Additional Mathematics Mark Scheme for the Unit June 2007 - OCR

FSMQ Additional Mathematics ADVANCED FSMQ 6993

Mark Scheme for the Unit June 2007

6993/MS/R/07

Oxford Cambridge and RSA Examinations

OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body, established by the University of Cambridge Local Examinations Syndicate and the RSA Examinations Board in January 1998. OCR provides a full range of GCSE, A level, GNVQ, Key Skills and other qualifications for schools and colleges in the United Kingdom, including those previously provided by MEG and OCEAC. It is also responsible for developing new syllabuses to meet national requirements and the needs of students and teachers. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by Examiners. It does not indicate the details of the discussions which took place at an Examiners’ meeting before marking commenced. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the Report on the Examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2007 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: Facsimile: E-mail:

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CONTENTS

Additional Mathematics FSMQ (6993)

MARK SCHEME FOR THE UNIT

Unit

Content

Page

6993

Additional Mathematics

1

*

Grade Thresholds

10

Mark Scheme 6993 June 2007

1

6993

Mark Scheme

June 2007

Q. Answer Section A 1 3( x + 2) > 2 − x

⇒ 3x + 6 > 2 − x ⇒ 4 x > −4

Mark

Notes

M1

Expand and collect

A1 A1

Only 2 terms 3

⇒ x > −1 2

v = 6 + 3t 2 ⇒ s = 6t + t 3 + c Take s = 0 when t = 1 ⇒ c = −7

M1 A1

Ignore c

DM1

Either sub to find c or sub and subtract from definite integral

A1

When t = 3, s = 18 + 27 − 7 = 38 Alternatively: 3

M1 int A1 DM1 sub and sub 4 A1

s = ∫ ( 6 + 3t 2 ) dt = ⎡⎣6t + t ⎤⎦ = (18 + 27 ) − ( 6 + 1) = 38 1 3 3

1

3

M1

x2 + y 2 − 4 x − 6 y + 3 = 0

Complete the square

⇒ x 2 − 4 x + y 2 − 6 y = −3 ⇒ x2 − 4x + 4 + y2 − 6 y + 9 = 4 + 9 − 3 ⇒ ( x − 2 ) + ( y − 3) = 10 2

2

B1 A1

⇒ Centre (2, 3), radius 10 ( ≈ 3.162...) SC: Penultimate line M1 A1 S.C. Centre Find a point on the circle and then use Pythagoras to find radius 4

B1 M1 A1

sin x = −4 cos x ⇒ tan x = −4 ⇒ x = ±75.96

B1 B1

0

⇒ x = 180 − 75.96 = 1040

M1 A1 A1

and x = 360 − 75.96 = 2840

Alternatively Use of s 2 + c 2 = 1 1 ⇒ cos 2 x = 17 ⇒ x = ±75.960 ⇒ x = 180 − 75.96 = 104

M1

A1 0

and x = 360 − 75.96 = 284

Centre Radius 3 Accept correct answers with no working

M1 A1 0

S.C. Graphical method ±20 tolerance B1 B1 S.C. Answers with no working B1 for both.

2

A1

For either value from calculator For method to find a correct answer from that given on 5 calculator -1 extra values Ignore values outside 3600

6993

5

(i)

Mark Scheme

Using v 2 = u 2 + 2as

M1

⇒ 102 = 302 + 2a.300

A1

⇒ 600a = −800 4 ⇒a=− 3 (ii)

June 2007

Got to be used!

A1

Ignore –ve sign. 3

Using v = u + at

M1

4 ⇒ 10 = 30 − t 3 3 ⇒ t = 20 × = 15 4

F1

From their a 2

u+v t 2 30 + 10 ⇒ 300 = t 2 600 ⇒t = = 15 40 Or: s =

6

This could be used in (i) to find t then a

dy = 3x 2 − 3 dx

B1 M1

Diff correctly Substitute in their gradient function

dy = 9 ⇒ y − 6 = 9( x − 2) dx ⇒ y = 9 x − 12

DM1

Set up equation with their gradient

dy = 3 x 2 − 4 x − 15 dx = 0 when 3x 2 − 4 x − 15 = 0

M1 A1 M1

At (2, 6)

7

A1 4

⇒ ( 3x + 5 )( x − 3) = 0

=0 and attempt to solve

A1

5 ⇒ x = 3, − 3 2 d y = 6 x − 4: dx 2 d2 y When x = 3, 2 > 0 dx ⇒ minimum ⇒ x=3

M1

Differentiate again and substitute

F1 A1 7 Providing all other marks earned

N.B. Any valid method is acceptable, but not that x = 3 is the right hand value or that the y value is lower then for the other value of x.

