AQA Further pure 1/FP1 exam 5th June 2015

i got k=38

What was the inequality you got with k in it?

My answers' other than show that questions:

1) a) a+b = -3, ab = 7/2
b) 4x^2 + 45 = 0
c) a^2 = 1 + 3/2 sqrt(5) i, b^2 = 1 - 3/2 sqrt(5) i
2) a) Undefined at one limit of integration
b) Does not have a finite value
3) a) 2-11i
b) i) q=20 (z^3-11z+20=0)
ii) z^2- 4z + 5
iii) z = -4
4) a) x = {-5 + 120n, 35 + 120n}
b) 235
5) a) c=4, d=-1
b) i) -16I
ii) enlargement scale factor 2, rotation 45 degrees clockwise
iii) Fromleft to right, top to bottom: [[65536sqrt(2), 65536sqrt(2)], [-65536sqrt(2), 65536sqrt(2)]]
6) a) Hyperbola crossing (3,0) and (-3,0)
b) y = 4/3 x + 4, y = -4/3 x -4
7) a) 39.60
b) n(n+1)(2n+3)
c) i) 2/3 r
ii) k = 39
8) a) y = 1
b) (1, -1/2) and (-3, 3/2)

Not sure if they're right, feel free to make corrections.

Edit: what happened to my indentation?

My 1c was what you got but squared
3a was 2+11i
How did you do 3)b)ii)?
for 5b)iii) I got 256sqrt(2), did I need to square 256?

How did you even get K?

Could you do an unofficial mark scheme thread because most of these look good

1/3(k+1)(k+2)(2(k+1)+3) < 44000

The final equation after substitution was the same as the original equation given in the start of the equation where you substituted 39 and 40 and saw a change in sign... The only difference was that instead of an "a" you had a "k" and because the inequality was less than zero the correct value was 39...

Wow didn't notice that at all!!! Great spot!!! Was wondering why it was so few marks :P

Hahah well don't worry if that's all you did wrong then you did great... I stared at the question for 5 minutes and then just looked at the question again and spotted it... all we can do now is wait...

Guys can anyone show me how u did the matrices in 5)a. The one where we had to work out c and d

I did find k=39 by trial and error after finding the cubic so hopefully that's an accepted method

You premultiply the original coordinates by A and let that equal to the transformed coordinates
Expand (?) the matrix and you get two simple equations

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Yes i tried to do that but we has to multiply out a 1 by 2 matrice withva 2by2 matrice to equal 1by2 matrices rite? Or did i use the wrong method

yeah but 2x2 should be before 1x2 like
Ax(coordinates)=(coordinates)

Yes hopefully.... I did a stupid thing on question one the last part though... ;( I found the equation and wrote it down, then found the roots on the calculator and forgot to right them! ;( it was only 2 marks but still that is very stupid of me... ;(