January 2005 - 6672 Pure P2 - Mark scheme
January 2005
6672 Pure Mathematics P2
Mark Scheme
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|Question Number |Scheme |Marks |
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|1a) |[pic] |B1,B1,B1 |
| | |(3) |
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| |[pic] | |
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|1b) | |M1 A1√ |
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| |[pic] | |
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| | |A1 |
| | |(3) |
| |[pic] | |
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|2a) | | |
| |[pic] | |
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| |[pic] |M1 |
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| | |A1 |
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| | |(2) |
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| | |B1 |
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| | |B1 |
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| | |(2) |
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| | |B1 |
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| | |B1 |
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| | |B1 |
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| | |(3) |
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|Question Number |Scheme |Marks |
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|3a) |u2 = (–1)(2) + d = –2 + d |B1 |
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| |u 3 = (–1)2(–2 + d) + d = –2 + 2d Attempting to find u 3 in |M1 |
| | | |
| |terms of d |A1 |
| |u 4 = (–1)3(–2 + 2d) + d = 2 – d u 3 and u 4 correct | |
| | |A1* |
| |u 5 = (–1)4(2 – d) + d = 2 * cso fully correct |(4) |
| | | |
|b) | |B1√ |
| |u10 = u2 = d – 2 o.e. their u2 must contain d |(1) |
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|c) | |M1 A1 |
| |–2 + 2d = 3(–2 + d) [pic] d = 4 M1 equating their u 3 to their 3u.2 |(2) |
| |must contain d | |
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|4a) | |B1 |
| |(0,4), or x =0 and y = 4 |(1) |
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|b) |[pic] attempt use of, must have pi |M1 |
| | | |
| |x2 = y – 4 or [pic] |B1 |
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| |[pic] attempt to integrate |M1 |
| |[pic] correct integration ignore pi | |
| | |A1 |
| |using limits in a changed form to give 8,4 either way but must subtract | |
| | | |
| |[pic]. (c.a.o) |M1 |
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| | |A1 |
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| | |(6) |
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|Question Number |Scheme |Marks |
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|5a) |log3x = log5 taking logs |M1 |
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| |[pic] | |
| | |A1 |
| |= 1.46 cao | |
| | |A1 |
| | |(3) |
| |2 = log2[pic] | |
|b) | | |
| |[pic] or equivalent; 4 |M1 |
| |2x +1 = 4x multiplying by x to get a linear equation | |
| |x =[pic] |B1 |
| | | |
| | |M1 |
| |sec x = 1/ cos x | |
| | |A1 |
| |sin x = cos x [pic] tan x = 1 x = 45 use of tan x |(4) |
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| | |B1 |
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|c) | |M1, A1 |
| | |(3) |
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|Question Number |Scheme |Marks |
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|6a) |I = 3x + 2ex |B1 |
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| |Using limits correctly to give 1 + 2e. (c.a.o.) must subst 0 and 1 and |M1 A1 |
| |subtract |(3) |
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|b) |A = (0, 5); y = 5 |B1 |
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| |[pic] |B1 |
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| |Equation of tangent: y = 2x +5; c = –2.5 | |
| | |M1; A1 |
| | |(4) |
| |[pic] [pic] | |
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|c) | |M1; A1 |
| |[pic] or equivalent must be in terms of x | |
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| | |A1 |
| |gf(0) = g(5); =3 att to put 0 into f and then their answer into g |(3) |
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|d) | | |
| | |M1; A1 |
| | |(2) |
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|Question Number |Scheme |Marks |
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|7a) |Complete method for DE [e.g. split triangle ADE and sin, or sine or cos rule] |M1 |
| | | |
| |DE = 4 sin θ * ( c.s.o.) | |
| | |A1* |
| | |(2) |
| |P = 2 DE +2EF or equivalent. With attempt at EF | |
|b) | |M1 |
| |= 8sin θ + 4cos θ * ( c.s.o.) | |
| | |A1* |
| | |(2) |
| |8sin θ + 4cos θ = R sin (θ + α) | |
|c) | | |
| |= R sin θ cos α + R cos θ sin α | |
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| | | |
| |Method for R, method for α need to use tan for 2nd M | |
| | |M1 M1 |
| |[R cos α = 8, R sin α = 4 tan α = 0.5, R = [pic]] | |
| | | |
| |R = 4[pic] or 8.94, α = 0.464 (allow 26.6), awrt 0.464 | |
| | | |
| | |A1 A1 |
| | |(4) |
| | | |
| |Using candidate’s R sin (θ + α) = 8.5 to give (θ + α) = sin-1[pic] | |
| | | |
| |Solving to give θ = sin-1[pic] – α, θ = 0.791 (allow 45.3) |M1 |
|d) | | |
| |Considering second angle: θ + α = π ( or 180) – sin-1[pic]; | |
| | |M1 A1 |
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| |θ = 1.42 (allow 81.6) |M1 |
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| | |A1 |
| | |(5) |
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|Question Number |Scheme |Marks |
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|8a) |[pic] |M1A1;A1 |
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| |[pic] | |
| | |M1A1 * cso |
| | |(5) |
| |[pic] | |
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| | |M1;A1 |
|b) |f(4.905) = < 0 (–0.000955), f(4.915) = > 0 ( + 0.000874) evaluate | |
| | |(2) |
| |Change of sign indicates root between and correct values to 1 sf) | |
| | |M1 |
|c) | | |
| | |A1 |
| |[pic] [pic] |(2) |
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| |[pic] * (c.s.o.) M1 for use of e to | |
|d) |the power on both sides | |
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| |x1 = 4.9192 | |
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| |x2= 4.9111, x3 = 4.9103, both, only lose one if not 4dp | |
| | |M1;A1 |
| | |(2) |
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| | |B1 |
|e) | | |
| | |B1 |
| | |(2) |
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-----------------------
B1 numerator, B1 denominator ;
B1 either form of answer
M1 for equating f(x) to x + 1 and forming quadratic.
A1 candidate’s correct quadratic
attempting to find eq. of tangent and subst in y = 0, must be linear equation
putting y = and att. to rearrange to find x .
M1 for evidence of differentiation. Final A –no extras
(or subst x = 0.5)
Subst 0.5 or their value for x in
Reflected in[pic]- axis 0 ................
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