Condensed. Mathematics. General Certificate of Education Advanced Subsidiary Examination January Unit Pure Core 2.

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1 General Certificate of Education Advanced Subsidiary Examination January 0 Mathematics MPC Unit Pure Core Monday January am to 0.0 am For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed 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), (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. 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. P567/Jan/MPC 6/6/ MPC

2 The diagram shows a sector OAB of a circle with centre O and radius r cm. r cm A O.5 r cm B The angle AOB is.5 radians. The perimeter of the sector is 9 cm. (a) Show that r ¼. ( marks) Calculate the area of the sector OAB. ( marks) (a) Use the trapezium rule with five ordinates (four strips) to find an approximate value for ð 5 x þ (i) (ii) giving your answer to three significant figures. ( marks) ð Find x þ 6x, giving the coefficient of each term in its simplest form. ( marks) Hence find the value of ð x þ 6x. ( marks) (0) P567/Jan/MPC

3 The diagram shows a triangle ABC. C 5cm 6cm A B The lengths of AC and BC are 5 cm and 6 cm respectively. The area of triangle ABC is.5 cm, and angle ACB is obtuse. (a) Find the size of angle ACB, giving your answer to the nearest 0.. ( marks) Find the length of AB, giving your answer to two significant figures. ( marks) Given that log a N log a x ¼ express x in terms of a and N, giving your answer in a form not involving logarithms. ( marks) 5 The point Pð, 8Þ lies on a curve, and the point M is the only stationary point of the curve. The curve has equation y ¼ 6 þ x 8 x. (a) Find dy. ( marks) Show that the normal to the curve at the point Pð, 8Þ has equation x þ y ¼. ( marks) (c) (i) Show that the stationary point M lies on the x-axis. ( marks) (ii) Hence write down the equation of the tangent to the curve at M. ( mark) (d) The tangent to the curve at M and the normal to the curve at P intersect at the point T. Find the coordinates of T. ( marks) Turn over s (0) P567/Jan/MPC

4 6 (a) A geometric series begins 0 þ 9 þ 05:8 þ :::: (i) Show that the common ratio of the series is 0.7. ( mark) (ii) Find the sum to infinity of the series. ( marks) (iii) Write the nth term of the series in the form p q n, where p and q are constants. ( marks) The first term of an arithmetic series is 0 and the common difference of the series is 8. The nth term of the series is u n. (i) Write down an expression for u n. ( mark) (ii) Given that u k ¼ 0, find the value of X k n¼ u n. ( marks) 7 (a) Describe a geometrical transformation that maps the graph of y ¼ x onto the graph of y ¼ x. ( marks) Sketch the curve with equation y ¼ x, indicating the value of the intercept on the y-axis. ( marks) (c) The curve with equation y ¼ x intersects the curve y ¼ x at the point P. Use logarithms to find the x-coordinate of P, giving your answer to three significant figures. (5 marks) 8 (a) Expand þ x. ( mark) 8 in ascending powers of The first four terms of the binomial expansion of þ x x are þ ax þ bx þ cx. Find the values of the constants a, b and c. ( marks) þ x 8. ( marks) (c) Hence find the coefficient of x in the expansion of þ x (0) P567/Jan/MPC

5 5 9 (a) Write down the two solutions of the equation tanðx þ 0 Þ ¼tan 79 in the interval 0 x 60. ( marks) Describe a single geometrical transformation that maps the graph of y ¼ tan x onto the graph of y ¼ tanðx þ 0 Þ. ( marks) (c) (i) Given that 5 þ sin y ¼ð5þcos yþ cos y, show that cos y ¼. (5 marks) (ii) Hence solve the equation 5 þ sin x ¼ð5 þ cos xþ cos x in the interval 0 < x < p, giving your values of x in radians to three significant figures. ( marks) Copyright ª 0 AQA and its licensors. All rights reserved. (05) P567/Jan/MPC

