um i think it was the first equation 5x +3y+3=0(Original post by BlackthornRose)
Do you happen to remember the simultaneous equations? I'd hate to lose marks on the first question and it's the one I remember least as I probably gave it the least thought.
and um think the second equatuion was 3x + or  2y or sumn +17 i think
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C1 Maths AS aqa 2016 (unofficial mark scheme new)
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 18052016 18:07
Post rating:2 
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 18052016 18:09
I got question 5a right but I didn't cross out my working before it which was wrong. Does that mean I get no marks?

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 18052016 18:13
(Original post by GabbytheGreek_48)
um i think it was the first equation 5x +3y+3=0
and um think the second equatuion was 3x + or  2y or sumn +17 i think
y=4
so x=3
(3,4) 
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 144
 18052016 18:15

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 18052016 18:16
Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?
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 146
 18052016 18:17
(Original post by BlackthornRose)
Do you happen to remember the simultaneous equations? I'd hate to lose marks on the first question and it's the one I remember least as I probably gave it the least thought. 
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 147
 18052016 18:18
(Original post by Jin99)
Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?
Posted from TSR Mobile 
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 148
 18052016 18:19
(Original post by Jin99)
Do you know for the area of shaded region, do you subtract the triangle from the area under the curve?
Posted from TSR Mobile 
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 149
 18052016 18:20
For question 7 I think...
(a) You had to find the derivative of the equation 4x^23x^3 using point P(2,24)?
dy/dx= (2x9x^2) so subbing in 2 gives 32.
y24=32(x+2)
y=32x64+24
y=32x40
(b) then you'd use that to find coordinate Q by subbing in y=0
into y=32x40
0=32x40 to get
x=5/4
(c)? Then you had to find that integral.
(d)?
To find the area of the triangle you use Coordinate Q 5/4
You had to subtract 5/4 from the limit (2)
So 5/4   2 = 3/4
Base of the triangle is then 3/4
So the area of the triangle is 0.5 * 0.75 * 24 = 9Last edited by Pentaquark; 18052016 at 18:22. 
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 18052016 18:20
I'm pretty sure the answer for 1b was (3,4).

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 18052016 18:20
(Original post by Applebananaaaa)
(x5)^225+(y+3)^29=0
= (x5)^2+(y+3)^2=34 therefore radius=Square root of 34
It is definitely 65.
(Original post by GabbytheGreek_48)
um i think it was the first equation 5x +3y+3=0
and um think the second equatuion was 3x + or  2y or sumn +17 i think
Once again thank youLast edited by BlackthornRose; 18052016 at 18:22. Reason: Extra info 
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 152
 18052016 18:20
(Original post by Pentaquark)
For question 7 I think...
(a) You had to find the derivative of the equation 4x^23x^3 using point (2,24)?
dy/dx= (2x9x^2) so subbing in 2 gives 32.
y24=32(x+2)
y=32x64+24
y=32x40
(b) then you'd use that to find coordinate Q by subbing in y=0
into y=32x40
0=32x40 to get x=5/4
Then you had to find that integral.
(b) (ii)
To find the area of the triangle you use Coordinate Q 5/4
You had to subtract 5/4 from the limit (2)
So 5/4   2 = 3/4
Base of the triangle is then 3/4
So the area of the triangle is 0.5 * 0.75 * 24 = 9 
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 153
 18052016 18:21
In question 8 they made you calculate (d^2y) / (dx^2) in part a) i) for 2 marks i think. They then asked you in part a) ii) to verify that p (x = 3/2) was a minimum point, for which there were 4 marks available. You had to first prove dy/dx = 0 at P and then calculate (d^2y) / (dx^2) which > 0 > minimum point at P.
In part b) you had to show that when y was increasing, dy/dx > 0 with an equation of something like 2x^2 + 4x + 8 or whatever
In part c) you had to find the values of k or something for which the final answer was k < 3/2 and k > 6
Hope this helps, good luck w/ mark scheme and hope everyone did okPost rating:1 
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 154
 18052016 18:24
i got 801/36 for the integral would that be okay

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 18052016 18:26
(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 [5]
3) a) y=(x7/2)^2  41/4
b) min value = 41/4
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^23x+14) + 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) i think there was 4 parts to this question i cant remember part a and b
b)
c) definite integral = 81/4 [5]
d) area of shaded region = 45/4 [3]
8) a) 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
 d^2y/dx^2=  2x  9x^2 sub in xcoord of p to get 45, 45>0 therefore minimum 
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 156
 18052016 18:26
(Original post by Rager6amer)
In question 8 they made you calculate (d^2y) / (dx^2) in part a) i) for 2 marks i think. They then asked you in part a) ii) to verify that p (x = 3/2) was a minimum point, for which there were 4 marks available. You had to first prove dy/dx = 0 at P and then calculate (d^2y) / (dx^2) which > 0 > minimum point at P.
In part b) you had to show that when y was decreasing, dy/dx < 0 with an equation of something like 2x^2 + 4x + 8 or whatever
In part c) you had to find the values of k or something for which the final answer was k < 3/2 and k > 6
Hope this helps, good luck w/ mark scheme and hope everyone did ok 
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 157
 18052016 18:27
(Original post by BlackthornRose)
No, you have to understand that you only minus the 'extra numbers' when you go from the 'multiplied out' version of the equation of the circle and are asked to complete the square to get the 'factorised/completed square' version. Here they already gave you the coordinates for centre, you simply have to work out the radius and put it all into the general equation of a circle which is (xa)^2+(yb)^2 = r^2 where (a,b) are the coordinates of the centre.
It is definitely 65.
Oh yes! Thank you very much, i didn't in fact get it right. The two simultaneous equations were 5x+3y+3=0 and 3x2y+17=0, and I solved them to get 19x=57 hence x=3, then sub that into either on of equations to find y which equals to 4, hence coordinates B are (3,4) (for anyone who wanted to know).
Once again thank you 
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 158
 18052016 18:28
(Original post by Sniperdon227)
See bold writingPost rating:1 
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 18052016 18:29
https://www.youtube.com/watch?v=JGdwNwi4G80 reaction to that paper

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 18052016 18:34
Did the question where you had to complete the square for 84x2x^2 have two parts where you had to complete the square then find the roots or was it one question?
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Updated: May 27, 2016
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