June 2005 - 6676 Pure P6 - Mark scheme
June 2005
6676 Pure P6
Mark Scheme
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|Question Number|Scheme |Marks |
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|1 |(a) [pic] | |
| | |B1 (1) |
| |(b) [pic] | |
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| |If result true for n = k , i.e. [pic] |B1 |
| |[pic] | |
| |If [pic] then [pic] so [pic] | |
| |Hence result is true for n = k + 1 Conclusion and no wrong working seen | |
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| | |M1A1 |
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| |(a) (i) b [pic]a is perpendicular to a (and b) | |
| | |A1 (4) |
| |a. b [pic] a = |a| | b [pic] a| cos 90( = 0 or equivalent | |
| | |[5] |
| |(ii) a [pic]b = a [pic]c ( a [pic] (b – c) = 0 | |
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|2 |As a [pic] 0 and b [pic] c, |B1 |
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| |a is parallel to (b – c), so b – c = [pic]a |B1 (2) |
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| |(b) (i) If A non-singular, then A–1 AB = A–1 AC ( B = C (*)AG |M1 |
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| |(ii) [pic] | |
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| |Set [pic]and finding two equations |A1 (2) |
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| |Any non-zero values of a, b, c and d such that |M1A1 (2) |
| |a + 2c = 1 and b + 2d = 7. | |
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| | |B1 |
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| | |M1 |
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| | |A1 (3) |
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| | |[9] |
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|Question Number|Scheme |Marks |
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| |(a) Normal to plane is [pic] = 6i + j – 4k (or any multiple) | |
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|3 |(b) Equation of plane is 6x + y –4z = d |M1A1 (2) |
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| |Substituting appropriate point in equation to give 6x + y – 4z = 16 | |
| |[e.g. (1, 6,–1), (3, –2, 0), (3, 6, 2) etc.] |M1 |
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| |p = –2 |A1 (2) |
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| |Direction of line is perpendicular to both normals | |
| | |B1 (1) |
| |[pic] = –9i + 10j –11k | |
| |[Planes are: 6x + y –4z = 16, x + 2y + z = 2] | |
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| |Finding a point on line | |
| | |M1 |
| |a and b identified | |
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| |Any correct equation of correct form e.g.[pic]x [pic]= 0. | |
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| | |M1A1 |
| |Alternative: Using equations of planes to find general point on line | |
| | |M1 |
| |Using equations of planes to form any two of | |
| |10x + 9y = 24, 11x – 9z = 30, 11y + 10z = –4 M1 | |
| |Putting in parametric form M1 | |
| | |A1 (5) [10] |
| |e.g. [pic] A1 | |
| |a and b identified M1 | |
| |Writing in required form; a correct equation A1 | |
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| |(a) | |
| |Circle A | |
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| |3 Correct circle. | |
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| |0 (centre (0, 3), radius 3) | |
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| |(b) Drawing correct half-line passing as shown | |
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| |Find either x or y coord of A. | |
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| |z = – [pic] + (3 + [pic]) i | |
| |[Algebraic approach, i.e. using [pic]and equation of circle |M1 |
| |will only gain M1A1, unless the second solution is ruled out, | |
| |when B1 can be given by implication, and final A1, if correct] | |
| | |A1 (2) |
|4 |(c) |z – 3i| = 3 ( [pic] = 3 | |
| |( | 2i – 3i( | = 3 | |
| ||(| | |
| |( | ( – 2/3 | = | ( | |B1 |
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| |Line with equation u = 1/3 (x = 1/3) |M1A1 |
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| |Some alternatives: | |
| | |A1 (4) |
| |(i) [pic][pic] M1A1 | |
| |As [pic], [pic] M1,A1A1 | |
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| |(ii) [pic] M1A1 | |
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| |= [pic][pic], M1A1 |M1 |
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| |So locus is line u = [pic] A1 | |
| | |A1 |
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| | |M1A1 |
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| | |A1 (5) |
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| | |[11] |
| |(a) [pic]+ i sin n( ), [pic] – i sin n() | |
| |Completion (needs to be convincing) [pic] [pic] (*)AG | |
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| |[pic]= [pic] | |
| |= [pic] | |
| |([pic])5 = 32i sin5 ( = 2i sin 5( – 10i sin 3( + 20i sin ( | |
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| |( sin5 ( = [pic]( sin 5( – 5sin 3( + 10 sin ( ) (*) AG | |
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| |(c) Finding sin5 ( = ¼ sin( | |
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| |[pic] (both) | |
| |(sin4 ( = ¼) ( sin ( = ± [pic] | |
| |[pic] | |
| |(a) [pic] | |
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| |(b) [pic] | |
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| |[pic] ( [pic][pic][pic] | |
| |( [pic] | |
| |Adding to give [pic] [pic] | |
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| |(c) Diff: [pic] | |
| |Substituting appropriate vales ( 4[pic] ( [pic] | |
| |(d) y = [pic] = 1 + [pic] x + [pic] x2 – [pic] x3 + … |M1 |
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| |(e) 1.05119 |A1 (2) |
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| |(a) Det = –12 –2(2k – 8) + 16 = 20 – 4k (*) AG | |
| | |M1A1 |
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|5 |(b) Cofactors [pic] [A1 each error] | |
| |A–1 = [pic][pic] | |
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| |(c) Setting [pic] | |
| | |M1A1 |
| |[pic] = –1 | |
| | |A1 (5) |
| |(d) Forming equations in x, y and z: [pic][pic] | |
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| |–5x + 2y + 4z = 0, 2x + 2z = 8y, 4x + 2y – 5z = 0 |M1 |
| |Establishing ratio x: y: z : [x = 2y, x = z] | |
| |Eigenvector ([pic] |B1 |
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| | |M1 |
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| | |A1;A1 (5) |
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| | |[12] |
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| | |B1 (1) |
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| | |M1A1 |
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| | |M1 |
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| | |A1 |
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|6 | |M1A1 (6) |
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| | |M1A1 |
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| | |M1A1 (4) |
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| | |M1A1√(2) |
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| | |A1 (1) |
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| | |[14] |
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| | |M1A1 (2) |
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| | |M1A3 |
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| | |M1A1√ (6) |
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|7 | | |
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| | |M1 |
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| | |A1 (2) |
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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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| | |[14] |
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