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, or (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 MPC - AQA GCE Mark Scheme 0 Jan series Q Solution Marks Total Comments (a) Arc = r (=.5r) Within (a), r or 5 used for the arc length PI P = r + r + r =9 m Use of r + r + r for the perimeter. m0 if no indication that 5 comes from r.5r = 9 9 r =.5 A CSO AG {Area of sector =} r Within, r stated or used for the sector area. =. 5 = 90 (cm ) A NMS: 90 scores marks Total 5 (a) h = B PI f(x) = x h I {f()+f(5)+[f()+f()+f()]} h {f()+f(5)+[f()+f()+f()]} OE summing of areas of the four trapezia h with { } = = (8...)+[ (88...)] = 0.58(6 )+[0.58(8...)]=.56(08 ) A OE Accept dp (rounded or truncated) for non-terminating decs. equiv. (I ) = = 0.68 (to sf) A CAO Must be 0.68 SC for those who use 5 strips, max possible is B0AA0 (i) x 6x x 6x = / / 0.5 = x x.5 (+c) A One term correct (even unsimplified) Both terms correct (even unsimplified) (+c) A Must be simplified. (ii) x 6x = [ ( 0.5 )+(.5 )] [ ( 0.5 )+(.5 )] Attempt to calculate F() F() where F(x) follows integration and is not just the integrand = ( +) ( +) = 9 A Since Hence NMS scores 0/ Total 9

8 MPC - AQA GCE Mark Scheme 0 Jan series Q Solution Marks Total Comments (a) 5 6sin C.5 (Area=) 5 6sin C sin C =0.8(..) A AWRT 0.8 or 5/6 OE PI by e.g. seeing 56 or better (C is obtuse) C =.6º A AWRT.6 { AB } cosc RHS of cosine rule used = 6 60 ( 0.55 ) = 9.(66 ) m Correct ft evaluation, to at least sf, of AB or AB using c s value of C. (AB =) 9.7 (cm to sf) A If not 9.7 accept AWRT 9.70 or AWRT 9.7 log log a a N log N x a x Total 6 A log law used correctly. PI by next line. N a x m Logarithm(s) eliminated correctly x a N A ACF of RHS Total

9 MPC - AQA GCE Mark Scheme 0 Jan series Q Solution Marks Total Comments 5(a) 8 B 8x PI by its derivative as x dy 6x 6x or x 6 Differentiating either 6+x correctly or differentiating 8/x correctly. A 6x OE At P(, 8), dy 6 (= ) Attempt to find d y when x Gradient of normal at P = m m m = used Eqn. of normal at P: y 8 x x + y = A CSO AG (c)(i) dy At St. Pt 0, 6x 0 dy Equating c s to 0 d y Accept =0 so x= stated with no errors seen (6x = ) x = A x = When x =, y = 6 =0; M(,0) lies on x-axis A Need statement and correct coords. (c)(ii) Tangent at M has equation y = 0 B y = 0 OE (d) Intersects normal at P when x + 0 = PI Solving c s eqn. of tangent with ans as far as correctly eliminating one variable. T (, 0) A Accept x =, y = 0 Total

10 MPC - AQA GCE Mark Scheme 0 Jan series 6(a)(i) 9 r = Solution Marks Total Comments AG. Accept any valid justification to B the given answer (ii) { S =} a 0 r 0.7 a r used { S =} 00 A 00 NMS mark as / or 0/ (iii) nth term = 600 (0.7) n B If not B award B for 0 (0.7) n OE (i) { u n }8 8n B Accept ACF (ii) u k 0 8k 8 8 8k=0 OE e.g. 0+(k )( 8)=0 ft if no recovery, on c s (i) answer k n u n k = k = = 0 0 A k k For 0 0 or for c's u 0 k OE e.g. c's u k 8 k n u n (= 5.5 0) = 70 A 70 Total 0

