AQA A Level Maths Core 2 - 25th May 2016 [Exam Discussion]

I thought I got that one, but apparently its like 13.7 and I don't remember writing that.. Almost forgot to add the arc length on but spotted it with 10 mins to go :P

I changed stretch s.f. 1/3 in x-direction after writing what you did. I don't think s.f. 3 in y-direction is right because when you expanded both it was y=X+3 changes to y=3X+3 or something

f(x) was √x^2 + 9 , or something similar.

and the transformation was 3√x^2 + 9 which is 3f(x) so answer is stretch s.f. 3 vertical i think

From BBC Bitesize
y = af(x) is a stretch, parallel to the y axis, scale factor a

Reply 221

akpbrooks

1

Original post by Alfie Oliver

Did anyone get k (in the translation question) as 5.5?

I did!!!

Reply 222

Excuse Me!

19

For the binomial I used the c4 method to fully expand whatever the quadratic was - (2+x)^7 ?

Problem is I completely forgot to times each answer by 2^7 although I did take it out the brackets and write it.

I then used the indices pairings from both expansions and multiplied any thats powers would add to 10.

How many marks would I lose?

Reply 223

Excuse Me!

19

On 8bii if you got up to this bit 5+3cosx=0

how many marks would you lose?

Reply 224

TheGreatPumpkin

15

Original post by akpbrooks

I did!!!

****, I got 11

Reply 225

MLW12345

7

hi whats the deal with resitting c1 - c4 seeing as this is the last type of this exam paper?

Reply 226

luckystarr

Original post by Excuse Me!

f(x) was √x^2 + 9 , or something similar.

and the transformation was 3√x^2 + 9 which is 3f(x) so answer is stretch s.f. 3 vertical i think

From BBC Bitesize
y = af(x) is a stretch, parallel to the y axis, scale factor a

The new transformation was actually 3√(x^2+1), so I got the answer as stretch SF 1/9 in x direction...

Reply 227

Crespo0008

1

Original post by Excuse Me!

On 8bii if you got up to this bit 5+3cosx=0

how many marks would you lose?

I'd say 2-3 because you had to say the minimum answer is 2 as cos anything can range between between -1 and 1 and then inverse cos -1 gave you 0 and 2 pie

Reply 228

tomdavis.

Updated - Unofficial Mark Scheme.
Here's what I got

1. b) a=-2
2. b) x=-0.861
c) reflection in y-axis
3. a) 3x^-1/2 - 1
b) y=6
c) y + 2x = 13
d) k = 5.5
4. b) d = -2, a = 28
c) Sum to 21 - sum to 3. 168-78 = 90
5. b) area = 19.9
c) perimeter = 13.8
6. a) 65.6
b) i) translation (0,5)
ii) stretch in X-direction 1/3 sf
7. a) p= -10 q= 40 r= -80
b) -1648
8. b) Tan(X) = 1 and tan(X) =-5/4 (to the nearest degree) X = 45, 225, 129, 309
c) lowest = 2, when theta = pi
9. a) c^(m/2 -6n)
b) X = 5/2

I don't exactly remember the answer to 7 b) if anyone could help.

Reply 229

jackpriceburns

Thats exactly the same as what I got so hopefully we both got it :tongue:

Reply 230

ItsMJ

1

Original post by M.N1997

For the 3(x^3 + 1)^1/2 translation, I took the 3 into the square root and times it all by factor nine, this gave me stretch in X direction sf 1/9??

Your first bit was right. You do take the 3 inside the square to make it a 9. Then you get (9x^2+ 9)^1/2. Which is basically ((3x)^2+9)^1/2. So it's a stretch along x-axis, scale factor (1/3).

Reply 231

bheaton

1

Anyone know if there's an unofficial mark scheme for statistics 1 Aqa - 25th may 2016

Reply 232

jackpriceburns

Original post by tomdavis.

Updated - Unofficial Mark Scheme.
Here's what I got

1. b) a=-2
2. b) x=-0.861
c) reflection in y-axis
3. a) 3x^-1/2 - 1
b) y=6
c) y + 2x = 13
d) k = 5.5
4. b) d = -2, a = 28
c) Sum to 21 - sum to 3. 168-78 = 90
5. b) area = 19.9
c) perimeter = 13.8
6. a) 65.6
b) i) translation (0,5)
ii) stretch in X-direction 1/3 sf
7. a) p= -10 q= 40 r= -80
b) -1648
8. b) Tan(X) = 1 and tan(X) =-5/4 (to the nearest degree) X = 45, 225, 129, 309
c) lowest = 2, when theta = pi
9. a) c^(m/2 -6n)
b) X = 5/2

I don't exactly remember the answer to 7 b) if anyone could help.

9 a) was 3^(m/2 -6n) :tongue:

Reply 233

SaltandSugar

2

For 2c, instead of putting reflection in the y axis I put stretch scale factor -1? from y=0.2^x to y=5^x

Reply 234

jackpriceburns

Original post by bheaton

Anyone know if there's an unofficial mark scheme for statistics 1 Aqa - 25th may 2016

Isn't S1 on June 01?

Reply 235

OfficialChemist

3

Original post by tomdavis.

Updated - Unofficial Mark Scheme.
Here's what I got

1. b) a=-2
2. b) x=-0.861
c) reflection in y-axis
3. a) 3x^-1/2 - 1
b) y=6
c) y + 2x = 13
d) k = 5.5
4. b) d = -2, a = 28
c) Sum to 21 - sum to 3. 168-78 = 90
5. b) area = 19.9
c) perimeter = 13.8
6. a) 65.6
b) i) translation (0,5)
ii) stretch in X-direction 1/3 sf
7. a) p= -10 q= 40 r= -80
b) -1648
8. b) Tan(X) = 1 and tan(X) =-5/4 (to the nearest degree) X = 45, 225, 129, 309
c) lowest = 2, when theta = pi
9. a) c^(m/2 -6n)
b) X = 5/2

I don't exactly remember the answer to 7 b) if anyone could help.

Was K not 6.5??? for question 3)d)

Reply 236

goofyygoober

8

Original post by Excuse Me!

f(x) was √x^2 + 9 , or something similar.

and the transformation was 3√x^2 + 9 which is 3f(x) so answer is stretch s.f. 3 vertical i think

From BBC Bitesize
y = af(x) is a stretch, parallel to the y axis, scale factor a

Yep, I got the same as you did.
But they said the transformation was 3√x^2 + 1 not 3√x^2 + 9 which totally confused me.

I remember when I looked at the question it was 3√x^2 + 9, right??

I just really didn't know whether I read the question wrong or not.