C1 MS Jan 2006c

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

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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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