11 MPC - AQA GCE Mark Scheme 0 Jan series Q Solution Marks Total Comments 7(a) Stretch(I) in y-direction(ii) scale factor (III) OE Need (I) and either (II) or (III) A All correct. Need (I) and (II) and (III) [> transformation scores 0/] B Shape with indication of correct asymptotic behaviour in nd quadrant below pt of intersection with y-axis B Only intersection is with y-axis, and only intercept is stated/indicated (c) x x OE eqn. in x x x log log log m Log Law (or Law applied x to or OE) used correctly or x correct rearrangement to x =/ OE simplified e.g.6 x = or x =(/ ) log x log x log m Log Law applied correctly twice (dependent on both & m) or a correct method using logs to solve an eqn. of form a kx =b, b>0 (including case k=) (dependent on and valid method to a kx ) x log log log log6 A Correct expression for x or for x e.g. x = log PI by correct sf value or better x = 0.96(06 ) = 0.96 (to sf) A 5 If logs not used explicitly then max of mm0. Total 9

12 MPC - AQA GCE Mark Scheme 0 Jan series Q Solution Marks Total Comments 8(a) Unsimplified equivalent answers, 8 6 = x x x (or 8x 6x ) B e.g. etc. must be x x x correctly simplified in part (c) to one of the two forms in solution to retrospectively score the B here 8 x 8 x 8 x 8 x Any valid method. PI by a correct { } value for either a or b or c 7 7 = { } x x x... 8 {a =, b =.75 OE, c = OE} AAA A for each of a, b, c SC a = 8, b = 8, c = 56 or a =, b = 8, c = 58 either explicitly or within expn (A0) (c) 8 6 ( x x ) 7 7 x x x 8 Product of c s two expansions either stated explicitly or used x x terms from expansion of x are ax and 8 bx and 6 cx m 8 Any two of the three, ft from products of non-zero terms using c s two expansions. May just use the coefficients. ax + 8 bx + 6 cx AF Ft on c s non-zero values for a, b and c and also ft on c s non-zero coeffs. of /x and /x in part (a). Accept x s missing i.e. sum of coeffs. PI by the correct final answer. Coefficient of x is ++ = 0 A OE Condone answer left as 0x. Ignore terms in other powers of x in the expansion. Total 9

13 MPC - AQA GCE Mark Scheme 0 Jan series Q Solution Marks Total Comments 9(a) x + 0º = 79º, x + 0º = 80º +79º x = 9º B 9 as the only solution in the interval 0 x 90 x = 9º B AWRT 9. Not given if any other soln. in the interval 90 x 60. Ignore anything outside 0º x 60º Translation; B Accept translat as equivalent. [T or Tr is NOT sufficient] (c)(i) 0 B 0 5 sin (5 cos ) cos 5 sin 5cos cos B OE Accept full equivalent to vector in words provided linked to translation/ move/shift and correct direction. (0/ if > transformation). Correct RHS. 5 cos 5cos cos sin = cos used to get a quadratic in cos. 6 5cos cos or cos + 5cos 6 (= 0) A ACF with like terms collected. ( cos )(cos + ) (= 0) m Correct quadratic and (c±)(c±) or by formula OE PI by correct values for cos. Since cos, cos A 5 CSO AG. Must show that the soln cos = has been considered and rejected (ii) 5 sin x (5 cos x) cos x x Using (c)(i) to reach cosx = ¾ or finding at least solutions of cos and dividing them by. cos x = 0.7(7 ), 0.7(7 ), +0.7(7 ), 0.7(7...), m Valid method to find all four positions of solutions. x = 0.6,.78,.50, 5.9 A CAO Must be these four sf values but ignore any values outside the interval 0 x. Total TOTAL 75

14 Scaled mark unit grade boundaries - January 0 exams A-level Maximum Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E LAW0 LAW UNIT MD0 MATHEMATICS UNIT MD MD0 MATHEMATICS UNIT MD MFP MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MB MATHEMATICS UNIT MB MMB MATHEMATICS UNIT MMB MPC MATHEMATICS UNIT MPC MPC MATHEMATICS UNIT MPC MPC MATHEMATICS UNIT MPC MPC MATHEMATICS UNIT MPC MSA MATHEMATICS UNIT MSA MS/SSA/W MATHEMATICS UNIT SA - WRITTEN MS/SSA/C MATHEMATICS UNIT SA - COURSEWORK MSB MATHEMATICS UNIT MSB MSB MATHEMATICS UNIT MSB MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT