WJEC FP2 Summer 2011 Post-Exam Discussion

Q7 was just WTF. I haven't seen anything like that before. At least it was only worth 3 marks.

But I thought the rest of the paper was pretty standard. What did everyone have for the interval of f(A)?

Reply 5

I made a few mistakes, did f(A) wrong, I forgot to prove the stat points were maxima and minima, couldn't do question 7 and sorta cheated my way through the partial fraction bit in question (8) :L

Reply 6

JosephRedfern

OP

7

Original post by MooMaster

Q7 was just WTF. I haven't seen anything like that before. At least it was only worth 3 marks.

But I thought the rest of the paper was pretty standard. What did everyone have for the interval of f(A)?

I think it was [4, 7]? I might have got that wrong though... I probably did, in fact :biggrin:

Original post by Pronitron

I made a few mistakes, did f(A) wrong, I forgot to prove the stat points were maxima and minima, couldn't do question 7 and sorta cheated my way through the partial fraction bit in question (8) :L

I did the same! I ended up showing that what they gave us was the same as the original formula, buy putting them over the Lowest common denominator, and factoring :biggrin:

(edited 12 years ago)

Reply 7

Original post by JosephRedfern

I think it was [4, 7]? I might have got that wrong though... I probably did, in fact :biggrin:

I did the same! I ended up showing that what they gave us was the same as the original formula, buy putting them over the Lowest common denominator, and factoring :biggrin:

You had to do (something) ? (something else) because the graph was defined for two separate domains, so you had to find f(A) for x<2 and f(a) for x>2 (as the graph was not continuous) :/

Reply 8

The question mark is a union symbol btw -_-

Reply 9

Original post by JosephRedfern

I think it was [4, 7]? I might have got that wrong though... I probably did, in fact :biggrin:

I did the same! I ended up showing that what they gave us was the same as the original formula, buy putting them over the Lowest common denominator, and factoring :biggrin:

Ha I didn't even do that, I just wrote some bull**** about how because the top was x^2 and the bottom x^2 you needed a constant also which would be 1, pretending I knew what I was doing and that the 1 there was obvious haha :/

Reply 10

Does f(A) = (-2, 2) u (3, 7) look familiar?

If anyone has reliable answers for this paper it would be much appreciated! :h:

Reply 11

Original post by MooMaster

Does f(A) = (-2, 2) u (3, 7) look familiar?

If anyone has reliable answers for this paper it would be much appreciated! :h:

Yeh I forgot about the discontinuity thing so I just had [-2,7], was A an interval of round brackets :/?

Reply 12

JosephRedfern

OP

7

Original post by Pronitron

You had to do (something) ? (something else) because the graph was defined for two separate domains, so you had to find f(A) for x<2 and f(a) for x>2 (as the graph was not continuous) :/

Original post by MooMaster

Does f(A) = (-2, 2) u (3, 7) look familiar?

If anyone has reliable answers for this paper it would be much appreciated! :h:

Ah, damn! I just took the lower value into the first equation, and the higher into the second. Oh well :frown:

I found it NOT to be continuous, and for it NOT to be strictly increasing (or strictly decreasing).

I also got something like 0.794 + 0.794i for my FIRST cubic root of the complex number... I can't remember the other two though, but I know that for one, the real part was negative, and the imaginary part was positive, and for the other it was swapped around.

Reply 13

Original post by JosephRedfern

Ah, damn! I just took the lower value into the first equation, and the higher into the second. Oh well :frown:

I found it NOT to be continuous, and for it NOT to be strictly increasing (or strictly decreasing).

I also got something like 0.794 + 0.794i for my FIRST cubic root of the complex number... I can't remember the other two though, but I know that for one, the real part was negative, and the imaginary part was positive, and for the other it was swapped around.

Yeh you did the same as I did then, oh well, we might maybe get one mark I guess.
I had the same number twice for my first root also, but I got the function to be strictly increasing across all of it's domain, as f'(x)>0 for both bits :L

Reply 14

Original post by Pronitron

Yeh I forgot about the discontinuity thing so I just had [-2,7], was A an interval of round brackets :/?

No, they were square brackets. Crap. :argh:

Reply 15

Original post by MooMaster

No, they were square brackets. Crap. :argh:

I guess you'd have curly brackets around the 2 bit though on each side wouldn't you as there would be a big of nothing there between the 1 and the 4 or whatever it was when x=2. I see you're going to Imperial too btw :smile:

? My friend has got an offer for maths there and has to get 2A*'s and 2A's, what's your offer like?

Reply 16

Original post by Pronitron

I guess you'd have curly brackets around the 2 bit though on each side wouldn't you as there would be a big of nothing there between the 1 and the 4 or whatever it was when x=2. I see you're going to Imperial too btw :smile:

? My friend has got an offer for maths there and has to get 2A*'s and 2A's, what's your offer like?

Ahhh man I'm gutted about that! I spent ages going over functions, goddamit! :mad:

They want an A* in Further Maths and an A in Physics, including the A I've already got in Maths :^_^:

But it all depends on how well M3 and S3 go :s-smilie:

not looking hopeful though tbh

Reply 17

Original post by MooMaster

Ahhh man I'm gutted about that! I spent ages going over functions, goddamit! :mad:

They want an A* in Further Maths and an A in Physics, including the A I've already got in Maths :^_^:

But it all depends on how well M3 and S3 go :s-smilie:

not looking hopeful though tbh

I hope M3 is okay to make up for any stupidity I did on FP2, and I hope nothing comes up that our teacher hasn't taught us. (question 7 I'm looking at you!) How come you're doing S3? Are you doing it instead of FP3 then? If so that might be a good call as I have FP3 on Friday and it's a bitch -_-

Reply 18

Joord

Original post by MooMaster

Does f(A) = (-2, 2) u (3, 7) look familiar?

If anyone has reliable answers for this paper it would be much appreciated! :h:

That looks familiar to me! :biggrin:

(I think)

Original post by JosephRedfern

Ah, damn! I just took the lower value into the first equation, and the higher into the second. Oh well :frown:

I found it NOT to be continuous, and for it NOT to be strictly increasing (or strictly decreasing).

I also got something like 0.794 + 0.794i for my FIRST cubic root of the complex number... I can't remember the other two though, but I know that for one, the real part was negative, and the imaginary part was positive, and for the other it was swapped around.

I found it to be not continuous and strictly increasing :/

Also that cubic sounds familiar..

Reply 19