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# AQA AS Core 2 maths unofficial MARK SCHEME (25/05/16) Watch

1. (Original post by mattdeeee1)
ahhh so you squared out the whole bracket here : c^1/2 / d^2 --> (3^m) ^1/2 / (27^n)^2

I used rules of indices and presumed that you could change that to 27^2n
That makes more sense. Thanks
2. What would the marks be to get an A out of 75, just for one paper
3. For the most part a straight forward paper, but if you didn't understand the perimeter question and the binomial expansion then that's a lot of marks to lose. My prediction is between 54 and 59, depending on how many people didn't answer Q8,Q9 and the perimeter question correctly.
4. (Original post by plower)
What would the marks be to get an A out of 75, just for one paper
For the core 2 paper last year it was 58 for an A, 55 the year before and 61 the year before that. IMO the paper was quite straightforward, although i did run out of time, so i reckon you're looking at 58-61 for an A. Hopefully by tomorrow we will have made a proper mark scheme so you can predict how well you think you did. I wish you the very best of luck!!
5. (Original post by plower)
Still gonna cry 😂😭😭
Im sorry to hear that. Given that you have just done an exam, i think you deserve a session of looking at some memes to wind down! :P
6. (Original post by Rager6amer)
For the core 2 paper last year it was 58 for an A, 55 the year before and 61 the year before that. IMO the paper was quite straightforward, although i did run out of time, so i reckon you're looking at 58-61 for an A. Hopefully by tomorrow we will have made a proper mark scheme so you can predict how well you think you did. I wish you the very best of luck!!
That gives me a little bit of hope, thank you!! And to you too!! Btw what was the perimeter one, did you had to halve the area of the triangle to get the area of sector?
7. (Original post by plower)
That gives me a little bit of hope, thank you!! And to you too!! Btw what was the perimeter one, did you had to halve the area of the triangle to get the area of sector?
Yeah, you're right. Ended up with 13.8 for the perimeter
8. (Original post by Sirelis)
I got x = 0.8 for the last question...? and when i put 0.8 back into the log thing they both equalled the same thing so I assumed it was right..? seemed too easy for 4 marks though so i musta got it wrong
I did the same. To get the final mark or two you have to do the discriminant of the quadratic
I did the same. To get the final mark or two you have to do the discriminant of the quadratic
wut I didn't get a quadratic hahaha oh dear I was retaking C2, already got a B in it so it's not the end of the world if I don't improve I guess
10. what did people get for the transformation onto y=3√x^2+1? I'm not sure if that was the question asked correct me if I'm wrong??

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11. (Original post by plower)
That gives me a little bit of hope, thank you!! And to you too!! Btw what was the perimeter one, did you had to halve the area of the triangle to get the area of sector?
Thanks very much. So in the previous part you calculated the area of the triangle (was it 19.9 cm^2 or something??) and then it told you that the sector area = area of shaded region. SO therefore 1/2(θr^2) = 19.9 - 1/2(θr^2)
so therefore add 1/2(θr^2) to both sides to get 2/2(θr^2) = 19.9 and hence θr^2 = 19.9. From the previous part you already know that θ = 0.586 or something so divide both sides by θ to get r^2 = 19.9/0.586 and then root both sides to get r = whatever it was. Then you use r and θ to calculate arc length and then do (side AC - r) + (side AB - r) + arc length + 5 (5 is length BC)

so therefore perimeter if i remember correctly was about 13.6???
12. (Original post by mattdeeee1)
ahhh so you squared out the whole bracket here : c^1/2 / d^2 --> (3^m) ^1/2 / (27^n)^2

I used rules of indices and presumed that you could change that to 27^2n
I also got m/2 -6n

Because c^1/2 / d^2 = (3^m)^1/2 / (3^3n)^2
Then you multiply the indicies to give
3^m/2 / 3^6n
= 3^(m/2-6n)
13. I thought the final question was 1/2 but you're probably right as my friend got 5/2 too
14. What did everyone get for the 16+9sin^2(x)/5-3cosx
15. (Original post by Sirelis)
wut I didn't get a quadratic hahaha oh dear I was retaking C2, already got a B in it so it's not the end of the world if I don't improve I guess
As in, you got 5/2 twice, by solving a quadratic (that's how I did it anyway)
So then you do the discriminant of that quadratic to prove there's only one solution.
16. (Original post by Rager6amer)
Thanks very much. So in the previous part you calculated the area of the triangle (was it 19.9 cm^2 or something??) and then it told you that the sector area = area of shaded region. SO therefore 1/2(θr^2) = 19.9 - 1/2(θr^2)
so therefore add 1/2(θr^2) to both sides to get 2/2(θr^2) = 19.9 and hence θr^2 = 19.9. From the previous part you already know that θ = 0.586 or something so divide both sides by θ to get r^2 = 19.9/0.586 and then root both sides to get r = whatever it was. Then you use r and θ to calculate arc length and then do (side AC - r) + (side AB - r) + arc length + 5 (5 is length BC)

so therefore perimeter if i remember correctly was about 13.6???

thankgod for that!!that was a big one, otherwise I was ready to go pack my bag and go live as a goat in Slovakia , that option is still open 😴
17. (Original post by Kayleigh216)
What did everyone get for the 16+9sin^2(x)/5-3cosx
Ditto, what was it
18. (Original post by BanterBus)
Binomial expansion:
Part a) (1-2x)^5 = -32x^5+80x^4-80x^3+40x^2-10x+1, and in ascending powers of x: 1-10x+40x^2-80x^3+80x^4-32x^5
Part b) Find (2+x)^7. Find the sum of the coefficients of (ax^5)(ax^5)+(ax^4)(ax^6)+(ax^3) (ax^7); which should give (-32x84)+(14x80)+(-80x1) = -1648
wasn't the question (2-x)^7
19. So noone else got 10.7 as the perimeter? I got 4.12 as the value of r??
20. (Original post by Kayleigh216)
What did everyone get for the 16+9sin^2(x)/5-3cosx
i substituted (1- cos^2(x)) to get 16 + 9 - 9cos^2
= 25 - 9cos^2(x) which is a difference of two squares, so you can change it to (5+3cosx)(5-3cosx) which you can divide by 5-3cosx to get just 5+3cosx on its own

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