General Certificate of Education Advanced Subsidiary Examination June 2011

Mathematics

MPC1

Unit Pure Core 1 Wednesday 18 May 2011

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.

P38883/Jun11/MPC1 6/6/6/

MPC1

2

The line AB has equation 7x þ 3y ¼ 13 .

1 (a)

Find the gradient of AB .

(b)

The point C has coordinates ð1, 3Þ . (i)

(2 marks)

Find an equation of the line which passes through the point C and which is parallel to AB . (2 marks) 1 2

(ii) The point ð1 , 1Þ is the mid-point of AC . Find the coordinates of the point A.

(2 marks)

The line AB intersects the line with equation 3x þ 2y ¼ 12 at the point B. Find the coordinates of B. (3 marks)

(c)

2 (a) (i)

Express

pffiffiffiffiffi pffiffiffi 48 in the form k 3 , where k is an integer.

pffiffiffiffiffi pffiffiffiffiffi 48 þ 2 27 pffiffiffiffiffi (ii) Simplify , giving your answer as an integer. 12 pffiffiffi pffiffiffi 15 5 pffiffiffi in the form m þ n 5 , where m and n are integers. Express 3þ 5

(b)

(1 mark) (3 marks)

(4 marks)

The volume, V m3 , of water in a tank after time t seconds is given by

3

V¼

t3  3t þ 5 4

dV . dt

(a)

Find

(b) (i)

Find the rate of change of volume, in m3 s1 , when t ¼ 1 .

(2 marks) (2 marks)

(ii) Hence determine, with a reason, whether the volume is increasing or decreasing

when t ¼ 1 . (c) (i)

Find the positive value of t for which V has a stationary value.

(1 mark) (3 marks)

d 2V , and hence determine whether this stationary value is a maximum value or dt 2 a minimum value. (3 marks)

(ii) Find

(02)

P38883/Jun11/MPC1

3

Express x 2 þ 5x þ 7 in the form ðx þ pÞ2 þ q , where p and q are rational numbers. (3 marks)

4 (a)

A curve has equation y ¼ x 2 þ 5x þ 7 .

(b) (i)

Find the coordinates of the vertex of the curve.

(ii) State the equation of the line of symmetry of the curve. (iii) Sketch the curve, stating the value of the intercept on the y-axis.

(2 marks) (1 mark) (3 marks)

Describe the geometrical transformation that maps the graph of y ¼ x 2 onto the (3 marks) graph of y ¼ x 2 þ 5x þ 7 .

(c)

The polynomial pðxÞ is given by pðxÞ ¼ x 3  2x 2 þ 3 .

5 (a)

Use the Remainder Theorem to find the remainder when pðxÞ is divided by x  3 . (2 marks)

(b)

Use the Factor Theorem to show that x þ 1 is a factor of pðxÞ.

(c) (i)

Express pðxÞ ¼ x 3  2x 2 þ 3 in the form ðx þ 1Þðx 2 þ bx þ cÞ , where b and c are integers. (2 marks)

(ii) Hence show that the equation pðxÞ ¼ 0 has exactly one real root.

(2 marks)

Turn over

s

(03)

(2 marks)

P38883/Jun11/MPC1

4

The curve with equation y ¼ x 3  2x 2 þ 3 is sketched below.

6

y

Bð1, 2Þ

A 1

O

x

The curve cuts the x-axis at the point Að1, 0Þ and passes through the point Bð1, 2Þ . ð1 (a)

Find

ðx 3  2x 2 þ 3Þ dx .

(5 marks)

1

(b)

Hence find the area of the shaded region bounded by the curve y ¼ x 3  2x 2 þ 3 and the line AB . (3 marks)

Solve each of the following inequalities:

7 (a)

2ð4  3xÞ > 5  4ðx þ 2Þ ;

(2 marks)

(b)

2x 2 þ 5x 5 12 .

(4 marks)

(04)

P38883/Jun11/MPC1

5

A circle has centre Cð3, 8Þ and radius 10 .

