OCR Mathematics Advanced Subsidiary GCE Core 4 (4724) 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, VIEW ALL DOCUMENTS, PAST PAPERS JUNE SERIES 2010, QUESTION PAPER UNIT 4724/01 CORE MATHEMATICS 4 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. Question 1 Now using the formula for the binomial expansion: 1+ =1+! + +! We need to replace x with 3x n = ( x 3x) + 1-5x + x 9x x + 20x 2 - x3 + x(3x) 2 + x(3x) 3 + x 27x 3 + Valid for 3 < 1 divide both sides by 3 < Page 1

2 Question 2 i) the quotient rule is = where y = let u = cos x = -sin x let v = 1 sin x = -cos x = = = = Page 2

3 Question 3 Partial fractions: = + + Turning the RHS into a single fraction = Comparing the numerator = Let x = 2 4 = C C = 4 Let x = 1 1 = 0 - B + 0 B = -1 Let x = 0 0 = 2A 2B + C We already know B = -1 and C = 4 0 = 2A A = -3 = Page 3

4 Question 4 Let u = +2 = (x + 2) 1/2 = (x + 2)-1/2 Multiply both sides by dx = (x + 2)-1/2 dx Multiply both sides by 2 2du = (x + 2) -1/2 dx Multiply both sides by (x + 2) 1/2 2(x + 2) 1/2 du = dx Replace (x + 2) 1/2 with u 2udu = dx Limits: When x = 7, u = 7+2 = 3 When x = -1, u = 1+2 = 1 if u = +2 then u 2 = x + 2 u 2 2 = x now replace all the x terms with u terms x 2u du = 2 2 = = ( ) ( ) = + 56 = = Page 4

5 Question 5 i) to find the stationary points we need to differentiate and set equal to 0 this is using implicit differentiation as the x and y terms are mixed up together to differentiate 4xy we need to use the product rule the product rule is = v+ u let u = 4x v = y = 4 = = 4y + 4x Differentiating the whole equation = = 0 subtract 4y and subtract 2x from both sides 4 + 4= 4 2 Factorise 4+4= 4 2 Divide both sides by 4+4 = = = Set this to 0 We want the numerator to = 0 2+=0 So we have = Page 5

6 we now have simultaneous equations to solve = 2 and =0 Substitute the x value =0 Expand the brackets =0 Group terms =0 Add 2y 2 to both sides 2 = 18 Divide both sides by 2 =9 Square root both sides y = ±3 we know that = 2 so when y = +3, x = -6 when y = -3, x = +6 two coordinates are: (-6, 3) and (6, -3) Page 6

7 Question 6 i) a.b = rearrange to make cos the subject cos=. Now if the two lines are perpendicular then θ = 90ᵒ and cos θ = 0 This will mean that a.b = 0 we use the direction vector for both a and b 2 2 a =, b = a.b = (2 x 2) + (a x 2) + (1 x -6) = 4 + 2a 6 = a = 0 2a = 2 a = Page 7

8 ii) a) if the two lines do meet then there will be a solution to the following 3 simultaneous equations 0 + 2t = 3 + 2s eqn at = 0 + 2s eqn t = -1-6s eqn 3 From eqn 1 2t = 3 + 2s 2s = 2t 3 6s = 6t 9 Substitute this into eqn t = -1 (6t 9) 1 + t = -1 6t + 9 Add 6t to both sides 1 + 7t = 8 7t = 7 t = 1 2s = 2t 3 2s = 2(1) 3 = -1 s = - now substitute these values for s and t into eqn a(1) = 2(- ) 1 + a = -1 a = Page 8

9 b) a.b = rearrange to make cos the subject cos=. we use the direction vector for both a and b 2 2 a = 2, b = a.b = (2 x 2) + (-2 x 2) + (1 x -6) = = -6 = = = 9 = 3 = = = 44 cos= = =cos =107.5ᵒ Page 9

10 Question 7 i) we need to differentiate the x equation and the y equation separately to differentiate the x equation we need the quotient rule the quotient rule is = let u = t + 2 let v = t + 1 = 1 = 1 where y = = = y = 2(t + 3) -1 = -2(t + 3)-2 x 1 = -2(t + 3) -2 = Using the chain rule = x = x = The numerator and the denominator are both square numbers so must be positive hence > Page 10

11 ii) A Cartesian equation is just an equation that links x and y (and involves no other parameter such as t) we can rearrange the y equation so that t is the subject Multiply both sides by t + 3 y(t + 3) = 2 Divide both sides by y t + 3 = Subtract 3 from both sides t = 3 = substitute this value for t into the equation for x x = multiply the top and bottom by y x = = multiply both sides by 2-2 y x(2 2y) = 2 y 2x 2xy = 2 y add y to both sides y + 2x 2xy = Page 11

12 Question 8 i) x - 4 x - 1 x 2-5x + 6 x 2 - x Quotient is x 4 Remainder is 2-4x + 6-4x Page 12

13 ii) a) swap the x and y around so that all the y terms are on the left and all the x terms are on the right multiply both sides by dx dy = 5 dx Multiply both sides by dy = dx Divide both sides by 1 and by 5 dy = dx In part i) we know that when we divide 5+6 by 1 the quotient is x 4 and the remainder is 2 rewriting the RHS dy = 4+ dx Add the integral sign to both sides dy = 4+ d Now integrate both sides ln 5 = ln( 1) + c we need to work out what c equals we know that y = 7, x = 8 ln 7 5 = 8 4 x 8 + 2ln(8 1) + c ln 2 = ln(7) + c ln 2 = ln(7 ) + c c = ln 2 ln 49 = ln( ) Page 13

14 now use this value of c together with x = 6 to find y ln 5 = 6 4 x 6 + 2ln(6 1) + ln( ) ln 5 = ln(5) + ln( ) ln 5 = 6 + 2ln(5) + ln( ) = =. y = = 5.00 (3 significant figures) Page 14

15 Question 9 i) First expand the brackets +cos2 2 = +cos2+cos2= The first term will be a straightforward integration but we will have to use integration by parts for 2cos2 And cos 2 will need to be rearranged and put in a different form = + c 2cos2 integrate by parts From the formulae sheet Let u = 2 =cos2 = uv - =2 = sin2 2cos2 = 2-2 x sin2 = 2 - sin2 = 2 + cos2 + c From the double angle formulae cos4=cos 2 sin 2=cos 2 1 cos 2=2cos 2 1 Rearrange to make cos 2 the subject cos 2= cos4+ Now we are able to integrate cos 2 = cos4+ = sin4+ + Putting the three terms back together we have +cos2 2 = cos2 + sin Page 15

16 ii) for volume of revolution we want = +cos2 2 We have already done this integration in part i) Now just multiply by π and put in the limits cos2 + sin4 + + π( cos20 + sin cos + sin2 + π( cos = + = + If you found these solutions helpful and would like to see some more then visit our website It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by OCR. In addition these solutions may not necessarily constitute the only possible solutions Page 16

OCR Mathematics Advanced Subsidiary GCE Core 4 (4724) January 2010

OCR Mathematics Advanced Subsidiary GCE Core 4 (4724) January 2010 Link to past paper on OCR website: www.ocr.org.uk The above link takes you to OCR s website. From there you click QUALIFICATIONS, QUALIFICATIONS BY TYPE, AS/A LEVEL GCE, MATHEMATICS, VIEW ALL DOCUMENTS,