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C4 maths mark schemes, June 2014 back to January 2010

GCE Core Mathematics C4 (6666) January 2011 1

Question Number

C4 JUNE 2014 Marks

1.

(a)

dM1

A1 cso

[5]

(b)At

see notes M1and either T:

or ,

T: or A1 cso

[2]

7

2.

(a)Either or

see notes M1

leading to A1

[2]

(b)

Either or or M1

Either or

M1


or 22.5A1

[3]

5

3.x 1 2 3 4

y 1.42857 0.90326 0.682116... 0.55556

(a) 0.68212B1 cao

[1]

(b)

Outside brackets or B1 aef

For structure of M1

anything that rounds to 2.5774 A1

[3]

(c) Overestimate

B1

and a reason such as

{top of} trapezia lie above the curve a diagram which gives reference to the extra

area concave or convex

(can be implied) bends inwards curves downwards

[1]

(d) or B1


Either or M1

with no other terms.

M1

A1 cso

Substitutes limits of 2 and 1 in u

(or 4 and 1 in x) and subtracts the correct way

round.

M1

or or

A1 oe cso

[6]

11

4. ,

M1

A1

M1 oe

or

When dependent on the previous M1

see notesdM1

or orA1 oe


[5]

5

5.

(a)M1 oe

So, Adds their expanded x (which is in terms of t) to dM

1

* Correct proofA1 *

[3]

(b)Applies to achieve

an equation containing only x’s and y’s.

M1

A1

[2]

5

6. (i)

M1

A1

A1

[3]

(ii)M1


or equivalent.

A1

{Ignore subsequent working}.

[2]

(iii)

at

or

B1 oe

Applying

M1

Integrates to give M1

A1

B1

or Use of and

in an integrated equation containing c

M1

giving A1

[7]

12

7.

(a) ,


their divided by their M1

Correct A1 oe

At ,

Some evidence of

substituting into

their

M1

applies M1

Either N: see notes M1

or

{At Q, so, } giving or or awrt

1.67

A1 cso

[6]

(b) see notes

M1

So, see notes A1

A1

Applies

M1


Dependent on the first method mark. For dM

1

A1

Dependent on

the third method

mark.

dM1

{So }

M1

A1

[9]

15

8.

(a) M1; A1

[2]

(b)

or

B1ft

[1]

(c) M1


Applies dot product formula between

their

and their

M1

Correct proof

A1 cso

[3]

(d)

or , with

either or , or

a multiple of .

M1

Correct vector equation. A1 ft

[2]

(e)

Either

or

M1

At least one set of coordinates are correct.

A1 ft

Both sets of coordinates are correct.

A1 ft

[3]

(f)M1

or

or or awrt 4.9 or equivalent

A1 oe


3 3

3023

P

82 6 3 3. 4.8989...3

h

2l114

C

3

138

B

1l

247

A

132

D

3

dM1

A1 cao

[4]

15

8. (f) Helpful Diagram!


Candidates do not need to prove this result for part (f)


Question Number

C4 JUNE 2014 (R) Marks

1. (a) B1

B1

M1

A1; A1

[5]

(b)

M1

M1

A1

[3]

8

2. (a) B1; M1

(2 dp)A1 cao

[3]


(b) Any one of

Increase the number of strips Use more trapezia Make h smaller Increase the number of x and/or y values used Shorter /smaller intervals for x More values of y. More intervals of x Increase n

B1

[1]

(c)

,

M1

A1

A1 oe

dM1

A1 oe

[5]

9

3.

(a)

dM1

Simplifying gives A1 cso oe

[5]


(b) M1

So ,

M1

gives or A1 oe

ddM1

A1 cao

[5]

10

4. (a)

B1

B1 cso

M1

leading to A1

[4]

(b) B1

For their partial fraction


M1

A1ft

A1ft

dM1

So, A1 oe

[6]

10

5. (a)From question, ,

B1 oe

M1 oe

When , dM1

Hence, (cm2 s-1)A1


[4]

(b) M1; oe

When ,

Hence, (cm2 s-1)A1 cso

[2]

6

6.

