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AQA A Level Maths Core 1 - 18th May 2016 [Exam Discussion]

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Unofficial mark scheme please
Original post by medicine71012
Unofficial mark scheme please


http://www.thestudentroom.co.uk/showthread.php?t=4095491
its not complete but it has the most answers
Original post by EwanatTSR
That should be a B. Normally an A would be around the 63-65+ mark so you should have a strong B :smile:.


Im kinda nervous though, like in june 2012 it was 51 for a B but in june 2015 it was 58! Surely the boundaries cant egt any higher its ridiculous, for people like me where it actually trakes so much effort to get good at the exams.
Original post by Fambox
I got the same I hope we are right !!! Mine was in fraction form like 135/12


I got 135/12 as well but after exam I found the simplified form is 45/4, do you think we will lose mark just because our answers are not simplified????
Mark for A predictions? How did everyone find it as a whole
Original post by money-for-all
(X-5)^2 - 25 + (Y+3)^2 -9 = radius

(x-5)^2 + (y+3)^2 -34 = radius

(x-5)^2 + (y+3)^2 = 34

Radius = ROOOOOOOT 34 BRUV


thats wrong because we didnt originally complete the square
you are jumping to the second half of completing the square which is not what we were asked to do

completing the square
1) make sure coefficient of x^2 = 1
2) take out a factor of x from X^2 and bx
3) half the coefficient of bx
4) raise the new B and x to the power of 2
5) AND THEN MINUS THE SQUARE OF THE NEW B

you jumped to number 5 without doing all the before, how is that right
thats like doing this

y=x^3 +5x
dy/dx = 3x^2
you see where its wrong? the 5x is differentiated twice this is the same as what u done, jumped the gun
Original post by money-for-all
(X-5)^2 - 25 + (Y+3)^2 -9 = radius

(x-5)^2 + (y+3)^2 -34 = radius

(x-5)^2 + (y+3)^2 = 34

Radius = ROOOOOOOT 34 BRUV


You only complete the square when you are given x^2 + y^2 + ax + by + c = 0. You don't when you are given the Centre and a point.

You simply put (x-a)^2 + (y-b)^2 = r^2 where Centre = C(a,b) and r = radius.

To calculate R use Pythagoras.
Original post by tajtsracc
Was wondering where you got this from? I too messed up the integral and it's reassuring to know previous errors supposedly don't carry on, could you confirm where you got this from?


Sure, I've attached an image (tangent.png) to make it clearer.

The area of the integral (dark grey and light grey) was 81/4 or 20.25.

But we don't want the light grey section (because we want it bounded by the tangent). So, using the coordinate of Q(-5/4,0) we determine that the base of the triangle is 3/4. The height it 24, and (24 x 0.75) / 2 = 9 (second image)

For my workings, I wrote that 9 = 36/4, and then 81/4 - 36/4 = 45/4 or 11.25
tangent.png50.5KB
Screenshot from 2016-05-18 14-39-16.png6.0KB
Original post by medicine71012
Mark for A predictions? How did everyone find it as a whole


It was okay. Lots of big scawee numbers in the surds question - and I used all but 4 minutes of the time - I normally have 30 minutes left on past papers.

For an A... Hmm, I'm guessing 62, but maybe the same as last year (64).
Reply 289
Avatar for moe889
moe889
7
was the length of ct root 65?
omdays, you guys are giving such mixed answers your scaring the hell outaa meh >;'(((
Original post by timtjtim
Sure, I've attached an image (tangent.png) to make it clearer.

The area of the integral (dark grey and light grey) was 81/4 or 20.25.

But we don't want the light grey section (because we want it bounded by the tangent). So, using the coordinate of Q(-5/4,0) we determine that the base of the triangle is 3/4. The height it 24, and (24 x 0.75) / 2 = 9 (second image)

For my workings, I wrote that 9 = 36/4, and then 81/4 - 36/4 = 45/4 or 11.25


No no no, I was referring to what you said about the marks. The entire question was out of 8, I got the area of the triangle correct but I messed up the integral, you said only 2 marks are lost but shouldn't more be lost?
Original post by timtjtim
you only complete the square when you are given x^2 + y^2 + ax + by + c = 0. You don't when you are given the centre and a point.

You simply put (x-a)^2 + (y-b)^2 = r^2 where centre = c(a,b) and r = radius.

To calculate r use pythagoras.


oh my days tim tim , you are clapped fam.

Why you bringing pythagoras into this ? Thats just the same as saying you use logarithms to integrate...
Original post by Not a Genius :(
omdays, you guys are giving such mixed answers your scaring the hell outaa meh >;'(((


here are the right answers
http://www.thestudentroom.co.uk/showthread.php?t=4095491
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! :biggrin:
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=0tangent 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

For x I got <-3/2 and >6
Also k=20 and 4
Original post by moe889
was the length of ct root 65?


no it was 9
Original post by money-for-all
oh my days tim tim , you are clapped fam.

Why you bringing pythagoras into this ? Thats just the same as saying you use logarithms to integrate...


lol because there is a right angle triangle?

distance of AC = radius
we have co ordinates of A and C
you can work it out from there
Original post by money-for-all
oh my days tim tim , you are clapped fam.

Why are you bringing Pythagoras into this ? That's just the same as saying you use logarithms to integrate...


Take a look at my image I've attached for you bruv. As you can see, you go to the side 7 and you're uppin and downin by 4 - well 4^2 + 7^2 = 16+49 = 65.

I've attached a second image, of desmos, showing that the only circle which intersects A has radius of 65 - waddya think? If you're still not getting the hang of these circles, feel free to drop me a private message bro.
circle.png23.9KB
Screenshot from 2016-05-18 15-43-35.png119.1KB
How did u find the coordinates of B
Original post by timtjtim
Sure, I've attached an image (tangent.png) to make it clearer.

The area of the integral (dark grey and light grey) was 81/4 or 20.25.

But we don't want the light grey section (because we want it bounded by the tangent). So, using the coordinate of Q(-5/4,0) we determine that the base of the triangle is 3/4. The height it 24, and (24 x 0.75) / 2 = 9 (second image)

For my workings, I wrote that 9 = 36/4, and then 81/4 - 36/4 = 45/4 or 11.25

I got 243/12 for first part?

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