# Concepts for Advanced Mathematics (C2) THURSDAY 15 MAY 2008

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2 Section A (36 marks) Express 7π 6 radians in degrees. [] The first term of a geometric series is.4 and the common ratio is 0.. (i) Find the fourth term of the series. [] (ii) Find the sum to infinity of the series. [] 3 State the transformation which maps the graph of y = x + onto the graph of y = 3x +. [] 4 Use calculus to find the set of values of x for which f(x) =x x 3 is an increasing function. [3] In Fig., A and B are the points on the curve y = x with x-coordinates 3 and 3. respectively. y Not to scale B A O 3 3. Fig. x (i) Find the gradient of the chord AB. Give your answer correct to decimal places. [] (ii) Stating the points you use, find the gradient of another chord which will give a closer approximation to the gradient of the tangent to y = x at A. [] 6 A curve has gradient given by dy dx = 6 x. Find the equation of the curve, given that it passes through the point (9, 0). [4] OCR /0 Jun08

3 3 7 Not to scale 6cm.6 Fig. 7 A sector of a circle of radius 6 cm has angle.6 radians, as shown in Fig. 7. Find the area of the sector. Hencefindtheareaoftheshadedsegment. [] 8 The th term of an arithmetic progression is. The sum of the first 0 terms is 0. Find the 4th term. [] 9 Use logarithms to solve the equation x = 3, giving your answer correct to decimal places. [3] 0 Showing your method, solve the equation sin θ = cos θ + forvaluesofθ between 0 and 360. [] OCR /0 Jun08 [Turn over

4 4 Section B (36 marks) y 0 x 0 Fig. Fig. shows a sketch of the cubic curve y = f(x). The values of x where it crosses the x-axis are, and, and it crosses the y-axis at (0, 0). (i) Express f(x) in factorised form. [] (ii) Show that the equation of the curve may be written as y = x 3 + x 4x 0. [] (iii) Use calculus to show that, correct to decimal place, the x-coordinate of the minimum point on the curve is 0.4. Find also the coordinates of the maximum point on the curve, giving your answers correct to decimal place. [6] (iv) State, correct to decimal place, the coordinates of the maximum point on the curve y = f(x). [] OCR /0 Jun08

5 y x Fig. A water trough is a prism. m long. Fig. shows the cross-section of the trough, with the depths in metres at 0. m intervals across the trough. The trough is full of water. (i) Use the trapezium rule with strips to calculate an estimate of the area of cross-section of the trough. Hence estimate the volume of water in the trough. [] (ii) A computer program models the curve of the base of the trough, with axes as shown and units in metres, using the equation y = 8x 3 3x 0.x 0., for 0 x Calculate (8x 3 3x 0.x 0.) dx and state what this represents. 0 Hence find the volume of water in the trough as given by this model. [7] [Question 3 is printed overleaf.] OCR /0 Jun08

6 6 3 The percentage of the adult population visiting the cinema in Great Britain has tended to increase since the 980s. The table shows the results of surveys in various years. Year 986/87 99/9 996/97 999/00 000/0 00/0 Percentage of the adult population visiting the cinema This growth may be modelled by an equation of the form Source: Department of National Statistics, P = at b, where P is the percentage of the adult population visiting the cinema, t is the number of years after the year 98/86 and a and b are constants to be determined. (i) Show that, according to this model, the graph of log 0 P against log 0 t should be a straight line of gradient b. State, in terms of a, the intercept on the vertical axis. [3] Answer part (ii) of this question on the insert provided. (ii) Complete the table of values on the insert, and plot log 0 P against log 0 t.drawbyeyealineof best fit for the data. [4] (iii) Use your graph to find the equation for P in terms of t. [4] (iv) Predict the percentage of the adult population visiting the cinema in the year 007/008 (i.e. when t = ), according to this model. [] OCR /0 Jun08

7 3 (ii) Year 986/87 99/9 996/97 999/00 000/0 00/0 t P log 0 t.04 log 0 P.73 log 0 P log 0 t Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. OCR is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. OCR /0 Ins Jun08

8 47 Mark Scheme June (C) Concepts for Advanced Mathematics Section A 0 c.a.o. for π rads = 80 soi 7 (i).4 0-3, or 000 for S =.4 / ( 0.) 3 (ii) 6 www 3 stretch, parallel to the y axis, sf 3 for stretch plus one other element correct 4 [f (x) =] 3x their f (x) > 0 or = 0 soi < x < B 3 (i) grad of chord = ( 3. 3 )/0. o.e. =.74 c.a.o. (ii) correct use of A and C where for C,.9 < x < 3. answer in range (.36,.74) 6 [y =] kx 3/ [+ c] k = 4 subst of (9, 0) in their eqn with c or c = 3 condone x or between - and or chord with ends x = 3 ± h, where 0 < h 0. s.c. for consistent use of reciprocal of gradient formula in parts (i) and (ii) may appear at any stage must have c; must have attempted integration sector area = 8.8 or [cm ] c.a.o. area of triangle = ½ 6 sin.6 o.e. their sector their triangle s.o.i. 0.8 to 0.8 [cm ] for ½ 6.6 must both be areas leading to a positive answer 8 a + 0d = or =.(a+0d) (a + 9d) = 0 o.e. a = s.o.i. www and d = s.o.i. www 4th term is 9 log 3 x log = log3 or x = log ( cos θ) = cos θ + cos θ = cos θ s.o.i. valid attempt at solving their quadratic in cos θ cos θ = - ½ www θ = 90, 70, 0, 40 A D or =.(a + ) gets M eg a + 9d = 4 or x = log 3 for 3.4 or versions of for - cos θ = sin θ substituted graphic calc method: allow M3 for intersection of y = sin θ and y = cos θ + and A for all four roots. All four answers correct but unsupported scores B. 0 and 40 only: B. 8 6

9 47 Mark Scheme June 008 Section B i (x + )(x )(x + ) for a (x + )(x )(x + ) ii [(x + )](x + 3x 0) x 3 + 3x 0x + x + 6x 0 o.e. for correct expansion of one pair of their brackets for clear expansion of correct factors accept given answer from (x + )(x 4) as first step iii iv y = 3x + 0x 4 their 3x + 0x 4 = 0 s.o.i. x = 0.36 from formula o.e. ( 3.7,.6) (-.8,.6) M B+ B+ if one error or for substitution of 0.4 if trying to obtain 0, and for correct demonstration of sign change accept (.9,.6) or f.t.( ½ their max x, their max y) 6 i Area = (-)0.36 seen [m ] www Volume = 0.34 [m 3 ] or ft from their area. 4 M3 for 0./ ( [ ]) M for one slip; for two slips must be positive ii x 4 x 3 0. x 0.x o.e. value at 0. [ value at 0] = 0.37 area of cross section (of trough) or area between curve and x-axis r.o.t. to 3 or more sf [m 3 ] m 3 seen in (i) or (ii) M E B U for terms correct dep on integral attempted must have neg sign 7 3 i log P = log a + b log t www comparison with y = mx + c intercept = log 0 a must be with correct equation condone omission of base 3 ii log t log P plots f.t. ruled line of best fit iii gradient rounding to 0. or 0.3 a = 0.49 s.o.i. P = 3t m allow the form P = 0 0.logt +.49 accept to or more dp for y step / x-step accept.47.0 for intercept accept answers that round to 30 3, their positive m 4 iv answer rounds in range 60 to

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