June 2005 6663 Core Mathematics C1 Mark Scheme Question
edexcel
Scheme
Marks
Number 1.
(a)
Ml
2
Penalise +
Bl
(1) (b)
--
1
1
11
V64
(a)2
8 3 = -7= or —7 or —;=or—
V82
Allow ±
81
Ml Al
= - or 0.25 4
(2) (3)
(b)
Ml
for understanding that"-" power means reciprocal
2
1
83 = 4 is M0A0 and - - is Ml A0 4
2.
(a)
—- = 6 + ox dx
x
—> x
both
(b)
f 2 6x (ox-4jc )dx =
J
2
r-4*
i
Ml Al
(2) Ml Al Al
+c
(3) (5)
(b)
1st A1 for one correct term inx:
or + 4x~}
(or better simplified versions)
2nd Al for all 3 terms as printed or better in one line.
-
6663 Core Mathematics C1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
I
Number
3.
(a)
Marks
Scheme
Question
x2-$x-29 = (x-4)2-45
(x±4)2 (x-4)2-16 + (-29) (x±4)2-45
Ml Al
Al
(3)
ALT
Compare coefficients
- 8 = 2a
a = -4 AND
equation for a
Ml
Al
az+b = -29
Al
6 =-45
(3) (b)
(jc - 4)2 = 45
(follow through their a and b from (a))
=>jc-4 = +V45 x = 4±3S
c=4 d=3
Ml Al
Al
(3)
(6)
(a)
(b)
Ml for (x ± 4)2 or an equation for a.
Ml for a full method leading to x - 4 = ... or x -...
Al for c and Al for d
Note Use of formula that ends with —=
scores Ml Al AO fbut must be v5 }
i.e. only penalise non-integers by one mark.
6663 Core Mathematics C1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
Marks
Scheme
Question
Number
4.
(a)
Shape
Bl
Points
Bl
(2) 6
"7
\
(b)
Ml
/ -2 and 4 max
Al
Al
(5)
Marks for shape: graphs must have curved sides and round top.
(a)
(3)
1st Bl for n shape through (0,0) and ((fc,0) where k>0)
2nd B1 for max at (3,15) and 6 labelled or (6, 0) seen
Condone (15,3) if 3 and 15 are correct on axes. Similarly (5,1) in (b)
(b)
Ml for r> shape NOT through (0,0) but must cut x-axis twice.
1st Al for -2 and 4 labelled or (-2,0) and (4,0) seen
2nd Al for max at (1, 5). Must be clearly in 1st quadrant 5.
Ml
x = l + 27andsub -+(l + 2yf+y2 =29 =>572+47-28(=0)
Ml
i.e. (57 + 14)(7-2) = 0
(7=)2or-—
v = 2=>x = l + 4 = 5;
1st Ml 1st Al 2nd Ml 2nd Al 3rd Ml
(o.e.)
7 =
(both) 14 5
=> x =
23 5
.
.
(o.e)
Attempt to sub leading to equation in 1 variable Correct 3TQ (condone = 0 missing) Attempt to solve 3TQ leading to 2 values for 7. Condone mislabelling x = for 7 = ... but then M0A0 in part (c). Attempt to find at least one x value
3rd Al f.t.
f.t. only in x = 1 + 27
(3sf if not exact) Both values
N.B. False squaring (e.g. 7 = x2 +472 = 1) can only score the last 2 marks.
6663 Core Mathematics C1
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
Al
MlAl f.t.
(6)
edexcel Question Number 6. (a)
(2x-l)(*-3)
Ml
=>8x>2 x > 4
(b)
Marks
Scheme
6x + 3>5-2x
Al
or 0.25 or 8
(2) Ml
(>0)
Critical values x = —, 2
•s;
3
(both)
Choosing "outside" region x>3
or
x < — 2
Al
Ml
Al f.t.
(4) (c)
x>3
or
- < jc< 4 2
Blf.t. Blf.t
(2)
(8)
(a)
Ml
(b)
1st Ml
Multiply out and collect terms (allow one slip and allow use of = here)
Attempting to factorise 3TQ -> x =...
2nd Ml Choosing the outside region 2nd Al f.t.
ft. their critical values
N.B.(x>3, x > - is M0A0)
For p<x<q where/? > q penalise the final Al in (b).
(c)
f.t. their answers to (a) and (b)
1st Bl
a correct f.t. leading to an infinite region
2nd Bl a correct f.t. leading to a finite region Penalise < or > once only at first offence.
e.g.
(a)
x>4 x>4
(b)
-<x<3 2 x>3, x>2
(c)
-<x<3 2 x>3
6663 Core Mathematics C1 June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
Mark
BOBl B1B0
edexcel 7. (a)
Mil
mmm
(3-Vx)2 =9Al c.s.o.
-+9x~2-6 +
(2) (b)
9x2
Ml A2/1/0
3^ 2
2
use y = â&#x20AC;&#x201D; and x = 1
2
Ml
- = 18 3 c = -
-x2 -12 3
So
Al
C.S..O.
