2017 AQA Level 2 Further Maths Paper 2 Unofficial Mark Scheme

This poll is closed

How many marks do you think you got?

0-30 3%

31-50 6%

51-60 9%

61-70 9%

71-80 14%

81-90 16%

91-100 30%

101-10512%

Total votes: 180

etothepiiplusone

15

1a.

30

(and

-30

is probably allowed) Marks=3?
b.

-1.5

Marks=?

2a.

\begin{pmatrix} -13 & 2 \\ 6 & 1 \end{pmatrix}

Marks=?
b.

\frac{11}{8}

Marks=?
c. Because the number of columns of B does not equal the number of rows of A Marks=1

3a. Since

1365 = 3*5*7*13

, choose any two factors, multiply those together, and write the other two factors: e.g

3, 5, 91

Marks=?
b. Factorised,

ab-11b = b(a-11)

so

b

is an arbitrary square e.g

4

, and

a=36

Marks=?

4.

2744

Marks=?

5.

\frac{4}{5}

Marks=?

6.

12.31

Marks=?

7.

16

Marks=?

8.

106.1

Marks=?

9.

x=-2

Marks=? www.desmos.com/calculator/um4a23ffwt .
Otherwise it is clear that

\{-2, -1, 0, 1\} \cap \{-2, -3, -4, ...\} = \{-2\}

, the first two sets being the integer solutions to the two inequalities

10.

15.2

Marks=?

11a. Draw the thing. Marks=?
11b.

-5 <= f(x) <= 7

Marks=?

12a.

3(5-x)(5+x)

Marks=?
b.

12n

Marks=?

13.

\frac{50a^2}{3}

Marks=5?

14.

a=4

,

b=10

Marks=?

15.

y=\frac{8x}{w+1}

Marks=3?

16a.

3^{-2b}

Marks=1
b.

5^{x+2}

Marks=1
c.

2^{3m}

Marks=1

17a.

(4, -25)

Marks=?
c. The roots are

x=5.79

and

x=1.21

Marks=?
18.

\frac{32y^2}{5}

Marks=?

19a.

k

. Marks=1 Think of 19a. and 19b. as transformations: https://www.desmos.com/calculator/4kphkki8vl
b.

-k

Marks=1
c.

\sqrt{1-k^2}

. This is rearranging of

\sin^2 x + \cos^2 x = 1

with

\sin^2 x = k^2

Marks=2

20a. Angle at circumference is half angle at centre Marks=1
b. e.g Angle in triangle

90-x

Angle on straight line

90+x

Opposite angle in cyclic quadrilateral

90-x

Other base angle

90-x

Other angle in triangle

180-2(90-x) = 2x, QED

Marks=5?
There are obviously many other ways to do this. I think we all know if we got or did not get marks on this question

21a.

(0, 8)

Marks=?
b.

-2x+2

Marks=?
c. Normal equation

y= \frac{-1}{2}x + 8

So intersects x-axis at

(16, 0)

The rest is easy (show one length is twice the other) Marks=?

22.

(-6/5, -7/5)

and

(2, 5)

were the pairs of solutions Marks=?

23.

-1

is the constant, because

\frac{1}{\tan^2 x} - \frac{1}{\sin^2 x}

= \frac{1}{(\frac{\sin x}{\cos x})^2} - \frac{1}{\sin^2 x}

= \frac{1}{\frac{\sin^2 x}{\cos^2 x}} - \frac{1}{\sin^2 x}

= \frac{\cos^2 x}{\sin^2 x} - \frac{1}{\sin^2 x}

= \frac{\cos^2 x - 1}{\sin^2 x}

= \frac{- (-\cos^2 x + 1)}{\sin^2 x}

= \frac{- \sin^2 x}{\sin^2 x}

= -1

Marks=4?

24.

3(2x-5)^2 - 70

Marks=5

Not like anyone will ever read this with the official ms released, but I was one of 9 students in the country to get 100% marks in this GCSE.