(a) {B lies on B1

[1]

(b)

e.g. i: M1

So, A1

Point of intersection is B1

Finds and either

checks and is true for the third component.

substitutes into to give

and substitutes into to give

B1

[4]

(c) M1


M1

A1

[3]

(d)M1

A1

[2]

10

7.

(a) B1

M1 A1; A1

then eg either... or...

leading to with no incorrect working/statements. A1 * cso

[5]

(b) B1

So

and or

Eg. M1


So

A1

A1

[4]

(c)

(fish) (nearest 100) B1

[1]

10

8. lies on the curve,

(a)

M1

or

so A1

[2]

(b) ,

B1

B1

So,

At ,

M1;

A1

cao cso

[4]

(c) M1


gives A1

So or M1; A1

or dM1

A1

[6]

12

Question Number

C4 JUNE 2013 MARK SCHEME Marks

1. (a) , 1st Application: , 2nd Application:

, M1

A1 oe

Either

or for

M1

M1

Correct answer, with/without A1

(5)

(b)

Applies limits of 1 and 0 to an expression of the form

and and subtracts the correct way

round.

M1


csoA1 oe

(2)

[7]

2. (a)

B1

M1 A1 A1

M1

Answer is given in the question.

A1 *

(6)

(b)M1

ie: B1

so,

A1 cao

(3)

[9]

3. (a)1.154701 B1

cao

(1)

(b)

B1; M1

(4 dp)1.7787 or awrt 1.7787 A1


(3)

(c)

For .Ignore limits and .

Can be implied.

B1

M1

or equivalent

A1

A1 cao cso

(4)

[8]

4.

(a) , or At least one of or

correct.B1

Both and are correct.B1

So,

At ,

Applies their divided by

their and substitutes

into their .

M1;

Correct value for of 1

A1 cao cso

(4)

(b) M1

So, or or or equivalent.

A1 cso isw


Either or B1

(3)

(c) Range: or or B1 B1

(2)

[9]

5. (a) or or

B1

M1

A1 * cso

(3)

(b)

M1 A1

So

Integrates

to

obtain any one of or

M1

At least one term correctly followed through

A1 ft

. A1 cao

So,

Applies limits of 3 and 1 in u or 9 and 1 in x

in their integrated function and subtracts

M1


the correct way round.

A1 cso cao

(7)

[10]

6.

(a) or

B1

or

M1 A1;

M1 A1

M1

then either... or...

dddM1

A1 * leading to

(8)

(b) M1

Uses correct order of operations by moving

from

to give and

dM1


,

where

= 161 (s) (nearest second) awrt 161 A1

(3)

[11]

7.

(a)

dM1

A1 cso oe

(5)

(b) M1

A1

M1*

dM1*

A1

When , When , ddM1*

A1 cso

(7)

[12]


8. , ,

(a)

Finds the difference

between and .

Ignore labelling.

M1

Correct difference. A1

M1

A1 cso

(4)

(b) M1

So, caoA1 cao

It follows that, or B1 ft

{Note that }

or

and

Uses a correct method in order to find both possible

sets of coordinates of B.M1


Both coordinates are correct.

A1 cao

(5)

[9]

Question Number C4 JUNE 2013 (R) MARK SCHEME

Marks

1. At least one of “A”

or “C” are correct.

B1

Breaks up their partial

fraction correctly into

three terms and

both and .

B1 cso

1.

2. Writes down a correct

identity and attempts to

find the value of either one “A” or “ B”

or “C”.

M13.

Either

leading to

Correct value for “B” which is found using

a correct identity and

follows from their partial

fraction decompositio

n.

A1 cso


So,

[4]

2.

(ignore)

B1 oe

Differentiates implicitly to include

either

.

M1*

B1

A1

4. Substitutes into

their differentiated equation or expression.

dM1*

5.

dM1*

Uses to

achieve A1 cso

[7]

3.

