OCR (MEI) Mathematics Advanced Subsidiary GCE Core 4 (4754) January 2011

Transcription

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 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 x n y n 1+ = = = = = Page 1

2 h((y 0 + y n ) + 2(y 1 + y y n )) h = 1 = strip width x 1 x (( ) + 2( )) x ((3.9619) + 2(4.5121)) x ( ) = 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 Page 2

3 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 = = Page 3

4 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 < Page 4

5 Question i) = + = + = = = + = + = = 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 = 3. 0 = (2 x 5) + (3 x 0) + (-5 x 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) = = 38 length BC = = (5) +(0) +(2) = = 29 area = ½ x 38 x 29 = 16.6 units squared Page 5

6 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 Page 6

7 Question i) = 2 6+ substitute these values into the equation for the plane 2(-8-3λ) 3(-2) + (6 + λ) = λ λ = λ = 11 Add 5λ to both sides -4 = λ 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 ( 3) 1 = 2 = 2 = Page 7

8 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) = = -5 = (2) +( 3) +(1) = = 14 = ( 3) +(0) +(1) = = 10 cos=. = = =cos = ᵒ However question asked for the acute angle and this is obtuse so simply subtract from 180ᵒ = 65.0ᵒ Page 8

9 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) = v2 = 10 - v2 express RHS as a single fraction = Page 9

10 Multiply both sides by = 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) ()() = Page 10

11 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 = = B B = 1 ()() = Page 11

12 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 ( ) Page 12

13 iv) using the rearranged version v = ( ) as t, 0 v = () = 5ms-1 when t = 0.5 v = ( ) = 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 produced by the second model. First model is better Page 13

14 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 = Page 14

15 A 5 cos β adj C opp F tan= / Multiply both sides by =tan x = 5sectan GF is double CF GF = 10sectan Page 15

16 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 Page 16

17 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 Page 17

18 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 = tan= Take the inverse tan of both sides =tan =28.4ᵒ Page 18

19 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 MEI. In addition these solutions may not necessarily constitute the only possible solutions Page 19