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OCR MEI AS Mathematics 2017 (C1, C2,S1/M1) [Official Thread] (Unofficial Mark Shemes)

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Yeah I got n=26!Always nice to know someone got the same
(edited 6 years ago)
Decent paper apart from showing that last logs one..
I also didn't get that final Trapezium question where it asked you to show that the area for that shape was more than 90% of the whole field
Original post by FZMalek
I think i got something like that I can't remember.
What did you get on the logaW one? I didn't even get time to finish it. :frown:


Yeah dw, I didnt have time for the very last one....Cause I went back and did the trapezium one again :wink:
Unofficial markscheme please!!!
Original post by FailedMyMocks
Decent paper apart from showing that last logs one..
I also didn't get that final Trapezium question where it asked you to show that the area for that shape was more than 90% of the whole field


I got around 92%? Not sure though....Did whole of that question in last 2 mins.
I got W=a^3+log something
What do you guys rreckon grade boundaries will be? It was harder.. well trickier than last years and GB were 56.. I'm hoping they drop to 54/55 but idk it just seems too easy to get an A then..
Original post by AdamWallisjr
I got around 92%? Not sure though....Did whole of that question in last 2 mins.


Yeah I think 92.5 was the answer - not that I got it lol..
Reply 68
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FZMalek
2
I managed the last one but didn't get to finish off the logaW and trig. Identities one..😒

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Original post by charlotte285
n = 26 for us here


I originally put 26, but it said in which year does he first get 1.2 million so surely as it was during the 25th year n = 25. I put 25.
Will this paper have a low grade boundary?
Original post by callumhowieson
Will this paper have a low grade boundary?


I think they'll be same as last year at the very least, if not sightly lower 2 at the most.. so 54 if you're very lucky
Original post by callumhowieson
Will this paper have a low grade boundary?


I hope....Mei ocr exams have got harder this year....And even if not harder they are different from all the past exams (C1)....
Anyone remember what they got for question 8, the trig identities one?
I remember getting 4 answers, two in terms of pi and two in terms of radians. I think they was something like
Pi/6, 5pi/6, 3.something radians and 5.something radians
i got that
This is what I can remember, random order :/

1)
a) Integral 5 to 1: 4xdx = [2x^2] = 50 – 2 = 48.
b) Integral: 6x^1/2 = 4x^3/2 + c

2)
a) (log0.2)-(log0.1)/(0.2-0.1) = 3.01
b) Plot c in between a and b

3)
Logaw = 3 + logax^5 + loga6 – loga2x
Logaw = logaa^3 + logax^5 + loga6 – loga2x
Logaw = loga(6a^3x^5) – loga2x
Logaw = loga(6a^3x^5/2x)
Logaw = loga(3a^3x^4)
W = 3a^3x^4

4) Y=2x^3. Normal at x = 2. y = 16. Dy/dx = 6x^2. Dy/dx = 24. Normal = -1/24. Equation: x + 24y = 386.

5)
a) Cosine rule. a^2 = 32^2 + 15^2 - (2 x 32 x 15 x cos116). AE = 40.86….
b)
Perp. Distance from AE to D.
Area ADE = 1/2 absinC = 1/2 x 32 x 15 x sin116 = 215.7.
Area ADE = b x h x 1/2.
215.7... = 40.86... x h x 1/2
h = 10.55… > 10. Therefore pond lies in triangle.
c)
Area of pond = 116/360 x Pi x 10^2 = 101.22…
Area of ADE = 215.7…
Area of meadow = 114.48…
d)
Angle C = 360 - 90 - 90 - 116 = 64.
Use that to work out total length of base of trapezium = 70 something.
Area of trapezium = 1/2 x (32 + 70 something) x 80 = 4120.74…
Area of car park = trapezium – ADE = 3905.033…
90% of trapezium = 3708.66…
3905>3708 so carpark takes up more than 90% of field.

6)
6cosx^2 = 5 - sinx
6(1-sinx^2) = 5 - sinx
6 - 6sinx^2 = 5 - sinx
6sinx^2 - sinx - 1 = 0
(3sinx + 1)(2sinx - 1) = 0
sinx = -1/3 or 1/2
x = 3.48, 5.94, 1/6PI, 5/6Pi

7)
a)
Asif = 30000 + (n-1)1000. Bettina = 25000 x 1.05^n-1.
10th year – Asif = £39000. Bettina = £38783. Asif had more.
11th year – Asif = £40000. Bettina = £40722. Bettina had more
b)
Total after 17 years.
Asif = 646000. Bettina = 646000 (to the nearest hundred). The same
c)
Total > £M.
25000(1.05^n – 1)/1.05-1 > M
25000(1.05^n – 1)/0.05 > M
25000(1.05^n – 1) > 0.05M
500000(1.05^n – 1) > M
(500000 x 1.05^n) – 500000 > M
500000 x 1.05^n > M + 500000
1.05^n > (M + 500000)/500000
log1.05^n > log (M + 500000) – log500000
nlog1.05 > log (M + 500000) – log500000
n > (log (M + 500000) – log500000)/log1.05
n = 25 or 26 :/

8)
a)
Stretch parallel to y-axis SF 2.
b)
Translation (3 0)

9)
Curve goes through (2,10)
Dy/dx = 12x^3 - 7
Equation of curve: y = 3x^4 -7x + c
c = -24
y = 3x^4 -7x - 24

10)
a)
V = 400
400 = Pi r^2 h
h = 400/Pir^2
A = 2Pir^2 + 2Pirh
A = 2Pir^2 + 800/r
b)
dA/dr = 4Pir -800/r^2
d2A/dr2 = 4Pi +1600/r^3
c)
Min value of r = 3.992..
d2A/dr2 = 37.6... > 0 Therefore minimum.
A with this value = 300.53.. = 301

11)
a) Sketch graph y = 2^x. Goes through (0,1)
b) ?

12)
a)
Sum of (3r + 2) from 1 to 5 = 5 + 8 + 11 + 14 + 17 = 55
b)
AP First term = 4.2. Sixth term = 1.8.
4.2 + 5d = 1.8
5d = -2.4
d = -0.48
(edited 6 years ago)
Reply 76
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FZMalek
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I've lost around 10/12 marks already :frown:

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Reply 77
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FZMalek
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Did any body else get around 646000 or something(?) for the total of 17 years

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Original post by FZMalek
Did any body else get around 646000 or something(?) for the total of 17 years

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Yes :smile:
For last question i am 99.9999999999999999999999999998%sure that its 25

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