January 2006
6663 Core Mathematics C1 Mark Scheme Question
Scheme
number
1.
Marks
x(x2-4x + 3)
Factor of x. (Allow (x - 0))
Ml
= x(x-3)(x-l)
Factorise 3 term quadratic
Ml Al
(3) Total 3 marks
Alternative:
(x2 - 3x)(x -1) or (x2 - x)(x - 3) scores the second Ml (allow Âą for each sign), then x(x - 3)(x -1) scores the first M1, and A1 if correct. Alternative:
Finding factor (x -1) or (x - 3) by the factor theorem scores the second Ml, then completing, using factor x, scores the first M1, and A1 if correct.
Factors "split": e.g. x(x2 - 4x + 3) => (x - 3)(x -1). Allow full marks. Factor x not seen: e.g. Dividing by x => (x - 3)(x -1). MO Ml A0. If an equation is solved, i.s.w.
Question
Scheme
Marks
number 2.
Bl
(a)u2=(-2)2=4 «3 = 1, «4 = 4
For u3, ft (u2 - 3)2
'
Blft,Bl
(3) (b) u2Q = 4
Blft
(1) Total 4 marks
(b) ft only if sequence is "oscillating".
Do not give marks if answers have clearly been obtained from wrong working,
e.g. w2 = (3-3)2=0
w3 = (4-3)2=l
«4=(5-3)2=4
Question
Scheme
number 3.
(a) y = 5 - (2 x 3) = -1
Marks
(or equivalent verification)
(*)
Bl
0) (b) Gradient of L is —
Bl
2
y ~ (-1) = ~ (x ~ 3), x - 2y. - 5 = 0
(ft from a changed gradient") (or equiv. with integer coefficients)
Ml Alft Al
(4)
Total 5 marks
(a) y -(-1) = -2(x-3) => y = 5 -2x is fine for Bl.
Just a table of values including x = 3, j = -1 is insufficient.
(b) Ml: eqn of a line through (3, -1), with any numerical gradient (except 0 or oo).
For the Ml A lft, the equation may be in any form eg y~^~^ x-3
1
2'
Alternatively, the Ml may be scored by using y = mx + c with a numerical gradient and substituting (3,-1) to find the value of c, with A lft if the value of c follows through correctly from a changed gradient.
Allow x - 2y = 5 or equiv., but must be integer coefficients. The "= 0" can be implied if correct working precedes.
Question
Scheme
Marks
number
4.
(a) *¥- = 4x + 18x"4 dx
Ml: x2 -> x or x"3 -» x"4
MIA
(2) ~
(b)—
3
/-
-2
^+C
Ml: x2-> x3 or x'3 ^ x"2 or + C
Ml Al Al
(3)
[-— + 3x"2 + C
FirstAl: 1%\c Second Al:
*—
-2
Total 5 marks
In both parts, accept any correct version, simplified or not.
Accept 4x' for Ax. + C in part (a) instead of part (b):
Penalise only once, so if otherwise correct scores Ml AO, Ml Al Al.
Question
Scheme
number 5.
Marks 111
(a) 3V5
(or a = 3)
Bl
(1)
2(3 + V5)J3 + V5) (3-V5) "(3 + V5)
(3 - V 5)(3 + V 5) = 9 - 5
Ml
(=4)
(Used as or intended as denominator)
(3 + yJ5)(p Âą qa/5) = ... 4 terms (p * 0,q * 0) or
(Independent)
Bl
Ml
(6 + 2^l5)(pÂąq^5) = ... 4 terms (p*0,q*0)
[Correct version: (3 + V5)(3 + V5) = 9 + 3V5 + 3V5 + 5,or double this.]
2(14 + 6V5)=7 + 3V5
,.A1:i.7> 2-A1:c = 3
Al Al
(5) Total 6 marks
(b) 2nd M mark for attempting (3 + ^5)(p + q^5) is generous. Condone errors.
L
'
J Question
Scheme
Marks
number
6.
(a)
,i
\
1)
(See below)
Clearly through origin (or (0, 0) seen)
/ \
3\
„ ^
3 labelled Cor (3. 0) seen^
Ml Al Al
(3)
(b)
.
\ / \ i
4\
w
Al
1 and 4 labelled (or (1, 0) and (4, 0) seen)
Ml
Stretch parallel to >>-axis
b labelled (or (U, b) seen)
Al
(3) 5) v
-cs
/-\ 2
8\
r
Al
2 and 8 labelled (or (2, 0) and (8, 0) seen)
Ml
Stretch parallel to x-axis
3 labelled <^or (U, i) seen)
Al
(3)
Total 9 marks
(a)Ml:\
(b) Ml: \
\
(c) M1:
with at least two of:
/^\
\y
\
(1, 0) unchanged
(4, 0) unchanged (0,3) changed
\
with at least two of:
\ / \
\
(1, 0) changed
(4, 0) changed
(0, 3) unchanged
Beware: CanLdidates may sometimes re-label the parts of their solution.
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Question
Scheme
number
7.
(a) 500+ (500+ 200) =1200 or
Marks
S2 = — 2{l000 + 200} = 1200
(b) Using a = 500, d = 200 with n = 7, 8 or 9
(*)
a + (« - l)rf or "listing"
500 + (7x200) = (£)1900
Bl
Ml Al
(c) Using -«{2a + (n - \)d} or -n{a + /}, or listing and "summing" terms
= (£) 9600
Al
Al
Ml: General Sn, equated to 32000
Ml Al
n2 + An - 320 = 0 (or equiv.)
