mark scheme june 2007 6666 core mathematics c4

Mark scheme - June 2007 - 6666 - Core Mathematics C4

Comparison of key skills specifications 2000/2002 with 2004 standards

ask CODE "WHAT CODE?" X015461

ask DATE "WHAT DATE?" July 2004

ask ISSUE "WHAT ISSUE?" Issue 1

GCE Mathematics (6666/01)

June 2007

6666 Core Mathematics C4

Mark SchemeQuestion NumberScheme Marks

** represents a constant

1. (a)

Takes 3 outside the bracket to give any of or .

See note below.B1

with

Expands to give a simplified or an un-simplified

;

A correct simplified or an un-simplified expansion with candidates followed thro

M1;

A1

Anything that cancels to

Simplified

A1;

A1

[5]

5 marks

Question NumberScheme Marks

Aliter

1.

Way 2

with

or(See note )Expands to give an un-simplified or simplified ;

A correct un-simplified or simplified expansion with candidates followed thro

B1

M1

A1

Anything that cancels to

Simplified

A1;

A1

[5]

5 marks

Attempts using Maclaurin expansions need to be escalated up to your team leader.

If you feel the mark scheme does not apply fairly to a candidate please escalate the response up to your team leader.


2., with substitution

or

or B1

where k is constantM1

M1

A1

change limits: when x = 0 & x = 1 then u = 1 & u = 2

Correct use of limits

u = 1 and u = 2depM1

or or A1 aef

Exact value only![6]

Alternatively candidate can revert back to x

Correct use of limits

x = 0 and x = 1depM1

or or A1 aef

Exact value only!

6 marks


3. (a)

(see note below)

Use of integration by parts formula in the correct direction.

Correct expression. M1

A1

or with

dM1

Correct expression with +c A1

[4]

(b)

Substitutes correctly

for in the

given integralM1

or underlined expressionA1;

Completely correct expression with/without +cA1

[3]

7 marks

Notes:(b)

This is acceptable for M1M1

This is also

acceptable for M1M1


Aliter

3. (b) Way 2

Substitutes correctly for in the

given integral M1

or

and

or underlined expressionA1


[3]

Aliter (b)

Way 3

Substitutes correctly for in M1

or underlined expressionA1;


[3]

7 marks


4. (a)A method of long division gives,

Way 1

B1

or

Forming any one of these two identities. Can be implied.M1

Let

EMBED Equation.DSMT4

See note below

Let

either one of or

A1

both B and C correctA1

[4]

Aliter

4. (a)

Way 2

See below for the award of B1 decide to award B1 here!!

for

B1

Forming this identity.

Can be implied.M1

Equate x2,

Let


See note below

Let

either one of or

A1

both B and C correctA1

[4]


4. (b)

Either or

or either or

M1

B1

or

See note below.A1 cso & aef

Substitutes limits of 2 and 1

and subtracts the correct way round. (Invisible brackets okay.)depM1

Use of correct product (or power) and/or quotient laws for logarithms to obtain a single logarithmic term for their numerical expression.M1

A1

Or and k stated as .[6]

10 marks


5. (a)If l1 and l2 intersect then:

Any two of

Writes down any two of these equations correctly.M1

Solves two of the above equations to find

either one of or correctA1

both and correctA1

Either

Complete method of putting their values of and into a third equation to

show a contradiction.B1

or for example:

Lines l1 and l2 do not intersectthis type of explanation is also allowed for B1.

[4]

Aliter

5. (a)

Uses the k component to find

and substitutes their value of

into either one of the i or j component.

Way 2

M1

either one of the s correctA1

both of the s correctA1

Either: These equations are then inconsistent

Or:

Or: Lines l1 and l2 do not intersectComplete method giving rise to any one of these three explanations.B1

[4]


Aliter

5. (a)

Way 3If l1 and l2 intersect then:

Any two of

Writes down any two of these equationsM1

either one of the s correctA1

both of the s correctA1

Either: These equations are then inconsistent

Or:

Or: Lines l1 and l2 do not intersectComplete method giving rise to any one of these three explanations.B1

[4]

Aliter

5. (a)

Way 4 Any two of

Writes down any two of these equationsM1

A1

RHS of (3) = 3A1

(3) yields

Complete method giving rise to this explanation.B1

[4]


5. (b)

Only one of either

or or or .(can be implied) B1

or

Finding the difference between their and.