8

Express the equation of the circle in the form

(a)

ðx  aÞ2 þ ðy  bÞ2 ¼ k

(2 marks)

(b)

Find the x-coordinates of the points where the circle crosses the x-axis.

(3 marks)

(c)

The tangent to the circle at the point A has gradient 2 . Find an equation of the line CA , giving your answer in the form rx þ sy þ t ¼ 0 , where r, s and t are integers. (3 marks)

(d)

The line with equation y ¼ 2x þ 1 intersects the circle.

5

(i)

Show that the x-coordinates of the points of intersection satisfy the equation x 2 þ 6x  2 ¼ 0

(3 marks)

(ii) Hence that the x-coordinates of the points of intersection are of the form pffiffishow ffi

m

n , where m and n are integers.

(2 marks)

END OF QUESTIONS

Copyright ª 2011 AQA and its licensors. All rights reserved.

(05)

P38883/Jun11/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)

y=

13 7 – x 3 3

(gradient =) − (b)(i)

Marks

Total

M1

7 3

A1

y – 3 = ‘their grad’(x – – 1)

2

M1

or 7x + 3y = k and attempt at k using x = –1 and y = 3 or y = (their m)x + c and attempt at c using x = –1 and y = 3

7 ( x + 1) or 7 x + 3 y = 2 3 7 2 or y = − x + c, c = 3 3

A1cso

2

( 4, – 5)

B1,B1

2

y –3 = −

(ii)

(c) 7x + 3y = 13 and 3x + 2y = 12  equation in x or y only

A1 A1 Total

correct equation in any form and replacing – – with + sign

x = 4 , y = –5 withhold if clearly from incorrect working must use correct pair of equations and attempt to eliminate y (or x)

M1

x= –2 y =9

Comments attempt at y = a + bx Δy with 2 correct points or Δx condone slip in rearranging if gradient is correct

3 9

MPC1 (cont) Q 2(a)(i) 48 = 4 3 (ii)

Solution

Marks B1

4 3 +6 3 2 3

=5

Total 1

Comments condone k = 4 stated

M1

attempt to write each term in form k 3 with at least 2 terms correctly obtained

A1

correct unsimplified in terms of

A1cso

3

3 only

must simplify fraction to 5

 3 M1  or ×  3   24 + 36 correct with integer terms = A1 12 =5 A1cso Alternative 1 ×

Alternative 2

12 12

48 + 108 12 = 4+ 9 = 5

Alternative 3

M1 A1 A1cso

48 27 M1 +2 12 12 9 A1 =2+2 4 A1cso =5

if hybrid of methods used, award M1 and most appropriate first A1 NMS (answer =) 5 scores full marks (b)

1– 5 5 3 – 5 × 3+ 5 3– 5

M1

(numerator =) 3 − 5 – 15 5 + 25

m1

correct unsimplified but must write 5 5 5 = 25 PI by 28 seen later

(denominator = 9 – 5 =) 4

B1

must be seen as denominator

28 – 16 5 giving 4 (answer =) 7 – 4 5

A1 Total

4 8

m = 7 , n = –4

MPC1 (cont) Q

Solution

Marks M1 A1

2

3(a)

(b)(i)

 dV  3t = –3   dt  4

t =1 

dV 3 = –3 dt 4 1 = −2 4

Total

2

dV dt dV correct (–2.25 OE) BUT must have dt

substituting t = 1 into their

M1 A1cso

Comments one of these terms correct all correct (no + c etc)

2

must have used

(ii) Volume is decreasing when t = 1

dV in (b)(i) or starts dt

again because

(c)(i)

dV 0 dt 2 dt 2  minimum

PI by “correct” equation being solved

A1

obtaining t n = k correctly from their 3

(condone unsimplified) ft their

M1

ft their 3 11

dV dt

withhold if answer left as t = ± 2

B1

A1cso Total

1 dV < 0 (or − 2 < 0 etc) 4 dt ft increasing plus explanation dV >0 if their dt

must state that

M1

A1cso

t=2

1

dV dt

d 2V and value of t from (c)(i) dt 2

MPC1 (cont) Q 4(a)

