Mathematics (MEI) Advanced Subsidiary GCE Core 2 (4752) June 2010

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1 Link to past paper on OCR website: The above link takes you to OCR s website. From there you click QUALIFICATIONS, QUALIFICATIONS BY TYPE, AS/A LEVEL GCE, MATHEMATICS (MEI), VIEW ALL DOCUMENTS, PAST PAPERS JANUARY SERIES 2010, QUESTION PAPER UNIT 4752/01 CORE MATHEMATICS 2 These solutions are for your personal use only. DO NOT photocopy or pass on to third parties. If you are a school or an organisation and would like to purchase these solutions please contact Chatterton Tuition for further details. Section A Question 1 We have u n = 1 when n = 1 we can substitute u n and n into u n+1 to get the next term u 2 = u 3 = u 4 = = ½ ½ = ⅓ ⅓ = Page 1

2 Question 2 i) the means we want the sum from a value of r = 2 up to r = 5 that means = = 2 ii) if we set this out in vertical columns it is easier to see x x x x x 7 r r + 1 r x (r + 1) = r(r+1) f(r) = r(r + 1) we want this to go from r = 2 up to r = 6 we have r(r + 1) Page 2

3 Question 3 i)to differentiate we multiply by the power and then reduce the power by 1 3x 2 (2 x 6x) 15 = 3x 2-12x 15 ii) to find the stationary points we differentiate and then set the differential to 0 = 3x2-12x 15 = 0 we can divide both sides by 3 x 2-4x 5 = 0 factorise (find two numbers that multiply to give -5 and add to give -4) two numbers are -5 and +1 (x -5)(x + 1) = 0 So x = 5 or x = -1 question only asked for x coordinates so we are done Page 3

4 Question 4 y = f(2x) this is a stretch (as we are multiplying or dividing rather than adding or subtracting) The change is inside the brackets so affecting the x values When the x values are affected they are affected in the opposite way to what you think This is a stretch of scale factor ½ parallel to the x axis (all the x values are half what they were, and all the y values are unchanged) Page 4

5 y = 3f(x) this is a stretch (as we are multiplying or dividing rather than adding or subtracting) The change is outside the brackets so affecting the y values This is a stretch of scale factor 3 parallel to the y axis (all the y values are 3 times what they were, and all the x values are unchanged) Page 5

6 Question 5 first rewrite with the x to a negative power (1 6x -3 ) (1 6x-3 ) dx to integrate you first add a power to the x term and then divide by the new power as we have limits we don t need to worry about adding a constant c x put in the limits 5 2 = x (5 + ) (2 + ) = Page 6

7 Question 6 If we have the gradient of a curve then we can integrate to get the equation of the curve To integrate you first add a power to the x term and then divide by the new power y = (6x x 1/2 ) dx y = + + c = 2x c we know the curve passes through the point (4,10) so substitute these values for x and y (to get c) 10 = (2 x (4 3 )) + (8 x 4 ) + c simplifiying 10 = (2 x 64) + (8 x 8) + c 10 = c 10 = c subtract 192 from both sides c = -182 put this value for c back into the equation of the curve to get Y = 2x Page 7

8 Question 7 log a x 3 = 3log a x log a = log a = log ax so we have 3log a x + log ax = 3.5 log a x Page 8

9 Question 8 from the basic identity sin 2 + cos 2 = 1 we can get sin 2 = 1 - cos 2 now substitute this into our equation 4(1 - cos 2 )= 3 + cos 2 expand the brackets 4-4cos 2 = 3 + cos 2 add 4cos 2 to both sides of the equation 4 = 3 + 5cos 2 subtract 3 from both sides 1 = 5cos 2 divide both sides by 5 cos 2 = square root both sides cos = take the inverse cos (cos -1 ) of both sides = cos -1 ( ) = 63.4⁰ or 116.6⁰ Page 9

10 we need to see where each of these get repeated the cosine graph is symmetrical about the line x = 180⁰ so we need to subtract each of these values from 360⁰ = 296.6⁰ = 243.4⁰ so we have = 63.4⁰, 116.6⁰, 243.4⁰ or 296.6⁰ Page 10

11 Question 9 we can substitute the coordinates into the equation to get two simultaneous equations 6 = a x 2 n and 18 = a x 3 n divide the second equation by the first to eliminate the a = 3 = ( )n take logs of both sides log3 = log( )n log3 = n x log( ) divide both sides by log( ) n = substitute this value for n back into the first equation 6 = a x = a x divide both sides by a = 0.92 n = Page 11

