OCR MEI Core2 2015 Unofficial Mark Scheme Watch

Connorbwfc
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#1
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1i)4x^(-2/3)
1ii) -3x^(-2)+C

2) 10.00625

3) A=12 D=-0.1 Sum=477.5

4) 6+9+9=21cm

5) AND
6i) Sketch both through (0,1). Y^(3x) above Y(x) for positive x-values but below for negative x-values.
6ii) x=3 3^x=27

7) 0.66 & 5.63 radians

8) y=100x^3

9i) Angle=13.80 degrees Area=31.00 cm^2
9ii) 21.28+9.84=31.12cm^2

10i) y=11x-1310ii) y=2x^(2)+3X-5. (-2.5,0) (1,0) Min = (-0.75, -6.125)
10iii) y=8x^(2)+6x-5 Min=(-0.375, -6.125)

11i) 6561
11ii) 21523359
11iii) (3(3^(n)-1))/2 > 1,000,000. 3(3^(n)-1) > 2,000,000. 3^(n+1) – 3 >2,000,000. 3^(n+1) > 2,000,003. (n+1)*log(10)3 > log(10)2,000,0003. n+1> (log(10)2,000,003)/(log(10)3) n > (log(10)2,000,003)/(log(10)3) – 1
11iv) 21523359-65534=21457824
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ETRC
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10) iii) looks wrong. If you stretch by 1/2 then the x coordinate is doubled not halved
x coordinate is halved if you stretch by 2

it should be -1.5
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Connorbwfc
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(Original post by ETRC)
10) iii) looks wrong. If you stretch by 1/2 then the x coordinate is doubled not halved
x coordinate is halved if you stretch by 2

it should be -1.5
No. If you stretch by scale factor 0.5, it is actually a compression scale factor 2, so the X co-ordinate halves.

If you stretch by two (the graph becomes twice as stretched), the x-coordinate will become twice as far from the y-axis - think about it.
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ETRC
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(Original post by Connorbwfc)
No. If you stretch by scale factor 0.5, it is actually a compression scale factor 2, so the X co-ordinate halves.

If you stretch by two (the graph becomes twice as stretched), the x-coordinate will become twice as far from the y-axis - think about it.
yeahh i know i read x-axis as y-axis lol
i didn't even read the question just started doing it- lost 3 marks but oh well
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Hody421
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#5
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Put a poll for grade boundaries

and what do you think the A grade boundary will be?
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Qwertykeyboard15
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I've lost about 20 marks from just silly errors .. So annoyed at myself
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Connorbwfc
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(Original post by Hody421)
Put a poll for grade boundaries

and what do you think the A grade boundary will be?
I'd guess around 57/58
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Hody421
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(Original post by Connorbwfc)
I'd guess around 57/58
How many marks out of 7 was for finding the minimum?
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wednesday_adams
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#9
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are you sure that 11 i and 11 ii are right? Wasn't that the one about Junes decedents or whatever? I dont get how she could have 6561 decendents after 8 generations when after two she had 9..
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Connorbwfc
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(Original post by Hody421)
How many marks out of 7 was for finding the minimum?
Not sure, but I'd guess 2.

(Original post by wednesday_adams)
are you sure that 11 i and 11 ii are right? Wasn't that the one about Junes decedents or whatever? I dont get how she could have 6561 decendents after 8 generations when after two she had 9..
1 - 3
2 - 9
3 - 27
4 - 81
5 - 243
6 - 729
7 - 2187
8 - 6561
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Ecconomist
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#11
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For question 8 which was 4 marks, I got log10x in terms of log10y (2 marks?) and then 'attempted' to raise each side to the power of 10 and kind of got an answer (but wrong).

Do you think this will be 3 marks?
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Connorbwfc
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(Original post by Ecconomist)
For question 8 which was 4 marks, I got log10x in terms of log10y (2 marks?) and then 'attempted' to raise each side to the power of 10 and kind of got an answer (but wrong).

Do you think this will be 3 marks?
If you got log(10)y = 3log(10)x + 2 you will probably get 2.
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Ecconomist
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(Original post by Connorbwfc)
If you got log(10)y = 3log(10)x + 2 you will probably get 2.
However I then did 10 to the power of both sides to then get y=10^3log(10)x+2 and then got my own answer. Was that even the correct initial step though?
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Connorbwfc
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(Original post by Ecconomist)
However I then did 10 to the power of both sides to then get y=10^3log(10)x+2 and then got my own answer. Was that even the correct initial step though?
Yeah that step is correct (I think) even though I didn't do it that way. You must have gone wrong after there. You will get 2/3 marks still.
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ASH451
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Does anyone know the marks for each question?
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username1972843
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#16
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for the minimum point, its (-1.5,-6.125) after stretched by scale factor 1/2
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Connorbwfc
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(Original post by emre7eren)
for the minimum point, its (-1.5,-6.125) after stretched by scale factor 1/2
No. It (-0.375,-6.125). I think you may have stretched by scale factor 2.
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Natia.baj
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#18
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#18
(Original post by Connorbwfc)
1i)4x^(-2/3)
1ii) -3x^(-2)+C

2) 10.00625

3) A=12 D=-0.1 Sum=477.5

4) 6+9+9=21cm

5) AND
6i) Sketch both through (0,1). Y^(3x) above Y(x) for positive x-values but below for negative x-values.
6ii) x=3 3^x=27

7) 0.66 & 5.63 radians

8) y=100x^3

9i) Angle=13.80 degrees Area=31.00 cm^2
9ii) 21.28+9.84=31.12cm^2

10i) y=11x-1310ii) y=2x^(2)+3X-5. (-2.5,0) (1,0) Min = (-0.75, -6.125)
10iii) y=8x^(2)+6x-5 Min=(-0.375, -6.125)

11i) 6561
11ii) 21523359
11iii) (3(3^(n)-1))/2 > 1,000,000. 3(3^(n)-1) > 2,000,000. 3^(n+1) – 3 >2,000,000. 3^(n+1) > 2,000,003. (n+1)*log(10)3 > log(10)2,000,0003. n+1> (log(10)2,000,003)/(log(10)3) n > (log(10)2,000,003)/(log(10)3) – 1
11iv) 21523359-65534=21457824


9)i) 0.5*13*20*sinA i got 122.48
moreover logically triangle with 13,20,8 cannot really have area of 31
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papayarama
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#19
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Hi! for question 10 i) Are you sure it was asking for the tangent? For some reason, I looked for the normal I probably just misread the question.
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papayarama
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(Original post by Natia.baj)
9)i) 0.5*13*20*sinA i got 122.48
moreover logically triangle with 13,20,8 cannot really have area of 31

I got 31.00 too, its 0.5*13*20*sin13.8
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