Condensed. Mathematics. General Certificate of Education Advanced Subsidiary Examination June Unit Further Pure 1.

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General Certificate of Education Advanced Subsidiary Examination June 011 Mathematics MFP1 Unit Further Pure 1 Friday 0 May 011 1.0 pm to.00 pm For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics 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 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. 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. P9/Jun11/MFP1 6/6/6/ MFP1

1 A curve passes through the point ð, Þ and satisfies the differential equation dy dx ¼ pffiffiffiffiffiffiffiffiffiffi 1 þ x Starting at the point ð, Þ, use a step-by-step method with a step length of 0.5 to estimate the value of y at x ¼. Give your answer to four decimal places. (5 marks) The equation x þ 6x þ ¼ 0 has roots a and b. (a) Write down the values of a þ b and ab. ( marks) (b) Show that a þ b ¼. ( marks) (c) Find an equation, with integer coefficients, which has roots a b and b a (5 marks) It is given that z ¼ x þ iy, where x and y are real. (a) Find, in terms of x and y, the real and imaginary parts of ðz iþðz iþ ( marks) (b) Given that ðz iþðz iþ ¼ 8i find the two possible values of z. ( marks) (0) P9/Jun11/MFP1

The variables x and Y, where Y ¼ log 10 y, are related by the equation Y ¼ mx þ c where m and c are constants. (a) Given that y ¼ ab x, express a in terms of c, and b in terms of m. ( marks) (b) It is given that y ¼ 1 when x ¼ 1 and that y ¼ 7 when x ¼ 5. On the diagram below, draw a linear graph relating x and Y. ( marks) (c) Use your graph to estimate, to two significant figures: (i) the value of y when x ¼ ; ( marks) (ii) the value of a. ( marks) Y ~ 1 0 0 1 5 ~ x 5 (a) Find the general solution of the equation cos x p pffiffi ¼ 6 giving your answer in terms of p. (5 marks) (b) Use your general solution to find the smallest solution of this equation which is greater than 5p. ( marks) Turn over s (0) P9/Jun11/MFP1

6 (a) Expand ð5 þ hþ. (1 mark) (b) A curve has equation y ¼ x x. (i) Find the gradient of the line passing through the point ð5, 100Þ and the point on the curve for which x ¼ 5 þ h. Give your answer in the form p þ qh þ rh where p, q and r are integers. ( marks) (ii) Show how the answer to part (b)(i) can be used to find the gradient of the curve at the point ð5, 100Þ. State the value of this gradient. ( marks) 7 The matrix A is defined by " pffiffiffi # 1 A ¼ pffiffi 1 (a) (i) Calculate the matrix A. ( marks) (ii) Show that A ¼ ki, where k is an integer and I is the identity matrix. ( marks) (b) Describe the single geometrical transformation, or combination of two geometrical transformations, corresponding to each of the matrices: (i) A ; (ii) A. ( marks) ( marks) 8 A curve has equation y ¼ 1 x. (a) (i) Write down the equations of the three asymptotes of the curve. ( marks) (ii) Sketch the curve, showing the coordinates of any points of intersection with the coordinate axes. ( marks) (b) Hence, or otherwise, solve the inequality 1 x < ( marks) (0) P9/Jun11/MFP1

5 9 The diagram shows a parabola P which has equation y ¼ 1 8 x, and another parabola Q which is the image of P under a reflection in the line y ¼ x. The parabolas P and Q intersect at the origin and again at a point A. The line L is a tangent to both P and Q. y Q A P L O x (a) (i) Find the coordinates of the point A. ( marks) (ii) Write down an equation for Q. (1 mark) (iii) Give a reason why the gradient of L must be 1. (1 mark) (b) (i) Given that the line y ¼ x þ c intersects the parabola P at two distinct points, show that c > ( marks) (ii) Find the coordinates of the points at which the line L touches the parabolas P and Q. (No credit will be given for solutions based on differentiation.) ( marks) END OF QUESTIONS Copyright Ó 011 AQA and its licensors. All rights reserved. (05) P9/Jun11/MFP1

Q Solution Marks Total Comments 1 Attempt at 0.5 y () (= 0.5) M1 Other variations are allowed y(.5).5 y().5 + 0.5 y (.5) m1.5 + 0.57(0) A1F PI; OE; ft c s value for y(.5).857 A1 5 dp needed Total 5 (a) α + β =, αβ = B1B1 (b) + β = ( ) () = A1 α M1A1 AG; A0 if α + β has wrong sign (c) Sum = (α + β) = B1F ft wrong value for α + β Product = 10 ( ) = x Sx + P (= 0) M1 Signs must be correct for the M1 1 αβ α + β M1A1F ft wrong values Eqn is x + 1x + 1 = 0 A1 5 Integer coeffs and = 0 needed Total 9 (a) Use of z = x iy M1 (z i)(z i) = (x + y 1) ix m1a1 A1 may be earned in (b) (b) Equating R and I parts M1 x = 8 so x = A1 16 + y 1 = so y = ± (z = ± i) m1a1 A0 if x = used Total 7 (a) Use of one law of logs or exponentials M1 lg a = c and lg b = m A1 OE; both needed So a = 10 c and b = 10 m A1 (b) Points (1, 1.08), (5, 1.) plotted M1A1 M1 A0 if one point correct Straight line drawn through points A1F ft small inaccuracy (c)(i) Attempt at antilog of Y() M1 OE When x =, Y 1.5 so y 18 A1 Allow AWRT 18 (ii) Attempt at a as antilog of Y-intercept M1 OE a 9. to 10 A1 AWRT Total 10 5(a) π cos 6 = B1 OE stated or used; deg/dec penalised at 5th mark cos( ) 6 B1F OE; ft wrong first value Introduction of nπ M1 (or nπ) at any stage π Going from x to x 6 m1 incl division of all terms by GS: = π x 18 ± 18 + nπ A1F 5 ft wrong first value (b) n = 8 will give the required solution M1 GS must include nπ for this 16... which is π ( 16.755) A1 from correct GS; 8 allow π or dec approx Total 7 9