3

6993

8

(i)

Mark Scheme

June 2007

M1

Equate and attempt to collect terms

⇒ x2 − 4 x + 3 = 0

M1

Solve a quadratic

⇒ ( x − 3)( x − 1) = 0

A1

4 x − x2 = x2 − 4 x + 6 ⇒ 2 x2 − 8x + 6 = 0

⇒ x = 3,1 (ii)

Area =

Ans only seen - B1 3

∫ ( 4 x − x ) dx − ∫ ( x 3

3

2

1

1

2

)

− 4 x + 6 dx

3

3

⎡ ⎤ x3 ⎤ ⎡ x3 = ⎢2 x2 − ⎥ − ⎢ − 2 x2 + 6 x ⎥ 3 ⎦1 ⎣ 3 ⎣ ⎦1 1⎞ ⎛ ⎛1 ⎞ = (18 − 9 ) − ⎜ 2 − ⎟ − ( 9 − 18 + 18 ) + ⎜ − 2 + 6 ⎟ 3⎠ ⎝ ⎝3 ⎠ 2 1 2 = 9 −1 − 9 + 4 = 2 3 3 3

M1

Integrate

A1

All terms; condone one slip

DM1

Substitute and subtract (even if limits wrong)

A1 4

Alternatively: Area =

∫ (8x − 2 x 3

1

2

M1 integrate

)

− 6 dx

A1

3

⎡ 2 2 x3 ⎤ = ⎢4 x − − 6x⎥ 3 ⎣ ⎦1

DM1 sub and sub

2 ⎛ ⎞ ⎛ 2⎞ = ( 36 − 18 − 18 ) − ⎜ 4 − − 6 ⎟ = 0 − ⎜ −2 ⎟ 3 ⎝ ⎠ ⎝ 3⎠ 2 =2 3 9

(i)

(ii) (iii)

AB =

( 5 − −1) + (8 − 1) 2

2

= 85

AC =

(8 − −1) + ( 3 − 1)

2

= 85

2

A1

M1 A1 2 B1

⎛ 5 + 8 8 + 3 ⎞ ⎛ 13 11 ⎞ M= ⎜ , ⎟=⎜ , ⎟ 2 ⎠ ⎝ 2 2⎠ ⎝ 2 8−3 5 Grad BC = =− 5−8 3 11 − 1 9 9 3 Grad AM = 2 = 2 = = 13 + 1 15 15 5 2 2 5 3 ⇒ m1 .m2 = − . = −1 3 5

1 E1

Both gradients; AM ft from their M

B1

Both and demonstration 2

Allow a geometric argument with reference to M being midpoint and the triangle isosceles. (iv)

For sight of Pythagoras used at least once

M1

3 ( x + 1) 5 ⇒ 5 y = 3x + 8 y −1 =

Must use (−1, 1) or their M and their g

A1 2

4

6993 10

Mark Scheme

(i)

June 2007 B1 E1

One line Shading

B1 E1

2nd line Shading

B1

Other two lines and 5 shading

B1 B1

Allow B2 for 12

N.B. -1 no scales (ii)

Maximum value on y-axis (0, 4) giving 12

2

5

6993

Mark Scheme

June 2007

Section B 11

(a)(i)

x3 − 3x 2 − 4 x = 0

M1

⇒ x x 2 − 3x − 4 = 0

A1 A1

(

)

⇒ x ( x − 4 )( x + 1) = 0 ⇒ x = 0 , −1, 4

Accept any valid method

A1 4

S.C. just answers B2 (ii)

B1

1 Must have points on axes (b)(i)

Remainder theorem or long division G(−1) = 12

M1 A1

(ii)

g(2) = 0

B1

For sub −1 2 For sub x = 2 1

(iii)

By continued trial

M1

or by division and quadratic factorisation g(3) = 0, g( − 2) = 0 ⇒ x = 2, 3, − 2

A1 A1 A1

S.C. just answers B2 Alternatively: By division by (x − 2) and quadratic factorisation M1 (x − 2)(x2 − x − 6) = 0 A1 ⇒(x − 2)(x + 2)(x − 3) = 0 A1 ⇒x = 2, −2, 3. A1

6

3 −2 Final answer 4

6993

12

Mark Scheme

(i)

June 2007

8

⎛ 9 ⎞ P(All males) = ⎜ ⎟ = 0.00168 ⎝ 20 ⎠

M1 A1 2

(ii)

(iii)

3

⎛ 9 ⎞ ⎛ 11 ⎞ P(5 females) = 8 C5 ⎜ ⎟ ⎜ ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = 0.2568 ≈ 0.257

M1 M1 A1 A1

5

23 17 ⎞ ⎛ ⎜ Or P(PT) = ⎟ 40 40 ⎠ ⎝ P(at least two part-time) = 1 − P(all FT) − P(7FT,1PT)

P(full-time) =

8

7

⎛ 23 ⎞ ⎛ 23 ⎞ ⎛ 17 ⎞ = 1− ⎜ ⎟ − 8⎜ ⎟ ⎜ ⎟ ⎝ 40 ⎠ ⎝ 40 ⎠ ⎝ 40 ⎠ = 1 − 0.0119 − 0.0706 = 0.917

6

M1 2

8

A1 powers A1 Coeffs A1 Ans

= 0.917 S.C. Read “At least two” as “exactly two” 6

M1 A1

probability

M1

1−2correct terms

A1 A1 A1

Powers coefficient Ans 6

Alternatively: Add 7 terms ⎛ 23 ⎞ ⎛ 17 ⎞ ⎛ 17 ⎞ 28 ⎜ ⎟ ⎜ ⎟ + ........ ⎜ ⎟ ⎝ 40 ⎠ ⎝ 40 ⎠ ⎝ 40 ⎠

powers coefficient 56 (could be 4 implied)

2

⎛ 23 ⎞ ⎛ 17 ⎞ 28 ⎜ ⎟ ⎜ ⎟ = 28 × 0.00653 = 0.1828 ⎝ 40 ⎠ ⎝ 40 ⎠

7

B1

6993

13

(i)

Mark Scheme

June 2007

Pythagoras:

M1

OM2 = 372 − 122 ⇒ OM = 35 CM2 = 202 − 122 ⇒ CM = 16

A1 A1

Correct use of Pythagoras for at least one 3

(ii)

M1 M1 A1 A1

Use cosine rule on triangle OCM ⇒ cosC =

16 + 40 − 35 ⇒ C = 60.50 2 × 16 × 40 2

2

2

Correct angle Correct use of cosine formula Ans 4

(iii)

Sight of attempt to find base area 1 Area = × 16 × 24 = 192 2 Sight of attempt to find height h = 40sin60.5 = 34.8 1 ⇒ Volume = ×192 × 34.8 = 2228 ≈ 2230cm3 3

8

M1 A1

Can be implied

M1 A1

Can be implied

A1 5

6993

14

Mark Scheme

(i)

Apply Pythagoras to both triangles: x2 = y2 + 4 (x + 0.95)2 = (y + 1.05)2 + 4

June 2007

B1 B1 2

(ii)

(iii)

Subtract: 2×0.95x + 0.952 = 2×1.05y + 1.052 ⇒2.1y = 1.9x − (1.052 −0.952) ⇒2.1y = 1.9x − 0.2 Alternatively: Multiply out one of the brackets B1 Substitute for y2 M1 Correct result A1

M1 A1

Substitute for y:

M1

Get y as subject

M1

Sub expression for y

⎛ 1.9 x − 0.2 ⎞ x2 = ⎜ ⎟ +4 2.1 ⎠ ⎝ ⇒ 2.12 x 2 = 1.92 x 2 − 2 × 0.2 × 1.9 x + 0.22 + 4 × 2.12

A1 3

2

⇒ 0.8 x 2 + 0.76 x − 17.68 = 0 −0.76 ± 0.76 + 4 × 0.8 × 17.68 −0.76 + 7.56 = = 4.25 1.6 1 .6 Substitute : ⇒x=

2

⎛ 1.9 x − 0.2 ⎞ y=⎜ ⎟ = 3.75 2.1 ⎠ ⎝

A1 Correct quadratic Solve Ignore other root

DM1 A1 DM1 F1 7

Withhold last mark if more than one answer given The quadratic in y is 20y2 + 21y − 360 = 0 Integer coefficients for x equation gives 20x2 + 19x − 442 = 0

9

FSMQ Advanced Additional Mathematics 6993 June 2007 Assessment Session

Unit Threshold Marks Unit 6993

Maximum Mark 100

A

B

C

D

E

U

70

60

50

40

30

0

The cumulative percentage of candidates awarded each grade was as follows: A 6993

28.8

B 38.6

C

D

E

U

48.1

57.5

66.8

100

10

Total Number of Candidates 5500

11

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