Solution

( x + 2.5)

Marks

2

M1

( x + 2.5)

A1

2

+ 0.75 mark their final line as their answer

 5 3  – ,  2 4

(ii)

x= –

5 2 unsimplified attempt at q = 7 – 'their' p 2 25 3 q=7 – = 4 4 3

A1cao

5 2

B1

2

1

correct or ft “ x = – ' their ' p ”

B1

y intercept = 7 stated or seen in table as y = 7 when x = 0 or 7 marked as intercept on y-axis (any graph)

M1

∪ shape

A1

vertex above x-axis in correct quadrant and parabola extending beyond y-axis into first quadrant

y (iii)

5 cao found using calculus 2 condone correct coordinates stated x = –2.5, y = 0.75

or x = −

M1

or y = ‘their’ q

Comments

p=

B1

q = 7 – 'their' p2

(b)(i) x = – ‘their’ p

Total

x

(c) Translation – 5  through  3 2   4 

E1

and no other transformation

M1

ft either ‘their’ –p or ‘their’ q or one component correct for M1

A1cao Total

3

3 12

both components correct for A1; may describe in words or use a vector

MPC1 (cont) Q Solution 3 2 5(a) p ( 3) = 3 − 2 × 3 + 3 ( = 27 – 18 + 3) = 12 (b)

Marks M1 A1

p ( –1) = ( –1) – 2 ( −1) + 3 3

2

2 (c)(i) Quadratic factor ( x – 3x + 3)

2

A1cso

– 3 x + 3)

A1

Total 3

correctly shown = 0 plus statement b = –3 or c = 3 by inspection

2

must see correct product ‘their’ discriminant considered possibly within quadratic equation formula

M1

b 2 − 4ac < 0  no real roots from quadratic   A1cso  only one real root 

( x

2

or full long division attempt or comparing coefficients

b2 − 4ac = (−3)2 − 4 × 3

6(a)

p(–1) attempted; not long division

M1

(ii) Discriminant of quadratic

1

Comments p(3) attempted; not long division

2

M1

p ( –1) = – 1 – 2 + 3 = 0  x + 1 is a factor

( p ( x ) = ) ( x +1) ( x

Total

2

8

)

– 2 x 2 + 3 dx

–1

M1 A1 A1

1

 x 4 2 x3  = – + 3x  3 4  –1

=  1 – 2 + 3  4 3

– 1 + 2 – 3       4 3 

one term correct another term correct all correct (condone + c)

B1

= 42 3

A1cso

 1  (b) Area of ∆  = × 2 × 2   2  = 2

Shaded region has area 4 2 – 2 3

= 22 3

B1

PI

M1

± their (a) ± their ∆ area

A1cso Total

5

‘their’ F(1) – F(–1) with (–1)3 etc evaluated correctly but must have earned M1 14 , 56 etc 3 12 but combined as single fraction

3 8

8 , 32 etc 3 12 but combined as single fraction

MPC1 (cont) Q Solution 8 – 6 x > 5 – 4 x – 8 7(a) 11 > 2 x 11   x < 51  or x <  2 2   (b)

Marks M1

Total

A1cso

2

Comments multiplying out correctly and > sign used

accept 5.5 > x OE

2 x2 + 5x –12  0 ( x + 4)(2 x − 3)

M1

correct factors

−5 ± 121 4 6 16 etc both CVs correct; condone , − 4 4 here but must be single fractions (or roots unsimplified)

Critical values are – 4 and 3 2

A1 M1

y

sketch or sign diagram including values +

–4

3 2

x

–4

x  −4 , x  3 2 take their final line as their answer

A1 Total

4 6

+

– 3 2

fractions must be simplified condone use of OR but not AND

MPC1 (cont) Q Solution 2 2 8(a) ( x − 3) + ( y + 8) = 100 (b)

