Condensed. Mathematics. General Certificate of Education Advanced Subsidiary Examination January 2012

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1 General Certificate of Education Advanced Subsidiary Examination January 01 Mathematics MPC1 Unit Pure Core 1 Friday 13 January am to am For this paper you must have: the blue AQA booklet of formulae and statistical tables. You must not use a calculator. Time allowed 1 hour 30 minutes 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. Information The marks for questions are shown in brackets. The maximum mark for this paper is 75. Condensed 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/Jan1/MPC1 6/6/6/ MPC1

2 1 The point A has coordinates ð6, 4Þ and the point B has coordinates ð, 7Þ. (a) Given that the point O has coordinates ð0, 0Þ, show that the length of OA is less than the length of OB. (b) (i) Find the gradient of AB. ( marks) (ii) Find an equation of the line AB in the form px þ qy ¼ r, where p, q and r are integers. (c) The point C has coordinates ðk, 0Þ. The line AC is perpendicular to the line AB. Find the value of the constant k. (a) Factorise x 4x 1. (1 mark) (b) (c) (i) Sketch the graph with equation y ¼ x 4x 1, stating the values where the curve crosses the coordinate axes. (4 marks) Express x 4x 1 in the form ðx pþ q, where p and q are positive integers. ( marks) (d) (ii) Hence find the minimum value of x 4x 1. (1 mark) The curve with equation y ¼ x 3 4x 1 is translated by the vector. Find an equation of the new curve. You need not simplify your answer. ( marks) p 3 (a) (i) Simplify 3 ffiffiffi. (1 mark) p (ii) Show that 3 ffiffi pffiffi 1 þ 3 þ is an integer and find its value. (4 marks) (b) Express 4 p ffiffiffi pffiffiffi 5 7 ffiffiffi p p pffiffi in the form m ffiffiffi n, where m and n are integers. 5 þ (4 marks) (0) P45841/Jan1/MPC1

3 3 4 The curve with equation y ¼ x 5 3x 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 O x (a) Given that y ¼ x 5 3x þ x þ 5, find: (i) (ii) dy dx ; d y dx. (1 mark) (b) Find an equation of the tangent to the curve at the point Að1, 0Þ. ( marks) (c) Verify that the point B, where x ¼ 1, is a minimum point of the curve. (d) The curve with equation y ¼ x 5 3x 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 O x (i) Find ð 1 1 ðx 5 3x þ x þ 5Þ dx. (5 marks) (ii) Hence find the area of the shaded region bounded by the curve between A and B and the line segments AO and OB. ( marks) Turn over s (03) P45841/Jan1/MPC1

4 4 5 The polynomial pðxþ is given by pðxþ ¼x 3 þ cx þ dx 1, where c and d are constants. (a) When pðxþ is divided by x þ, the remainder is 150. Show that c d þ 65 ¼ 0. (b) Given that x 3 is a factor of pðxþ, find another equation involving c and d. ( marks) (c) By solving these two equations, find the value of c and the value of d. 6 A rectangular garden is to have width x metres and length ðx þ 4Þ metres. (a) The perimeter of the garden needs to be greater than 30 metres. Show that x > 11. (1 mark) (b) The area of the garden needs to be less than 96 square metres. Show that x þ 4x 96 < 0. (1 mark) (c) Solve the inequality x þ 4x 96 < 0. (4 marks) (d) Hence determine the possible values of the width of the garden. (1 mark) (04) P45841/Jan1/MPC1

5 5 7 A circle with centre C has equation x þ y þ 14x 10y þ 49 ¼ 0. (a) Express this equation in the form ðx aþ þðy bþ ¼ r (b) Write down: (i) the coordinates of C ; (ii) the radius of the circle. ( marks) (c) Sketch the circle. ( marks) (d) A line has equation y ¼ kx þ 6, where k is a constant. (i) (ii) Show that the x-coordinates of any points of intersection of the line and the circle satisfy the equation ðk þ 1Þx þ ðk þ 7Þx þ 5 ¼ 0. ( marks) The equation ðk þ 1Þx þ ðk þ 7Þx þ 5 ¼ 0 has equal roots. Show that 1k 7k 1 ¼ 0 (iii) Hence find the values of k for which the line is a tangent to the circle. ( marks) Copyright ª 01 AQA and its licensors. All rights reserved. (05) P45841/Jan1/MPC1

6 Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A,1 or 1 (or 0) accuracy marks x EE deduct x marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf significant figure(s) dp 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.

