Mark Scheme (Results) Summer 2007 - Edexcel

Mark Scheme (Results)
Summer 2007
GCE
GCE Mathematics
Further Pure Mathematics FP1 (6674)
Edexcel Limited. Registered in England and Wales No. 4496750
Registered Office: One90 High Holborn, London WC1V 7BH

June 2007
6674 Further Pure Mathematics FP1
Mark Scheme
Question Scheme Marks
number
1.
1
1 and 3 are ‘critical values’, e.g. used in solution, or both seen as asymptotes B1
2
( x + 1)( x − 3) = 2x
− 3 ⇒ x(
x − 4) = 0
x = 4, x = 0 M1: attempt to find at least one other critical value M1 A1, A1
1
1
0 < x < 1 , 3 < x < 4 M1: An inequality using 1 or 3 M1 A1, A1 (7)
2
2
First M mark can be implied by the two correct values, but otherwise a method
must be seen. (The method may be graphical, but either (x = ) 4 or (x =) 0 needs
to be clearly written or used in this case).
Ignore ‘extra values’ which might arise through ‘squaring both sides’ methods.
≤ appearing: maximum one A mark penalty (final mark).
7

Question Scheme Marks
number
2. Integrating factor
∫
−tan
x dx
ln(cos x)
−ln(sec
x)
⎛ ⎞
( ) ⎟
⎠
e
⎛ dy
⎜cos
x − y sin x =
⎝ dx
∫
y cos x = 2sec x dx
y cos x = 2 tan x ( + C)
2
2sec
= e
⎞
x⎟
⎠
(or equiv.)
or
e
,
= cos x ⎜or
⎝
2 ⎛
⎞
( ) ⎟
⎠
(or equiv.)
1
sec x
M1, A1
d
2
⎜Or : y cos x = 2sec x
M1 A1(ft)
⎝ dx
y = 3 at x = 0: C = 3 M1
2 tan x + 3
y = (Or equiv. in the form y = f(x)) A1 (7)
cos x
1 st tan d ln(cos ) ln(sec )
M: Also scored for e
∫
x x − x
x
= e ( or e )
, then A0 for sec x .
2 nd M: Attempt to use their integrating factor (requires one side of the equation
‘correct’ for their integrating factor).
2 nd A: The follow-through is allowed only in the case where the integrating
=
4
factor used is sec x or − sec x .
⎛
y sec x 2sec x dx
⎞
⎜
∫
⎟
⎝ ⎠
3 rd M: Using y = 3 at x = 0 to find a value for C (dependent on an integration
attempt, however poor, on the RHS).
Alternative
1 st M: Multiply through the given equation by cos x .
1 st dy
2
A: Achieving cos x − y sin x = 2sec x . (Allowing the possibility of
dx
integrating by inspection).
A1
7

Question Scheme Marks
number
3
3
2
3. (a) ( r + 1) = r + 3r
+ 3r
+ 1 and ( r −1)
= r − 3r
+ 3r
−1
M1
3
3
2
3
( r + 1) − ( r −1)
= 6r
3
2
+ 2
(*) A1cso (2)
3
(b) r = 1: 2 − 0 = 6(1 ) + 2
3
3
3
2
2
r = 2 : 3 −1
= 6(2 ) + 2
: : : : : : : : : : :
r = n :
3
3
( n + 1) − ( n −1)
2
= 6n
+ 2 M: Differences: at least first, last M1 A1
and one other.
3 3
2
Sum: ( n + 1) + n −1
= 6∑
r + 2n
M: Attempt to sum at least one side. M1 A1
2 3 2
( 6 ∑ r = 2n
+ 3n
+ n)
n
∑
r=
1
r
2
1
= n(
n + 1)(2n
+ 1)
6
(Intermediate steps are not required) (*) A1cso (5)
2n
2n
n−1
2
2
2 1
1
(c) ∑ r = ∑ r −∑
r , = (2n)(2n
+ 1)(4n
+ 1) − ( n −1)
n(2n
−1)
6
6
r=
n
r=
1
r=
1
2
( 16n
+ 12n
+ 2) − ( 2n
− 3 + 1
1 2
= n n )) M1
6
M1, A1
1
= n ( n + 1)(14n
+ 1)
A1 (4)
6
(b) 1 st A: Requires first, last and one other term correct on both LHS and RHS
(but condone ‘omissions’ if following work is convincing).
2n
∑
∑
∑
(c) 1 st M: Allow also for r = r − r .
r=
n
2
2n
r=
1
2
2 nd 1
M: Taking out (at some stage) factor n , and multiplying out brackets to
6
reach an expression involving terms.
n
r=
1
2
n
2
11

