January 2006 - 6675 Pure P5 and Further Pure FP2 - Mark …



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|Question Number|Scheme |Marks |

| 1 |[pic] |B1 |

| |I = [pic] = [ [pic] or equiv. = arsinh ¾ [pic][pic] | |

| |[ M1 does not require limits; A1 f.t. on completing square, providing arsinh ] |M1 A1√ |

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| |Into ln form [pic] ; = ln2 | |

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| |[ If straight to ln form : B1, ln[pic] M1 |M1A1 |

| |Using limits correctly M1A1√ , ln2 A1 ] | |

| | |[5] |

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| |(a) Using [pic] [ 4 = 16 [pic] ] e = [pic] or equiv. (1.12) | |

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| |(b) Distance between foci = 2 a e [ 2 x 4 x [pic] ] ; = 4 [pic] | |

| |[ A1√ dependent on both Ms ] | |

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| |(c) Ellipse, centred on origin | |

| |Hyperbola, both branches | |

| |Totally correct, touching, with |M1A1 |

| |correct intercepts |(2) |

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|2 | | |

| | |M1A1√ |

| | |(2) |

| |[pic] (both) | |

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| |[pic] ; = [pic] | |

| |Use of “half-angle formula” [pic]; s = [pic] | |

| |Using limits correctly and surd form; = 2 √2 (allow [pic] |B1 |

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| | |B1 |

| |(a) Using cosh [pic] and attempt to progress | |

| |Correct intermediate step as far as [pic] |B1 |

| |= [pic] | |

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| |(b) Using part (a) to reduce to cosh2x = [ 2 ] |(3) |

| |Correct method to form ln x or find e X or e 2 x | |

| |x = ln [pic] , ln [pic] or equivalent |[7] |

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| |or ½ ln [pic], ½ ln [pic], ( after finding e 2 x = ……) | |

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| | |B1 |

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| | |M1A1 |

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|3 | | |

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| | |M1A1√ |

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| | |M1A1 |

| | |[7] |

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| | |M1 |

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| | |A1 |

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|4 | | |

| | |A1 (3) |

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| | |M1 |

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| | |M1 |

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| | |A1 A1√ |

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| | |(4) |

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| | |[7] |

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|Question Number|Scheme |Marks |

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| |[pic] ; [pic] |B1B1 |

|5 | | |

| |[pic] | |

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| |= [pic] or equivalent |M1 |

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| |Method to find “c”: [pic] | |

| |c = [pic]or [pic] or – 1.32 | |

| | |A1A1 |

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| |[pic][pic][pic] or equivalent | |

| | |M1 |

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| | |A1 |

| |(a) [pic][pic] | |

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| |[pic][pic] or equivalent |A1√ |

| | |[8] |

| |Application of formula for radius of curvature [[pic] = [pic]] | |

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| |Use of sinh2 x = [pic]1 | |

| | |B1 |

| |[pic][pic][pic] = [pic] AG | |

|6 | | |

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| |(b) cosh x = [pic] |M1A1 |

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| |Using found value of cosh x in formula for [pic]; [pic] = 4.8 | |

| | |M1 |

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| |(a) [pic] |M1 |

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| |= [pic] |A1* |

| | |(6) |

| |= [pic][pic] | |

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| |[pic] | |

| | |B1 |

| |[ [pic]] [pic] AG | |

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| |(b) Relating [pic] to [pic] using result from (a) |M1A1 |

| | |(3) |

| |[pic] = [pic] [pic] = [pic] [pic] |[9] |

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| | |M1A1 |

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|7 | | |

| | |A1√ |

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| | |M1A1 |

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| | |A1* |

| | |(6) |

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| | |M1 |

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| |(a) ar tanh [pic] or equivalent |A1A1 |

| | |(3) |

| |= [pic] or equivalent | |

| | |[9] |

| |= [pic] AG | |

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| |Alternative Approach | |

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| |If using [pic]ar tanh (sin x) [pic] [ or cosh2 [pic]2 ] | |

| |and then use exponentials: | |

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| |Progression as far as [pic] M1 | |

| |Converting to ln form M1 | |

| |Answer as given A1* | |

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| |Note: [pic] can earn M1M1 but for A1* there must be a | |

| |convincing further step. | |

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| |(b) [pic] | |

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| |Note: If [pic] is differentiated M1 requires [pic] | |

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| |(c) Attempt at by parts and use of result in (b) | |

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| |= [pic] | |

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| |= [pic] | |

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| |Using limits correctly : = [pic] or exact equivalent | |

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| | |M1 |

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| |(a) Correct method for finding [pic] [pic] | |

| | |M1 |

| |Gradient of normal = – p | |

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| |Equation of normal: [pic] |A1* |

|8 | |(3) |

| |[pic] AG | |

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| |(b) Using both equations and eliminating x or y | |

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| |[pic] may be unsimplified | |

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| |[pic] | |

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| |Finding the other coordinate | |

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| |[pic] | |

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| |(c) Using pq = 3 in both x and y (in any form) | |

| | |M1A1 |

| |[[pic][pic]] |(2) |

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| |Complete method for relating x and y, independent of p and q | |

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| |A correct equation, in any form | |

| | |M1 |

| |[ e.g. : [pic]= [pic]] | |

| | |A1 |

| |[pic] | |

| | |M1 |

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| | |M1A1 |

| | |(5) |

| | |[10 ] |

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| | |M1 |

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| | |A1 |

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| | |M1 |

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| | |A1* |

| | |(4) |

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| | |M1 |

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| | |A1 |

|9 | | |

| | |A1 |

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| | |M1 |

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| | |A1 |

| | |(5) |

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| | |M1 |

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| | |M1 |

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| | |A1 |

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| | |A1√ |

| | |(4) |

| | |[13] |

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