AQA maths core 1, 13th May 2015

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1a. -3/5 [2]
1b. 4x-3y+1=0 (I thought it was this form) [3]
1c. A(9,-4) [3?]
2. 7 + √15 [5]
3a. y-6=-10(x+1) [4]
3bi. 108/5 [5]
3bii. 162/5 [3]
4a. (x+1)^2 + (y-3)^2 = 50 [2?]
4bi. C(-1,3) [1]
4bii. 5√2 [2]
4c. k=-8, 2 [2]
4d. minimum distance =7 [2]
5a. p = 3/2, q = -¼ [3]
5bi. (-3/2, -¼) [2]
5bii. x=-3/2 [1]
5c. y = x^2 - x + 4 [3?]
6ai. h = 24/r -r/2 [3]
6aii. show V = 24πr -π/2 r^3 [3]
6bi. 24π -3π/2 r^2 [2]
6bii. =-12π therefore max (r=+4) [4]
7a. draw x^2(x-2) [3]
7bi. R = 36 [2]
7bii. R= 0 therefore root [2]
7biii.(x-2)(x^2-5x+10) [2]
7biv. x=2 [3]
8a. Show that x^2 + 3(k-2)x -13-k=0 [1]
8b. 9k^2 -32k -16 < 0 [3]
8c. -4/9 < k <4 [4]

Edit: added marks, not sure about a few of them.

if this is right then ive probably only lost around 3-4 marks :smile:

Reply 441

Lobster37

Original post by SaadMuhammad

I think my paper did not go as well as I was expecting. I have definitely lost 2 marks on the question where you had to work out the shortest distance between the centre and QR. By the looks of other answers on TSR, I think I have lost the last question, which is out of 4 I think. I believe I may have made a mistake on the integration question - though I am unsure about this; hopefully full marks on this question. So in total, I definitely lost 6 and maybe another 4-5 just to be on the safe side.

In the end, a 60/75 is still 80% - an A I hope! I will work harder for C2 and Decision and should get higher marks to compensate for this paper.

Good luck for your other exams!.

You too 😄

Reply 442

-jordan-

Original post by FS492

Its not completing the square though? [+] value is taken away from x and +[y] is added to integer?

You have to complete the square to find the new equation. Completing the square gives you the original vertex and then you can slot in the numbers for the translation on the vertex and multiply out.

You can't simply add 4 to the y-intercept of 2 because the x-coordinate moves too and if you imagine the quadratic on a graph and moving it along -2 in the x-direction , this will change where it crosses the y-axis. And then you apply the translation in the y-direction and it changes again. So the y-intercept is the new position when translated by x + the translation on y

(edited 8 years ago)

Reply 443

apache324

Original post by Lobster37

-2***

Yes, X=-2

Reply 444

RhyaCeri

Original post by AbbiC

Can someone do an unofficial mark scheme for this? :smile:

http://www.thestudentroom.co.uk/showthread.php?t=3322639

Reply 445

pottermaniac1998

Yyaaaaaay i got that too!!!

Reply 446

tanyapotter

Original post by RhyaCeri

if this is right then ive probably only lost around 3-4 marks :smile:

in the first one, shouldn't it be 5x, not 4x? or can i not remember properly?

Reply 447

kas69

Easy paper if u did most/all the other past papers! Hard questions were the minimum length, cylinder question and finding the equation after vector 4,2...
If u want to no any of the methods msg me...

Reply 448

Urvs

I got x^2 + 3x + 6 when I applied the vector (minus 2 and add 4)

(x+1.5)^2 - 0.25

(x+ (1.5-2))^2 - (0.25+4)

when you multiply out its x^2 + 3x + 6

Reply 449

SunDun111

Original post by student0042

Was gradient -10 how it was -7?

Reply 450

student0042

Original post by RhyaCeri

if this is right then ive probably only lost around 3-4 marks :smile:

Even though I've put 162/5 for 6, in the exam I put the wrong answer so I think I've lost around 3-4 marks too. This should still be 95+ UMS however. (hopefully.)

Reply 451

RhyaCeri

Original post by tanyapotter

in the first one, shouldn't it be 5x, not 4x? or can i not remember properly?

its 5x? i think, im not sure, pretty sure im gonna have nightmares about that cylinder question tonight though

Reply 452

RhyaCeri

Original post by student0042

Even though I've put 162/5 for 6, in the exam I put the wrong answer so I think I've lost around 3-4 marks too. This should still be 95+ UMS however. (hopefully.)