎝⎝
∫
1 ⎞ ⎛ ⎜ x + ⎟ dx 4x ⎠ ⎝
M1 A1
A1 2
⎡ x 2 ln x ⎤ ln 2 1 ln 0.5 15 1 − − = + ln 2 ⎢ + ⎥ = 2+ 4 8 4 8 2 2 4 ⎣ ⎦ 0.5 15 1⎞ ⎛ ⎜a = , b = ⎟ 8 2⎠ ⎝
M1 A1
(7) [7]
Question Number 4.
Scheme
Marks
(a) ⎛ e A + e− A ⎞ ⎛ e B + e− B ⎞ ⎛ e A − e− A ⎞ ⎛ e B − e− B ⎞ cosh A cosh B − sinh A sinh B = ⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟ 2 2 2 2 ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ 1 = e A+ B + e− A+ B + e A− B + e− A− B − e A+ B + e− A+ B + e A− B − e− A− B 4 1 e A− B + e−( A− B ) = 2e − A + B + 2e A − B = = cosh( A − B) ¿ cso 4 2
(
)
(
)
cosh x cosh1 − sinh x sinh 1 = sinh x
(b)
M1
M1
cosh x cosh 1 = sinh x(1 + sinh 1) ⇒ tanh x =
e + e−1 e + e−1 e2 + 1 2 tanh x = = = e − e−1 2 + e − e −1 e 2 + 2e − 1 1+ 2
M1 A1 (3)
cosh 1 1 + sinh 1
M1
¿
M1 A1 (4)
cso [7]
Alternative for (b) e x −1 + e 2
Leading to
−( x −1)
e x − e− x 2 2 e +e e2 x = e −1 =
e x − e − x e 2 x − 1 e 2 + e − (e − 1) e2 + 1 tanh x = x = = = ¿ e + e − x e 2 x + 1 e 2 + e + (e − 1) e 2 + 2e − 1
M1 M1 cso
M1 A1 (4)
Question Number 5.
Scheme
Marks
(a) x = t − sin 2t ⇒ x& = 1 − 2 cos 2t ⇒ && x = 4 sin 2t y = cos 2t ⇒ y& = −2 sin 2t ⇒ && y = − 4 cos 2t
either both
&&& − &&& Obtaining x& 2 + y& 2 and xy xy in terms of t 2 2 2 x& + y& = 1 + 4 cos 2t − 4 cos 2t + 4sin 2 2t = 5 − 4 cos 2t &&& − &&& xy xy = −4 cos 2t (1 − 2 cos 2t ) − 4sin 2t ( −2sin 2t ) = 8 − 4 cos 2t
( 5 − 4 cos 2t ) ρ=
3
M1 A1 A1
2
8 − 4 cos 2t
(b)
M1 A1
A1
The least value of y ( cos 2t ) is −1
B1
(5 + 4) ρ=
B1
8+ 4
3
2
=
9 4
accept equivalent fractions or 2.25
(6)
(2) [8]
Question Number
Scheme 8
6.
Marks
4 4 3 I n = − 3 ⎡ x n (8 − x) 3 ⎤ + nx n −1 (8 − x) 3 dx 4 ⎣⎢ ⎥⎦ 0 4
(a)
=
∫ nx
n −1
3 4
∫
∫
4
nx n −1 (8 − x) 3 dx
M1 A1
ft numeric constants only A1ft
(8 − x )(8 − x) 3 dx = ∫ nx n −1 8(8 − x) 3 dx − ∫ nx n −1 x (8 − x) 3 dx 1
1
1
24n 3 I n = 6nI n −1 − nI n ⇒ I n = I n −1 ¿ 4 3n + 4
(b)
I0 =
∫
8 0
I=
(8 − x )
∫
8 0
1
cso
A1
(6)
8
3
4 ⎤ 3 4 ⎡ 3 dx = ⎢ − (8 − x) 3 ⎥ = × 8 3 = 12 ⎣ 4 ⎦0 4
M1 A1
1
x( x + 5)(8 − x) 3 dx = I 2 + 5I1
⎛ 576 ⎞ I0 ⎟ ⎜= ⎝ 35 ⎠ 168 ⎞ ⎛ I0 ⎟ ⎜ The previous line can be implied by I = I 2 + 5 I1 = 5 ⎝ ⎠ 24 ⎞ 2016 ⎛ 576 I =⎜ + 5 × ⎟ × 12 = ( = 403.2 ) 7 ⎠ 5 ⎝ 35 I1 =
M1 A1
48 48 24 24 I1 = × I 0 I0 , I2 = 10 10 7 7
M1 M1 A1
A1
(6) [12]
Question Number
7.
