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 2010, QUESTION PAPER UNIT 4753/01 CORE MATHEMATICS 3 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 Factorise out e x e x (e x 5) = 0 e x = 0 or e x = 5 Take natural logs of both sides (ln) x = ln(0) which is not defined so this leads to no solution or x = ln (5) = 1.609 www.chattertontuition.co.uk 0775 950 1629 Page 1
Question 2 i) we are given some information, we can use these to set up some equations initially (when t = 0) T = 100 100 = 20 + be -k(0) 100 = 20 + b b = 80 now we can use b = 80 in the equation after 5 minutes, T = 60 60 = 20 + 80e -5k Subtract 20 from each side 40 = 80e -5k Divide both sides by 80 0.5 = e -5k Take natural logs (ln) of both sides ln(0.5) = -5k Divide both sides by -5 - ln (0.5) = k This is the same as k = ln (2) = 0.139 ii) now set T = 50, to find t 50 = 20 + 80e -1/5tln2 Subtract 20 from each side 30 = 80e -1/5tln2 Divide both sides by 80 = e-1/5xtln2 Take natural logs (ln) of both sides ln( ) = - tln(2) multiply both sides by -5-5 ln( ) = tln(2) Divide both sides by ln(2) t = ( ) () = 7.075 minutes www.chattertontuition.co.uk 0775 950 1629 Page 2
Question 3 i) y = (1 + 3x 2 ) 1/3 We can use the chain rule here Let u = 1 + 3x 2 Then y = u 1/3 = 6x = u-2/3 = x = u-2/3 x 6x Substitute back u = 1 + 3x 2 = (1 + 3x2 ) -2/3 x 6x = 2(1 + 3x2 ) -2/3 = ( ) ii) Differentiate both sides with respect to x 3y 2 = 6x Divide both sides by 3y 2 = = Now replace y with (1 + 3x 2 ) 1/3 so that y 2 = (1 + 3x 2 ) 2/3 = = same as part i) ( ) www.chattertontuition.co.uk 0775 950 1629 Page 3
Question 4 i) This is the type of integral where the top is the differential of the bottom () = ln (f(x)) + c () = ln( + 1) 1 0 = ln 2 ln 1 = ln 2 0 = ln 2 ii) integrate by substitution let u = 1 + x = 1 Multiply both sides by dx du = dx If u = 1 + x x = u - 1 Limits When x = 1, u = 1 + 1 = 2 When x = 0, u = 1 + 0 = 1 New limits are 2 and 1 Now substitute all the x terms for u terms () dx = () du = (2 ) = 2 2ln 2 1 4 2ln 2 (2 2ln 1) = 2 2ln 2 www.chattertontuition.co.uk 0775 950 1629 Page 4
Question 5 It would help here to see the original sine curve i) Looking at the x axis first on the original sine curve the distance between a peak and trough should be π, for the transformed graph the distance is now π/2, this implies that the transformed graph (so far) is y = sin 2x (c = 2) (always affects the x in the opposite way to what we think) now looking at the y axis: on the original sine curve the vertical distance between peak and trough is 2, on the transformed graph this is 6 (3 times bigger) so we have (so far) y = 3 sin 2x (b = 3 and c = 2) on the original sine curve the height above the x axis is the same as the height below the x axis. This is still the same on the transformed curve. So we have y = 0 + 3 sin 2x a = 0, b = 3, c = 2 ii) Looking at the x axis first the distance between a peak and trough should be π, for the transformed graph the distance is still π, this implies that the transformed graph (so far) is y = sin x (c = 1) now looking at the y axis: on the original sine curve the vertical distance between peak and trough is 2, on the transformed graph this is still 2. However the graph is a reflection of the sine curve in the x axis so we have (so far) y = - sin x (b = -1 and c = 1) on the original sine curve the height above the x axis is the same as the height below the x axis. This has now all been shifted up by 1 unit so we have y = 1 + -1sin x a = 1, b = -1, c = 1 www.chattertontuition.co.uk 0775 950 1629 Page 5
