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 (MEI), VIEW ALL DOCUMENTS, PAST PAPERS JANUARY SERIES 2011, QUESTION PAPER UNIT 4754/01 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. Section A Question 1 i) The trapezium rule is given in the formulae booklet First set up a table for each of the terms n 0 1 2 3 4 x n -2-1 0 1 2 y n 1+ = 1.0655 1+ = 1.1696 1+ = 1.4142 1+ = 1.9283 1+ = 2.8964 www.chattertontuition.co.uk 0775 950 1629 Page 1
h((y 0 + y n ) + 2(y 1 + y 2 + + y n )) h = 1 = strip width x 1 x ((1.0655 + 2.8964) + 2(1.1696 + 1.4142 + 1.9283)) x ((3.9619) + 2(4.5121)) x (3.9619 + 9.0242) = 6.493 ii) because of the shape of this curve the trapezium rule overestimates the true value more strips would therefore lead to a more accurate estimate and would therefore give a smaller estimate www.chattertontuition.co.uk 0775 950 1629 Page 2
Question 2 A Cartesian equation is just an equation that links x and y (and involves no other parameter such as t) Multiply both sides by 1 + t x(1 + t) = 1 Divide both sides by x 1 + t = Subtract 1 from both sides t = 1 = Substitute this value of t into the equation for y y = ( ) ( ) Multiply the top and bottom by x y = () () Expand brackets y = = www.chattertontuition.co.uk 0775 950 1629 Page 3
Question 3 (3 2x) -3 = (3(1 - x))-3 = 3-3 (1 - x)-3 = (1 - x)-3 Now using the formula for the binomial expansion: (1+) =1+! +() +! We need to replace x with - x n = -3 (1 + (-3 x - () x) + x(- x)2 + ) (1 + 2x + x x2 + ) (1 + 2x + x2 + ) + x + x2 + Valid for < 1 Multiply both sides by 3 and divide both sides by 2 < www.chattertontuition.co.uk 0775 950 1629 Page 4
Question 4 2 4 2 i) = + = + = 0 + 3 = 3 1 6 5 4 9 5 = + = + = 3 + 3 = 0 6 4 2 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 2 5. = 3. 0 = (2 x 5) + (3 x 0) + (-5 x 2) = 10 + 0 10 = 0 5 2 The dot product is 0 so AB is perpendicular to BC ii) we know we have a right angled triangle so we can just multiply base by height and divide by 2 let AB be base and BC be height length AB = = (2) +(3) +( 5) = 4+9+25 = 38 length BC = = (5) +(0) +(2) = 25+0+4 = 29 area = ½ x 38 x 29 = 16.6 units squared www.chattertontuition.co.uk 0775 950 1629 Page 5
Question 5 from formulae sheet sin 2θ = sin (θ + θ) = sinθcosθ + cosθsinθ = 2sinθcosθ cos 2θ = cos (θ + θ) = cosθcosθ - sinθsinθ = cos 2 θ sin 2 θ working on the LHS we have From the basic identity sin 2 θ + cos 2 θ = 1 we can rearrange to get cos 2 θ = 1 - sin 2 θ ( ) = = = = = tanθ www.chattertontuition.co.uk 0775 950 1629 Page 6
Question 6 8 3 i) = 2 6+ substitute these values into the equation for the plane 2(-8-3λ) 3(-2) + (6 + λ) = 11-16 - 6λ + 6 + 6 + λ = 11-4 - 5λ = 11 Add 5λ to both sides -4 = 11 + 5λ Subtract 11 from both sides -15 = 5λ Divide both sides by 5-3 = λ Now substitute this back into the equation of the line to find the point of intersection 8 3 8 3( 3) 1 = 2 = 2 = 2 6+ 6+ 3 3 www.chattertontuition.co.uk 0775 950 1629 Page 7
ii) a.b = rearrange to make cos the subject cos=. 2 The normal vector to the plane is 3 1 To find the angle we can use this as a and we use the direction vector for the line as b 3 b = 0, 1 a.b = (2 x -3) + (-3 x 0) + (1 x 1) = -6 + 0 + 1 = -5 = (2) +( 3) +(1) = 4+9+1 = 14 = ( 3) +(0) +(1) = 9+0+1 = 10 cos=. = = -0.422577 =cos 0.422577=114.997ᵒ However question asked for the acute angle and this is obtuse so simply subtract from 180ᵒ 180 114.997 = 65.0ᵒ www.chattertontuition.co.uk 0775 950 1629 Page 8
