Condensed. Mathematics. General Certificate of Education Advanced Subsidiary Examination June Unit Pure Core 1. Time allowed * 1 hour 30 minutes

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1 General Certificate of Education Advanced Subsidiary Examination June 01 Mathematics Unit Pure Core 1 Wednesday 16 May am to 10.0 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 0 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 each question in the space provided for that question. If you require extra space, use an AQA supplementary answer book; do not use the space provided for a different question. 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. Condensed 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. P50000/Jun1/ 6/6/6/

2 1 Express 5 p ffiffiffi 6 ffiffiffi p p in the form m þ n ffiffi, where m and n are integers. þ ( marks) The line AB has equation x y ¼ 7. (a) (i) Find the gradient of AB. ( marks) Find an equation of the straight line that is parallel to AB and which passes through the point Cð, 5Þ, giving your answer in the form px þ qy ¼ r, where p, q and r are integers. ( marks) (b) (c) The line AB intersects the line with equation x y ¼ at the point D. Find the coordinates of D. ( marks) The point E with coordinates ðk, k Þ lies on the line AB. Find the value of the constant k. ( marks) The polynomial pðxþ is given by pðxþ ¼x þ x 5x 6 (a) (i) Use the Factor Theorem to show that x þ 1 is a factor of pðxþ. ( marks) Express pðxþ as the product of three linear factors. ( marks) (b) Verify that pð0þ > pð1þ. ( marks) (c) Sketch the curve with equation y ¼ x þ x 5x 6, indicating the values where the curve crosses the x-axis. ( marks) (0) P50000/Jun1/

3 The diagram shows a solid cuboid with sides of lengths x cm, x cm and y cm. y x x The total surface area of the cuboid is cm. (a) (i) Show that x þ xy ¼ 16. ( marks) Hence show that the volume, V cm, of the cuboid is given by V ¼ 1x 9x ( marks) (b) Find dv dx. ( marks) (c) (i) Verify that a stationary value of V occurs when x ¼. ( marks) Find d V dx value when x ¼. and hence determine whether V has a maximum value or a minimum ( marks) Turn over s (0) P50000/Jun1/

4 5 (a) (i) Express x x þ 5 in the form ðx pþ þ q. ( marks) Hence write down the equation of the line of symmetry of the curve with equation y ¼ x x þ 5. (1 mark) (b) The curve C with equation y ¼ x x þ 5 and the straight line y ¼ x þ 5 intersect at the point Að0, 5Þ and at the point B, as shown in the diagram below. y B A R O x (i) Find the coordinates of the point B. ( marks) Find ð ðx x þ 5Þ dx. ( marks) (iii) Find the area of the shaded region R bounded by the curve C and the line segment AB. ( marks) (0) P50000/Jun1/

5 5 6 The circle with centre Cð5, 8Þ touches the y-axis, as shown in the diagram. y C O x (a) Express the equation of the circle in the form ðx aþ þðy bþ ¼ k ( marks) (b) (i) Verify that the point Að, 1Þ lies on the circle. (1 mark) Find an equation of the tangent to the circle at the point A, giving your answer in the form sx þ ty þ u ¼ 0, where s, t and u are integers. (5 marks) (c) The points P and Q lie on the circle, and the mid-point of PQ is Mð7, 1Þ. p (i) Show that the length of CM is n ffiffi 5, where n is an integer. ( marks) Hence find the area of triangle PCQ. ( marks) 7 The gradient, dy, of a curve C at the point ðx, yþ is given by dx dy dx ¼ 0x 6x 16 (a) (i) Show that y is increasing when x 10x þ 8 < 0. ( marks) Solve the inequality x 10x þ 8 < 0. ( marks) (b) The curve C passes through the point Pð, Þ. (i) Verify that the tangent to the curve at P is parallel to the x-axis. ( marks) The point Qð, 1Þ also lies on the curve. The normal to the curve at Q and the tangent to the curve at P intersect at the point R. Find the coordinates of R. (7 marks) Copyright Ó 01 AQA and its licensors. All rights reserved. (05) P50000/Jun1/

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 - AQA GCE Mark Scheme 01 June series Q Solution Marks Total Comments m1 correct 8 7 (Numerator ) (Denominator = 1 9 ) B1 must be seen as denominator (a)(i) CSO; accept 16 9 Total 7 y x grad AB = y 'their grad' x c y xk y or x with correct points condone slip in rearranging if gradient is correct; condone 1. or better and attempt to use x, y 5 or y5 'theirgrad AB' x or x y = k and attempt to find k using x = and y = 5 y5 x 7 or y x x y = 7 correct equation in any form but must simplify to + integer coefficients in required form eg 8x + 6y = 5 (b) x y 7 and x y must use correct pair of equations and 8x9x1 1 etc attempt to eliminate x or y (generous) x 5, 5 (c) 7 y or D k k sub x k, y k into x y 7 k 8 6k 9 7 k Total 10 and attempt to multiply out with all k terms on one side (condone one slip)