Q Solution Marks Total Comments 6(a) (5 + h) = 15 + 75h + 15h + h B1 1 Accept unsimplified coefficients (b)(i) y(5 + h) = 100 + 65h + 1h + h B1F PI; ft numerical error in (a) Use of correct formula for gradient M1 Gradient is 65 + 1h + h A,1F A1 if one numerical error made; ft numerical error already penalised (ii) As h 0 this 65 E,1F E1 for h = 0 ; ft wrong values for p, q, r Total 7 7(a)(i) A = M1A1 M1 if at least two entries correct (ii) A 8 0 M1 if at least two entries correct = 0 8..= 8I A1 (b)(i) A gives enlargement with SF 8 (centre the origin) M1A1F M1 for enlargement (only); ft wrong value for k (ii) Enlargement and rotation M1 Some detail needed Enlargement scale factor A1 Rotation through 10 (antic wise) A1 Total 9 8(a)(i) Asymptotes x =, x =, y = 0 B1 (ii) Middle branch generally correct B1 Allow if max pt not in right place Other branches generally correct B1 All branches approaching asymps B1 Asymps must be shown correctly 1 Intersection at ( 0, ) indicated B1 on diagram or elsewhere; B0 if any other intersections are shown (b) y = when x = ±. 5 B1 Allow NMS Sol n < x <.5,.5 < x < B,1 Condone dec approx n for. 5 ; B1 if used instead of < Total 10 9(a)(i) Elimination to give x = 1 8 x M1 OE A is (8, 8) A1 NMS / (ii) Equation of Q is x = B1 1 OE; condone y = 8x 1 8 y (iii) Points of contact are images in y = x E1 1 (b)(i) Eliminating y to give x + c = 1 8 x M1 (ie x + 8x 8c = 0) Distinct roots if Δ > 0 E1 stated or implied Δ = 6 + c, so c > A1 convincingly shown (AG) (ii) For tangent c =, so x + 8x + 16 = 0 M1 OE... and x =, y = A1 Reflection in y = x M1 or other complete method x =, y = A1F ft wrong answer for first point; allow NMS / Total 11 TOTAL 75

Scaled mark unit grade boundaries - June 011 exams A-level Max. Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E HBIO GCE HUMAN BIOLOGY UNIT 80-56 51 6 1 7 HBIT GCE HUMAN BIOLOGY UNIT T 50-1 8 5 1 HBIX GCE HUMAN BIOLOGY UNIT X 50-0 6 19 HBIO GCE HUMAN BIOLOGY UNIT 90 61 56 51 6 1 6 HBIO5 GCE HUMAN BIOLOGY UNIT 5 90 65 60 55 50 5 1 HBI6T GCE HUMAN BIOLOGY UNIT 6T 50 0 7 1 8 HBI6X GCE HUMAN BIOLOGY UNIT 6X 50 9 5 1 7 INFO1 GCE INFO AND COMM TECH UNIT 1 80-5 8 8 INFO GCE INFO AND COMM TECH UNIT 80-5 7 1 5 0 INFO GCE INFO AND COMM TECH UNIT 100 7 66 60 5 8 INFO GCE INFO AND COMM TECH UNIT 70 6 57 50 6 0 LAW01 GCE LAW UNIT 1 96-7 66 59 5 6 LAW0 GCE LAW UNIT 9-70 61 5 5 LAW0 GCE LAW UNIT 80 69 6 55 8 6 LAW0 GCE LAW UNIT 85 7 66 59 5 7 1 XMCAS GCE MATHEMATICS UNIT XMCAS 15-99 89 79 70 61 MD01 GCE MATHEMATICS UNIT D01 75-59 5 5 9 MFP1 GCE MATHEMATICS UNIT FP1 75-61 5 8 6 MM1A GCE MATHEMATICS UNIT M1A 100 No candidates were entered for this unit MM1B GCE MATHEMATICS UNIT M1B 75-6 55 8 1 MPC1 GCE MATHEMATICS UNIT PC1 75-6 55 8 1 MS1A GCE MATHEMATICS UNIT S1A 100-7 6 5 5 6 MS/SS1A/W GCE MATHEMATICS UNIT S1A - WRITTEN 75 5 6 MS/SS1A/C GCE MATHEMATICS UNIT S1A - COURSEWORK 5 0 10