Mathematics Edexcel Advanced Subsidiary GCE Core 3 (6665) January 2010

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Cassandra Armstrong
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1 Link to past paper on Edexcel website: The above link takes you to Edexcel s website. From there you click QUALIFICATIONS, GCE from 2008, under the subject list click MATHEMATICS, click QUESTION PAPERS, click JANUARY 2010 If you have any difficulties downloading the question paper call us for help. 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 To put this as a single fraction we need to have a common denominator. Before we do this though see if either fractions simplify. For the first fraction we can factorise the denominator. = () = ( ) ()() () We now have () Page 1

2 To get the common denominator multiply both denominators together 3(x -1)(3x + 1) If we multiply the first fraction by then we will have the denominator we want If we multiply the second fraction by () then we will have the denominator () we want () () = () ()() ()() ()() Tidying up the numerator = ()() ()() Page 2

3 Question 2 a) x 3 + 2x 2 3x 11 = 0 we need to pick one of the x terms and make that the subject. The trick is which one? I can see that the equation we want has 3x + 11 as the numerator so if we add both sides by 3x + 11 then we might get close x 3 + 2x 2 = 3x + 11 factorise the LHS x 2 (x + 2) = 3x + 11 divide both sides by (x + 2) x 2 = square root both sides x = b) if you set up your calculator correctly then you can do this without having to keep keying the whole lot in. Key in 0 = and then AC. Now key in the equation using ans in place of the x n value. Every time you press equal you will get the next x value ( ) The first iteration gives x 2 = Pressing the = button again gives x 3 = Pressing the = button again gives x 4 = Page 3

4 c) first rearrange the equation so that we have the equation equal to 0 - x = 0 to show that = to 3 decimal places we must look at and and if we get a sign change then that shows that must be between those two values (ie is to 3 decimal places) when x = we get (.) = when x = we get (.) = Therefore we do have a sign change so = Page 4

5 Question 3 a) From the formulae booklet we know that cos(x + ) = cosxcosα sinxsinα So set 5cosx 3sinx = Rcosxcosα Rsinxsinα Comparing coefficients 5 = Rcosα -3 = -Rsinα Using the identity that sin 2 α + cos 2 α = 1 Square both equations and add them (-3) 2 = R 2 cos 2 α + R 2 sin 2 α = R 2 (sin 2 α + cos 2 α) = R = R 2 34 = R 2 R = 34 Using the fact that = tanα Divide the second equation by the first = = -tanα tanα = inverse tan (arctan) both sides α = c so we have 34 cos(x ) Page 5

6 b) we can use part a) to say that 34 cos(x ) = 4 Divide both sides by 34 cos(x ) = inverse cos (cos -1 ) both sides x = before we make the final adjustment to get x we need to find where else cosy = cos gets repeated at by subtracting from 2π 2π = x = or Now subtract from both sides x = 0.27 c or 4.93 c Page 6

7 Question 4 i) we need to use the quotient rule here = Let u = ln (x 2 + 1) = Let v = x = 1 = ( ) = ( ) ii) x = tany = we need to take care here as the x and y have been interchanged from their normal positions again we can use the quotient rule (but note how I have changed the x and y) = Let u = siny = cosy Let v = cosy = -siny = = = = sec2 y Page 7

8 If we start with the basic identity cos 2 y + sin 2 y = 1 and divide throughout by cos 2 y we get + = This is the same as 1 + tan 2 y = sec 2 y So we have = sec2 y = 1 + tan 2 y = 1 + x 2 So = Page 8

9 Question 5 The graph should be symmetrical about the y axis. It crosses the x axis when x = -1 and Page 9

10 Question 6 i) there are two transformations happening here. The first is inside the bracket so is just affecting the x values. This one is a reflection in the y axis. The other change is outside of the brackets so affects the y values. This is a translation of 1 unit in the positive y direction ii) there are two transformations happening here. The first is inside the bracket so is just affecting the x values. When the x values are affected they do the opposite to what you would think. We are adding 2 units so we move the curve 2 units in the negative x direction. The other change is outside of the brackets so affects the y values. This is a translation of 3 units in the positive y direction Overall we have vector translation of Page 10

11 iii) Again we have two transformations here. This time they are both stretches. The first is affecting the x values as it is inside the brackets. When the x values are affected then the effect is the opposite to what you would expect. The curve is squashed by a scale factor of ½ parallel to the x axis. The other stretch is of scale factor 2 parallel to the y axis Page 11