Marks B1 B1

y = 0  'their ' ( x – a ) + b 2 = k 2

2

Total

2

( x − 3) = 36 or x − 6 x − 27 (= 0) (PI)  x = – 3, 9

accept ( y − −8)2 condone RHS = 10 2 or k = 10 2 Alternative

M1

2

Comments

A1 A1

3

(d 2 =) 102 – 82 d = 36  x = – 3, 9

A1

2 x + 5 y + 34 = 0 2

their ( x – 3) + ( 2 x + 1 + 8 ) 2

2

2

A1cso

3

or

x + (2 x + 1) − 6 x + 16(2 x + 1)

M1 A1

or d = 6

A1

any form of correct equation 2 34 eg y = − x + c, c = − 5 5 integer coefficients - all terms on 1 side substituting y = 2x + 1 correctly into LHS of “their” circle equation and

(+ 73) M1

2

10

M1

CA has equation ( y + 8) = – 2 ( x − 3) 5

(d)(i)

8

C 2

2 (c) Line CA has gradient – 5

d

attempt to expand in terms of x only

2

x – 6 x + 9 + 4 x + 36x + 81 = 100 or x2 + 4 x2 + 4 x +1 − 6 x + 32 x + 16 + 73 = 100

any correct equation (with brackets expanded) must see this line or equivalent

A1

 5x2 + 30 x –10 = 0  x2 + 6 x – 2 = 0 (ii)

A1cso

( x + 3)2 = 11

3

AG; all algebra must be correct or correct use of formula

M1

must get as far as x =

x = – 3 ± 11

A1cso Total TOTAL

2 13 75

exactly this

– 6 ± 44 2



Scaled mark unit grade boundaries - June 2011 exams A-level Max. Scaled Mark

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

Code

Title

HBIO2 HBI3T HBI3X HBIO4 HBIO5 HBI6T HBI6X

GCE HUMAN BIOLOGY UNIT 2 GCE HUMAN BIOLOGY UNIT 3T GCE HUMAN BIOLOGY UNIT 3X GCE HUMAN BIOLOGY UNIT 4 GCE HUMAN BIOLOGY UNIT 5 GCE HUMAN BIOLOGY UNIT 6T GCE HUMAN BIOLOGY UNIT 6X

80 50 50 90 90 50 50

61 65 43 43

56 41 34 56 60 40 39

51 38 30 51 55 37 35

46 35 26 46 50 34 31

41 33 22 41 45 31 27

37 31 19 36 41 28 24

INFO1 INFO2 INFO3 INFO4

GCE INFO AND COMM TECH UNIT 1 GCE INFO AND COMM TECH UNIT 2 GCE INFO AND COMM TECH UNIT 3 GCE INFO AND COMM TECH UNIT 4

80 80 100 70

72 63

53 53 66 57

48 47 60 50

43 41 54 43

38 35 48 36

33 30 43 30

LAW01 LAW02 LAW03 LAW04

GCE LAW GCE LAW GCE LAW GCE LAW

96 94 80 85

69 73

73 70 62 66

66 61 55 59

59 52 48 53

52 43 42 47

46 35 36 41

XMCAS MD01 MFP1 MM1A MM1B MPC1

GCE MATHEMATICS UNIT XMCAS GCE MATHEMATICS UNIT D01 GCE MATHEMATICS UNIT FP1 GCE MATHEMATICS UNIT M1A GCE MATHEMATICS UNIT M1B GCE MATHEMATICS UNIT PC1

125 75 75 100 75 75

-

99 89 79 70 59 52 45 39 61 54 48 42 No candidates were entered for this unit 62 55 48 41 62 55 48 41

61 33 36

GCE MATHEMATICS UNIT S1A GCE MATHEMATICS UNIT S1A - WRITTEN GCE MATHEMATICS UNIT S1A - COURSEWORK

100 75 25

-

73 53 20

36 26 10

MS1A MS/SS1A/W MS/SS1A/C

UNIT 1 UNIT 2 UNIT 3 UNIT 4

-

63

54

45

34 34