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Aqa gcse further maths unofficial markscheme 14/06/19

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Reply 1
Avatar for D.p.005
D.p.005
2
Circle theorem question, x was 26 degrees
Height = 4sqrt(10)
Circle theorem, x=26°
Gradient question, a=-3.5
Last question, x=2, x=-3
Prove it’s >/0 , one solution x=2.5 , graph
Point B on line (11,-2.5)
Matrix transformation (-1,0,0,1)
Smallest value of x inequality x = -3
Perimeter 6sqrt(2) 4sqrt(2)
Factorise 2(3x-2y)(x 5y)
Cube root question = 2, answer was 8
Matrix question 5 and -2
(edited 4 years ago)
The matrixes question was something like 5 and -2
Reply 4
Avatar for D.p.005
D.p.005
2
the coordinates for the lines that was ab:bc is 1:3 or something like that was 11 and -2.5 or something like that. Also, can't remember which way round
last question I got x= 1/2 and -3 (I think)
(edited 4 years ago)
Original post by D.p.005
the coordinates was 11 and -2.5 or something like that. Also, can't remember which way round


Yeah I think it was that. Definitely got 11
Original post by runawaytrolley
last question I got x= 1/2 and 3 (I think)


Yeah I got that I think, it was like (2x-4)(x+3)
Original post by Maximus 190
Yeah I got that I think, it was like (2x-4)(x+3)


Yeah that's it, I meant -3
Original post by runawaytrolley
Yeah that's it, I meant -3


I accidentally did a plus, I can’t remember if it was negative or positive.
Original post by Maximus 190
I accidentally did a plus, I can’t remember if it was negative or positive.


one of them was definitely negative
What about the cone question? I got 4 root ten but I’m not sure wether it is right
Original post by Emma0603ii
What about the cone question? I got 4 root ten but I’m not sure wether it is right


I think I got that
Original post by runawaytrolley
I think I got that


Okay good, hopefully it is right, or at least some method marks
Can someone explain their working for the x circle theorems question? Only question I struggled with :/
Yeah. It was 4 sqrt(20) for height

R was sqrt(20),
So l= 3r = 3sqrt(20)

It’s super long to type. But if you square root (L^2-r^2) = 4sqrt(20)
(edited 4 years ago)
Original post by sdr981
Can someone explain their working for the x circle theorems question? Only question I struggled with :/


I did angles on the circumference are half those at the centre so it is x+23, then I used base angles in an isosceles triangle are equal to get 37° I think for one of them. I then did a tangent meets a radius at 90° so I added x + 23 to 3x and then took away the 37 and put it equal to 90
There may have been an easier way and I’m not great at explaining sorry
Original post by sdr981
Can someone explain their working for the x circle theorems question? Only question I struggled with :/


Work out the isosceles triangle angle in terms of x. Add it to the corner angle. Then that equals 3x as it’s the alternate segment theorem.
****, knew it was going to be that easy.
Original post by Maximus 190
Work out the isosceles triangle angle in terms of x. Add it to the corner angle. Then that equals 3x as it’s the alternate segment theorem.
Original post by Maximus 190
Yeah. It was 4 sqrt(20) for height

R was sqrt(20),
So l= 3r = 3sqrt(20)

It’s super long to type. But if you square root (L^2-r^2) = 4sqrt(20)


I thought it had to be in the form a root 10?

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