12 Question 7 a) there are two ways to do this; we could either use the quotient rule with u as 1 and v as cosx, or we can set sec x = cos -1 x and then use the chain rule Quotient Rule = let u = 1 = 0 let v = cos x = - sin x = ( )( ) Chain Rule Let u = cos x y = u -1 = -u-2 = -sin x = = = x = secx tanx = x =-u-2 x -sin x = ( ) x (-sin x) = = x = secx tanx Page 12

13 b) we need to use the product rule here = + u = e 2x = 2e2x v = sec 3x = 3 sec 3x tan 3x (from part a) = 2e sec3+3 e 2x sec 3x tan 3x = e 2x sec 3x (2 + 3tan 3x) Page 13

14 c) turning points are where the gradient is equal to 0 set equal to 0 and solve e 2x sec 3x (2 + 3tan 3x) = 0 e 2x will not equal 0 sec 3x = 0 also cannot equal tan 3x = 0 (this can be solved) subtract 2 from both sides 3tan 3x = -2 divide both sides by 3 tan 3x = take inverse tan (arctan) of both sides 3x = arctan( ) = c Before we adjust for the 3x we need to find where else this gets repeated For the tan curve we just keep adding or subtracting π 3x = c, c divide both sides by 3 x = c (any more would be out of the range) When x = c, we can substitute into y = e 2x sec 3x y = a = and b = (-0.196, 0.812) Page 14

15 Question 8 Starting with the basic identity sin 2 x + cos 2 x = 1 Divide each term by sin 2 x + = We have 1 + cot 2 x = cosec 2 x Replace cosec 2 2x with 1 + cot 2 2x 1 + cot 2 2x cot 2x = 1 We now have a quadratic involving cot 2x Rearrange and set the quadratic equal to 0 cot 2 2x cot 2x = 0 Factorise out the cot 2x cot 2x (cot 2x 1) = 0 So either cot 2x = 0 or cot 2x 1 = 0 in which case cot 2x = 1 cot 2x = 0 cot 2x is the same as cot 2x = 0 means tan 2x = 2x = 90⁰ or 270⁰ x = 45⁰ or 135⁰ cot 2x = 1 cot 2x is the same as cot 2x = 1 means tan 2x = 1 2x = 45⁰ or 225⁰ x = 22.5⁰ or 112.5⁰ x = 22.5⁰, 45⁰, 112.5⁰, 135⁰ Page 15

16 Question 9 i) a) take e of both sides 3x 7 = e 5 Add 7 to both sides 3x = e Divide both sides by 3 x = Question asked for exact answer so don t put in your calculator, just leave it as it is b) take ln of both sides ln(3 x e 7x+2 ) = ln 15 split the log ln (3 x ) + ln (e 7x+2 ) = ln 15 bring the powers to the front xln 3 + (7x + 2)ln e = ln 15 lne = 1 xln 3 + (7x + 2) = ln 15 xln 3 + 7x + 2 = ln 15 subtract 2 from both sides xln3 + 7x = -2 + ln 15 factorise x(ln 3 + 7) = -2 + ln 15 divide both sides by ln3 + 7 x = Question asked for exact answer so don t put in your calculator, just leave it as it is Page 16

17 ii) a) f(x) = e 2x + 3 y = e 2x + 3 to find the inverse function (f -1 ) swap the x and y around and then rearrange to make y the subject x = e 2y + 3 subtract 3 from both sides x 3 = e 2y take ln of both sides ln (x 3) = 2y divide both sides by 2 y = () now switch y back to f -1 (x) f -1 (x) = () the domain of the inverse function is the same as the range of the original function the range of e 2x is > 0 the range of e 2x + 3 is > 3 domain of f -1 (x) will be x > 3 b) when we have two functions fg(x) we work from the inside outwards g(x) = ln (x 1) so fg(x) = f(ln(x 1)) = e 2ln(x 1) + 3 the 2 can move to be a power of (x 1) fg(x) = + 3 fg(x) = (x 1) anything squared is always 0 but we also know that from g(x) that x > 1 so (x 1) 2 > 0 and fg(x) > range is y > Page 17

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OCR (MEI) Mathematics Advanced Subsidiary GCE Core 3 (4753) January 2010

OCR (MEI) Mathematics Advanced Subsidiary GCE Core 3 (4753) January 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 (MEI), VIEW ALL DOCUMENTS,