6.

or

Either M1

Either A1

8. Correct substitution

(Ignore integral sign and A1

An attempt to divide each term by u.

dM1


ddM1

A1 ft

Applies limits of 5 and 3 in u

or 4 and 0 in x in their integrated function and

subtracts the correct way round.

M1

A1

cao cso

[8]

4. (a)Power

of

M1

9.

or B1

10.

M1 A1

A1; A1

(6)

(b)

Writes down or

uses B1

When M1


So, 19.2201

csoA1 cao

(3)

[9]

5. (a) 6.248046798... = 6.248 (3dp) 6.248 or awrt 6.248 B1

(1)

(b)B1; M1

(2 dp) 49.37 or awrt 49.37 A1

(3)

(c)

M1

A1

B1

A1

Substitutes limits of 8 and 0 into an

integrated function of the form of either

or

and subtracts the

correct way round.

dM1


A1

(6)

(d)

Difference

1.46 or awrt 1.46 B1

(1)

[11]

6.

, ,

(a)

A is on l, so

B1

Substitutes their value of into M1

A1 cao

(3)

(b)

Finds the difference

between and .

Ignore labelling.

M1

M1; A1 ft


ddM1;

A1 cso cao

(5)

(c) M1

So, A1 cao

(2)

(d) M1;A1 cao

(2)

[12]

7.

(a) ,

At least one of

or correct.

B1

Both and are correct.

B1

At ,

Applies their

divided by their

M1;

A1 cao cso

(4)

(b) M1

A1 * cso

or B1


(3)

(c)

For

or Ignore limits and .

Can be implied.

B1

Either or

oe

M1

oeA1

Substitutes limits of 125 and 27 into an integrated function

and subtracts the correct way round.

dM1

or or A1

(5)

[12]

8. where M is a constant

(a)

is the rate of increase of the mass of waste products.

M is the total mass of unburned fuel and waste fuel

(or the initial mass of unburned fuel)

Any one correct explanation.

B1

Both explanations are correct.

B1

(2)

(b) or B1


or M1 A1

M1

then either... or...

ddM1

A1 * cso leading to or

oe

(6)

(c)

M1

So A1

dM1

A1 cso

(4)

[12]


Question Number

C4 JANUARY 2013 MARK SCHEME Marks

1.or

B1

see notesM1

A1

See notes

below!

A1;

A1

[5]

5

2. (a)

,

In the form

M1

simplified or

un-simplified.

A1

simplified

or un-simplified.


dM1

Correct answer,

with/without A1

[5]

(b)

Applies limits of 2

and 1 to their part

(a) answer and

subtracts the correct

way round.

M1

or

equivalent

.

A1

[2]

7

3. Method 1: Using one identity

their constant term B1

Forming a correct

identity.B1

Either

or

Attempts to find the

value of either one of

their B or their C from

their identity.

M1

Correct values for

their B and their C,

which are found using

a correct identity.

A1


[4]

Method 2: Long Division

their constant term B1

So,

Forming a correct

identity.B1

Either

or

Attempts to find the

value of either one of

their B or their C from

their identity.

M1

Correct values for

their B and their C,

which are found usingA1

So, [4]

4

4. (a)1.0981 B1

cao

[1]

(b)B1;

M1

(3 dp)

2.843 or awrt 2.843 A1


[3]

(c) or

B1

M1

A1

Expands to give a “four

term” cubic in u.

Eg: M1

An attempt to divide at least three terms in their

cubic by u. See notes.M1

A1

=

Applies limits of 3 and 2 in u or 4 and 1 in x and subtracts

either way round.

M1

Correct exact answer

or equivalent.A1

[8]

12

5. Working parametrically:


(a)Applies to obtain a

value for t.M1

When , Correct value for y.A1

[2]

(b)

Applies to obtain a

value for t.

(Must be seen in part (b)).

M1

When ,

A1

[2]

(c) and either

or

B1

Attempts their divided by

their

M1

At A, so

Applies and M1

or

or equivalent.

