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AQA maths core 1, 13th May 2015

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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...

For 3a did you get gradient as -7 or -10

Reply 461

JackC3001

Just thought id correct my previous answer - you are correct that it is (x+2) when x=-2. I had to try and remember my answers...

Reply 462

vj_5005

What do you lots think the grade boundaries will be for an A?

Reply 463

apache324

Original post by Heffalump .

x^2-x+4

Oh yeah, I got that

Reply 464

tanyapotter

Original post by student0042

(y+3) = 5/3(x+2)
3y + 9 = 5x + 10
5x - 3y + 1 = 0

Yes, working through that you are right, I will change my answer. Thanks.

got a bit scared then! thank you so much, i think i am now certain of my mark (or at least a confidence interval within which it definitely lies). will make results day less of a shock.

also, in the integration question, i must have added something up wrong because i got 107/5 for the first part, not 108/5. this means that the next part of the question (54-107/5= 163/5) was also wrong, as it was carried forward. how many marks would i lose in total? do i get penalised twice for that one mistake, or can i still the full 3 marks for the second part of calculating the area, even though it is wrong? thank you

Reply 465

RhyaCeri

Original post by SunDun111

For 3a did you get gradient as -7 or -10

-10 i think

Reply 466

ProNoob

Original post by Heffalump .

x^2-x+4

Got that!!!
Lets hope it is correct

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Reply 467

JackC3001

Original post by Heffalump .

I put
-5x + 3y -1 =0
It's the same thing, but did the question specifically ask for it to be in the form ax - by + r = 0 ?

That should also be correct. I actually wrote it in both formats to be on the safe side, though it didn't ask for it to be written in any specific order format so long as all of the parts are on one side of the equation.

Reply 468

ProNoob

Original post by SunDun111

For 3a did you get gradient as -7 or -10

I got -10 as the gradient

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Reply 469

SunDun111

Original post by ProNoob

I got -10 as the gradient

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So did I but my teacher got -7? How what was the dy/dx equation again

Reply 470

student0042

Original post by tanyapotter

I made a mistake in the first bit too, so maybe -1 or -2 marks there and in the second bit I think -1 mark as you didn't get the final correct answer. (I presume one for 54 is area of trapezium, 1 for follow through? and one for correct answer.)

Reply 471

omarharoun3

Question 1

(a) -3/5
(b)(i)perp. gradient = 5/3
in the form mx + ny + p = 0
5x - 3y + 1 = 0

Question 2

7+root15
n=7 (Had to be stated separately)

Question 3

a) y-6 = -10(x+1)
y = -10x - 4

bi) 108/5 or 21.6
bii) 162/5 or 32.4

Question 4

a) (x+1)^2 + (y-3)^2 = 50
bi) C is (-1, 3)
bii) r = √50 = 5√2
c) k= 8, -2
d) min. distance = √49 = 7

Question 5

a) p = 3/2, q = -¼
bi) (-3/2, -1/4)
bii) x=-3/2
c) y = x^2 - x + 4

Question 6

ai) Surface Area = 2πrh
48π = 2πrh - 2πr^2
h = 24/r - r/2

aii) V = πr^2h
V = πr^2 * (24/r - r/2)
V = 24πr - (π/2 * r^3)

bi) 24π - (3πr^2) / 2
bii) 24π - (3πr^2) / 2 = 0 when r = ±4
r = 4 as positive value
d^2V/dr^2 = 3πr
when r = 4, second derivative = -12π, therefore maximum

Question 7

a) draw x^2(x-3) where line touches at 0 and crosses at 3
bi) Remainder = 36
bii) R = 0 therefore is a factor
biii) (x-2)(x^2-5x+10)
biv) b^2 - 4ac < 0 therefore no real roots for (x^2-5x+10)
(x-2) only root when x = 2

Question 8

a) show that = x^2 + 3(k-2)x -13-k=0 by setting two equations equal to each other
b) b^2 - 4ac < 0 therefore 9k^2 -32k -16 < 0
c) -4/9 < k <4

Reply 472

ProNoob

Original post by SunDun111

So did I but my teacher got -7? How what was the dy/dx equation again

I have forgotten sorry :frown:

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Reply 473

-jordan-

Original post by SunDun111

So did I but my teacher got -7? How what was the dy/dx equation again

Remember my dy/dx being 4x^3 + 6x so substituting in -1 gives you -10. Your teacher is wrong judging by the fact everyone else got -10.

Reply 474

Sekaib

Original post by Lobster37

my exact words were 'x=2 is only real root' would i get the mark for this. Also are there any oher statements you can think of that we should have written?

Yeah I think you will get a mark for that. I think you get marks for saying dy/dx =0 when talking about a stationary point
And d2y/dx2 is less than zero so it was a maximum point

Reply 475

student0042

Original post by JackC3001

I always assume that when it wants it in the form ax + ... = 0, that a should be positive. Someone did say earlier that in the mark scheme it says condone -ax + ... =0.

Reply 476

apache324

Original post by SunDun111

For 3a did you get gradient as -7 or -10

-10 i think.

Reply 477

JackC3001

Original post by student0042

I always assume that when it wants it in the form ax + ... = 0, that a should be positive. Someone did say earlier that in the mark scheme it says condone -ax + ... =0.

It definitely didn't ask for a certain format other than grouping the terms, so judging by other mark schemes, they will allow either. :smile:

Reply 478

aoxa

Original post by SunDun111

So did I but my teacher got -7? How what was the dy/dx equation again

You got -10 by using the dy/dx method.

If you used the method of y-y/x-x to get the gradient, you got -7 or something. I don't think you were supposed to se that method though.

Reply 479

SunDun111

Original post by aoxa

You got -10 by using the dy/dx method.

If you used the method of y-y/x-x to get the gradient, you got -7 or something. I don't think you were supposed to se that method though.