Mark scheme - June 2007 - 6676 - Further Pure …



Comparison of key skills specifications 2000/2002 with 2004 standardsX015461July 2004Issue 1

GCE Mathematics (6676/01)

June 2007

6676 Further Pure Mathematics FP3

Mark Scheme

|Question |Scheme |Marks |

|Number | | |

| | | |

| 1. |(a) [pic] |M1 |

| | [pic] |A1 (2) |

| | | |

| |(b) [pic] … |B1 |

| | [pic]… |M1 |

| | [pic] cao |A1 (3) |

| | | [5] |

| | | |

| | | |

| 2. | (a) [pic] |M1 |

| | At [pic] [pic] |M1 A1 cso (3) |

| | | |

| |(b) [pic] Allow anywhere |B1 |

| | [pic] … | |

| | [pic] + … |M1 A1ft, A1 (dep)(4) |

| | | [7] |

| | | |

| | | |

| | | |

| | | |

| | | |

|Question |Scheme |Marks |

|Number | | |

| | | |

| 3. |(a) [pic] | |

| | Third row [pic] |M1 A1 (2) |

| | | |

| |(b) [pic] |M1 A1 |

| | First row [pic] Method for either |M1 |

| | Second row [pic] Both correct |A1 ft (4) |

| | | |

| |(c) [pic] | |

| | [pic] | |

| | [pic] | |

| | [pic] Obtaining 3 linear equations |M1 |

| | | |

| | [pic] | |

| | [pic] Reducing to a pair of equations and |M1 |

| | solving for one variable | |

| | [pic] Solving for all three variables. |M1 A1 (4) |

| | | [10] |

| | | |

| |Alternative to (c) | |

| | [pic] |M1 M1 |

| | [pic] |M1 A1 (4) |

| | | |

| | | |

|Question |Scheme |Marks |

|Number | | |

| | | |

| 4. |(a) [pic] | |

| | [pic] both |M1 |

| |Adding [pic]( cso |A1 (2) |

| | | |

| |(b) [pic] |M1 |

| | [pic] |M1 |

| | [pic] |M1 |

| | [pic] |A1, A1 |

| | [pic] A1 any two correct | (5) |

| | | |

| |(c) [pic] | |

| | [pic] |M1 A1ft |

| | [pic] or exact equivalent |M1 A1 (4) |

| | | |

| | | [11] |

| | | |

| | | |

|Question |Scheme |Marks |

|Number | | |

| | | |

| 5. | [pic]: [pic] |B1 |

| | (Hence result is true for [pic].) | |

| | [pic] | |

| | [pic], by induction hypothesis |M1 |

| | [pic] | |

| | [pic] | |

| | [pic] |M1 A1 |

| | [pic] [pic] | |

| | (Hence, if result is true for [pic], then it is true for [pic]) | |

| | By Mathematical Induction, above implies the result is true for all [pic]( | |

| | |A1 (5) |

| |cso | |

| | | [5] |

| | | |

| | | |

| 6. | (a) [pic] | |

| | [pic] |M1 |

| | [pic] can be implied |A1 |

| | [pic] |M1 |

| | Hence [pic] ( cso |A1 (4) |

| |Note: [pic] is divisible by 240 and other appropriate multiples of 15 lead to the required result. | |

| | | |

| |(b) [pic]: [pic] |B1 |

| | (Hence result is true for [pic].) | |

| | From (a) [pic] say. By induction hypothesis [pic] say. | |

| | [pic] |M1 |

| | (Hence, if result is true for [pic], then it is true for [pic]) | |

| | By Mathematical Induction, above implies the result is true for all [pic]( | |

| | Accept equivalent arguments cso |A1 (3) |

| | | |

| |(c) [pic] is not divisible by 15, so result is not true for all [pic]. |B1 (1) |

| |Note: There is no integer for which [pic] is divisible by 15 and any specific example | [8] |

| | should be accepted. | |

| | | |

|Question |Scheme |Marks |

|Number | | |

| | | |

| 7. |(a) [pic] any two |B1 |

| | [pic] |M1 A1 A1 |

| |Give A1 for any two components correct or the negative of the correct answer. | (4) |

| | | |

| |(b) Cartesian equation has form [pic] | |

| | [pic] or use of another point |M1 |

| | [pic] ( or any multiple |A1 (2) |

| | | |

| |(c) Parametric form of line is [pic] or equivalent form |M1 A1 |

| |Substituting into equation of plane | |

| | [pic] |M1 |

| |Leading to [pic] |A1 |

| | [pic] |A1 (5) |

| | | |

| |(d) [pic] both |M1 |

| | These are parallel and hence A, B and T are collinear ( (by the axiom of parallels) |M1 A1 (3) |

| | | [14] |

| |Alternative to (d) | |

| | The equation of AB: [pic] or equivalent | |

| | i: [pic] |M1 |

| | [pic] |M1 |

| | Hence A, B and T are collinear ( cso |A1 (3) |

| | | |

| |Note: Column vectors or bold-faced vectors may be used at any stage. | |

| | | |

|Question |Scheme |Marks |

|Number | | |

| | | |

| 8. |(a) Let [pic]; [pic] |M1 |

| | [pic] |M1 |

| | [pic] |A1 |

| | [pic] |M1 |

| | Eliminating [pic] gives a line with equation [pic] or equivalent |A1 (5) |

| | | |

| |(b) Let [pic]; [pic] |M1 |

| | [pic] |M1 |

| | [pic] |A1 |

| | [pic], [pic] |M1 |

| | [pic] | |

| | [pic] |M1 |

| | Reducing to the circle with equation [pic]( cso |M1 A1 (7) |

| | | |

| |(c) | |

| | | |

| | v ft |B1ft |

| |their line | |

| | Circle through origin, centre in correct quadrant |B1 |

| | Intersections correctly placed |B1 (3) |

| | u | [15] |

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

| | | |

|Question |Scheme |Marks |

|Number | | |

| | | |

| 8. |Alternative for (b) | |

| | Let [pic]; [pic] |M1 |

| | [pic] |M1 |

| | [pic] |A1 |

| | [pic], [pic] |M1 |

| | | |

| |[pic][pic] | |

| | [pic] |M1 |

| | [pic] ( |M1 A1 (7) |

| | | |

| 8. |Alternative for (b) | |

| | Let [pic]; [pic] |M1 |

| | [pic] |M1 |

| | u( + v(( + 1) + [v( – u(( + 1)]i = ( – (i |A1 |

| |Equating real & imaginary parts | |

| | u( + v(( + 1) = ( (i) v( – (u – u = –( (ii) |M1 |

| |From (i) [pic] From (ii) [pic] | |

| | [pic] |M1 |

| | Reducing to the circle with equation [pic]( |M1 A1 (7) |

| | | |

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Mark Scheme (Post-Standardisation)

Summer 2007

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GCE

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