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, 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 1 + 6 4 + 5+2 + 11 + 28 + 3+1 + 5 + 2 6 + 26 + 3 6 + 30 + 12 4 9 + 1 4 20 8 11+9 Quotient is + 6 4 Remainder is 11+9 www.chattertontuition.co.uk 0775 950 1629 Page 1
Question 2 i) given that angle ABC = 90ᵒ, that means that AB is perpendicular to BC 5 1 6 = + = + = 10 + 2 = 12 12 3 15 1 3 2 = + = + = 2 + 6 = 4 3 3+ 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 0 6 2. = 12. 4 = (6 x 2) + (12 x 4) + (-15 x (3 + p)) = 0 15 3+ 12 + 48 45 15p = 0 15 15p = 0 Add 15p to both sides 15 = 15p Divide both sides by 15 p = 1 www.chattertontuition.co.uk 0775 950 1629 Page 2
ii) if ABC is a straight line then AB is parallel to BC there will be a constant k such that 6 2 12 = k 4 15 3+ 6 = 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 = -8 www.chattertontuition.co.uk 0775 950 1629 Page 3
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 3 + 1 = 3 + 1 3 + 1 www.chattertontuition.co.uk 0775 950 1629 Page 4
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 + = 2 + 1 = 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 = = = = www.chattertontuition.co.uk 0775 950 1629 Page 5
Question 5 i) (1 + x) 1/3 Now using the formula for the binomial expansion: 1+ =1+! + +! n = 1/3 1 + ( x x) + 1 + x + x x2 + 1 + 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 < www.chattertontuition.co.uk 0775 950 1629 Page 6
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 = +3 www.chattertontuition.co.uk 0775 950 1629 Page 7
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 +4+2 = 3 3 + 4+2 = 3 subtract 2 from both sides 3 + 4= 3 2 Factorise 3 2 = 3 + 4 Divide both sides by 3 2 = www.chattertontuition.co.uk 0775 950 1629 Page 8
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= 4+2 4+2=0 www.chattertontuition.co.uk 0775 950 1629 Page 9
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 = 1 www.chattertontuition.co.uk 0775 950 1629 Page 10
Question 9 i) First find the position vector of P, replace t with 1 3+11 4 P = 1+1 1 = 0 1+12 3 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 1 3 2 a.b = (4 x 1) + (0 x -1) + (3 x 2) = 4 + 0 + 6 = 10 = 4 +0 +3 = 16+0+9 = 25 = 5 = 1 + 1 +2 = 1+1+4 = 6 cos=. = = 0.816497 =cos 0.816497=35.3ᵒ This is already acute so no need to adjust www.chattertontuition.co.uk 0775 950 1629 Page 11
3+ ii) Position vector Q = 1 1+2 If OQ is perpendicular to l then the dot product of them will be 0 3+ 1 1. 1 = 0 1+2 2 (3 + t) (1 t) + 2(1 + 2t) = 0 3 + t 1 + t + 2 + 4t = 0 4 + 6t = 0 Subtract 4 from both sides 6t = -4 Divide both sides by 6 t = - substitute this back into Q 3 3+ Q = 1 = 1+ = 1+2 1 iii) length of OQ = = + + = + + = = 2.89 units www.chattertontuition.co.uk 0775 950 1629 Page 12
Question 10 i) Partial fractions: = + Turning the RHS into a single fraction = Comparing the numerator 1= 6 + 3 Let x = 3 1 = 3A A = Let x = 6 1 = -3B B = - = - www.chattertontuition.co.uk 0775 950 1629 Page 13
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 www.chattertontuition.co.uk 0775 950 1629 Page 14
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 2 www.chattertontuition.co.uk 0775 950 1629 Page 15
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 48 + 17x = 75 Subtract 48 from both sides 17x = 27 Divide both sides by 17 x = = 1.59 grams www.chattertontuition.co.uk 0775 950 1629 Page 16
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