Question 6 if f(x) is an odd function then this means that -f(x) = f(-x) if g(x) is an even function then g(x) = g(-x) f(x) = -f(-x) g(x) = g(-x) f odd implies gf(x) = g(-f(-x)) g even implies g(-f(-x)) = g(f(-x)) = gf(-x) so we have gf(x) = gf(-x) so gf is an even function www.chattertontuition.co.uk 0775 950 1629 Page 6
Question 7 Let arcsin x = θ So x = sin θ If arcsin x = θ then so does arccos y = θ (because given that arcsin x = arcos y) Arccos y = θ So y = cos θ We know the basic identity that sin 2 θ + cos 2 θ = 1 So replacing x = sin θ and y = cos θ We have x 2 + y 2 = 1 www.chattertontuition.co.uk 0775 950 1629 Page 7
Section B Question 8 i) At P and Q the y coordinate is 0 0 = xcos 3x So either x = 0 or cos 3x = 0 x = 0 is not P or Q cos 3x = 0 3x = arcos 0 =,,,, P and Q are the first two of these Divide both sides by 3 x =, P = (, 0) Q = (, 0) ii) to find the gradient we need to first differentiate we need to use the product rule = + where y = let u = x and v = cos 3x = 1 = 3sin3 = cos 3x + 3sin3 Now substitute the x value at P (x = ) cos ( ) + sin( ) = 0 + = - gradient at P is - www.chattertontuition.co.uk 0775 950 1629 Page 8
turning points occur when = 0 cos 3x + 3sin3 = 0 divide both sides by cos 3x 1 3x tan x = 0 Add 3x tan x to both sides 1 = 3x tan x Divide both sides by 3 x tan x = iii) we need to integrate the curve y = xcos 3x between the limits of 0 and cos3 We need to integrate by parts dx = b a - dx Let u = x = 1 and = cos3x v = sin 3x cos3 0 cos3π/6 0 + cos - cos0 = - = sin3π/6 0 - sin3 www.chattertontuition.co.uk 0775 950 1629 Page 9
Question 9 i) the quotient rule is = let u = 2x 2-1 let v = x 2 + 1 = 4x = 2x where y = = f (x) = ( ) ( ) = = ( ) ( ) ( ) If x > 0 then 6x will also be positive The denominator is always positive anyway as it is a square So if x > 0 then f (x) is always positive so has a positive gradient and is an increasing function ii) we can see from the diagram shown that the smallest y value is at -1 and the largest y value is when x = 2 when x = 2, y = = () () = range is -1 y www.chattertontuition.co.uk 0775 950 1629 Page 10
iii) to maximise f (x) we want to differentiate and set equal to 0 f (x) differentiated is f (x) which is given to us ( ) = 0 We just need to set the numerator to 0 6 18x 2 = 0 Add 18x 2 to both sides 6 = 18x 2 Divide both sides by 18 x 2 = x is positive as domain was 0 x 2 x = = now substitute into f (x) f (x) = ( ) = = www.chattertontuition.co.uk 0775 950 1629 Page 11
iv) the domain of the inverse function will be the same as the range for the original function the range of the inverse function will be the same as the domain for the original function domain: -1 x range: 0 y 2 g(x) will be the reflection of f(x) in the line y = x g(x) is shown here in red www.chattertontuition.co.uk 0775 950 1629 Page 12
v) to find the inverse we swap the x and y around and then rearrange to get y as the subject y = x = multiply both sides by (y 2 + 1) x(y 2 + 1) = 2y 2 1 expand xy 2 + x = 2y 2 1 subtract xy 2 from both sides x = 2y 2 xy 2 1 add 1 to both sides x + 1 = 2y 2 xy 2 factorise x + 1 = y 2 (2 x) divide both sides by (2 x) y 2 = square root both sides (only want the positive root) y = g(x) = www.chattertontuition.co.uk 0775 950 1629 Page 13
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