Alf.t.
(6) (8)
(a)
Ml Attempt to multiply out (3 - Vjc)2 . Must have 3 or 4 terms, allow one sign error Al cso Fully correct solution to printed answer. Penalise wrong working.
(b)
1st Ml
Al
Some correct integration:
,n+\
xn -> x
At least 2 correct unsimplified terms Ignore + c
A2
All 3 terms correct (unsimplified)
2nd Ml Alc.s.o.
Use of y = - and x = 1 to find c. No + c is MO.
for -12. (o.e.)
Award this mark if " c = -12 " stated i.e. not as part
of an expression for y
Alf.t. for 3 simplified x terms with y = ... and a numerical value for c. Follow through their value of c but it must be a number.
6663 Core Mathematics C1
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
edexcel .
Question
Scheme
Number
8.
•11 *"w>
Marks
(a)
Ml Al
3y - x + 21 = 0
(o.e.) (condone 3 terms with integer coefficients e.g. 3y+21=x)
Al
(3' (b)
Equation of /2 is:
Solving lx and /2:
y = -2x
Bl
(o.e.)
\/f1
JV11
- 6x - x + 21 = 0
p is point where xp = 3,
yp = -6
x
or y
yP or xp
(c)
(lx isy = -x-l)
C is (0,-7)
or
OC = 7
11 21 Area of AOCP = — OCxxp, =—x 7x3 = 10.5 or —
Al
Alf.t.
(4)
Blf.t.
Ml Alc.a.o.
(3) (10)
(a)
Ml
for full method to find equation of lx
1 stA1
(b)
Ml
any unsimplified form
Attempt to solve two linear equations leading to linear equation in one variable
2nd Al ft.
(c)
only ft. their xp or yp in y~ -2x
Blft.
Either a correct OC or ft. from their lx
Ml
for correct attempt in letters or symbols for AOCP
Al c.a.o. — x 7 x 3 scores Ml A0 2
6663 Core Mathematics C1
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics
edexcel Question
Scheme
Marks
Number 9 (a)
SSI
Bl
(S=)[a + (n-\)d]+ 2S = [2a + (n- l)d]+
+a + [2a + (n - l)d]
Ml
} either
dMl
Al c.s.o
(4)
2
(a = 149, d = -2)
(b)
(c)
MlAl
m2]=149 + 20(-2) =£109
Sn = -[2 x 149 + (n -1)(-2)]
(2)
(= n(l 50 - nj)
Sn = 5000 => n2 -150n + 5000 = 0
(*)
MlAl Al c.s.o
(3) (d)
Ml
(«-100)(n-50) = 0
A2/1/0
n = 50 or 100
(e)
Mioo < 0
(3) Bl ft.
•'• n = 100 not sensible
0)
(13)
(a)
Bl
requires at least 3 terms, must include first and last terms, an adjacent term dots and + signs.
1st Ml
for reversing series. Must be arithmetic with a, d (or a, I) and n.
(NB
Allow first 3 marks for use of / for last term but as given for final mark)
(b)
Ml
for using a = 149 and d= ±2 in a + {n -1)d formula.
(c)
Ml
for using their a, Jin Sn
Al cso
for putting Sn =5000 and simplifying to given expression. No wrong work
Ml
Attempt to solve leading to n -...
2nd dMl for adding, must have 25 and be a genuine attempt." Either line is sufficient. Dependent on 1st Ml
(d)
(e)
A1 any correct expression
All \ 10
Give A1A0 for 1 correct value and A1A1 for both correct
Bl f.t.
Must mention 100 and state w100 < 0 (or loan paid or equivalent) If giving f.t. then must have n > 76.
6663 Core Mathematics C1 June 2005 Advanced Subsidiary /Advanced Level in GCE Mathematics
edexcel Question Number
Scheme
Marks
10 (a)
( 9 - 36 + 27=0 is OK)
Bl
(b)
— = -x2 -2x4xx + dx
Ml Al
(=x2-8x + 8)
3
Whenx = 3, ^- = 9-24 + 8 =>m = -7
Ml
etc
Equation of tangent:
Mi
y - 0 = -7(x - 3)
Ml
Al c.a.o
(c)
—=m
gives
jc -8jc + 8 = -7
Ml
or 5
x=5
-53-4x52+8x5 + 3 ,.1 -15— 3
(b)
MR
46 Al
3
some correct differentiation (x" -> jc"
2nd Ml
substituting xp(= 3) in their — clear evidence
1st Al
Al
Ml
1st Ml
rd Ml 3rd
(c)
or
Ml
correct unsimplified (all 3 terms)
for one term)
dx
using their m to find tangent at p.
1st Ml
forming a correct equation " their — = gradient of their tangent''
2nd Ml
for solving a quadratic based on their — leading to x =
3rd Ml
for using their x value inj> to obtain y coordinate
dx
dx
For misreading (0, 3) for (3, 0) award B0 and then Ml Al as in scheme. Then allow all M marks but no A ft. (Max 7)
6663 Core Mathematics C1
June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics