General Certificate of Education Advanced Subsidiary Examination January 2012

Mathematics

MPC1

Unit Pure Core 1 Friday 13 January 2012

9.00 am to 10.30 am

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For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You must not use a calculator.

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Time allowed * 1 hour 30 minutes

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Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Fill in the boxes at the top of this page. * Answer all questions. * Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. * You must answer the questions in the spaces provided. Do not write outside the box around each page. * Show all necessary working; otherwise marks for method may be lost. * Do all rough work in this book. Cross through any work that you do not want to be marked. * The use of calculators is not permitted.

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Information * The marks for questions are shown in brackets. * The maximum mark for this paper is 75. Advice * Unless stated otherwise, you may quote formulae, without proof, from the booklet. * You do not necessarily need to use all the space provided.

P45841/Jan12/MPC1 6/6/6/

MPC1

2

The point A has coordinates ð6, 4Þ and the point B has coordinates ð2, 7Þ .

1 (a)

Given that the point O has coordinates ð0, 0Þ , show that the length of OA is less than the length of OB. (3 marks)

(b) (i)

Find the gradient of AB.

(2 marks)

(ii) Find an equation of the line AB in the form px þ qy ¼ r , where p, q and r are

integers.

(3 marks)

The point C has coordinates ðk, 0Þ . The line AC is perpendicular to the line AB. Find the value of the constant k. (3 marks)

(c)

Factorise x 2  4x  12 .

2 (a)

(1 mark)

(b)

Sketch the graph with equation y ¼ x 2  4x  12 , stating the values where the curve crosses the coordinate axes. (4 marks)

(c) (i)

Express x 2  4x  12 in the form ðx  pÞ2  q , where p and q are positive integers. (2 marks)

(ii) Hence find the minimum value of x 2  4x  12 .

(1 mark) 

 3 The curve with equation y ¼ x 2  4x  12 is translated by the vector . 2 (2 marks) Find an equation of the new curve. You need not simplify your answer.

(d)

3 (a) (i)

 pffiffiffi 2 Simplify 3 2 .

(ii) Show that

(b)

(02)

pffiffiffi 2  pffiffiffi 2  3 2  1 þ 3 þ 2 is an integer and find its value.

pffiffiffi pffiffiffi pffiffiffi 4 57 2 Express pffiffiffi pffiffiffi in the form m  n , where m and n are integers. 2 5þ 2

(1 mark) (4 marks)

(4 marks)

P45841/Jan12/MPC1

3

The curve with equation y ¼ x 5  3x 2 þ x þ 5 is sketched below. The point O is at the origin and the curve passes through the points Að1, 0Þ and Bð1, 4Þ .

4

y

Bð1, 4Þ

A 1

x

O

Given that y ¼ x 5  3x 2 þ x þ 5 , find:

(a) (i)

dy ; dx

(3 marks)

d2 y (ii) . dx 2

(1 mark)

(b)

Find an equation of the tangent to the curve at the point Að1, 0Þ .

(2 marks)

(c)

Verify that the point B, where x ¼ 1 , is a minimum point of the curve.

(3 marks)

(d)

The curve with equation y ¼ x 5  3x 2 þ x þ 5 is sketched below. The point O is at the origin and the curve passes through the points Að1, 0Þ and Bð1, 4Þ . y

Bð1, 4Þ

A 1

ð1 (i)

Find

O

ðx 5  3x 2 þ x þ 5Þ dx .

x

(5 marks)

1

(ii) Hence find the area of the shaded region bounded by the curve between A and B and

the line segments AO and OB.

Turn over

s

(03)

(2 marks)

P45841/Jan12/MPC1

4

The polynomial pðxÞ is given by pðxÞ ¼ x 3 þ cx 2 þ dx  12 , where c and d are constants.

5

(a)

When pðxÞ is divided by x þ 2 , the remainder is 150 . Show that 2c  d þ 65 ¼ 0 .

(3 marks)

(b)

Given that x  3 is a factor of pðxÞ, find another equation involving c and d. (2 marks)

(c)

By solving these two equations, find the value of c and the value of d.

(3 marks)

A rectangular garden is to have width x metres and length ðx þ 4Þ metres.

6 (a)

The perimeter of the garden needs to be greater than 30 metres. Show that 2x > 11 .

(b)

The area of the garden needs to be less than 96 square metres. Show that x 2 þ 4x  96 < 0 .

(c)

Solve the inequality x 2 þ 4x  96 < 0 .

(d)

Hence determine the possible values of the width of the garden.

(04)

(1 mark)

(1 mark) (4 marks) (1 mark)

P45841/Jan12/MPC1

5

A circle with centre C has equation x 2 þ y 2 þ 14x  10y þ 49 ¼ 0 .

7

Express this equation in the form

(a)

ðx  aÞ2 þ ðy  bÞ2 ¼ r 2

(3 marks)

Write down:

(b) (i)

the coordinates of C ;

(ii) the radius of the circle.

(2 marks)

(c)

Sketch the circle.

(d)

A line has equation y ¼ kx þ 6 , where k is a constant. (i)

(2 marks)

Show that the x-coordinates of any points of intersection of the line and the circle (2 marks) satisfy the equation ðk 2 þ 1Þx 2 þ 2ðk þ 7Þx þ 25 ¼ 0 .

(ii) The equation ðk 2 þ 1Þx 2 þ 2ðk þ 7Þx þ 25 ¼ 0 has equal roots. Show that

12k 2  7k  12 ¼ 0 (iii) Hence find the values of k for which the line is a tangent to the circle.

(3 marks) (2 marks)

Copyright ª 2012 AQA and its licensors. All rights reserved.

(05)

P45841/Jan12/MPC1

Key to mark scheme abbreviations M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA c sf dp

mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach candidate significant figure(s) decimal place(s)

No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.

MPC1 Q

Solution

1(a)

 OA   6  (4)  OB   (2)  OA   52 and  OB   53 2

2

2

2

;

2

or

Marks 2

7

2

 OA  

52 and

 OB  

grad AB 

(ii)

or OA2  52 and OB 2  53 A1

3

M1 11 8

11  x – 6 8

52

53 seen

and 53 or

53

A1

y – –4 = ‘their grad AB’(x – 6) or y – 7 = ‘their grad AB’(x – –2)

y4 –

both correct values 52 or

A1

74 –2  6 –

Comments either correct PI by 52 or 53 seen

M1

2

OA  52 and OB  53  OA  OB

(b)(i)

Total

condone one sign error 2 or

y = ‘their grad AB’ x + c and attempt to find c using x  6 , y  – 4 or x  – 2, y  7

M1

OE

correct working + concluding statement involving OA and/or OB

any correct form eg

A1

y–

11 34 x 8 8

but must simplify – – to +

 11x  8 y  34

(c) (grad AC =)

A1

8 11

3

FT

B1

4 8  ' their ' OE k –6 11

condone 8 y  11x  34 or any multiple of these equations –1 / ‘their grad AB’

equating gradients; LHS must be correct and RHS is “attempt” at perp grad to AB

M1

 2k  12  11  k 

23 2

A1cso Total

3

k = 11.5 OE

11

8 (c) Alternative: Eqn AC :  y  4   ' their '  x – 6  B1 ( 11y  8 x  92 ) AND must sub y  0 for M1 11 8 or  y  0   ' their '  x – k  B1 AND must sub x = 6 , y  4 for M1 11

MPC1 (cont) Q 2(a)

Solution

 x – 6  x  2 

(b)

Marks

Total

B1

1

Comments ISW for x  6, x  2 etc

y

–2

x

6

–12

x  – 2  x6  y  –12  – shaped curve

(c)(i)

(stated or –12 marked on sketch)

M1

approximately

A1

 x  2

M1

2 2

(ii) (Minimum value is ) (d)

B1

“correct” shape in all 4 quadrants with minimum to right of y-axis

 x  2

correct x values or FT ‘their’ factors (x-intercepts stated or marked on sketch) may be seen in (a)

B1

A1

– 16 –16

B1

Replacing each x by x + 3 OR adding 2 to their quadratic

4

p 2 2

p  2 and q 16

1

FT ‘ their – q ’ in original equation or ‘their’ completed square or factorised form or replacing y by y – 2

M1

y   x  3  – 4  x  3  – 12   2   2

or

y   x  1 –14

or or

y  x  2 x –13 y  2  ( x  3)( x  5)

2

A1

2

2

Total

10

OE any correct equation in x and y unsimplified

MPC1 (cont) Q 3(a)(i)

(ii)

3 2  3

Solution 2

Marks

Total

 18

B1

1



M1

FT their 3 2

A1

 19  6 2 

B1

 11  6 2 

 18 – 3 2 – 3 2  1

3  2   9  3 2

2 3 2 2

 Sum  30

(b)



2

2  1  ' their 18'– 3 2 – 3 2  1

A1cso

4 5–7 2 2 5 2  2 5 2 2 5– 2

Comments



2

4

M1

Numerator  8

 5

2

– 4 5 2 – 14 5 2  7



Denominator = 2 5

 –  2 2

 2

2

  54  18 10 

m1

correct unsimplified

B1

must be seen as denominator

2

 18  Answer = 3 – 10

A1cso

Total

4 9

MPC1 (cont) Q 4(a)(i)

(ii)

(b)

(c)

Solution

Marks M1 A1 A1

 dy    5x4 – 6x  1   dx 

 d2 y  3  2   20 x – 6 d x  

B1

dy  5( 1) 4  6( 1)  1 (  12) dx  y  12  x  1 x 1 

x1 

dy  5– 6 1 dx

Total

3

Comments one term correct another term correct all correct (no + c etc)

1

FT ‘their’

must sub x = –1 into 'their'

M1 A1cso

2

when x  1,

or E1

3

d2 y  20  6 > 0 dx 2  (B is a) minimum (point)

must have correct

(d)(i)

dy dx

shown  0 plus correct statement

A1cso

d2 y  14 dx 2  (B is a ) minimum (point)

dy dx

any correct form with ( x  1) simplified condone y  12 x  c, c  12 sub x  1 into their

M1

dy  0  stationary point dx

dy dx

d2 y dy and 2 for E1 dx dx

x6 3x3 x 2   5x – 6 3 2

M1 A1 A1

one term correct another term correct all correct (may have + c)

1 1 1  1   6 – 1  2  5 –  6  1  2 – 5    

m1

‘their’ F(1) – F(–1) with powers of 1 and –1 evaluated correctly

= 8 (ii) ‘their answer to part (i)’ – 2

 Area  6 Total

A1cso

5

M1 A1cso

2 16

MPC1 (cont) Q Solution 3 5(a) p( – 2)  (–2)  (2) 2 c  (–2)d – 12

Marks M1

‘their’ 8  4c – 2d – 12  – 150

m1

 2c – d  65  0

(b)

(c)

Total

A1cso

p(3)  33  32 c  3d – 12

M1

9c  3d  15  0

A1

2c – d  65  0    5c   70 3c  d  5  0 

putting expression for remainder = –150 3

A1 A1 Total

6(a) Sides are x and x  4  x  x  x  4  x  4  30 or 2 x  2 x  8  30

AG terms all on one side in any order (check that there are no errors in working)

or p(3) attempted long division by x–3 as far as remainder 2

any correct equation with terms collected eg 3c  d   5

Elimination of c or d

M1

 c  – 14 , d  37 OE

Comments p(–2) attempted or long division by x+2 as far as remainder

3 8

value of c or d correct unsimplified both c and d correct unsimplified

must see this line OE

or 2  2 x  4   30 or 4 x  8  30

  4 x  22   2 x  11 (b)

B1

x  x  4   96

 x  4 x – 96  0

B1

 x  12  x – 8 

M1

Critical values 8, –12

A1

8

x

+

AG

correct factors or correct quadratic equation formula

–

+ sketch or sign diagram

M1 –12

8

 – 12  x  8

(d)

1

or

y –12

AG (be convinced) condone 11 < 2x

must see this line OE 2

(c)

1

1 5  x8 2 Total

A1cso

4

B1

1 7

accept but not nor

x  8 AND x  – 12 x  8 OR x  – 12 x  8 , x  – 12

MPC1 (cont) Q 7(a)

Solution

 x  7    y – 5 2

2

 x  7    y – 5 2

2

= 52

(b)(i) C  –7, 5 

Marks

Total

M1

one term correct ; condone ( x – – 7)2

A1

both terms correct with squares and plus sign between terms

A1cao

3

r=5

B1

condone 25 for 52 correct or FT ‘their’ circle equation

B1

(ii)

Comments

2

correct or FT ‘their’ r2 > 0 condone 25 etc but not  25

(c) must draw axes

freehand circle with C correct or FT ‘their C’ for quadrant of centre

M1

A1

2

–7 (d)(i)

clear attempt to sub y  kx  6 into original or ‘their’ circle equation …

x 2  (kx  6)2  14 x  10(kx  6)  49  0 x 2  k 2 x 2 12kx  36  14 x –10kx – 60  49  0

circle touching x-axis at –7 with –7 marked (need not show 5 on y-axis ) but circle must not touch y-axis

…and attempt to multiply out

M1

(1  k 2 ) x 2  2 kx  14 x  25  0  ( k 2  1) x 2  2( k  7) x  25  0

(ii) Equal roots ‘ b2 – 4ac  0 ’

 2  k  7   – 4  25  k 2 1 2

A1cso

2

AG condone x 2 (1  k 2 )  2 x(7  k )  ... etc

B1

allow statement alone if discriminant in terms of k attempted

M1

discriminant (condone one slip)

4k 2 14k  49 – 25k 2  25  0

– 24k 2 14k  24  0

(iii)

 12k 2 – 7k –12  0

A1

 4k  3 3k – 4 

M1

3

AG all working correct but = 0 must appear before last line

correct factors or correct use of formula as far as k 

k 

3 4 , k  4 3

OE

are values of k for which line is a tangent Total TOTAL

A1

2 14 75

7  49  576 24



Scaled mark unit grade boundaries - January 2012 exams A-level Max. Scaled Mark

Scaled Mark Grade Boundaries and A* Conversion Points A* A B C D E

Code

Title

LAW02 LAW03

LAW UNIT 2 LAW UNIT 3

94 80

69

MD01 MFP1 MM1A MM1B MPC1

MATHEMATICS UNIT MD01 MATHEMATICS UNIT MFP1 MATHEMATICS UNIT MM1A MATHEMATICS UNIT MM1B MATHEMATICS UNIT MPC1

75 75 100 75 75

-

MS1A MATHEMATICS UNIT MS1A MS/SS1A/W MATHEMATICS UNIT S1A - WRITTEN MS/SS1A/C MATHEMATICS UNIT S1A - COURSEWORK

-

73 63

66 57

59 51

52 45

46 40

62 56 50 44 67 60 53 46 no candidates were entered for this unit 59 52 46 40 61 55 49 43

39 39 34 37

100 75 25

-

74 54 20

65

56

47

38 28 10

MS1B MD02 MFP2 MM2B MPC2 MS2B MFP3 MPC3 MFP4 MPC4

MATHEMATICS UNIT MS1B MATHEMATICS UNIT MD02 MATHEMATICS UNIT MFP2 MATHEMATICS UNIT MM2B MATHEMATICS UNIT MPC2 MATHEMATICS UNIT MS2B MATHEMATICS UNIT MFP3 MATHEMATICS UNIT MPC3 MATHEMATICS UNIT MFP4 MATHEMATICS UNIT MPC4

75 75 75 75 75 75 75 75 75 75

69 59 69 69 67 64 60 63

56 64 52 63 66 63 60 57 54 57

49 57 45 55 59 55 52 50 48 51

42 50 38 47 52 47 44 43 42 45

36 44 31 39 46 40 37 37 37 39

30 38 25 32 40 33 30 31 32 33

MEST1 MEST2 MEST3 MEST4

MEDIA STUDIES UNIT 1 MEDIA STUDIES UNIT 2 MEDIA STUDIES UNIT 3 MEDIA STUDIES UNIT 4

80 80 80 80

67 74

55 63 57 68

47 54 47 56

40 45 37 45

33 36 27 34

26 28 18 23

PHIL1

PHILOSOPHY UNIT 1

90

-

55

49

43

37

32