Lol!!! According to ur answers I prob failed this exam. I don't know how u guys managed to get x=1+_ root5. Because I got root5, I know that is wrong.
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C1 Maths AS aqa 2016 (unofficial mark scheme new)
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Why bother with a post grad? Are they even worth it? Have your say!  26102016 

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 19052016 07:19

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 19052016 08:12
(Original post by Alexcor)
My brain stopped working on question 7 😠😢😠😢 because i made a mistake somewhere so i spent about 20 minutes on it(mainly staring blankly...and crying inside 😥) so I'm pretty sure it was as follows:
a) answer was y=32x  40
b)find x coordinate of axis intersect (of Q i think?)
c) the 5 marker integration question: 81/4
d) shaded area (I believe you used the x coordinate you found of on part b to find the area of the triangle) .: area= 45/4
The point is, i somehow got y=32x  26 for part a so i got part b and c wrong too. Anybody know how many marks was it for QUESTION 7 PART A, B & D?!?
Thank you 😄 
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 263
 19052016 08:17
(Original post by timtjtim)
Mark Scheme with some workings etc. 
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 19052016 08:26
How many mark do you think I will lose for not simplifying final answers/ simplifying wrong??
Thanks 
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 265
 19052016 08:37
U guys think the grade boundary will low!!! I don't think so. overall, I think I did alright, except few parts of Qs. I found the paper challenging but not too hard though. I hope the grade boundry will be low as u guys say.

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 19052016 08:38
Can some1 explain 6d. I got my signs wrong.

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 19052016 09:25
It's root65, you need to work out the length of AC. Which is root65 and that's the radius. I got that wrong too so I went to my teacher and described the question fully and he said no that's not the way to do it. A lot of people got root34 as the radius though

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 19052016 09:31
For question 8 I'm pretty sure they gave DY over dx as 54 + 27x  6x^2 then you had to find the second derivative and from that prove x = 1.5 had a minimum. This right?

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 19052016 09:53
(Original post by Azi16)
Lol!!! According to ur answers I prob failed this exam. I don't know how u guys managed to get x=1+_ root5. Because I got root5, I know that is wrong. 
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 270
 19052016 10:01
Can anyone remember what they asking for question 1, I think I got the gradient wrong for the first part will this affect the rest of the answer on this question? Thanks

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 19052016 10:23
I think 7)a) asked you to find the tangent at x=2 on the curve and b) asked you to find the intersection with x axis of this tangent

