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Further Maths GCSE AQA Paper 1 (Unofficial mark scheme)

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What was that cubic graph question where we had to find k? I got 9?
Original post by Chittesh14
We both ****ed up lol.



That's weird though, what I did and probably NiamhM too:


(3 - sqrt(x))1/3 = -2

(31/3 - x 1/6) = -2
9 - x = 64
x = -55



Nah I got 45 for that, and -55 for this - I'm wrong, you're right. I'm dead.


Oh don't worry! Remember this is not even an entirely legitimate qualification for people our age; it is a certificate, not a GCSE. You'll be fine!

In your method, you can not apply the power of a third to each individual term in your second step. (3 - sqrt(x))1/3 = -2 is equal to cube root(3 - sqrt(x)) = -2, not cube root(3) - cube root( x1/2) = -2
Original post by Wolfram Alpha
What was that cubic graph question where we had to find k? I got 9?


Correct, good job; most people struggled with that!
Original post by Redcoats
Oh don't worry! Remember this is not even an entirely legitimate qualification for people our age; it is a certificate, not a GCSE. You'll be fine!

In your method, you can not apply the power of a third to each individual term in your second step. (3 - sqrt(x))1/3 = -2 is equal to cube root(3 - sqrt(x)) = -2, not cube root(3) - cube root( x1/2) = -2


Oh right, I get it lol.
So, its the whole thing in the bracket that the power is applied to - ffs I knew that as well.
Oh well - learn from mistakes.
I know it's a certificate, but I'm so lazy and I never revise, I want to change and I'm going nuts because I want to now onwards destroy everything, especially A-level. So, getting an A*^ would be a great start, don't worry - that's what the second paper is made for - comebacks.
Original post by Redcoats
I'm not too sure about your method because:

First, x cannot equal a negative number as you're square rooting it.

Secondly, when you squared both sides of the equation in 3 - sqrt(x) = -8, you square rooted incorrectly:

Square rooting both sides gives: (3-sqrt(x))2 = (-8)2

This is : (9 - 6sqrt(x) +x) = 64 as it is a double bracket, you can't just square each term as you have done.

Really sorry to tell you this! I'm sure you've still got your A^ - remember the boundaries for such are really low


Oh god whoops haha

I honestly should've known, of course you can't square root a negative number - how could I not have realised!!

Thank you. I'm just hoping now that I'll be able to pull it back in paper 2 (my stronger paper) - do you reckon if I manage to get around 95 in that I should get an A^ still? Bearing in mind I think I got anywhere between 50 and 60 in this paper?
Original post by Wolfram Alpha
What was that cubic graph question where we had to find k? I got 9?


Oh yes I got this too!! Took me bloody ages as I kept forgetting the - sign in y=-3x+k when x=3, so I kept getting -9 haha
Original post by NiamhM1801
Oh god whoops haha

I honestly should've known, of course you can't square root a negative number - how could I not have realised!!

Thank you. I'm just hoping now that I'll be able to pull it back in paper 2 (my stronger paper) - do you reckon if I manage to get around 95 in that I should get an A^ still? Bearing in mind I think I got anywhere between 50 and 60 in this paper?


Of course! The highest the grade boundary has ever been is 149 / 175 so if you get 54 in this paper, and 95 in the next, you've got an A^ - and, if the next paper is like the paper today, we're looking at a boundary lower than 149.
What was the question for m = 8, and p = -1 answer?
Original post by Redcoats
Correct, good job; most people struggled with that!


That's funny considering how little I know about further maths. We hardly did any in class and I'll only have one full day to revise for paper two since I have other exams that are top priority.
Original post by Ishan_2000
What was the question for m = 8, and p = -1 answer?


I remember it so vividly as I spent 25 minutes on it only to guess 12 and 2. Lmao

It was:
mx - 4 - 2(x + p) =(triple) 6(x + 1)
Find m and p

(Actually it might have been + 4 not -)
predicted grade boundaries anyone?

I think I got 50 from the unofficial mark scheme.
Original post by Redcoats
Of course! The highest the grade boundary has ever been is 149 / 175 so if you get 54 in this paper, and 95 in the next, you've got an A^ - and, if the next paper is like the paper today, we're looking at a boundary lower than 149.


The highest?? Really? That has GOT to have been 2014 - absolute dream of a paper! Today's was much harder than that. I'd say it was on par with, or maybe slightly easier than last year's. Although I've done worse due to silly mistakes.

