mark scheme june 2007 6666 core mathematics c4
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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.
Question NumberScheme Marks
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
Question NumberScheme 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
Question NumberScheme Marks
Aliter
3. (b) Way 2
Substitutes correctly for in the
given integral M1
or
and
or underlined expressionA1
Completely correct expression with/without +cA1
[3]
Aliter (b)
Way 3
Substitutes correctly for in M1
or underlined expressionA1;
Completely correct expression with/without +cA1
[3]
7 marks
Question NumberScheme 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
EMBED Equation.DSMT4
See note below
Let
either one of or
A1
both B and C correctA1
[4]
Question NumberScheme Marks
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
Question NumberScheme 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]
Question NumberScheme Marks
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]
Question NumberScheme Marks
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)
Question NumberScheme Marks
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]
Question NumberScheme Marks
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]
Question NumberScheme Marks
Aliter
6. (c)
Way 3
Uses M1
Uses
M1
Hence,
Eliminates t to write an equation involving x and y.ddM1
Hence,
A1
[4]
Aliter
6. (c)
Uses M1
Way 4
Uses
M1
then uses
ddM1
Hence,
A1
[4]
Question NumberScheme Marks
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,
Eliminates t to write an equation involving x and y.ddM1
Hence,
A1
[4]
12 marks
Question NumberScheme 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]
Question NumberScheme Marks
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:
Question NumberScheme Marks
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]
Question NumberScheme Marks
8. (c) 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]
(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
Question NumberScheme Marks
and
Aliter8. (a)
Way 2
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]
Aliter8. (a)
Way 3
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]
Question NumberScheme Marks
and
Aliter8. (c)
Way 2
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]
Question NumberScheme Marks
and
Aliter8. (c)
Way 3
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]
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
EMBED Equation.DSMT4
EMBED Equation.DSMT4
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
EMBED Equation.DSMT4
Case 3: EMBED Equation.DSMT4
and EMBED Equation.DSMT4
EMBED Equation.DSMT4
Case 4: their ft EMBED Equation.DSMT4
and EMBED Equation.DSMT4
EMBED Equation.DSMT4
Case 1: their ft EMBED Equation.DSMT4
and EMBED Equation.DSMT4
EMBED Equation.DSMT4
Case 5: their EMBED Equation.DSMT4
and their EMBED Equation.DSMT4
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
EMBED Equation.DSMT4
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.
EMBED Equation.DSMT4 is an acceptable response for the final accuracy A1 mark.
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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