8 - AQA GCE Mark Scheme 01 June series Q Solution Marks Total Comments p 1 ( 1) ( 1) 5( 1) 6 p 1 attempted not long division (a)(i) p x 1is a factor CSO; correctly shown = 0 plus statement long division as far as constant term Quad factor in this form: x bx c or comparing coefficients, or b 1 or c 6 by inspection x x x x x x 6 correct quadratic factor p 1 must see correct product (b) p0 6;p1 8 p0 p1 both p0 andp(1)attempted and at least one value correct AG both values correct plus correct statement involving p(0) and p(1) (c) y 1 x cubic with one max and one min with, 1, marked correct with minimum to right of y-axis AND going beyond and Total 10

9 - AQA GCE Mark Scheme 01 June series Q Solution Marks Total Comments (a)(i) x x + xyxyxy xy correct expression for surface area 6x 8xy (x xy xy) etc x xy 16 AG be convinced V x y OE correct volume in terms of x and y 16 x x 16 x or x x 9x 1 x OE CSO AG be convinced that all working is correct (b) dv 7 1 x dx one of these terms correct all correct with 9 evaluated (no + c etc) (c)(i) dv 7 dv x 1 attempt to sub x into ' their' dx dx dv dx 9 or or 1 0 etc 6 dv 0 stationary value CSO; shown = 0 plus statement dx d V 7x dv OE B1 FT for their a bx dx dx d V d V or sub of x into their when x, 0 maximum E1 dx dx maximum d V E0 if numerical error seen FT "minimum" if their 0 dx Total 10 5

10 - AQA GCE Mark Scheme 01 June series Q Solution Marks Total Comments 5(a)(i) x or p = 1.5 stated 11 x Mark their final line as their answer x x B1 1 correct or FT their x = p (b)(i) x x 5 x 5 x x x0 x y 9 eliminating x or y and collecting like terms (condone one slip) or y 5 y5 5 y y y x x 5x c one of these terms correct another term correct all correct (need not have + c) (iii) Area trapezium x 5 y Area of shaded region must have earned in part(b) their x F 0 correctly sub d F B condone 17. but not 16 etc or 10 6 etc 1 B B B1 FT their numerical values of xb, y Area = 1 1 (= 8) CSO; Total 1, accept 10.7 or better B 6

11 - AQA GCE Mark Scheme 01 June series Q Solution Marks Total Comments x 5 y 8 B1 6(a) (b)(i) = 5 B1 condone or 5 AC = 9+16 = 5 hence AC = 5 ; ( also radius = 5) A lies on circle B1 1 CSO radius = AC A lies on circle (must have concluding statement and circle equation correct if using equation) (must have concluding statement & RHS of circle equation correct or r =5 stated if Pythagoras is used) grad AC B1 Gradient of tangent is y 1 ' their tangent grad' x B1 FT their 1/ grad AC or y = their tangent grad x + c & attempt to find c using x =, y = 1 1 y 1 x or y x etc correct equation in any form x y 0 CSO; must have integer coefficients with 5 all terms on one side of equation accept 0 8 y 6 x 8 etc (c)(i) CM (7 5) (1 8) or CM CM 0 CM 5 0 PM PC CM 5 0 Pythagoras used correctly PM 5 Area PCQ CSO Total 1 d 5 5 eg 7

12 - AQA GCE Mark Scheme 01 June series Q Solution Marks Total Comments 7(a)(i) d Increasing y 0 dx either 0 x 6 x 16 0 correct interpretation of y increasing 6 x 0x16 0 must see at least one of these steps before or () (10 x x 8) 0 final answer for x 10x8 0 CSO AG no errors in working x x CVs are and correct factors or correct use of quadratic equation formula as far as 10 6 condone 8 6 and 1 6 line + + sketch or sign diagram x or x accept x AND x Mark their final line as their answer but not x OR x nor x, x here but not in final 8

13 - AQA GCE Mark Scheme 01 June series Q Solution Marks Total Comments 7(b)(i) x ; dy dy 0 16 sub x into and simplify terms dx d x dy 0 dx tangent at P is parallel to the x-axis must be all correct working plus statement x dy ; must attempt to sub x = into d dx d ( = ) 10 Gradient of normal = 1 10 Normal: y 1 'theirgrad' x 1 " their 10" m1 normal attempted with correct coordinates used and gradient obtained from their d y dx value 1 any correct form, eg 10 y x 1 but y1 x 10 must simplify to + (Equation of tangent at P is ) y B1 x 7 CSO; R, y x Total 15 TOTAL 75 9

14 Scaled mark unit grade boundaries - June 01 exams A-level Max. Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E MD01 MATHEMATICS UNIT MD MD0 MATHEMATICS UNIT MD MFP1 MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MA MATHEMATICS UNIT MA MA/W MATHEMATICS UNIT MA - WRITTEN MA/C MATHEMATICS UNIT MA - COURSEWORK MB MATHEMATICS UNIT MB MMB MATHEMATICS UNIT MMB MM0 MATHEMATICS UNIT MM MM0 MATHEMATICS UNIT MM MM05 MATHEMATICS UNIT MM MATHEMATICS UNIT MPC MATHEMATICS UNIT MPC MPC MATHEMATICS UNIT MPC MPC MATHEMATICS UNIT MPC MS1A MATHEMATICS UNIT MS1A

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