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

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1. (Original post by goofyygoober)
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.
I believe the answer was stretch in x s.f. 1/3, just checked on a graph. I believe it was from y = √x^2+9 to y = 3√x^2+1, which is a stretch in x s.f. 1/3.
2. (Original post by goofyygoober)
then stretch in x-direction by sf 1/3 is correct
I think it's 3:
3(x^2 +1)^1/2 =
= ( 9(x^2+1) )^1/2 = (9x^2 + 9)^1/2 =
( (3x)^2 + 9 ) ^1/2
So it has basically changed from (x)^2 in the original to (3x)^2
That means it's been timesd by 1/3 Inside the bracet, and so it has a sf of 3 (because with x the transformations are opposite )
3. (Original post by goofyygoober)
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.
tbh i totally didn't see that what was in the square root changed.

it was definitely 2 marks which is always one named transformation and then the scale factor / vector to get 2 marks so since there was definetly a stretch we'll get the marks.

Maybe it was to confuse people?
4. 1)b) was definitely -2
9)a) ^m/2 - 6n
5. (Original post by salome12)
I think it's 3:
3(x^2 +1)^1/2 =
= ( 9(x^2+1) )^1/2 = (9x^2 + 9)^1/2 =
( (3x)^2 + 9 ) ^1/2
So it has basically changed from (x)^2 in the original to (3x)^2
That means it's been timesd by 1/3 Inside the bracet, and so it has a sf of 3 (because with x the transformations are opposite )
if you type the equation on your graphic calculator, you will see that f(x) has become narrower in x-direction which means it is 1/3
6. Stretch s.f. 1/3 in the x direction was correct. If you do a stretch scale factor a, you replace the variable (in this case x) with x/a, giving you x/(1/3) which left you with 3x. Then it was (3x)^2 to give you 9x^2, so you could take 9 out as a factor (of 9x^2 + 9) and then square root it to get y = 3sqrt.(x^2 + 1.)
7. (Original post by Excuse Me!)
tbh i totally didn't see that what was in the square root changed.

it was definitely 2 marks which is always one named transformation and then the scale factor / vector to get 2 marks so since there was definetly a stretch we'll get the marks.

Maybe it was to confuse people?
I hope so. Really ran out of time to check the questions.
8. i left the normal equation in the form y-y1 = m(x-x1)

how many marks willl i lose
9. (Original post by money-for-all)
i left the normal equation in the form y-y1 = m(x-x1)

how many marks willl i lose
Im guessing you subbed numbers in to y1 and x1? If so, none. They normally accept all forms, unless they state otherwise. I, myself, simplified it to y + 2x = 13 which made it easier in the next question but you should be okay.
10. (Original post by BenAtkinson)
Im guessing you subbed numbers in to y1 and x1? If so, none. They normally accept all forms, unless they state otherwise. I, myself, simplified it to y + 2x = 13 which made it easier in the next question but you should be okay.
Woooooo i love you buddy!!!!
remind me how many marks this was again
thanks
11. (Original post by TIF141)
Somehow I got 4096 for the binomial expansion, think I must have forgot to put in the minuses or something!! D:
I got the same, i think i forgot minuses too well i got Methoda mark ehh
12. For the perimeter I got 15 .1 I used the area 19.9 then divided by 2 and than times by 2 and than used the angle and divided it and I get radius 34 I think and than I square rooted ani got 5 .82 and than I times it by two and foung the arc lengh and than I got 15.1
IMO that's very harsh. But at the end of the day I think I may have got 67 so probably 90 UMS, which I'm happy with
how much UMS would around 60/63 / 75 be?

also how many universities look at individual UMS for modules?
(Original post by Zohrayasin)
For the perimeter I got 15 .1 I used the area 19.9 then divided by 2 and than times by 2 and than used the angle and divided it and I get radius 34 I think and than I square rooted ani got 5 .82 and than I times it by two and foung the arc lengh and than I got 15.1

do you remember how many marks this question was worth in total?
14. for sketch y=2^x

i put (1,0) by mistake

but the sketch was correct,
and above the sketch i said 'when x=0 , y=1'

but i put the corordinate the other way around

will i still get the full marks ?
15. How did you find u12 for the arithmetic series. I got a = -12 and d= 2 but that's wrong. I couldn't work out how to do it.
16. (Original post by Jackmilne95)
How did you find u12 for the arithmetic series. I got a = -12 and d= 2 but that's wrong. I couldn't work out how to do it.
i got a and d but didnt have time to find u12 what is did was use the equation in the question before and use n/2 = (2a+(n-1)d) to find an equation for u12 and do simultaneous equations to find a= 28 d= -2
17. i got about 53? anyone any idea what grade this is.
18. (Original post by alkaline.)
how much UMS would around 60/63 / 75 be?

also how many universities look at individual UMS for modules?

do you remember how many marks this question was worth in total?
Should just get you an A
Only really Oxford and Cambridge
6 marks for perimeter I believe
19. (Original post by MotorboatMyGoat)
Should just get you an A
Only really Oxford and Cambridge
6 marks for perimeter I believe
honestly I think I GOT LIKE 55/75.....
**** my entire life.

i think I got near perfect on c1 (As expected) but *if* I get near perfect on stats do you still think overall I can get an A?
20. (Original post by Zohrayasin)
For the perimeter I got 15 .1 I used the area 19.9 then divided by 2 and than times by 2 and than used the angle and divided it and I get radius 34 I think and than I square rooted ani got 5 .82 and than I times it by two and foung the arc lengh and than I got 15.1
Perimeter was 13.8 to 3.s.f

Note premature rounding may have skewed accuracy

r=√(19.9/0.586) = 5.82743892

Side 1→ 9 – 5.82743892 = 3.17256108
Side 2 → 8 – 5.82743892 = 2.17256108
Arc Length = rθ = 3.414879207
Unaffected length of 5

Perimeter = [Side 1] + [Side 2] + [Arc Length] + [Unaffected Length]

Perimeter = 13.76000137

To 3 s.f = 13.8

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