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

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Oswin Hunter
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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 JANUARY 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 Quotient is Remainder is Page 1

2 Question 2 i) given that angle ABC = 90ᵒ, that means that AB is perpendicular to BC = + = + = = = + = + = = To find the angle between two vectors a.b = rearrange to make cos the subject. cos= Now if the two vectors are perpendicular then the angle between them will be 90ᵒ cos 90 = 0 So a.b will also need to be = = (6 x 2) + (12 x 4) + (-15 x (3 + p)) = p = p = 0 Add 15p to both sides 15 = 15p Divide both sides by 15 p = Page 2

3 ii) if ABC is a straight line then AB is parallel to BC there will be a constant k such that = k = 2k and 12 = 4k k = 3-15 = k(3 + p) Substitute k = 3-15 = 9 + 3p Subtract 9 from both sides -24 = 3p Divide both sides by 3 p = Page 3

4 Question 3 From the formulae sheet cos2=cos +=cos sin =cos 1 cos cos2=2cos 1 So we have = 2 = 2 sec 2 π/3 π/4 = tan tan = Page 4

5 Question 4 Let u = 2+ = Multiply both sides by dt = dt Multiply both sides by t t du = dt Limits: When t = e, u = 2 + = = 3 When t = 1, u =2+1=2+0=2 now replace all the t terms with u terms (though ok to leave some of the t terms as they will cancel each other out) x t du = = 3 2 = = = = Page 5

6 Question 5 i) (1 + x) 1/3 Now using the formula for the binomial expansion: 1+ =1+! + +! n = 1/3 1 + ( x x) x + x x x - x2 + x(x) 2 + ii) a) (8 + 16x) 1/3 = (8(1 + 2x)) 1/3 = 8 1/3 (1 + 2x) 1/3 = 2(1 + 2x) 1/3 We need to replace x with 2x And multiply the whole result by 2 2(1 + (2x) - (2x)2 + ) 2 + x - x2 + b) Valid for 2 < 1 divide both sides by 2 < Page 6

7 Question 6 = 9 - = 9 - y = t 3 3lnt = 3t2 - = x = (3t2 - ) x Set = 3 and solve = 3 = Multiply both sides by 9-3 =39 3 =27 Add to both sides 3 =27 Divide both sides by 3 =9 t = ± 3 But t must be positive otherwise ln 9t would be invalid t = Page 7

8 Question 7 i) to find the gradient of the normal we need to differentiate and then find the negative reciprocal the differentiation will be implicit differentiation as the x and y terms are mixed up together to differentiate 2 we need to use the product rule the product rule is = v+ u let u = 2x 2 v = y = 4x = = 4xy + 2x2 Differentiating the whole equation = = 3 subtract 2 from both sides 3 + 4= 3 2 Factorise 3 2 = Divide both sides by 3 2 = Page 8

9 To find the gradient of the tangent at the point (2, 1) substitute in x = 2 and y = 1 = = = = -4 To find the gradient of the normal we take the negative reciprocal of -4 Gradient of normal is Equation of a line is given by y y 1 = m(x x 1 ) substitute in x = 2, y = 1 and m = 1= 2 Multiply both sides by 4 4 4= 2 Subtract 4y from both sides and add 4 0= = Page 9

10 Question 8 i) sin ii) from formulae sheet = Let u = cos =sin = sin = cossin = cos π/2 / 0 - sin cos ) ( cos0 / ) + sin 0 ( 1 ) + = = e + 1 e = Page 10

11 Question 9 i) First find the position vector of P, replace t with P = = To find the angle between two vectors a.b = rearrange to make cos the subject cos=. 4 1 We use the two vectors 0 and a.b = (4 x 1) + (0 x -1) + (3 x 2) = = 10 = = = 25 = 5 = = = 6 cos=. = = =cos =35.3ᵒ This is already acute so no need to adjust Page 11

12 3+ ii) Position vector Q = If OQ is perpendicular to l then the dot product of them will be = (3 + t) (1 t) + 2(1 + 2t) = t 1 + t t = t = 0 Subtract 4 from both sides 6t = -4 Divide both sides by 6 t = - substitute this back into Q 3 3+ Q = 1 = 1+ = iii) length of OQ = = + + = + + = = 2.89 units Page 12

13 Question 10 i) Partial fractions: = + Turning the RHS into a single fraction = Comparing the numerator 1= Let x = 3 1 = 3A A = Let x = 6 1 = -3B B = - = Page 13

14 ii)a) first we need to solve this differential equation swap the x and t around so that all the t terms are on the right and all the x terms are on the left multiply both sides by dt dx = 3 6 dt divide both sides by 3 6 dx = k dt Add the integral sign to both sides dx = k dt Replace the single fraction with the partial fractions dx = k dt Now integrate both sides ln 3 + ln 6 = kt + c Express the logs on the LHS as a single log ln = kt + c we need to work out what c equals we know that when t = 0, x = 0 ln = 0 + c ln 2 = c Page 14

15 We need to work out what k equals We know that when t = 1, x = 1 ln = k + ln 2 ln = k + ln 2 Subtract ln 2 from both sides ln - ln 2 = k Express as a single log ln = k k = ln b) ln = kt + c replace k and c with the values we worked out ln = ln + ln 2 Now substitute t = 2 and solve for x ln = ln + ln 2 Multiply all terms by 3 ln = 2ln + ln Page 15

16 ln = ln + ln 2 express RHS as a single log ln = ln2 x take e of both sides = 2 x = Multiply both sides by 3 x and by 8 8(6 x) = 25(3 x) Expand 48 8x = 75 25x Add 25x to both sides x = 75 Subtract 48 from both sides 17x = 27 Divide both sides by 17 x = = 1.59 grams Page 16

17 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 17

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

OCR Mathematics Advanced Subsidiary GCE Core 4 (4724) June 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,