M1 A1 oe cso

[5]

(d)Complete substitution for both

andM1

B1


Either

or

or

M1*

A1

Depends on the previous

method mark.

Substitutes their changed limits

in t and subtracts either way

round.

dM1*

or equivalent.A1

[6]

15

6. (a), seen or implied.

M1

At least one correct value of x. (See notes).

A1

Both A1 cso

[3]

(b)

V

For .Ignore limits and

B1

See notes.

M1


Attempts to give any two of or

.

M1

Correct integration. A1

Applying limits the

correct way round. Ignore

M1

Two term exact answer.

A1

[6]

9

7. (a)

Any two equations.

(Allow one slip).M1

Eg: or An attempt to eliminate one of the parameters.

M1

Leading to Either A1

or M1 A1

[5]

(b)

,

Realisation that the dot product is required

between and

M1


.

Correct equation. A1

awrt 69.1 A1

[3]

(c)

,

M1 A1

M1

leading to A1

Position vector

M1 A1

[6]

14

8. (a) or

B1

or See notes. M1 A1

Correct completion to

.


A1 *

[4]

(b) ; See notes. M1; A1

Substitutes into an equation

of the form

or equivalent.

M1

Correct algebra to ,

where k is a positive value.M1

awrt 77 A1

[5]

9

Question Number C4 JUNE 2012 MARK SCHEME Marks

1. (a) B1 M1 any two constants correct A1 Coefficients of all three constants correct A1 (4)

(b) (i) GCE Core Mathematics C4 (6666) January 2011 42

M1 A1ft A1ft

(ii) M1

M1

A1 (6) [10]

2. (a) csoB1 (1)

(b) M1

At x = 8, A1 (2)(c) B1

M1

At x = 8 A1 (3)


[6]Question Number Scheme Marks

3. (a) M1 , , 2 or equivalent B1 M1; A1ft

or A1

A1 (6)

(b) B1ft (1)

(c) M1 A1 (2)

[9]

4. Can be implied. Ignore integral signs B1

=GCE Core Mathematics C4 (6666) January 2011 44

M1 A1

M1Leading to or equivalent

A1 (5) [5]


Question Number Scheme Marks

5. (a) Differentiating implicitly to obtain and/or M1

A1

or equivalent B1

M1

A1 (5)

(b) M1 Using or or M1Leading to

or M1 or or A1 A1


Substituting either of their values into to obtain a value of the other variable. M1

both A1 (7) [12]


6. (a) B1

M1 A1

M1

A1 (5)

(b) When can be implied B1

M1

M1GCE Core Mathematics C4 (6666) January 2011 47

A1 (4)

(c) M1 or equivalent

M1 A1 (3) [12]


7. (a) x 1 2 3 4y ln2 2ln8

0.6931 1.9605 3.1034 4.1589M1

B1 M1 7.49 cao

A1 (4)(b) M1 A1


M1 A1 (4)

(c) M1 Using or implying M1

A1 (3) [11]


8. (a) M1 A1 (2)

(b)

M1 A1ft (2)

(c) M1 A1

M1


Leading to A1

Position vector of P is M1 A1 (6) [10]

Question Number

C4 JANUARY 2012 MARK SCHEME Marks

1. (a)

not necessari

ly required.

At dM1 A1 cso

[5]

(b)So, m(N) = M1

N: M1

N: A1

[3]

(8

marks)

2. (a) M1 A1


A1

[3]

(b) M1 A1

A1 isw

Ignore subseque

nt working

[3]

(6

marks)

3. (a)or

B1

M1 A1ft

A1; A1

[5]

(b)Can be implied by

later work even in part (c).

M1

x terms:

giving, A1

[2]

(c) terms:

M1


So, or or A1

[2]

(9

marks

)

4.

Volume Use of . B1

M1

A1

Substitutes limits of 2 and 0

and subtracts the correct way round.

dM1

So Volume or

A1 oe isw

[5]

(5

marks

)

5. (a) ,

B1 B1

So,

B1

oe

(3)

(b)M1 oe

M1


A1 A1 A1

(5)

(8

mark

s)

6. (a)0.73508 B1

cao (1)

(b)B1 M1

(4 dp)

awrt 1.1504A1 (3)

(c) B1

B1

M1

dM1

AG

A1 cso (5)

(d)

Applying limits

and either way round.

