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2017 AQA Level 2 Further Maths Paper 2 Unofficial Mark Scheme

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Reply 100
Avatar for HollyCo
HollyCo
Does anyone know the working for the the last question?
Original post by etothepiiplusone
Come on Steven, these are basic exponent laws!

Please can you explain how it's wrong? Sorry i don't get it😂
Original post by Stevenscheese
Please can you explain how it's wrong? Sorry i don't get it😂


If m = 2
2^m = 4
So (2^m)^3 = 64
64 / 8 = 8
8 = 2^3
Sorry I can't see what I did wrong!
Original post by Stevenscheese
Please can you explain how it's wrong? Sorry i don't get it😂


Ok

(2m)3 (2^{m})^3
=(2m)(2m)(2m) = (2^{m})(2^{m})(2^{m})
=2m+m+m = 2^{m+m+m}
=23m = 2^{3m}

To use an alternative exponent law. You should know (ab)c=abc (a^b)^c = a^{bc}
Original post by jamal.academy
lol, i made the same exact mistake, how much marks would you lose?


As unfortunate as it is, it's good to know others made the same mistake. The boundaries will probably be lower as it was a hard paper.

It was out of 4 marks I think:

So 1 mark by substituting in x and y into Pythagoras
1 mark for simplifying Pythagoras theorem
1 mark for finding x in terms of y by square rooting both sides
1 mark for finding the area

We probably got at least 2 or 3 marks. Did you get 16/5? That's what I got.
Original post by etothepiiplusone
Ok

(2m)3 (2^{m})^3
=(2m)(2m)(2m) = (2^{m})(2^{m})(2^{m})
=2m+m+m = 2^{m+m+m}
=23m = 2^{3m}

To use an alternative exponent law. You should know (ab)c=abc (a^b)^c = a^{bc}


Oops... thank you!
Original post by HollyCo
Does anyone know the working for the the last question?


It's really pot luck or lots of bashing to factor out the three.

12x2−60x+5 12x^2 - 60x + 5

3(4x2−20x+53) 3(4x^2 - 20x + \frac{5}{3})

Now we complete the square on the inside as (2x−5)2−25=4x2−20x (2x-5)^2 - 25 = 4x^2 - 20x

So 3((2x−5)2−25+53) 3((2x-5)^2 - 25 + \frac{5}{3})


3((2x−5)2−703) 3((2x-5)^2 - \frac{70}{3})

3(2x−5)2−70 3(2x-5)^2 - 70

Done!
Original post by Stevenscheese
Surely
(2^m)^3
= 8^3m not 2^3m?
Correct me if I'm wrong!


I nearly did this mate hahahaha.

Where i went wrong was thinking of it as (2x^m)^3
This would have then been 8x^3m
This might have been what you thought like me.
Fortuantely i was able to spot it XD

Hope it helps
Reply 108
Avatar for Boil
Boil
2
Original post by Npt512
im pretty sure q8 was 106.1 (since angle = 106.06...)


yes exactly
Original post by Boil
yes exactly


I originally put 106.7. I then edited.
Reply 110
Avatar for Boil
Boil
2
Original post by etothepiiplusone
I originally put 106.7. I then edited.


ok thanks, could u remind me what question 9: x=-2 was again? and question 10 please
Original post by Boil
ok thanks, could u remind me what question 9: x=-2 was again? and question 10 please


I'd love to but I can't remember myself. Anyone?
Original post by Loci Pi
I think it will be around 137-141. Most of the time with the past papers, I knew what to do immediately. In today's paper though, I really had to think about what to do, especially for the last question and the trig one where you had to find k. I thought Thursday's paper was alright though. I guess that was AQA's way of tricking people into thinking today's paper was going to be good.


That's exactly what I found too. This paper required so much more thinking about actually what you had to do to get the answer, and I didn't really get that on Thursday or on any other past paper. I do hope the A^ grade boundary is low though, I messed up some easy questions and made silly mistakes throughout!
Original post by Tomazza123
That's exactly what I found too. This paper required so much more thinking about actually what you had to do to get the answer, and I didn't really get that on Thursday or on any other past paper. I do hope the A^ grade boundary is low though, I messed up some easy questions and made silly mistakes throughout!


Yeah, I messed up some easy questions too. For the find the area of the rectangle question I found the area of the triangle instead. I made a stupid mistake on the circle theorem question. I made a substitution error on the quadratic simultaneous equation. I got stuck on the 'find the value of cos in terms of k' and guessed incorrectly. Hopefully I still get method marks lol. Really hope I scrape enough marks for A^.
Original post by Loci Pi
Yeah, I messed up some easy questions too. For the find the area of the rectangle question I found the area of the triangle instead. I made a stupid mistake on the circle theorem question. I made a substitution error on the quadratic simultaneous equation. I got stuck on the 'find the value of cos in terms of k' and guessed incorrectly. Hopefully I still get method marks lol. Really hope I scrape enough marks for A^.


Wait what was the find the area of the rectangle question
Original post by gcse0
I got 19.2 units squared for the area of the triangle - is that right?


I also got 19.2
Original post by gunnafail123
but it was asking for the difference from 20th term to 8th term not 8th to 20th


But I am pretty sure it said between rather than from...to
Original post by etothepiiplusone
I'd love to but I can't remember myself. Anyone?


Original post by Boil
ok thanks, could u remind me what question 9: x=-2 was again? and question 10 please


it was the weird question where you were given two inequalities and you had to show that there was only one solution for both of them
the inequalities were -11<5x<=5 and the other simplified inequality was 2x+3<0
Reply 118
Avatar for gcse0
gcse0
10
Original post by CraigBackner
it was the weird question where you were given two inequalities and you had to show that there was only one solution for both of them
the inequalities were -11<5x<=5 and the other simplified inequality was 2x+3<0


Dammit, I misread the question since I thought they asked for 1 solution that cannot be in both inequalities.
-11/5 < x <= 1
x < -3/2

How many method marks will I get since I got to the end and wrote the wrong answer?
Original post by CraigBackner
it was the weird question where you were given two inequalities and you had to show that there was only one solution for both of them
the inequalities were -11<5x<=5 and the other simplified inequality was 2x+3<0


Thanks. @Boil if you plot the inequalities on a number line (yeah I know) it's clear to see -2 is the only answer: https://www.desmos.com/calculator/um4a23ffwt

Otherwise it is clear that {−2,−1,0,1}∩{−2,−3,−4,...}={−2} \{-2, -1, 0, 1\} \cap \{-2, -3, -4, ...\} = \{-2\} , the first two sets being the integer solutions to the two inequalities
(edited 6 years ago)

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