7 MPC1 Q Solution Marks Total Comments 1(a) OA ; OB OA 5 and OB 53 6 ( 4) ( ) 7 either correct PI by 5 or 53 seen or OA 5 and OB 53 A1 both correct values 5 or 5 and 53 or 53 seen OA 5 and OB 53 OA OB A1 3 or OA 5 and OB 53 correct working + concluding statement involving OA and/or OB (b)(i) grad AB A1 8 condone one sign error (ii) y 4 = their grad AB (x 6) or y 7 = their grad AB (x ) or y = their grad AB x + c and attempt to find c using x 6, y 4 or x, y y 4 x 6 OE A1 any correct form eg y x but must simplify to + 11x 8y 34 A1 3 condone 8y11x34or any multiple of these equations 8 (c) (grad AC =) B1 FT 1 / their grad AB equating gradients; LHS must be correct ' their ' OE k 6 11 and RHS is attempt at perp grad to AB k k A1cso 3 k = 11.5 OE Total 11 y 4 ' their 8 ' x 6 B1 (11y 8x 9 ) AND must sub y 0 11 y 0 ' their 8 ' x k B1 AND must sub x = 6, y 4 for 11 (c) Alternative: Eqn AC : or for

8 MPC1 (cont) Q Solution Marks Total Comments x x B1 1 ISW for x6, xetc (a) 6 (b) y 6 x 1 x x 6 B1 correct x values or FT their factors (x-intercepts stated or marked on sketch) may be seen in (a) y 1 B1 (stated or 1 marked on sketch) shaped curve approximately correct shape in all 4 quadrants with minimum to right of y-axis A1 4 (c)(i) x p x 16 A1 p and q 16 (ii) (Minimum value is ) 16 B1 1 FT their q (d) Replacing each x by x + 3 OR adding to their quadratic y x3 4 x3 1 or yx 1 14 or y x x 13 or y ( x3)( x 5) in original equation or their completed square or factorised form or replacing y by y A1 OE any correct equation in x and y unsimplified Total 10

9 MPC1 (cont) Q Solution Marks Total Comments 3(a)(i) 3 18 B1 1 (ii) 3 1 ' their 18' FT their A B Sum 30 A1cso 4 (b) Numerator m1 correct unsimplified Denominator = 5 18 B1 must be seen as denominator Answer = 3 10 A1cso 4 Total 9

10 MPC1 (cont) Q Solution Marks Total Comments 4(a)(i) dy one term correct 4 5 x 6x 1 A1 another term correct dx A1 3 all correct (no + c etc) (ii) d y dx x B1 1 FT their d y dx (b) (c) dy 4 dy x 1 5( 1) 6( 1) 1 ( 1) must sub x = 1 into 'their' dx d x y 1 x 1 A1cso any correct form with ( x 1) simplified condone y1 xc, c 1 dy x sub x 1 into their d y dx dx dy 0 stationary point dx A1cso shown 0 plus correct statement when x 1, d y d y 14 or 0 6 dx dx (B is a ) minimum (point) E1 3 (B is a) minimum (point) must have correct d y dx d y for E1 dx (d)(i) 6 3 x 3x x 5x 6 3 A1 A1 one term correct another term correct all correct (may have + c) m1 their F(1) F( 1) with powers of 1 and 1 evaluated correctly = 8 A1cso 5 (ii) their answer to part (i) Area 6 A1cso Total 16

