C1 MS Jun 2005c

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