Thank you!! :biggrin:
Original post by NiamhM1801
I remember it so vividly as I spent 25 minutes on it only to guess 12 and 2. Lmao

It was:
mx + 4 - 2(x + p) =(triple) 6(x + 1)
Find m and p

(Actually it might have been + 4 not -)


You have to expand brackets, collect like terms and compare coefficients (it was +4) :

mx + 4 - 2x - p = 6x + 6

(m - 2)x + (4 - p) = 6x + 6

So, if we compare coefficients: m - 2 = 6 and 4 - p = 6

Therefore, m = 8 and p = -2 or something like that
(edited 7 years ago)
Original post by NiamhM1801
I remember it so vividly as I spent 25 minutes on it only to guess 12 and 2. Lmao

It was:
mx - 4 - 2(x + p) =(triple) 6(x + 1)
Find m and p

(Actually it might have been + 4 not -)


Thank you!

I have more confidence now in my paper :wink:
Original post by Chittesh14
Unofficial Markscheme:

Correct me if I'm wrong. The gaps are for answers I don't remember or questions.

3x^2 - 20x
a = 3, b = -20
12th term
3/5 limiting factor

-5, 8 - basically it was opposite the numbers in the brackets
Root 10 - radius

m = 8, p = -1
8 < x < 12 (I put 9, 10, 11 - gonna assume it's correct)

x = 121
x^3 - 15x^2 + 75x - 125
-80, x = 16 and y = -1/5

h = 3, k = -7
-3, -7
-3 +/- root 7
Circle theorem:
180 - 3x
2x + 58 = 2(180 - 3x)
2x + 48 = 360 - 6x
8x = 312
x = 39
Another question: x = 45
When 3 is put into equation and a is 1, the equation is equal to 0.
(x-3)(x-7)(x+2) - use long division first to get quadratic then factorise it

2root7 - 4
5/6 for the cos and sine question?

k = 9

Use pythagoras theorem, 9 + 9 = 18, side = root 18 = 3 root 2 - did it a slightly different way
Trapezium - Idk, but I got the answer somehow: the bottom side of the triangle was root 3 and the hypotenuse was 2root3.
Add it all together: 3 + 3 + 3 + 2 root 3 + root 3 = 9 + 3 root 3, thats what u get when u expanded what they gave so it's correct.
x^4 - 81 (I think it was this) = (x^2 + 9)(x+3)(x-3)
Cosine rule.
Cos P = 1/3.
Use the rule, you find u have to do 4/12 to get 1/3.
So, you do 13n^2 - 4n^2 = 9n^2
That's w^2 so w = 3n.
w = 3n same as the other side, so the triangle is isoceles.


What I got right - well happy with this :biggrin:
Original post by Redcoats
You have to expand brackets, collect like terms and compare coefficients (it was +4) :

mx + 4 - 2x - p = 6x + 6

(m - 2)x + (4 - p) = 6x + 6

So, if we compare coefficients: m - 2 = 6 and 4 - p = 6

Therefore, m = 8 and p = -2 or something like that


Wow I went about that in COMPLETELY the wrong way 😂😂
Although I did expand, so if the examiner's feeling nice they might give me one mark of 4.
Still, it was only one question so I'm not gonna linger on it too long.

Thanks for the explanation though as I was well and truly clueless! (It wasn't even meant to be a hard one :redface:)
Original post by NiamhM1801
What I got right - well happy with this :biggrin:


Don't worry, the correct answer is 9, 10 and 11 lol.
Original post by Chittesh14
Don't worry, the correct answer is 9, 10 and 11 lol.


Ah thanks, don't you worry either - it took me a solid 5 minutes of trying to find factors of 96 that added to -20 before I even realised you had to complete the square!
Original post by Chittesh14
Don't worry, the correct answer is 9, 10 and 11 lol.


I almost put the inequality 8 < x < 12 but then God showed me mercy
Original post by NiamhM1801
Ah thanks, don't you worry either - it took me a solid 5 minutes of trying to find factors of 96 that added to -20 before I even realised you had to complete the square!


Did you?
I just factorised it lol... (x-8)(x-12)

Original post by Redcoats
I almost put the inequality 8 < x < 12 but then God showed me mercy


Lol, I'm a complete imbecile. I do this for every GCSE - just my lazy personality, I finish or don't finish. I never check the paper.

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