Year 12s: what will be the first way you research your future options? >>

AQA AS Core 2 maths unofficial MARK SCHEME (25/05/16)

EDIT: This is for the 2016 exam. You can find the unofficial markscheme for 2017 HERE.

Given that the core 2 exam is tomorrow, i wanted to make this thread so that by the time people are hopefully reading this, we can all help to build the core 2 unofficial mark scheme.

With the help of others i was able to create a more 'official' unofficial mark scheme for core 1 which you can view at the bottom of this post but as soon as core 2 has finished, its time to get the next mark scheme up!

If you can remember specific questions with their marks and answers post them below and i can eventually add it together in a more official document. Hopefully core 2 has gone well for people! If not, the grade boundaries will most likely be lower :smile:

(edited 7 years ago)

AQA Core 1 Unofficial Mark Scheme 2016.pdf622.6KB

Scroll to see replies

Reply 1

timothygrealish

I have contracted AIDS from that paper

Reply 2

rainbowtwist

The exam was ridiculous 😂
But for the mark scheme - Q1 (a) = -32x^-1 + ax^2 /2
and (b) a=-2

Reply 3

Rager6amer

2) a) sketch y = (0.2)^x [2] (negative gradient where the y value decreases exponentially as x increases, with y-intercept at 1 and x-axis being the curve's asymptote)

b) something about solving a log question e.g. (0.2)^x = 8 ? so log that to get x and provide answer to 3sf [3?]

c) transformation that maps the graph y =(0.2)^x onto y = (5)^x [1] answer: y = (1/5)^x there fore replace x with -x to get y = (5)^x so therefore transformation is reflection in y axis.

Reply 4

plower

Pretty sure it was stretch in y direction?? You know what you got for coefficient of x^10??

Reply 5

voltz

I got -1648 for the binomial question.

x = 5/2 for the last question.

a = -2 as the first answer (checked using integration)

Coorindates of M = (9,6)

k = 5 (guessed this one)

For the trig question I got 308 or something as one of them - all I remember.

For the trig with radians I got 1.15 but im unsure abour this.

Let me know if you agree/disagree with any :smile:

Reply 6

AdamSd

I think b) was (0.2)^x = 4. Use logs to find out x to 3 sign fig

Reply 7

Rager6amer

Original post by plower

Pretty sure it was stretch in y direction?? You know what you got for coefficient of x^10??

Yeah i put that but when speaking to others they said it was a reflection in y axis which makes sense :smile:

9a. 3^m/2 - 6n

I got x = 0.8 for the last question...? and when i put 0.8 back into the log thing they both equalled the same thing so I assumed it was right..? seemed too easy for 4 marks though so i musta got it wrong

Reply 10

Rager6amer

To recap / jog people's memories:
1) Indefinite and definite integration
2) exponential graph
3) curve with max point M
4) Arithmetic series
5) Trapezium rule and 2 graph transformations
6) Binomial expansion
7) Triangles and sectors
8) Trig equations
9) Logs.

(edited 7 years ago)

Reply 11

gnolhs

Stretch Y direction S.F 5 ?

Reply 12

mattdeeee1

Original post by Rager6amer

To recap / jog people's memories:
1) Indefinite and definite integration
2) exponential graph
3) curve with max point M
4) Arithmetic series
5) Binomial expansion
6) Triangles and sectors
7) ??
8) Trig equations
9) Logs.

Think 7 was binomial expansion.

Reply 13

Rager6amer

Original post by mattdeeee1

Think 7 was binomial expansion.

Hmm not sure but then what was 5?

Reply 14

plower

I got -192 for binomial ... Craaap

Reply 15

BanterBus

Binomial expansion:
Part a) (1-2x)^5 = -32x^5+80x^4-80x^3+40x^2-10x+1, and in ascending powers of x: 1-10x+40x^2-80x^3+80x^4-32x^5
Part b) Find (2+x)^7. Find the sum of the coefficients of (ax^5)(ax^5)+(ax^4)(ax^6)+(ax^3)(ax^7); which should give (-32x84)+(14x80)+(-80x1) = -1648

Reply 16

Rager6amer

Original post by plower

I got -192 for binomial ... Craaap

you will most likely get method marks if you wrote out the steps properly. It was like 5 marks so 2 or 3 is still possible :smile:

Reply 17

Thousand_Bird

Original post by mattdeeee1

9a. 3^m/2 - 6n

I got a slightly different answer:

9)a) log(27) d = n log(3)c = m

d = 27^n (3^3n), c = 3^m

c^1/2 / d^2 --> (3^m) ^1/2 / (27^n)^2

= 3^m/2 / 729^2n

=3^m/2 / (3^6)^2n

=3^m/2 - 12n

Reply 18

mattdeeee1

Original post by Thousand_Bird

I got a slightly different answer:

9)a) log(27) d = n log(3)c = m

d = 27^n (3^3n), c = 3^m

c^1/2 / d^2 --> (3^m) ^1/2 / (27^n)^2

= 3^m/2 / 729^2n

=3^m/2 / (3^6)^2n

=3^m/2 - 12n

ahhh so you squared out the whole bracket here : c^1/2 / d^2 --> (3^m) ^1/2 / (27^n)^2

I used rules of indices and presumed that you could change that to 27^2n