M1

or

or awrt

A1


14

4371

d

l

C

109

D

BA

ˆBAD

awrt 0.077

or awrt 6.3(%)

A1 cso (3)

(12

mark

s)

Question Number

Scheme Marks

7.

(a) M1; A1

[2]

(b)

or

M1 A1ft

[2]

Let d be the shortest distance from C to l.

(c) M1

Applies dot product formula

between

their

M1


and their

Correct followed through expression

or equation.A1

awrt 109

A1 cso AG

[4]

(d) M1

So, A1

[2]

(e)M1; dM1 A1

[3]

(f) or M1

awrt 3.54 A1

[2]

(15

marks)

Question Number

Scheme Marks

8. (a) Can be implied. M1

Either one. A1

giving

A1 cao, aef


[3]

(b) B1

M1*

A1ft

dM1*

eg:

Using any of the subtraction (or

addition) laws for logarithms

CORRECTLY

dM1*

eg: or eg: Eliminate ln’s correctly.

dM1*

gives

Make P the subject.

dM1*

or

etc.

A1

[8]

(c) . So population cannot exceed 5000.

B1

[1]

(12

marks)



1. B1

M1

Any two of A, B, C A1

terms All three correct A1 (4) [4]

2. M1 , or

B1

n not a natural number, M1

ft their A1 ft

A1GCE Core Mathematics C4 (6666) January 2011 57

A1 (6) [6]

3. (a) or equivalent M1 A1

At , M1 A1 (4)

(b) or M1

At , awrt 0.031A1 (2) [6]


4. (a) , awrt , B1 B1 (2)GCE Core Mathematics C4 (6666) January 2011 58

(b) B1 M1 Accept 1.3 A1 (3)

(c) B1 B1 M1Hence cso

A1 (4)

(d) M1 A1

M1 A1


= M1 A1 (6)

[15]


5. … B1

… M1 A1At , M1leading to Accept A1

At M1

A1 (7) [7]



6. (a) i: j: Any two equations M1 leading to , M1 A1

or M1 A1

k: , B1 (6) (As LHS = RHS, lines intersect)

(b) M1 A1 Acute angle is awrt 69.1 A1 (3)

(c) B1 (1)

(d) Let d be shortest distance from B to

M1

= awrt 7.5 A1


M1 awrt 6.99 A1 (4)

[14]


7. (a) or M1 awrt 1.05

A1 (2)

(b) , M1 A1At P, Can be implied A1 Using , M1 For normal M1 At Q, leading to 1.0625 A1 (6)


(c) M1 A1

A1 M1 A1

M1 A1 (7)

[15]


(a) M1 A1 (2)

(b) B1

M1


Using M1leading to A1 M1

or equivalent A1 (6) [8]

C4 MARK SCHEME JANUARY 2011Questio

n Number

Scheme Marks

1. M1 A1 A1

M1

M1 A1

[6]

2. M1 A1


At M1

M1 A1

[5]

Question

NumberScheme Marks

3.

(a) M1 A1

A1 (3)

(b)

ft constants M1 A1ft A1ft

(3)

(c) M1

M1 A1

depends on first two Ms in (c) M1 dep


Using depends on first two Ms in (c) M1 dep

A1 (6)

[12]

Question

NumberScheme Marks

4.

(a) M1 A1 (2)

(b) M1 A1ft (2)

or (c)

or B1

M1

Leading to M1 A1

(4)

(d) M1

accept A1


awrt 6.8 (2)

[10]

Question

NumberScheme Marks

5.

(a) B1

M1 A1

M1 A1

(5)

(b)

Coefficient of x; M1

Coefficient of ; A1 either M1 A1


correct Leading to M1 A1 (5)

(c) Coefficient of is M1 A1ft

cao A1 (3)

[13]


Question

NumberScheme Marks

6.