(can be implied) M1

Applying the dot product formula between allowable vectors. See notes below.M1

, & is angle

Applies dot product formula between and their

M1

Correct expression.A1

or 0.7 or

A1 cao

but not [6]

10 marks

Note: If candidate use cases 2, 3, 4 and 5 they cannot gain the final three marks for this part.

Note: Candidate can only gain some/all of the final three marks if they use case 1.Examples of awarding of marks M1M1A1 in 5.(b)ExampleMarks

M1M1A1

(Case 1)

M1M0A0

(Case 2)

M1M0A0

(Case 3)


6. (a)

,

Correct and B1

M1

A1

[3]

(b) When (need values)The point or

These coordinates can be implied. B1, B1

( is not sufficient for B1)

When m(T) =

any of the five underlined expressions or awrt 0.18B1 aef

T:

Finding an equation of a tangent with their point and their tangent gradient or finds c by using .M1 aef

T: or

Correct simplified

EXACT equation of tangentA1 aef cso

or

Hence T: or

[5]


6. (c)

Way 1

Uses M1

Eliminates t to write an equation involving x and y.M1

Rearranging and factorising with an attempt to makethe subject.ddM1

A1

[4]

Aliter

6. (c)

Uses M1

Way 2

Uses

M1 implied

Hence,

Eliminates t to write an equation involving x and y.ddM1

Hence,

A1

[4]


Aliter

6. (c)

Way 3

Uses M1

Uses

M1

Hence,


Hence,

A1

[4]

Aliter

6. (c)

Uses M1

Way 4

Uses

M1

then uses

ddM1

Hence,

A1

[4]


Aliter

6. (c)

Way 5

Draws a right-angled triangle and places bothand 1 on the triangle

Uses Pythagoras to deduce the hypotenuseM1

M1

Hence,


Hence,

A1

[4]

12 marks


7. (a)x0

y00.4459959270.6435942520.8174219461

Enter marks into ePEN in the correct order.0.446 or awrt 0.44600 B1

awrt 0.64359B1

awrt 0.81742B1

0 can be implied[3]

(b)

Way 1

Outside brackets

or

For structure of trapezium rule;

Correct expression

inside brackets which all must be multiplied by. B1

M1

A1

(4dp)for seeing 0.4726A1 cao

[4]

Aliter

(b)

Way 2

which is equivalent to:

and a divisor of 2 on all terms inside brackets.

One of first and last ordinates, two of the middle ordinates inside brackets ignoring the 2.

Correct expression inside brackets if was to be factorised out.B1

M1

A1

0.4726A1 cao

[4]


7. (c)Volume

or

Can be implied.

Ignore limits and M1

or

or A1

or

The correct use of limits on a function other than tan x; ie

minus . may be implied. Ignore dM1

or

or or or or

or

or or

or A1 aef

must be exact.[4]

11 marks

Beware: In part (c) the factor of is not needed for the first three marks.Beware: In part (b) a candidate can also add up individual trapezia in this way:


8. (a) and

Separates the variables with and on either side with integral signs not necessary.M1

Must see and ;

Correct equation with/without + c.A1

When

Use of boundary condition (1) to attempt to find the constant of integration.M1

Hence,

A1

[4]

(b)

Substitutesinto an expression involving P M1

or

or or

Eliminates and takes

ln of both sidesM1

or (to nearest minute)awrt or A1

[3]


8. (c) and


Must see and ;


When


Hence,

A1

[4]

(d)

or

Eliminates and makes or the subject by taking lnsM1

Then rearranges

to make t the subject.dM1

(must use sin-1)

or (to nearest minute)awrt or A1

[3]

14 marks


and

Aliter8. (a)

Way 2


Must see and ;


When


Hence,

A1

[4]