12 Section B Question 10 i) to find a tangent we first differentiate to get the gradient to differentiate we multiply by the power and then reduce the power by 1 = 4x3 substitute x = 2 into this to get the gradient at the particular point gradient = 4 x 2 3 = 4 x 8 = 32 we need to find the y coordinate at this point y = 2 4 = 16 (2, 16) we now have a gradient and a point that the tangent goes through y y 1 = m(x x 1 ) y 16 = 32(x 2) y 16 = 32x 64 add 16 to both sides y = 32x 48 ii) we need the coordinates at the two points then we can calculate the gradient of the line joining the two points when x = 2, y = 2 4 = 16 (2, 16) when x = 2.1, y = = (2.1, ) gradient of line is difference in y coordinates divided by difference in x coordinates gradient =.. =.. = Page 12

13 iii) (A) we can expand this by using the binomial expansion First set out pascal s triangle. The power is 4 so we only need 5 rows (one more than the power) The final row of the triangle corresponds to the coefficients that we need to use in our expansion I find it easier to work down the page. The first term (2) increases in powers from 0 to 4 whilst the second term ) decreases in powers from 4 to 0. On any row the sum of the powers should be 4. 1 x (2) 0 x h x (2) 1 x h x (2) 2 x h x (2) 3 x h x (2) 4 x h 0 now we need to tidy up each term h 4 + 8h h h Page 13

14 (B) = = h 3 + 8h h + 32 (C) the equation in B) gives the formula for the gradient of a chord joining the points (2, 2 4 ) and (2 + h, (2 + h) 4 ) as h gets smaller and smaller the above two points get closer and closer together so the gradient of the chord becomes closer to the gradient of the curve at the point (2, 16) so when h gets close to 0, we can substitute h = 0 into the equation in B) and see that the gradient will be 32 (this also agrees with the gradient we calculated in part i)) Page 14

15 Question 11 a) these questions are a mix of bearings and advanced trigonometry we need to calculate the angle PQR (shown as x in my diagram) and the angle shown as y we can then calculate the bearing as 360 (x + y) N y N 045⁰ 10.6 x Q P 68⁰ 9.2 R distance from R to Q is length RQ we can find this by using the cosine rule a 2 = b 2 + c 2 2bccosA where a = RQ, b = 10.6, c = 9.2, A = 68⁰ RQ 2 = (2 x 10.6 x 9.2 x cos 68⁰) = square root both sides RQ = = 11.1 km (3 sf) now we have RQ we can use the sine rule to work out angle Q (labelled x) = where Q = x, q = 9.2, P = 68⁰, p = Page 15

16 . =. multiply both sides by 9.2. sin x = = inverse sin (sin -1 ) both sides x = sin -1 ( ) = 50.02⁰ now to calculate angle y the two norths are parallel to each other and the line PQ crosses them this means that angle y = = 135⁰ bearing of R from Q is measured by going clockwise round from north at Q to meet the line QR 360 ( ) = = 175⁰ The distance and bearing of R from Q is 11.1km, 175⁰ Page 16

17 b) i) area of a sector is, r = 80, = area = x = = = 6702 cm 2 ii) triangle ADC a right angled triangle where angle A is ( - ) and AC is 80 we can use basic trigonometry to work out r opposite = r adjacent = don t have and don t need hypoteneuse = 80 we want to use sine (as have opposite and hypoteneuse SOH) sin = multiply both sides by 80 r = 80 x sin = 40 3 iii) consider DCE as a quarter of a circle of radius r (40 3) then the area of DCE = = = 1200 = area of triangle ADC = ½absinC where a = 40 3, b = 80, C = ( - - as angles in a triangle add up to ) triangle ADC = ½ x 40 3 x 80 x sin = = middle area (AEC) is quarter circle triangle ADC = = rudder area = sector ABC middle area (AEC) = = = 4318 cm Page 17

18 Question 12 this is all easier to see if we set up a table it is easy enough to complete the table for 10 stages Stage node Buds Stems Nodes Leaves i) A) looking at the bud column we can see by inspection that we are just multiplying by 2 each time eg, the 4 th term is 2 4, the 5 th term is 2 5 etc so the 10 th term will be 2 10 = 1024 B) looking at the stem column we can see by inspection that the number is always one less than a power of 2 eg, the 4 th term is = 31, the 5 th term is = 63 so the 10 th term will be = = 2047 ii) A) the number of buds is 2 n the number of stems is 2 n+1 1 the number of nodes is 2 n 1 the number of leaves is always 7 times the number of nodes so leaves = 7(2 n 1) Page 18

19 alternatively we can work out the number of nodes as being n-1 this is a geometric series with initial term 1 (a) and common ratio 2 (r) so the sum is = = 2 n 1 the number of leaves is always 7 times the number of nodes so leaves = 7(2 n 1) B) this means that 7(2 n 1) 200,000 expand the brackets (7 x 2 n ) 7 200,000 add 7 to both sides 7 x 2 n 200,007 take logs of both sides log 10 (7 x 2 n ) log ,007 log log 10 2 n log ,007 subtract log 10 7 from both sides log 10 2 n log ,007 - log 10 7 nlog 10 2 log ,007 - log 10 7 divide both sides by log 10 2 n, putting this in the calculator we get n but n has to be a whole number so the least possible value of n is Page 19

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