Mark Scheme (Results) Summer 2008
GCE
GCE Mathematics (6675/01)
Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH
June 2008 Further Pure Mathematics FP2 Mark Scheme Question number 1.
Scheme
Marks
d sech 2 x (ln(tanh x )) = dx tanh x
=
M1 A1
1 2 = = 2 cosech 2 x sinh x cosh x sinh 2 x
(*)
M1 A1
(4) 4
Notes −x
−x
1M1
Any valid differentiation attempt including ln(e − e ) − ln(e + e )
1A1
c.a.o. (o.e e.g.
2M1 2A1
Proceeding to a hyperbolic expression in 2x c.s.o.
x
cosh x sinh x ) − sinh x cosh x
2
x
June 2008 Further Pure FP2 Mark Scheme Question number 2.
Scheme
Marks
⎛ e x − e− x ⎞ ⎛ e x + e− x ⎞ ⎟⎟ = 13 ⎟⎟ − 4 ⎜⎜ 8 ⎜⎜ ⎝ 2 ⎠ ⎝ 2 ⎠
B1
4e x + 4e − x − 2e x + 2e − x = 13 2e 2 x − 13e x + 6 = 0
(or equiv.)
M1 A1
(2e x − 1) (e x − 6) = 0 ex =
1 , 2
x = ln
ex = 6
M1 A1ft
1 (or − ln 2), 2
x = ln 6
A1
(6) 6
Notes B1
Correctly substituting exponentials for all hyperbolics
1M1 1A1
To a three term quadratic in c.a.o. (o.e.)
2M1 2A1ft 3A1
Solving their equation to e = f.t. their equation. c.a.o.
ex
x
3
June 2008 Further Pure FP2 Mark Scheme Question number 3.
Scheme
∫
3 x2 − 9
dx +
∫
x x2 − 9
Marks
dx
B1
x ⎤ ⎡ = ⎢3 arcosh + x 2 − 9 ⎥ 3 ⎦ ⎣
M1 A1 A1 6
⎡ ⎛ x + x2 − 9 ⎞ ⎤ = ⎢3ln ⎜ ⎟ + x2 − 9 ⎥ ⎟ (3) ⎢⎣ ⎜⎝ ⎥⎦ 5 ⎠
⎛ ⎞ ⎛ 6 + 27 5+4 ⎞ = ⎜⎜ 3 ln ( ) + 27 ⎟⎟ − ⎜ 3 ln ( ) + 4⎟ 3 3 ⎠ ⎝ ⎠ ⎝ = 3 ln
M1 A1
6 + 27 2+ 3 + 27 − 4 = 3 ln +3 3−4 3 9
Notes B1 1M1 1A1 2A1 2DM1 3A1 4A1
(*)
A1
(7) 7
Correctly changing to an integrable form. Complete attempt to integrate at least one bit. One term correct All correct Substituting limits in all.Must have got first M1 Correctly (no follow through) c.s.o.
4
June 2008 Further Pure FP2 Mark Scheme Question number 4.
Scheme dy 3x 2 (a) = , dx 1 + x6
Marks dy 6 = =2 dx 3
At x = 2
M1 A1, A1
y − arsinh (2 2 ) = 2( x − 2 )
(
y = 2 x − 2 2 + ln 3 + 2 2 3a 2
(b)
1+ a
6
=2
M1
)
(*)
9a 4 = 4(1 + a 6 )
4a 6 − 9a 4 + 4 = 0
a2 =
A1
1 + 33 ≈ 0.92 8
a=
(5)
M1 A1
(a 2 − 2)(4a 4 − a 2 − 2) = 0
1 ± 1 + 32 8
A1
M1 A1
(5) 10
Notes (a)1M1 Attempt to differentiate need
(
1+ x
6
)
−1
2
at least
1A1 correct 2A1 c.a.o. 2M1 Substituting into straight line equation (linear). Must use x = 3A1 c.s.o. (b)1M1 Their derivative = their gradient (condone x throughout) 2M1= A mark cao, any form 1A1 quartic cao 3M1 Solving their quartic to ‘a’ = 2A1 c.a.o. (a.w.r.t. 0.92 to 2dp)
5
2
June 2008 Further Pure FP2 Mark Scheme Question number 5.
Scheme (a) I n =
π
∫
[
] ∫ e n sin
e x sin n x dx = e x sin n x −
0
Marks x
n −1
x cos x dx
M1 A1
[e
x
sin n x − ne x sin n −1 x cos x + n e x (− sin n x + (n − 1) cos x sin n − 2 x cos x) dx
] ∫
M1 A1
[e
x
sin n x − ne x sin n −1 x cos x
]
=0
B1
∫
M1
π 0
∫
I n = −n e x sin n x dx + n(n − 1) sin n − 2 x(1 − sin 2 x) dx I n = −nI n + n(n − 1) I n − 2 − n(n − 1) I n (b) I 4 = I0 =
4×3 I2 , 17
∫
π
=
[ ]
e x dx = e x
0
π 0
In =
n(n − 1) I n−2 n2 + 1
(*)
12 2 × I0 17 5
= ....
,
M1 A1
M1, A1
I4 =
(
)
24 π e −1 85
M1, A1 12
(a)1M1 1A1 2M1 2A1 1B1
n −1
Complete attempt to use parts once in the right direction need sin x cao Attempt to use parts again with sensible choice of parts, not reversing. Need to be differentiating a product. cao both = 0 at some point. (doesn’t need to be correct, must must =0)
= expressions in ∫ e x sin k x dx Depends on 2nd M
3DM1 I n
4DM1Expresssion in I n and I n − 2 to 3A1 c.s.o. (b)1M1 I 4 in terms of
I n = . Depends on 3rd M
I2
1A1 I 4 correctly in terms of I 0 [ o.e.] 2M1
(8)
∫ e dx x
2A1 c.a.o for I 4 .
