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EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question
Number
Scheme Marks
1.(a)
B1
or M1
awrt 1.287 A1
Hence, (3)
(b) Minimum value = or B1ft
(1)
(c)
M1
For applying M1
So, either or
M1
gives, awrt 3.84 OR 6.16 A1
awrt 3.84 AND 6.16 A1(5)[9]
1
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question
NumberScheme Marks
2.
(a)
An attempt to form a single fraction
M1
Simplifies to give a correct quadratic numerator over a
correct quadratic denominatorA1 aef
An attempt to factorise a 3 term quadratic numerator
M1
A1 (4)
(b)
An attempt to form a single fraction
M1
Correct result A1 (2)
(c)
M1
A1 aef
Either or A1
(3)[9]
2
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question
Number
Scheme Marks
3.
Substitutes either
or
or for
M1
Forms a “quadratic in sine” = 0 M1(*)
Applies the quadratic formulaSee notes for alternative methods.
M1
Any one correct answer A1
180-their pv dM1(*)
All four solutions correct. A1[6]
3
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question
NumberScheme Marks
4.(a) (eqn *)
Substitutes and into eqn * M1
A1(2)
(b)
Substitutes and into eqn * and rearranges eqn * to make e±5k the
subject.M1
Takes ‘lns’ and proceeds
to make ‘±5k’ the subject.dM1
Convincing proof that A1 (3)
(c)
where M1
A1 oe
When
Rate of decrease of (3 dp.) awrt A1(3)[8]
4
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question
NumberScheme Marks
5.(a)
Crosses x-axis
Either or Either one of {x}=1 OR x={8} B1
Coordinates are and Both and
B1
(2)
(b)
Apply product rule: M1
Any one term correct A1
Both terms correct A1(3)
(c)
Sign change (and as is continuous) therefore the x-coordinate of Q lies between 3.5 and 3.6.
Attempts to evaluate both and
M1
both values correct to at least 1 sf, sign change and conclusion
A1
(2)
(d) At Q, Setting . M1
Splitting up the numerator and proceeding to x=
M1
(as required)For correct proof.
No errors seen in working. A1
(3)
5
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question
NumberScheme Marks
(e) Iterative formula:
An attempt to substitute
into the iterative formula.Can be implied by
M1
Both and
A1
to 3 dp. all stated correctly to 3 dp
A1
(3)[13]
6
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question Number
Scheme Marks
6.
(a)Attempt to make x
(or swapped y) the subjectM1
Collect x terms together and factorise.
M1
A1 oe
(3)
(b) Range of g is -9≤ g(x)≤ 4 or -9≤ y ≤ 4 Correct Range B1(1)
(c) Deduces that is 0. Seen or implied.
M1
g g(2)= g (0) , from sketch. -6 A1(2)
(d)Correct order g followed by f
M1
5 A1
(2)(e)(i) Correct shape
B1
, B1
7
x
6
y
2
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question Number
Scheme Marks
(e)(ii)Correct shape
B1
Graph goes through and
which are marked.B1
(4)
(f) Domain of is -9≤ x ≤ 4 Either correct answer or a follow
through from part (b) answer B1
(1)[13]
8
-6
y
2
x
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question
NumberScheme Marks
7
(a)
Apply quotient rule:
Applying M1
Any one term correct on the numerator
A1
Fully correct (unsimplified). A1
For correct proof with an understanding that
No errors seen in working. (as required) A1*
(4)
(b) When , B1
At B1
Either T: or and
;
with ‘their TANGENT gradient’ and their y1;
or uses with ‘their TANGENT gradient’;
M1
T: A1(4)[8]
9
EDEXCEL CORE MATHEMATICS C3 (6665) – JANUARY 2011 FINAL MARK SCHEME
Question
Number
Scheme Marks
8.
(a)
Writes as and gives M1
or A1
Convincing proof.
Must see both A1 AG
(3)
(b)
M1
A1(2)
(c)Applies M1
Substitutes for M1
Attempts to use the identityM1
So
A1
(4)
[9]
10
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