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# AQA C1 Unofficial Markscheme watch

1. (Original post by BearackObama)
I agree with the K=-30 but still unsure how you came to root 65 for the radius?
I'm sure someone has already explained it, but since they gave you centre coordinates, you really don't have to complete the square at all.

I used the coordinates of the centre (5, -3) and coordinates point A (-2, 1) (if i recall the points correctly) to find the distance from C to A, which is essentially the radius.
Formula for distance^2= (x2-x1)^2 + (y2-y1)^2.
Put your x and y values from points A and C into the formula and you end up with
Distance^2= (7)^2 + (4)^2 (here I've left out whether it's minus or not as that makes no surrender whilst squaring it) which gives
Distance^2= 49+16=65
So distance = square root of 65.

Hope that helped.
2. (Original post by df97)
Hey, I can't remember all the questions but here are some of my answers. Let me know if you disagree with any!
1) m=-5/3 B(-3,4)k=-30
2) (3rt5)^2=45 75-32rt5
Can't remember the rest so here are some of my other answers I remembered in random order:
Area under curve=81/4 Area shaded region=45/4
(x-7/4)-41/4 (asked for rational form so use fractions) minimum point=-41/4 graph translated by vector [1/2, 41/4]
d^2y/dx^2=-2x-9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
B(12,-7) m=-4/7 m(tangent)=7/4 7x-4y+18=0
tangent for k (used b^2-4ac=0) k=20 k=4
(x-5)^2+(y+3)^2=65 length of CT=9
y=-32x-40 x-coord Q=-5/4
For graph I had n-shaped parabola intersect y-axis at 8 and x-axis at -1+-rt5
Three linear factors were (x-3)(x-4)^2
For the remainder theorem I think I got 20
I see a lot of similarities i thought i was the only one to get -1+_rt5
3. (Original post by money-for-all)
i got root 34 for radius , woooooo !!!!! i dont believe that it can be anything else.....peasants.....
lol same i got root 34, but no need to call others peasants...you peasant
4. (Original post by vyyzx)
Was the values for k ----> (-3/2<k<6 ) or (-3/2>k or k>6)

*Brackets show the 2 different answers*
**** I forgot to write them as inequalities
5. (Original post by Aw.9570)
There was a surd question at the start of a new question (Q3 maybe) (not the surd one which was question 2 I think)
Which said put the X values for the quadratic in the form m plus or minus root n

I put the quadratic into a completed the square form then rearranged to get X=

I ended up with X = 1 plus or minus root 5

But my friend had -1 and I can't remember which will be right (+/-) as I can't remember the quadratic

What did you guys get
I'm fairly sure it was -1 because the quadratic was like 2(X+1)^2 +10 when you completed the square so X+1= plus or minus root 5 then -1 plus or minus root is the answer
6. (Original post by Aw.9570)
There was a surd question at the start of a new question (Q3 maybe) (not the surd one which was question 2 I think)
Which said put the X values for the quadratic in the form m plus or minus root n

I put the quadratic into a completed the square form then rearranged to get X=

I ended up with X = 1 plus or minus root 5

But my friend had -1 and I can't remember which will be right (+/-) as I can't remember the quadratic

What did you guys get
-1 +- √5
7. (Original post by TheGreatPumpkin)
**** I forgot to write them as inequalities
what question was that?
8. Seem to be getting full marks at the moment . What king of graph did you all sketch? Was the maximum at y=15 ? Thanks
9. (Original post by tamcat)
lol same i got root 34, but no need to call others peasants...you peasant
tamcat ? more like tamrat or tamdog or tamcockroach or tam.... you get the jist
10. (Original post by Sniperdon227)
AQA were being ***** you did not even need R(1,0) for the whole question, yet still stated it in the q?
Well, they were showing that R was the point you were integrating up to.
11. (Original post by 28657)
Seem to be getting full marks at the moment . What king of graph did you all sketch? Was the maximum at y=15 ? Thanks
i got y=8 and my skecth was an upside down quadratic graph like an upside down 'U' is that right ? or nah
12. In case anyone is wondering about that surds question - see the image.

Attachment 533727
Attached Images

13. (Original post by money-for-all)
well done for remembering all of that bruv, well done.
good for you.
go get em' tiger
ROAR !
excellent.
I only put it so people can compare and see if I went wrong anywhere, there's no need to be sarcastic...
14. For the radius question i got root 65.
Geometrical transformation was translation by 1/2 - 41/4
CT - root 81 =9
Integration question got 81/4 & shaded area question was 45/4
I got B(12,-7) somewhere
K = -3/2 & 6
For (3 root 5)^2 i got 45
For this
(3√5)^2+√5
....7+3√5
I got 75+7 root 5 if i remember right
15. (Original post by 28657)
Seem to be getting full marks at the moment . What king of graph did you all sketch? Was the maximum at y=15 ? Thanks
If I'm thinking about the same question as you, you didn't have to work out the max point did you? It just said state clearly the y intercept
16. From what I can remember:

Q1 - find gradient of line AB. The second line is parallel to AB so it has the same gradient.
Circle radius was square root 65
K =30
Integration question was 243/12
Quadratic roots -1plus/minus square root 5
-3/2>k or k>6
17. For the radius question i got root 65.
Geometrical transformation was translation by 1/2 - 41/4
CT - root 81 =9
Integration question got 81/4 & shaded area question was 45/4
I got B(12,-7) somewhere
K = -3/2 & 6
For (3 root 5)^2 i got 45
For this
(3√5)^2+√5
....7+3√5
I got 75+7 root 5 if i remember right
18. (Original post by money-for-all)
i got y=8 and my skecth was an upside down quadratic graph like an upside down 'U' is that right ? or nah
Yep, I got that too.
19. (Original post by timtjtim)
In case anyone is wondering about that surds question - see the image.

Shouldn't it be 75-32rt5?
20. (Original post by timtjtim)
Well, they were showing that R was the point you were integrating up to.
Exactly what i was thinking.

(Original post by df97)
Hey, I can't remember all the questions but here are some of my answers. Let me know if you disagree with any!
1) m=-5/3 B(-3,4)k=-30
2) (3rt5)^2=45 75-32rt5
Can't remember the rest so here are some of my other answers I remembered in random order:
Area under curve=81/4 Area shaded region=45/4
(x-7/4)-41/4 (asked for rational form so use fractions) minimum point=-41/4 graph translated by vector [1/2, 41/4]
d^2y/dx^2=-2x-9x^2 sub in x-coord of p to get 45, 45>0 therefore minimum
B(12,-7) m=-4/7 m(tangent)=7/4 7x-4y+18=0
tangent for k (used b^2-4ac=0) k=20 k=4
(x-5)^2+(y+3)^2=65 length of CT=9
y=-32x-40 x-coord Q=-5/4
For graph I had n-shaped parabola intersect y-axis at 8 and x-axis at -1+-rt5
Three linear factors were (x-3)(x-4)^2
For the remainder theorem I think I got 20
So the translation was [1/2, 41/4] as opposed to [1/2, -41/4]? Please explain, i sort of understand and sort of don't as i can't completely remember which one it said is supposed to map onto the other.

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