Ml: Simplify to 3 term quadratic
Ml Al
(n + 20)(«-16) = 0
Ml: Attempt to solve 3 t.q.
Ml
«=16,
/» = ...
(2)
Ml
Ss = -8{2 x 500 + 7 x 200} or S, = ^-8(500 + 1900}, or all terms in list correct (d) -n{2 x 500 + (n -1) x 200} = 32000
(1)
Age is 26
(3)
Alcso,Alcso
(7) Total 13 marks
(b) Correct answer with no working: Allow both marks.
(c) Some working must be seen to score marks: Minimum working: 500 + 700 + 900 +...(+ 1900) =...
scores Ml (Al).
(d) Allow > or > throughout, apart from "Age 26".
A common misread here is 3200. This gives n = 4 and age 14, and can score Ml A0 Ml A0 Ml Al Al with the usual misread rule. Alternative: (Listing sumsl
(500, 1200, 2100, 3200, 4500, 6000, 7700, 9600,) 11700, 14000, 16500 19200,22100, 25200, 28500, 32000.
List at least up to 32000 All values correct
M3 A2
n= 16 (perhaps implied by age) Age 26
Alcso Alcso
If there is a mistake in the list, e.g. 16th sum = 32100, possible marks are: M3 A0 A0 A0
Alternative: (Trial and improvement)
Use of Sn formula with n = 16 (and perhaps other values)
M3
Accurately achieving 32000 for n = 16
A3
Age 26
A1
10
Question
Scheme
Marks
number
5x2
Ml: One term correct.
Ml Al
X1 Al: Both terms correct, and no extra terms.
(+ C not required here)
\2)
Ml Alft
\2)
6=3+2+4+C
Use of x = 1 and;; =6 to form eqn. in C
C = -3
Ml
Alcso 5
1
3x + 2x2+4x2-3
(simplified version required)
Al (ft Q
(7) [or: 3x + 2Vx +4Vx-3 or equiv.]
Total 7 marks
For the integration:
Ml requires evidence from just one term (e.g. 3 -> 3x), but not just "+C". Alft requires correct integration of at least 3 terms, with at least one of these terms having a fractional power.
For the final Al, follow through on C only.
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Question
Scheme
number 9.
(a)
(b)
-2 (P),
Q
Marks
2 (Q)
(± 2 scores B1 B1)
y = x* -x2-4x + 4 (May be seen earlier)
Bl, Bl
Multiply out, giving 4 terms
—-
(*)
.,
(2)
Ml Ml Alcso
(3)
(c)
Afx = -l:^ = 3(-l)2-2(-l)-4 = l ax
Eqn. of tangent:
y-6 = l(x-(-\)),
y = x+7
(*)
Ml Alcso
(2) (d)
3x2 - 2x - 4 = 1
(Equating to "gradient of tangent")
3x2-2x-5 = 0 5
x = -
(3jc-5)(jc + 1) = 0
Ml
x=...
Ml
or equiv.
(5
Y25
Al
1
2 r in
22
•y = l3"1AT~4J' = 3xr"9~J = ~I7 orequiv-
Ml, Al
(5) Total 12 marks
(b) Alternative:
Attempt to differentiate by product rule scores the second Ml: dv
( — MY
^ 4^ X 1 r 1
( Y"
1^ V ^ v^
Then multiplying out scores the first Ml, with Al if correct (cso)
(c) Ml requires full method: Evaluate — and use in eqn. of line through (-1,6), dx
(n.b. the gradient need not be 1 for this Ml).
Alternative:
Gradient of y = x + 7 is 1, so solve 3x2 - 2x - 4 = 1, as in (d)... to getx = -l.
(d)
2nd and 3rd M marks are dependent on starting with 3x2 - 2x - 4 = k , k is a constant.
12
Ml Alcso
where
Question
Scheme
Marks
number
10.
(2)
(3)
(a) x2+2x + 3 = (x + \f, +2
(a=l, b = 2)
B1,B1
(b)
"U"-shaped parabola
Ml
Vertex in correct quadrant (ft from (-a, b)
Alft
(0, 3) (or 3 ony-axis)
Bl
;r
(c)*'-4«-4-12 = -8
(3)
Bl
Negative, so curve does not cross x-axis
(d) b2 - 4ac = k2 -12 2)
(2)
(May be within the quadratic formula)
Bl
(2)
Ml
k2 -12 < 0
(Correct inequality expression in any form)
Al
-Vl2 <£<Vl2
(or -2V3 <£<2V3)
Ml Al
(4)
Total 11 marks
(b) The B mark can be scored independently of the sketch.
(3, 0) shown on they-axis scores the Bl, but if not shown on the axis, it is BO.
5) irks
(c) ".... no real roots" is insufficient for the 2nd B mark. ".... curve does not touch x-axis" is insufficient for the 2nd B mark.
(d) 2nd Ml: correct solution method for their quadratic inequality, e.g. k2 -12 < 0 gives k between the 2 critical values a <k < /? . whereas k2 -12 > 0 gives k < a, k > J3.
"k> -Vl2 and k < VT2 " scores the final Ml Al, but "k > -Vl2 or k < Vl2 " scores Ml AO, "k > -Vl2 , k<Jl2" scores Ml AO. N.B. k < ±Vl2 does not score the 2nd M mark. k < V12 does not score the 2nd M mark. < instead of <: Penalise only once, on first occurrence.
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