Aliter8. (a)

Way 3


Must see and ;


When


Hence,

A1

[4]


and

Aliter8. (c)

Way 2


Must see and ;


When


Hence,

A1

[4]


and

Aliter8. (c)

Way 3


Must see and ;


When


Hence,

A1

[4]

Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark.

ddM1 denotes a method mark which is dependent upon the award of the previous two method marks.

depM1 denotes a method mark which is dependent upon the award of .

ft denotes follow through

cao denotes correct answer only

aef denotes any equivalent form

If you see this integration applied anywhere in a candidates working then you can award M1, A1

To award this M1 mark, the candidate must use the appropriate law(s) of logarithms for their ln terms to give a one single logarithmic term. Any error in applying the laws of logarithms would then earn M0.

Note: The x and y coordinates must be the right way round.

A candidate who incorrectly differentiates EMBED Equation.DSMT4 to give EMBED Equation.DSMT4 or EMBED Equation.DSMT4 is then able to fluke the correct answer in part (b). Such candidates can potentially get: (a) B0M1A1 EMBED Equation.DSMT4 (b) B1B1B1M1A0 cso.

Note: cso means correct solution only.

Note: part (a) not fully correct implies candidate can achieve a maximum of 4 out of 5 marks in part (b).

t



1

Candidates can score this mark if there is a complete method for finding the dot product between their vectors in the following cases:

Case 2: EMBED Equation.DSMT4

and EMBED Equation.DSMT4


Case 3: EMBED Equation.DSMT4



Case 4: their ft EMBED Equation.DSMT4



Case 1: their ft EMBED Equation.DSMT4



Case 5: their EMBED Equation.DSMT4

and their EMBED Equation.DSMT4


Some candidates may find rational values for B and C. They may combine the denominator of their B or C with (2x +1) or (2x 1). Hence:

Either EMBED Equation.DSMT4 or EMBED Equation.DSMT4 is okay for M1.

Candidates are not allowed to fluke EMBED Equation.DSMT4 for A1. Hence cso. If they do fluke this, however, they can gain the final A1 mark for this part of the question.

Note: This is not a dependent method mark.

EMBED Equation.DSMT4 , gains B0M1A1A0

In (a) for EMBED Equation.DSMT4 writing 0.4459959 then 0.45600 gains B1 for awrt 0.44600 even though 0.45600 is incorrect.

In (b) you can follow though a candidates values from part (a) to award M1 ft, A1 ft

Special Case: If you see the constant EMBED Equation.DSMT4 in a candidates final binomial expression, then you can award B1

Special Case: If you see the constant EMBED Equation.DSMT4 in a candidates final binomial expression, then you can award B1

Note: You would award: B1M1A0 for


because EMBED Equation.DSMT4 is not consistent.

If a candidate states one of either B or C correctly then the method mark M1 can be implied.

If a candidate gives the correct exact answer and then writes 1.088779, then such a candidate can be awarded A1 (aef). The subsequent working would then be ignored. (isw)

There are other acceptable answers for A1, eg: EMBED Equation.DSMT4

NB: Use your calculator to check

eg. 0.240449

EMBED Equation.DSMT4 is an acceptable response for the final accuracy A1 mark.



There are so many ways that a candidate can proceed with part (c). If a candidate produces a correct solution then please award all four marks. If they use a method commensurate with the five ways as detailed on the mark scheme then award the marks appropriately. If you are unsure of how to apply the scheme please escalate your response up to your team leader.

EMBED Equation.DSMT4 written down without the first M1 mark given scores all four marks in part (a).

EMBED Equation.DSMT4 written down without the first M1 mark given scores all four marks in part (c).

EMBED Equation.DSMT4 written down without the first M1 mark given scores all four marks in part (a).

EMBED Equation.DSMT4 written down without the first M1 mark given scores all four marks in part (c).

Mark Scheme (Results)

Summer 2007

Edexcel Limited. Registered in England and Wales No. 4496750

Registered Office: One90 High Holborn, London WC1V 7BH

GCE

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mark scheme june 2007 6666 core mathematics c4

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