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AQA Maths Core 1+2 May 2013

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Reply 140
Avatar for ash9002
ash9002
Original post by Adam_Pritchard
How did you show that p was 3/2


I showed that p=2/3. I just visually mixed up 2/3 with 3/2 on the second part of the question. To show that p=2/3 you solved simultaneously; you had to sub in 24 for the limit to get 24=24p + q then use 72 = 96p + q or something from the info given. Not even 100% sure that those were the exact eqns but I did get 48=72p in the end to get 2/3=p.
For the question with root(8x^3 + 1) being translated from root(x^3 + 1)

I got stretch in the x-direction by scale factor 1/2, because I thought if you are replacing the x with 2x you are stretching it by a scale factor of 1/2 not 2 :s-smilie: Can someone explain please.
Original post by mike256
(Core 2) Here's what I can remember (question parts/numbers might be mixed up)

1 (b) sum to infinity = 160

3) (c) Integral = 39/2

For the question that was something like show that the x-coordinate of the stationary point can be written in the form 2p, where p is a rational number, I got p =6/5

For the long logarithm one where you had to show that the equation had only one solution, I ended up with (x + 3)^2 = 0, so x = -3 was the only solution.

For the last trig equation one, I got x = 16, x = 104, or x = 136.

That's all I can remember for now.



















I don't think x=-3 is the correct answer for logs question because the question says verify that the equation has only solution and for (x+3)^2, you get two x values even if both were the same, so one of the x values must be (zero) and the other x value could be any number so in this case the equation has only one solution as the other solution is zero
Original post by Equality4all
Unless I misread the question aha :P I had 50 minutes or so to check my work and the wording of the questions, so I do hope I haven't messed that one up. How do you think you did overall?

fingers crossed haha. I think it wasn't too bad but I cant be certain lol you'll be amazed how easily marks can be lost in C1 tho
Original post by Denacio
you had 2x^2+6x+5=0. 2(x^2+3x)+5=0
so complete the square for X^2+3x gives you (x+3)^2-9
so final is: 2(X+3)^2-4=0 as -9+5=-4.

should of gotten 1/2root5.


Completing the on x^2 +3x gives (x+1.5)^2 - 2.25. The complete answer was 2(x+1.5)^2 + 0.5, and the final answer was 0.5(10)^0.5.
for the stretch x8 was it a stretch to a scale factor 1/8, 8, or 64 not sure which one.
Original post by Alifff
fingers crossed haha. I think it wasn't too bad but I cant be certain lol you'll be amazed how easily marks can be lost in C1 tho


I thought it was tricky by C1 standards, but found it pretty straightforward. Yeah, I know about that too well. I got 97% last January and I was so confident I got everything right. Is this a retake for you?
Original post by Equality4all
Completing the on x^2 +3x gives (x+1.5)^2 - 2.25. The complete answer was 2(x+1.5)^2 + 0.5, and the final answer was 0.5(10)^0.5.

i can vouch for the first one
Reply 148
Avatar for The H
The H
1
Original post by Adam_Pritchard
for the stretch x8 was it a stretch to a scale factor 1/8, 8, or 64 not sure which one.


It was 2
Reply 149
Avatar for $oulja
$oulja
Original post by Equality4all
By merely stating translation you will get a mark, unless if they specified it was a translation in the question, which I don't think thy did. Hope you get the grade you want man.


i think i did put 6,-7 cause i even wrote next to it, 6 to the right, 7 down. hoping on getting near full marks on this. how did you find the paper overall?
Reply 150
Avatar for mike256
mike256
2
Original post by sohailkm96
For the question with root(8x^3 + 1) being translated from root(x^3 + 1)

I got stretch in the x-direction by scale factor 1/2, because I thought if you are replacing the x with 2x you are stretching it by a scale factor of 1/2 not 2 :s-smilie: Can someone explain please.


I think the question was actually the other way round: from sqrt(8x^3 + 1) to sqrt(x^3 + 1), so it was actually s.f. 2.

I got this question wrong; I put 1/2 :frown:
For C1

Can anyone remember AB^2 question?
how you got to your answers and how many marks was it out of?
Original post by sohailkm96
For the question with root(8x^3 + 1) being translated from root(x^3 + 1)

I got stretch in the x-direction by scale factor 1/2, because I thought if you are replacing the x with 2x you are stretching it by a scale factor of 1/2 not 2 :s-smilie: Can someone explain please.



the coefficient which is 8 has been added to x only so it is stretch in the x direction by scale factor of 1/8
Reply 153
Avatar for The H
The H
1
Original post by sophie phil
the coefficient which is 8 has been added to x only so it is stretch in the x direction by scale factor of 1/8


No that's the coefficient of x^3
The coefficient of x is 2
(2x)^3 = 8x^3
Original post by sophie phil
the coefficient which is 8 has been added to x only so it is stretch in the x direction by scale factor of 1/8


Original post by mike256
I think the question was actually the other way round: from sqrt(8x^3 + 1) to sqrt(x^3 + 1), so it was actually s.f. 2.

I got this question wrong; I put 1/2 :frown:



But surely it's scale factor 1/2 because it's (2x)^3 = 8x^3???
Reply 155
Avatar for mike256
mike256
2
Original post by sohailkm96
But surely it's scale factor 1/2 because it's (2x)^3 = 8x^3???


The translation was from 8x^3 to x^3.

If you translate by scale factor 2, you replace x by x/2, so you get

8(x / 2)^3 = 8 (x^3 / 8) / x^3.
Original post by $oulja
i think i did put 6,-7 cause i even wrote next to it, 6 to the right, 7 down. hoping on getting near full marks on this. how did you find the paper overall?


That's good then, I did the same thing :smile: I found the paper pretty straightforward, but tricky by C1 standards. I think the grade boundaries will be 60 for an A, 66 for 90 UMS, which means 72 for full UMS.
Reply 157
Avatar for Indiee
Indiee
1
I think question
1
The first term of a geometric series in 80 and the common ratio is 0.5
a) work out the third term in the series
b) work out the sum to infinity
c) work out the sum of the first 12 terms in the series to 2 decimal places.

there was one question to
a)draw the graph of y=9^x
b) work out x using logarithms to 3 significant places
c) y=f(x) has been reflected in the y axis write an expression for y=f(x)

the last question was
a) draw the graph of tanx on the axis given
b)work out the values of x for tanx= -1 between 0<x<360 to the nearest degree
c) given 6tanxsinx=5
how do you get
6cos^2x+5cos-6=0
d) i think it was something like 6tan3xsin3x=5
work out the values of x in between 0<x<180 to the nearest degree
Original post by mike256
The translation was from 8x^3 to x^3.

If you translate by scale factor 2, you replace x by x/2, so you get

8(x / 2)^3 = 8 (x^3 / 8) / x^3.




Ahhhh okay so I just mixed up what things were transforming thanks!
What did you all get for the question where it asks for 2^k only question that had me stumped

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