Scheme
(a)
(b)
Marks
)
(
1 d 1 1 −1 × x 2 arsinh x 2 = dx √ (1 + x ) 2 dy 1 At x = 4 , = dx 4 √ 5
⎛ ⎞ 1 ⎜⎜ = ⎟⎟ ⎝ 2 √ x √ (1 + x ) ⎠
accept equivalents
A1
(3)
dx = 2sinh θ cosh θ dθ
x = sinh 2 θ ,
∫ arsinh √ x dx = ∫ θ × 2 sinh θ cosh θ dθ
M1 A1
cosh 2θ dθ 2 2 sinh 2θ = ... − 4
= ∫ θ sinh 2θ dθ =
⎡θ cosh 2θ sinh 2θ ⎤ − ⎢⎣ 2 4 ⎥⎦ 0
M1 A1
θ cosh 2θ
−
∫
M1 A1 + A1 M1
arsinh 2
= …
attempt at substitution M1
⎡θ (1 + 2sinh 2 θ ) 2sinh θ cosh θ ⎤ 1 4 5 ⎥ = arsinh 2 × (1 + 8 ) − √ =⎢ − 2 4 4 ⎢⎣ ⎥⎦ 2 9 = ln 2 + √ 5 − √ 5 2
(
)
M1 A1 A1
(10) [13]
Alternative for (a) x = sinh 2 y,
2sinh y cosh y
dy =1 dx
M1
⎛ ⎞ dy 1 1 1 = = ⎜⎜ = ⎟ 2 dx 2sinh y cosh y 2sinh y √ ( sinh y + 1) ⎝ 2 √ x √ (1 + x ) ⎟⎠ At x = 4 ,
dy 1 = dx 4 √ 5
An alternative for (b) is given on the next page
accept equivalents
A1 A1
(3)
Question Number 7.
Scheme
Marks
Alternative for (b)
∫ 1× arsinh √ x dx = x arsinh √ x − ∫ x × 2 √ x √ (1 + x ) dx 1
∫
= x arsinh √ x −
Let x = sinh 2 θ ,
∫
√ x dx = √ (1 + x )
∫
arsinh 2
∫
4 0
√x dx 2 √ (1 + x )
dx = 2sinh θ cosh θ dθ
sinh θ × 2sinh θ cosh θ dθ cosh θ
= 2 ∫ sinh 2 θ dθ = 2 ⎡ sinh 2θ ⎤ ⎢⎣ 2 − θ ⎥⎦ 0
M1 A1 + A1
∫
M1 A1
cosh 2θ − 1 sinh 2θ −θ dθ , = 2 2
⎡ 2sinh θ cosh θ ⎤ =⎢ −θ ⎥ 2 ⎣ ⎦0
arsinh 2
=
2× 2× √ 5 − arsinh 2 2
⎞ 9 1 ⎛ 2× 2× √ 5 − arsinh 2 ⎟ = ln 2 + √ 5 − √ 5 arsinh √ x dx = 4 arsinh 2 − ⎜ 2⎝ 2 ⎠ 2
(
)
M1, M1 M1 A1 A1
The last 7 marks of the alternative solution can be gained as follows Let x = tan 2 θ ,
∫
√ x dx = √ (1 + x )
∫
dx = 2 tan θ sec 2 θ dθ
tan θ × 2 tan θ sec 2 θ dθ sec θ
dependent on first M1
M1 A1
= ∫ 2sec θ tan 2 θ dθ
∫ ( sec θ tan θ ) tan θ dθ = sec θ tan θ − ∫ sec θ dθ = sec θ tan θ − ∫ sec θ (1 + tan θ ) dθ 3
M1
2
∫
1 1 Hence ∫ sec θ tan 2 θ dθ = sec θ tan θ − sec θ dθ 2 2 1 1 = sec θ tan θ − ln ( sec θ + tan θ ) 2 2 1 1 arctan 2 [ ... ] 0 = × √ 5 × 2 − ln √ 5 + 2 2 2 4 1 9 ∫ 0 arsinh √ x dx = 4 arsinh 2 − √ 5 + 2 ln 2 + √ 5 = 2 ln 2 + √ 5 − √ 5
(
)
(
)
(
)
M1 M1 A1 A1
(10)
Question Number 8.
Scheme
Marks
2ap − 2aq 2 Can be implied = 2 2 ap − aq p+q Use of any correct method or formula to obtain an equation of PQ in any form. Leading to ( p + q ) y = 2 ( x + apq ) ¿ (a)
Gradient of PQ =
(b) Gradient of normal at P is − p . Given or implied at any stage Obtaining any correct form for normal at either point. Allow if just written down. y + px = 2ap + ap 3 y + qx = 2aq + aq 3 Using both normal equations and eliminating x or y. Allow in any unsimplified form. ( p − q ) x = 2a ( p − q ) + a ( p 3 − q 3 ) Any correct form for x or y 2 2 Leading to x = a( p + q + pq + 2) ¿ cso ¿ cso y = − apq ( p + q ) (c)
0 = 2 ( 5a + apq ) ⇒
pq = −5
Using pq = −5 in both x = a( p + q + pq + 2) and y = − apq ( p + q ) . x = a ( p 2 + q 2 − 3) y = 5a( p + q) Any complete method for relating x and y, independently of p and q. A correct equation in any form. ⎛ ⎛ y ⎞2 ⎞ 2 x = a ( p + q ) − 2 pq − 3 = a ⎜ ⎜ ⎟ + 10 − 3 ⎟ ⎜ ⎝ 5a ⎠ ⎟ ⎝ ⎠ 2 Leading to y = 25a ( x − 7a ) Accept equivalent forms of f ( x ) 2
(
2
B1 M1 A1
(3)
B1 M1 A1
M1 A1 A1 A1
(7)
B1 M1 M1 A1
)
The algebra above can be written in many alternative equivalent forms.
A1
(5) [15]