2017 AQA Level 2 Further Maths Paper 2 Unofficial Mark Scheme

im pretty sure q8 was 106.1 (since angle = 106.06...)

What do you think the grade boundaries will be then roughly? Because those papers both felt much harder than past papers... 137 for the A^?

I knew something seemed wrong about that question. I got 106.1 too, glad I'm not the only one.


Posted from TSR Mobile

Please add the marks sir I'll give you rep

For q14 you out 10 but that isn't a square number

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=?

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=4?

dude, i swear 18 is 16y^2/5, because the area x*2x/2, and x=4y/sqrt5, so x*2x/2 = (4y/sqrt5)^2

Isn't the area of the rectangle $x*2x=2x^2$? Why divide by two?

Surely
(2^m)^3
= 8^3m not 2^3m?
Correct me if I'm wrong!

I've just realised for the area of the rectangle question I found the area of the triangle I read the question wrong.
I found x in terms of y though. Will I get any method marks for this?

lol, it said find the area of the rectangle in terms of y

lol, it said find the area of the rectangle in terms of y

Can't really remember the question. If you could further detail it it would be great

nono, you are right, but can u tell me how much marks i will lose for that?

Tbh I have no idea. Anywhere between 1 and them all

i swear it was 16 square units

between them all, omg you must be joking, am sure 2 marks