Section B Question 7 i) when t = 0 v = 5(1 e 0 ) = 5(1 1) = 5 x 0 = 0 ms -1 consistent with initial velocity when t v = 5(1 ) = 5(1 0) = 5 ms -1 consistent with the terminal (long term) velocity when t = 0.5 v = 5(1 ) = 3.16 ms -1 ii) v = 5 5e -2t = 10e-2t From v = 5 5e -2t Add 5e -2t to both sides 5e -2t = 5 v Multiply by 2 10e -2t = 10 2v = 10e-2t = 10 2v iii) = 10 0.4v2 = 10 - v2 express RHS as a single fraction = www.chattertontuition.co.uk 0775 950 1629 Page 9
Multiply both sides by 10 10 = 100 4v2 Factorise the RHS 10 = 4(25 v2 ) Recognise that the LHS is the difference of two squares 10 = 4(5 v)(5 + v) Divide both sides by (5 v)(5 + v) ()() = 4 www.chattertontuition.co.uk 0775 950 1629 Page 10
Partial fractions: ()() = + Turning the RHS into a single fraction ()() = () () ()() Comparing the numerator 10 = A(5 + v) + B(5 v) Let v = 5 10 = 10A + 0 A = 1 Let v = -5 10 = 0 + 10B B = 1 ()() = + www.chattertontuition.co.uk 0775 950 1629 Page 11
Putting this back into the differential equation ( + ) = 4 Multiply both sides by dt and put in the integral signs ( + ) = 4 dt Now integrate both sides -ln (5 v) + ln (5 + v) = 4t + c Express LHS as a single log ln ( )= 4t + c We are told that when t = 0, v = 0 Substitute these values in to get c ln ( )= 0 + c ln 1 = 0 + c 0 = c So we have ln ( )= 4t Divide both sides by 4 t = ln ( ) www.chattertontuition.co.uk 0775 950 1629 Page 12
iv) using the rearranged version v = ( ) as t, 0 v = () = 5ms-1 when t = 0.5 v = ( ) = 3.808 ms-1 v)the first model gave velocity of 3.16 ms -1 after 0.5 seconds. This is closer to the true velocity of 3 ms -1 than the 3.808 produced by the second model. First model is better. www.chattertontuition.co.uk 0775 950 1629 Page 13
Question 8 i) SOHCAHTOA we have adjacent (adj) and we want to find AC (hypoteneuse hyp) so we need cosine (CAH) A adj 5 α hyp B C cos = Multiply both sides by AC cos=5 Divide both sides by cos = www.chattertontuition.co.uk 0775 950 1629 Page 14
A 5 cos β adj C opp F tan= / Multiply both sides by =tan x = 5sectan GF is double CF GF = 10sectan www.chattertontuition.co.uk 0775 950 1629 Page 15
ii) CE = BE - BC BE tan(+ )= Multiply both sides by 5 =5tan(+ ) BC tan= Multiply both sides by 5 =5tan CE = 5tan(+ ) - 5tan = 5(tan(+ ) - tan) From the formulae sheet tan(+ ) = So we have CE = 5( tan) = 5( ( ) CE = 5( ) = 5( ) CE = 5( ( ) ) = 5( ) CE = as required www.chattertontuition.co.uk 0775 950 1629 Page 16
iii) if α = 45ᵒ then tan=1 and sec= = 2 substituting these values in CE = CD = = = From the side view given DE = CD + CE = + We need a common denominator DE = () + () = + = ()() ()() ()() ()() ()() DE = ()() as required www.chattertontuition.co.uk 0775 950 1629 Page 17
iv) from part i) GF = 10sectan GF = 10 x 2 x t = 10 2 t v) if DE = 2GF then ()() = 2 x 10 2 t Divide both sides by 20t ()() = 2 Expanding the denominator = 2 Multiply both sides by 1 1= 2 2 Add 2 to both sides and subtract 1 from both sides 2 = 2 1 Divide both sides by 2 = - = 1 - = 1 - as required Square root both sides = 1 = 0.5412.. tan=0.5412 Take the inverse tan of both sides =tan 0.5412 =28.4ᵒ www.chattertontuition.co.uk 0775 950 1629 Page 18
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