(edited 6 years ago)

Reply 1

darude_sand

2

Original post by etothepiiplusone

1a. 30
b. -1.5
2a. 13 -2
6 1
b. 11/8
c. Because number of columns of B does not equal rows of A
3a. e.g 3, 5, 91
b. Factorised ab-11b = b(a-11) so b is arbitrary square e.g 4, a=36
4. 2744
5. 4/5
6. 12.31
7. 16
8. 106.7
9. x=-2
10. 15.2
11b. -5 <= f(x) <= 7
12a. 3(5-x)(5+x)
b. 12n
13. 50a^2 / 3
14. a=4, b=10
15. y=(8x)/(w+1)
16a.3^(-2b)
b. 5^(x+2)
c. 2^(3m)
17a. (4, -25)
c. 5.79 and 1.21
18. (32/5) y^2
19a.k
b. -k
c. sqrt(1-k^2)
20a. Angle at circumference is half angle at centre
b. e.g Angle in triangle 90-x
Angle on straight line 90+x
Opposite angle in cyclic quadrilateral 90-x
Other base angle 90-x
Other angle in triangle 180-2(90-x) = 2x QED
21a. (0, 8)
b. -2x+2
c. Normal equation y=-1/2 x + 8
So intersects at (16, 0)
2*6 = 16-4
22. (-6/5, -7/5) and (2, 5)
23. -1
24. 3(2x-5)^2 - 70

"Because number of columns of B does not equal rows of A"... Wut?

Reply 2

Frankiitch

6

How did you do the cos a question that was worth 2 marks

Reply 3

Shanpatel2000

3

How much do you think 148 would be

Original post by Shanpatel2000

How much do you think 148 would be

Probably an A^, but at least an A*.

Reply 5

DarkDay65

Original post by Frankiitch

How did you do the cos a question that was worth 2 marks

Square it so cos^2(a)= 1-sin^2(a)
Sina=k so Sin^2(a)=k^2
Thus cos^2(a)=1-k^2
Square Root it so cosa= root(1-k^2)

Reply 6

B0redBrioche

12

Because number of columns of B does not equal rows of A

^ For this question I put there are fewer columns in B than A. Mark or no mark?

Reply 7

LUSer21

3

Q1 difference between 20th and 8th terms = 20th term - 8th term = -30?

Reply 8

Shanpatel2000

3

Original post by _gcx

Probably an A^, but at least an A*.

Hoping for A^; thanks :smile:

Reply 9

Mermaidqueen

14

Original post by LUSer21

Q1 difference between 20th and 8th terms = 20th term - 8th term = -30?

Yeah but the values of the 20th term nd the 8th term were both negative numbers so you don't need to write -30 as a difference, just 30 is correct

Reply 10

gcse0

10

Can someone describe question 9 and 10 for me since I forgot?

Reply 11

YellowYellow2

16

I got around 87 in this and 55 in the other.... is this an A*

Reply 12

LUSer21

3

Original post by Mermaidqueen

Yeah but the values of the 20th term nd the 8th term were both negative numbers so you don't need to write -30 as a difference, just 30 is correct

Will I still get the mark if I write -30?

Reply 13

Oliver_2001

8

What was the answer for the question where you had to work out the area of OAB i got 24 but I don't see it in the ms

Reply 14

Mermaidqueen

14

Original post by LUSer21

Will I still get the mark if I write -30?

Im not sure, i don't remember how many marks it was but if it was out of 2 marks
you'd probably get one mark for calculating -30 and the second for writing 30

Reply 15

tom_6

3

Original post by Oliver_2001

What was the answer for the question where you had to work out the area of OAB i got 24 but I don't see it in the ms

It was the area of OPB (which was 16) not OAB

Reply 16

Mermaidqueen

14

Original post by Oliver_2001

What was the answer for the question where you had to work out the area of OAB i got 24 but I don't see it in the ms

What did u do for it? I don't remember my answer but what i did was:
calculate the coordinates of A and B, then found the distance AB
then i did 2/5 multiple by distance AB bc the ratio was 2;3 and the length we needed for the little triangle was the 2 part.
Anyways so i did 2/5 (distance AB) multiplied by 8 then divide by 2

Reply 17

LUSer21

3

Original post by Mermaidqueen

Im not sure, i don't remember how many marks it was but if it was out of 2 marks
you'd probably get one mark for calculating -30 and the second for writing 30

Idk, but I think the first mark is going to be substituting 20 and 9 into the nth term and the second mark is going to be the answer. So I think -30 will be in the accept column.