6030 06a c3 january 2011 mark scheme

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


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


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


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


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


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


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


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


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


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

6030 06a c3 january 2011 mark scheme

Documents