(a) , M1 A1

Using , at M1 A1

M1 A1 (6)

(b) B1

M1 A1 (3)

(c) M1

M1

M1 A1

M1

A1

(6)

[15]

Alternative to (c) using parameters

M1


M1

M1 A1

The limits are and M1

A1

(6)

Question

NumberScheme Marks

7.

(a) awrt B1

awrt or B1

(2)

(b) B1 M1 A1ft

0.542 or 0.543 A1 (4)

(c) B1

M1

A1


M1 A1

, B1

M1

A1

(8)

[14]


1. (a) accept awrt 4 d.p.B1 B1 (2)

(b)(i) B1 for

B1 M1 cao A1

(ii) B1 for B1 M1 cao A1 (6)

[8]GCE Core Mathematics C4 (6666) January 2011 71

2. B1 M1 A1 ft sign error A1ft or equivalent with u M1 cso A1 (6)

[6]

3. B1 M1 A1= A1Substituting M1 Accept exact equivalents

M1 A1 (7) [7]



4. (a) B1 B1

or equivalentM1 A1 (4)

(b) At , , B1

M1 A1

M1 M1 A1 (6)

[10]

5. (a) B1 M1 A1 A1 (4)

(b) M1 B1


B1

M1 ft their A1 ft stated or implied

A1 A1 (7) [11]


6. (a) M1 M1 cso

A1 (3)

(b) M1 A1 A1 M1 A1

M1


A1 (7) [10]

7. (a) j components M1 A1 Leading to accept vector forms A1 (3)

(b) Choosing correct directions or finding and M1

use of scalar productM1 A1

awrt A1 (4)

(c)

M1 A1 A1 , M1 A1 (5)


awrt 34 [12]

Question Number

Scheme Marks

8. (a) M1 A1GCE Core Mathematics C4 (6666) January 2011 76

B1 M1 Leading to cso

A1 (5)

(b) separating variables M1 M1 A1 When , M1 When awrt 10.4

M1 A1 (6)

[11]


Question

NumberC4 JANUARY 2010 MARK SCHEME Marks

Q1 (a) + … M1 A1

… A1; A1 (4)

(b) M1

csoA1 (2)

(c) M1

M1

cao A1 (3)

[9]

Q2 (a) 1.386, 2.291 awrt 1.386, 2.291 B1 B1 (2)

(b) B1

M1

ft their A1ft


(a) cao A1

(4)

(c)(i) M1 A1

M1 A1

(ii) M1

seen or implied M1

A1 (7)

[13]


Question

NumberScheme Marks

Q3 (a) M1 A1

Accept , A1 (3)

(b) At , M1

A1

awrt 0.349A1 (3)

(c) At , M1

M1

Leading to A1 (3)

[9]


A X

Y1l

2ld4 26

Question

Number

Scheme Marks


Q4 (a) A: Accept vector forms B1 (1)

(b) M1 A1

awrt 0.73A1 (3)

(c) X: Accept vector forms B1 (1)

(d) Either orderM1

caoA1 (2)

(e) M1

Do not penalise if consistent A1 (2)

incorrect signs in (d)(f) Use of correct right angled triangle M1


M1

awrt 27.9A1 (3)

[12]


Question

NumberScheme Marks

Q5 (a) M1

A1 (2)

(b) Integral signs not necessary B1

M1

ft their A1ft

, 3

28

23=9+61 n 1+C M1

A1

A1

(6)

[8]


Question

NumberScheme Marks

Q6 B1

B1

When M1

M1

awrt 0.299 A1

[5]


Question

NumberScheme Marks

Q7 (a) Any one correct value B1

At , Method for finding one value of x M1

At , At A, ; at B, Both A1

(3)

(b) Seen or implied B1

M1 A1

A1

M1

A1 (6)

[9]


Q8 (a) B1

M1

Use of M1

M1

M1

M1

A1 (7)

(b) M1

integral in (a) M1

their answer to A1ft (3)


part (a) [10]


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