11 MPC1 (cont) Q Solution Marks Total Comments 3 5(a) p( ) ( ) ( ) c ( ) d 1 p( ) attempted or long division by x+ as far as remainder their 8 4 c d m1 putting expression for remainder = 150 c d 65 0 A1cso 3 AG terms all on one side in any order (check that there are no errors in working) 3 (b) p(3) 3 3 c 3 d 1 p(3) attempted or long division by x 3 as far as remainder 9c3d 15 0 A1 any correct equation with terms collected eg 3c d 5 (c) c d 650 5c 70 3cd 50 Elimination of c or d c 14, d 37 OE A1 A1 3 Total 8 6(a) Sides are x and x 4 x x x4 x4 30 or x x 8 30 or x 4 30 or 4x x value of c or d correct unsimplified both c and d correct unsimplified must see this line OE x 11 B1 1 AG (be convinced) condone 11 < x (b) x x4 96 must see this line OE x 4 x 96 0 B1 1 AG (c) x 1 x 8 Critical values 8, 1 A1 y or 1 8 x correct factors or correct quadratic equation formula sketch or sign diagram 1 x 8 A1cso 4 accept x 8 AND x 1 but not x 8 OR x 1 nor x 8, x 1 (d) 1 5 x 8 B1 1 Total 7

12 MPC1 (cont) Q Solution Marks Total Comments 7(a) x 7 y 5 one term correct ; condone x 7 y 5 A1 both terms correct with squares and plus sign between terms = 5 A1cao 3 condone 5 for 5 ( x 7) (b)(i) C 7, 5 B1 correct or FT their circle equation (ii) r = 5 B1 correct or FT their r > 0 condone 5 etc but not 5 (c) must draw axes freehand circle with C correct or FT their C for quadrant of centre 7 A1 circle touching x-axis at 7 with 7 marked (need not show 5 on y-axis ) but circle must not touch y-axis (d)(i) x ( kx6) 14x10( kx6) 49 0 clear attempt to sub y kx6 into original or their circle equation x kx1kx36 14x 10 kx and attempt to multiply out (1 k ) x kx14x5 0 ( k 1) x ( k 7) x5 0 A1cso AG condone x k x k (1 ) (7 )... etc (ii) Equal roots b 4ac 0 B1 allow statement alone if discriminant in terms of k attempted k 7 4 5k 1 k k k discriminant (condone one slip) k 14k4 0 1 k 7 k 1 0 A1 3 (iii) 4k 33 k 4 AG all working correct but = 0 must appear before last line correct factors or correct use of 3 4 k, k OE 4 3 A1 are values of k for which line is a tangent Total 14 TOTAL 75 formula as far as k 4

13 Scaled mark unit grade boundaries - January 01 exams A-level Max. Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E LAW0 LAW UNIT LAW03 LAW UNIT MD01 MATHEMATICS UNIT MD MFP1 MATHEMATICS UNIT MFP MA MATHEMATICS UNIT MA 100 no candidates were entered for this unit MB MATHEMATICS UNIT MB MPC1 MATHEMATICS UNIT MPC MS1A MATHEMATICS UNIT MS1A MS/SS1A/W MATHEMATICS UNIT S1A - WRITTEN MS/SS1A/C MATHEMATICS UNIT S1A - COURSEWORK MS1B MATHEMATICS UNIT MS1B MD0 MATHEMATICS UNIT MD MFP MATHEMATICS UNIT MFP MMB MATHEMATICS UNIT MMB MPC MATHEMATICS UNIT MPC MSB MATHEMATICS UNIT MSB MFP3 MATHEMATICS UNIT MFP MPC3 MATHEMATICS UNIT MPC MFP4 MATHEMATICS UNIT MFP MPC4 MATHEMATICS UNIT MPC MEST1 MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT MEST3 MEDIA STUDIES UNIT MEST4 MEDIA STUDIES UNIT PHIL1 PHILOSOPHY UNIT

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