20
10
−2 −1 1 2
−10
−20
y
x
3
Question Scheme Marks
number
4. (a) f ′(
x ) = 3x
2 + 8 3x 2 + 8 = 0.......
or 3x
2 + 8 > 0.......
M1
Correct derivative and, e.g., ‘no turning points’ or ‘increasing function’. A1
Simple sketch, (increasing, crossing positive x-axis) B1 (3)
(or, if the M1 A1 has been scored, a reason such as ‘crosses
x-axis only once’).
(b) Calculate f (1) and f (2)
( Values must be seen) M1
f (1) = −10,
f (2) = 5 , Sign change, ∴Root A1 (2)
f (2)
5
(c) x 1 = 2 − ,
= 2 − ( = 1. 75) M1, A1
f ′(2)
20
f ( x1)
⎛ 0.359375 ⎞
x
2
= x1
− ,
⎜=
1.75 − ⎟ = 1.729 (ONLY) (α) M1, A1 (4)
f ′(
x )
⎝ 17.1875 ⎠
1
(d) Calculate f ( α − 0.0005)
and f ( α + 0.0005)
M1
(or a ‘tighter’ interval that gives a sign change).
f (1.7285) = −0.0077...
and f (1.7295) = 0.0092...
, ∴Accurate to 3 d.p. A1 (2)
11
(a) M: Differentiate and consider sign of f ′(
x)
, or equate f ′(
x)
to zero.
Alternative:
3
3
M1: Attempt to rearrange as x −19
= −8x
or x = 19 − 8x
(condone sign slips),
and to sketch a cubic graph and a straight line graph.
A1: Correct graphs (shape correct and intercepts ‘in the right place’).
B1: Comment such as “one intersection, therefore one root”).
(c) 1st A1 can be implied by an answer of 1.729, provided N.R. has been used.
Answer only: No marks. The Newton-Raphson method must be seen.
(d) For A1, correct values of f(1.7285) and f(1.7295) must be seen, together with a
conclusion. If only 1 s.f. is given in the values, allow rounded (e.g. − 0.008)
or truncated (e.g. − 0.007 ) values.

Question Scheme Marks
number
2
5. C.F. m + 3m
+ 2 = 0 m = −1
and m = −2
M1
y =
Ae
− x
+ Be
−2x
A1 (2)
P.I.
2
y = cx + dx + e
B1
d 2
2
y
d y
2
= 2cx
+ d,
= 2c
2c
+ 3(2cx
+ d)
+ 2( cx + dx + e)
≡ 2x
+ 6x
M1
2
dx
dx
2 c = 2
c = 1
(One correct value) A1
6 c + 2d
= 6
d = 0
2c + 3d
+ 2e
= 0
e = −1
(Other two correct values) A1
− x
−2x
2
General soln: y = Ae
+ Be
+ x −1
(Their C.F. + their P.I.) A1ft (5)
( A + 2)
x = 0 , y = 1: 1 = A + B −1
B =
M1
d −x
− 2 x
y
= −Ae
dx
− 2Be
dy
+ 2x,
x = 0, = 1:
1 = −A
− 2B
M1
dx
Solving simultaneously: A = 5 and B = −3
M1 A1
− x
−2x
2
Solution: y = 5e − 3e + x −1
A1 (5)
1 st M: Attempt to solve auxiliary equation.
12
2 nd 2
dy
d y
M: Substitute their and
2 into the D.E. to form an identity in x with
dx
dx
unknown constants. .
3 rd M: Using y = 1 at x = 0 in their general solution to find an equation in A and B.
4 th M: Differentiating their general solution (condone ‘slips’, but the powers of
d y
each term must be correct) and using = 1 at x = 0 to find an equation
dx
in A and B.
5 th M: Solving simultaneous equations to find both a value of A and a value of B.