June 2008 6
(4)
Further Pure FP2 Mark Scheme Question number 6.
Scheme (a)
∫
Marks
∫
cosh x arctan(sinh x) dx = sinh x arctan(sinh x) − sinh x
cosh x dx 1 + sinh 2 x
M1 A1 A1
1 = sinh x arctan(sinh x) − ln(1 + sinh 2 x) (+C ) 2
M1 A1
(5)
M1, A1
(2)
∫
............ − tanh x dx
Or:
= sinh x arctan(sinh x) − ln(cosh x) (+C )
M1 A1
Alternative: Let t = sinh x,
dt = cosh x, dx
t
∫ arctan t dt = t arctan t − ∫ 1 + t
2
dt
1 = ...... − ln (1 + t 2 ) 2 1 = sinh x arctan(sinh x) − ln(1 + sinh 2 x) (+C ) 2 1 [sinh x arctan(sinh x) − ln(cosh x)] 02 = ..... , (b) 10
(or equiv.)
M1 A1 A1 M1 A1
0.34
(*)
7
(a) Alternative: Let tan t = sinh x, sec2 t
dt = cosh x, dx
∫ t sec
2
∫
t dt = t tan t − tan t dt
M1 A1 A1
= ...... − ln (sec t )
M1
= sinh x arctan(sinh x) − ln 1 + sinh 2 x (+C ) Notes (a)1M1 1A1 2A1 2M1 3A1 (b)1M1 1A1
Complete attempt to use parts One term correct. All correct. All integration completed. Need a ln term. c.a.o. ( in x) o.e, any correct form, simplified or not Use of limits 0 and 2 and 1/10. c.s.o.
7
(or equiv.)
A1
June 2008 Further Pure FP2 Mark Scheme Question number 7.
Scheme (a)
Marks dy ⎡ dx ⎤ 4 sec t tan t , = = 3 sec 2 t ⎥ ⎢ dt dt ⎣ ⎦
2 x 2 y dy − =0 16 9 dx dy 9x 36 sec t 3 = = = dx 16 y 48 tan t 4 sin t y − 3 tan t =
M1 A1
− 4 sin t ( x − 4 sec t ) 3
M1
4 x sin t + 3 y = 25 tan t (b) Using b 2 = a 2 (e2 − 1) :
P: 4 sec t = 5
cos t =
(*) ae = a 2 + b 2 = 5
or e =
5 4
4 5
(6)
M1 A1
M1 A1
25 tan t 125 = 4 sin t 16
M1 A1 14
Notes (a)1M1 Differentitating 1A1 c.a.o. 2M1
dy in terms of t. dx
2A1 c.a.o. 3M1 Substituting gradient of normal into straight line equation. 3A1 c.s.o. (b)1M1 Use of b = a (e − 1) 1A1 c.a.o. for ae or for e 2M1 Using x coordinate of focus= x coordinate of P, to get single term f(t)= constant. (Allow recovery in (c) ) 3M1 Substituting into P coordinates to a number for x and for y. 2A1 c.a.o. (c)1M1 Attempt to find x coordinate of R. 2M1 Substituting into correct template i.e. ½ x |their Rx - their Hx| x their Py 1A1 c.a.o. 3 s.f. or better. 2
2
(5)
M1
1 (SR × SP ) = 1 × ⎛⎜ 125 − 5 ⎞⎟ × 9 = 405 ⎛⎜ = 3 21 ⎞⎟ 2 2 ⎝ 16 ⎠ 4 128 ⎝ 128 ⎠
Area of PRS:
A1
M1
⎛ 9⎞ Coordinates of P: (4 sec t , 3 tan t ) = ⎜ 5, ⎟ ⎝ 4⎠ (c) R: x =
M1 A1
2
8
(3)
June 2008 Further Pure FP2 Mark Scheme Question number 8.
Scheme
Marks
(a) x& = 3 + 3 cos t y& = 3 sin t
B1
2 sin t cos t dy y& sin t 2 2 = tan t = = = 2 dx x& 1 + cos t 2 2 cos t 2 (b) s =
∫
(*)
x& 2 + y& 2 dt = 3 2 ∫ 1 + cos t dt
t t t = 6 ∫ cos dt = 12sin 0 2 2
(c) tanψ = tan
∫
t 0
(Limits or establish C = 0 for A1)
(*)
2πy x& 2 + y& 2 dt = 18 2π ∫ (1 − cos t ) 1 + cos t dt
M1 A1
(4)
B1
(1)
M1 A1ft
t t = 72π ∫ sin 2 cos dt 2 2
M1
t⎞ ⎛2 = ...........⎜ sin 3 ⎟ 2⎠ ⎝3
M1 A1
But sin (a)1B1 both
t s L = = , 2 12 12
(3)
M1 A1ft
t t ⇒ ψ = ⇒ s = 12 sinψ 2 2
(d) Surface area =
M1 A1
so surface area =
144π L 3 π L 3 × 3 = 3 36 12
1M1 Attempt at y’/x’ 1A1 cso – on paper need to see half angles (b)1M1 Attempt at arc length, integral formula 1A1 cao follow through on their x’ and y’ one variable only 2M1 Integrating 2A1 cso – on paper (c) 1B1 cao (d) 1M1 Attempt at Surface area, integral formula.Condone lack of 2π. 1A1 cao follow through on their x’ and y’ condone lack of 2π. one variable only 2DM1Getting to integrable form condone lack of 2π. Depends on previous M mark. 3DM1integrating condone lack of 2π. Depends on previous M mark. 2A1 cao 4DM1Eliminating t to give expression in L only Depends on previous M mark. 3A1 cso – on paper.
9
(*)
M1 A1
(7)