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 19052016 10:30
I'm pretty sure that 7a) asked you to find the tangent to the curve at the point (2,24) and 7b) asked you to find the xintercept of this tangent which was 5/4
Post rating:1 
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 273
 19052016 10:55
(Original post by Konanabanana)
well you cant root a minus number so your obviously wrong so you made some amateur mistake there my g. 
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 19052016 11:08
well done
(Original post by moneyforall)
Love you bruv, i love people who make mark schemes, i imagine them as snails working away slowly on their computer, for hours on end.
You are mark scheme championPost rating:1 
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 275
 19052016 11:31
(Original post by beanigger)
Unoffical Mark scheme for C1 AQA 2016
It would help if you could link answers to questions as i cant remember them
Questions:
1) a)Asked to work out gradient of a tangent, m=  5/3 [2]
b)Asked to find coordinates of B B(  3,4) [3]
c)Asked to find K K=  30 [2]
2) a) simplify (3√5)^2 = 45 [1]
b) i cant remember the question but the answer was75  32√5 [4]
3) a) y=(x7/2)^2  41/4
b) min value = 41/4 [1]
c) Translation of (1/2 , 41/4)
4) a) show that (x+3) was a factor of x^3  5x^2 8x + 48, p(3) = 0, (x+3) is a factor of p(x).
b) three linear factors of p(x) = (x+3)(x4)(x4)
c) find remainder when p(x) was divided by (x+2) R=20
d) Factorise p(x) using (x+2) R = (x2)(x^23x14) + 20 [3]
5) a) asked to find equation of circle (x5)^2+(y+3)^2=65 (note 65 = r^2) [3]
b) asked to find coordinates of B (AB is diameter) B ( 12, 7) [2]
c) asked to find equation of tangent at A 7x4y+18=0 [5]
d) asked to find length of CT = 9 [2]
6) a) y=32x40
b)Q(5/4) [1]
c) upside down positive graph passing through y axis at 8 [2]
d) x = 1±√5
7) a)
b)
c) definite integral = = 81 /4 [5]
d) area of shaded region = = 45/4 [3]
8) a) d^2y/dx^2=  2x  9x^2 [2]
b) verify that P was a minimum point sub x coordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in xcoord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]
c)show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by 1 the inequality sign flips)
d) cant remember anything except the answers were k>6 and k<3/2
Answers that need a question to be assigned to
 k=4 and k=20
solve the inequality 2x^2 9x 18 >0
It was a four mark question. 
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 276
 19052016 12:38
(Original post by Cassette)
No, unfortunately you are expected to give things in the simplest form.
Does anyone remember how much of Q1 would be screwed up by giving the wrong gradient for 1a? I think every other answer I gave on the paper was correct, but I'm genuinely panicking about this one. I don't remember even doing Q1, which says it all really...Post rating:1 
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 277
 19052016 12:45
(Original post by beanigger)
Unoffical Mark scheme for C1 AQA 2016
It would help if you could link answers to questions as i cant remember them
Questions:
1) a)Asked to work out gradient of a tangent, m=  5/3 [2]
b)Asked to find coordinates of B B(  3,4) [3]
c)Asked to find K K=  30 [2]
2) a) simplify (3√5)^2 = 45 [1]
b) i cant remember the question but the answer was75  32√5 [4]
3) a) y=(x7/2)^2  41/4
b) min value = 41/4 [1]
c) Translation of (1/2 , 41/4)
4) a) show that (x+3) was a factor of x^3  5x^2 8x + 48, p(3) = 0, (x+3) is a factor of p(x).
b) three linear factors of p(x) = (x+3)(x4)(x4)
c) find remainder when p(x) was divided by (x+2) R=20
d) Factorise p(x) using (x+2) R = (x2)(x^23x14) + 20 [3]
5) a) asked to find equation of circle (x5)^2+(y+3)^2=65 (note 65 = r^2) [3]
b) asked to find coordinates of B (AB is diameter) B ( 12, 7) [2]
c) asked to find equation of tangent at A 7x4y+18=0 [5]
d) asked to find length of CT = 9 [2]
6) a) y=32x40
b)Q(5/4) [1]
c) upside down positive graph passing through y axis at 8 [2]
d) x = 1±√5
7) a)
b)
c) definite integral = = 81 /4 [5]
d) area of shaded region = = 45/4 [3]
8) a) d^2y/dx^2=  2x  9x^2 [2]
b) verify that P was a minimum point sub x coordinate of P into dy/dx (given in the question) to prove it was a stationary point, then sub in xcoord of p into d^2y/dx^2 to get +45, d^2y/dx^2>0 therefore minimum point [4]
c)show that y is decreasing, so make your dy/dx<0 and rearrange to get into the form they wanted (note when you multiply by 1 the inequality sign flips)
d) cant remember anything except the answers were k>6 and k<3/2
Answers that need a question to be assigned to
 k=4 and k=20
Also, what mark do you think we'll need for 100ums? The only part I missed was subbing xcoord of P into dy/dx in 8b, I only subbed it into d^2y/dx^2 and wrote statement about 45>0 therefore minimum. Do you think I'll still get 100ums since quite a few people weren't too happy with this paper? 
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 19052016 12:59
(Original post by Rager6amer)
a) was like 5 marks i think, (cos you had to find the gradient and the equation no?) and d) was only 3. But not to worry even if you made a mistake you should get some method marks e.g. for finding gradient etc.. and might get some follow through marks in part d) . Im sorry to hear about your issue with question 7, i think the grade boundaries will be a bit lower than usual and also a lot of people lost many marks on this question, including myself. Good luck though
😉thank you, and as you said, hopefully we pick up some method marks. Good luck to you too on C2 
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 279
 19052016 13:00
I know the number of marks for each question. I also know some questions. Let me know what question you want.
Marks:
q1) 6 mark total
q2) 5 mark total
q3) 6 mark total
q4) 10 mark total
q5) 13 mark total
q6) 8 mark total
q7)14 mark total
q8) 12 mark total 
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 280
 19052016 13:35
(Original post by Azi16)
As I said I know I got it wrong, but don't know how to get the answer.
bring 10 to other side and divide by 2 then square root both sides and move 1 to the other side. that gives x=1 /root 5
Last edited by Konanabanana; 19052016 at 13:41.
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Updated: May 27, 2016
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