Question Scheme Marks
number
*
6. (a) z = 3 + i
B1
z
z
*
=
( 3 − i)( 3 − i)
( 3 + i)( 3 − i)
3 − 2 3i −1
=
,
3 + 1
2
z ⎛ 1 ⎞ ⎛ ± 3 ⎞
(b) = ⎜ ⎟ , = 1
*
⎜ ⎟ +
z ⎝ 2 ⎠
2
⎝ ⎠
2
=
1
2
−
3
2
⎡
⎢Or
:
⎢
⎣
i
z
*
z
z 3 + 1 ⎤
= = , = 1⎥
*
z 3 + 1 ⎥
⎦
(c) arg ( w ) = arctan⎜
real( w)
⎟ or arg ( ) = arctan⎜
⎠
imag( w)
⎟ ⎠
(*) M1, A1cso (3)
M1, A1 (2)
⎛ imag ( w)
⎞
⎜ ±
w
⎛
⎜ ±
real ( w)
⎞
, M1
⎝
⎝
where w is z or
* z
z or
*
z
⎛ − 3 ⎞
⎛ z ⎞ ⎜ ⎟
⎜ ⎟ =
2
π
arg arctan⎜
⎟
= −
*
⎝ z ⎠ ⎜
1
⎟
3
⎝ 2 ⎠
A1
⎛ −1
⎞ π
⎛ 1 ⎞ π
*
arctan⎜
⎟ = − and arctan⎜
⎟ = (Ignore interchanged z and z ) A1
⎝ 3 ⎠ 6
⎝ 3 ⎠ 6
* π π π ⎛ z ⎞
arg z − arg z = − − = − = arg⎜
*
⎟
6 6 3 ⎝ z ⎠
A1 (4)
(d)
2 y
1
*
z
z and
*
z
(Correct quadrants, approx. symmetrical) B1
− 1 2 3
−
(e) ( − ( 3 − i
)( x − ( 3 + i
)
−
x
z
*
z z
(Strictly inside the triangle shown here) B1 (2)
x M1
⎛ − b ⎞
⎞
Or: Use sum of roots ⎜= ⎟ and product of roots ⎜
⎟ ⎝ a ⎠ ⎝ ⎠
2
x − 2 3x + 4
A1 (2)
13
(a) M: Multiplying both numerator and denominator by 3 − i , and multiplying
out brackets with some use of i 2 = −1.
(b) Answer 1 with no working scores both marks.
(c) Allow work in degrees: −60°, −30° and 30°
5π 11π π
Allow arg between 0 and 2π : , and (or 300°, 330° and 30°).
3 6 6
Decimals: Allow marks for awrt −1.05 (A1), −0.524 and 0.524 (A1), but then
A0 for final mark. (Similarly for 5.24 (A1), 5.76 and 0.524 (A1)).
(d) Condone wrong labelling (or lack of labelling), if the intention is clear.

6
4
2
Question Scheme Marks
number
7. (a) 5
(b)
2 4 6
Shape (closed curve, approx. symmetrical about
the initial line, in all ‘quadrants’ and
x
−4
5 − √3 5 + √3 ‘centred’ to the right of the pole/origin). B1
−2
5 Scale (at least one correct ‘intercept’ r value…
−6
shown on sketch or perhaps seen in a table). B1 (2)
(Also allow awrt 3.27 or awrt 6.73).
y = r sinθ
= 5sinθ
+ √ 3sinθ
cosθ
M1
d y
2
= 5cosθ
− √ 3sin θ + √ 3cos
θ
dθ
( = 5cosθ
+ √ 3cos 2θ
)
A1
2
2
5cosθ − √ 31−
cos θ + √ 3cos θ =
M1
2
−2
−4
y
( ) 0
2√
3cos θ + 5cosθ
− √ 3 = 0
2 √ 3cosθ
−1
cosθ
+ √ 3 = 0 cosθ
=
( )( ) ...
(0.288…)
⎛
1 ⎞
θ = 1.28 and 5.01 (awrt) (Allow ±1.28 awrt) ⎜Also allow ± arccos ⎟ A1
⎝
2√
3 ⎠
⎛ 1 ⎞ 11
r = 5 + √ 3⎜
⎟ = (Allow awrt 5.50) A1 (6)
⎝ 2√
3 ⎠ 2
2
2
(c) r = 25 + 10√
3cosθ
+ 3cos θ
B1
2 53θ
⎛ sin 2θ
⎞
25 + 10√
3cosθ + 3cos θ dθ
= + 10√
3sinθ
+ 3
∫
⎜ ⎟
M1 A1ft A1ft
2
⎝ 4 ⎠
(ft for integration of ( ) bcosθ
a + and c cos2θ
respectively)
2π
1 ⎡
3sin 2θ
3θ
⎤
25 10 3sin
2 ⎢ θ + √ θ + +
4 2 ⎥
⎣
⎦ 0
= ......
M1
53
= 1 ( 50π + 3π
) =
π or equiv. in terms of π.
2
2
A1 (6)
14
(b) 2 nd M: Forming a quadratic in cos θ .
3 rd M: Solving a 3 term quadratic to find a value of cos θ (even if called θ ).
Special case: Working with r cosθ
instead of r sinθ
:
1 st 2
M1 for r cosθ
= 5cosθ
+ √ 3cos θ
1 st A1 for derivative − 5sinθ
− 2√
3sinθ
cosθ
, then no further marks.
(c) 1 st M: Attempt to integrate at least one term.
2 nd M: Requires use of the 2
1 , correct limits (which could be 0 to 2π, or
−π to π, or ‘double’ 0 to π), and subtraction (which could be implied).
M1