Cambridge A Level Mathematics 9709 — 2001 Oct/Nov Paper 2 · Variant 1

CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level MATHEMATICS 8709/2 PAPER 2 Pure Mathematics 2 (P2) OCTOBER/NOVEMBER SESSION 2001 1 hour 15 minutes Additional materials: Answer paper Graph paper List of Formulae (MF9) TIME = 1: hour 15 minutes INSTRUCTIONS TO CANDIDATES Write your name, Centre number and candidate number in the spaces provided on the answer paper/answer booklet. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 50. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. The use of an electronic calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. This question paper consists of 3 printed pages and 1 blank page. UNIversITY of CAMBRIDGE Local Examinations Syndicate [Turn over © CIE 2001

nN 2 Solve the equation 2sec*x — tanx = 5, for 0° < x < 360°. [5] @ By using the substitution u = 2*, show that the equation 4* = 2*+! + 12 can be expressed as uw? —2u—12=0. W (ii) Hence find x, correct to 2 decimal places. [4] (@@) Sketch the graphs of 2y = x + 1 and 2y = |x — 4| on the same diagram. 3] (ii) Solve the simultaneous equations 2y=x4+1, 3 by = x4, 13] Iny (1, 2.4) (4, 0.6) oO Inx Variables x and y are related by the equation y=Ax’, where A and n are constants. When a graph of Iny against Inx is drawn, the resulting line passes through the points (1, 2.4) and (4, 0.6), as shown in the diagram. Find the values of n and A. [6] 2x (a) Variables x and y are related by the equation y = =F Find the rate of change of y with respect to x when x = 0. 3] (b) The equation of a curve is x* + y* = xy +7. Show that the equation of the tangent to the curve at the point (3, 2) is y + 4x = 14. {5}

The diagram shows the curve y = x cosx and a maximum point M. @) Show that the x-coordinate of M satisfies the equation x tanx = 2. [4] =tan~! { 2 4,41 = tan a)’ In with u, = 1, to find the x-coordinate of M correct to 2 decimal places, showing the values of u,, U3, ... aS appropriate. 13] (ii) Use the iteration formula (iii) Explain why the iteration formula, with the given value of u,, gives the required value for the x-coordinate of M. [2] at rd x @ Show that [' sin 2xdx = } and tha { cos*xdx = 4(@ +2). {61 0 0 in (ii) Use the results in part (i) to evaluate if Qsinx+3 cos .x)* dx. {5] 0 8709/2/0/N/01

CAMBRIDGE INTERNATIONAL EXAMINATIONS NOVEMBER 2001 ADVANCED SUBSIDIARY LEVEL MARK SCHEME SYLLABUS/COMPONENT : 8709/2 MATHEMATICS f§) University of CAMBRIDGE 8&9 Local Examinations Syndicate

Page 1 of 3 Mark Scheme Syllabus AS Level Examinations —- November 2001 Paper 2 8709 1 2sec*x —tanx =5 Use of sec?x = 1 + tan*x M41 Use of tan - sec link — 2tan’x — tanx- 3 =0 Al Correct only Solution of this DM1 Correct attempt to solve tanx =-1 or 1.5 x = 135° or 315° or 56.3° or 236.3° A1A1V | A1 for one pair correct. A1sq for other pair. 5 2 (i) | & =u and2***=2u B1 For both values u?=2u +12 (ii) | Leads to u = 4.6055 (or 1 + 13 ) B1 For correct value of u — even if other given Solution of 2” = “his value” by logs M1 Realises need to use logs (or TI if accurate) x = log 4.6055 + log2 M1 log + log x= 2.20 At Co to 3 sig figs (but allow 2.2) (Loses this A mark if 2 answers given) 5 3 (i) Graph of 2y =x +1 B1 Approx correct — no values needed Graph of 2y = |x-4| M1 Must be V-shape — no negatives — to x-axis At (2,0) y All gradients approx OK | A1 Two approx parallel, other with negative m 2y=xt+1 2y = |x-4 1 > 0 x (ii) | Solution occurs when 2y =x +1 and M1 Recognition of where solution lies 2y=4-x M1 Must be using (4 — x) not (x — 4) x= 1.5, y=1.25 Al Both needed

Page 2 of 3 Mark Scheme | Syllabus | Paper AS Level Examinations - November 2001 L__ 8709 2 4 Attempt at Y =mxX +c M1 Attempt at any y = mx + c eqn Y=-0.6X+c Al mand c correct Puts Y = Iny and X = Inx M1 Putting Y = Iny and X = Inx Iny = -0.6Inx + 3 y=ex® M1 Correct elimination of logs n=-0.6 and A=e°=20.1 A1A1 (a) _ eX _ (2x +3)2e%* -e7%.2 M1 Correct u/v formula — or uv y= dy/dx = ith e(2x + 3) 2x+3 (2x +3)? with e™(2x + 3) Al Correct unsimplified If x = 0, dy/dx = 4/9 At Co (b) Implicit differentiation. M1 Some evidence of implicit needed 2x + 2ydy/dx = y +xdy/dx A1A1 A1LHS, A1 RHS At (3,2), dy/dx =-4 Eqn of tangent y-2=-4(x-3) M1 Must have used calculus, not for normal ory + 4x = 14 Al Any form ok. 6 (i) y = x’cosx dy/dx = 2xcosx — x’sinx M1 Correct uv formula Al Unsimplified ok =0 whenx=0O or 2cosx =xsinx M1 Putting his dy/dx = 0 — xtanx =2. Al Co (ii) | u,= 1.107 Us = 1.065 u, = 1.081 M1 Correct manipulation of Up+1 from u, us=1.075 us = 1.078 ur = 1,077 At First two correct — Limit of 1.08 At Correct limit (iii) | Since a limit is reached (=L) Une1 = Un = Lb L = tan” (2/L) M1 Putting Une = Up =L LtanL =2. Al Co

Page 3 of 3 Mark Scheme Syllabus Paper AS Level Examinations — November 2001 8709 2 (i) (ii) 7 4 Jsinaxax = [=] =0-L%)= % ia t cos? cos2x , 1 J cos xdx = + zx = (om | = +X 42 4 =—(2+T7) a! ) J(2s + 3c) dx = (4s + 9c? +12sc)dx 12sc = 6sin2x Integral = 6 x ; =3 9c? Integral = 9 x ; X (1 + 2) 4s? = 4-4? Integral = 4x between 0 and in 4 x integral of c? from 0 to in = 9.36 or 1371/8 + 17/4 M1 Al M1 Al DM1 Al B1 B1 B1 M1 Al 11 Needs “—“ and cos 2x. Co Using double angles + attempt at integration Co Use of limits 0 to 11/4 Co beware of fortuitous answers. Correct squaring — needs all terms There could be alternatives to these marks. They could also be implied. Dealing correctly with Jas? Correct in either form.