Maths PP1

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Donald-Duck wrote:

Theres nothing strange about it at all, you do spend 2 hours looking at it and in reality there isn't that much on the exam paper so its not hard to believe that peopel can remember most of it, plus as said above, they could have it in front of them

i actually have the paper right in front of me, and am not bothered opening it.
from now till monday nothing but french and maths, who cares about business

CambridgeGuy

I know its asking a lot, but it would be great if someone could post a solution of Q6 9(c)(ii) of the ordinary paper. if it wasnt for that nasty bit i would have got 100%!

sheroo_05

i thought it was fine. easyer den past years. my hopes of gettin an A1 r still alive..

Lucas10101

rerm....some1 will post it up soon. Cambridgeman.

Gangsta

I think that paper was harder than last years?? But saying that i think I got about 97/98% got everything out except 6 c (iii)

genericgoon

analyse this wrote:

you got 18??!!! i got 20/3 and so did everyone else in my class!! damn i was sure I had got that write:(

I got a fraction too! ANd it was a third one. I created a triangle then got the area of the curve enclosed by the curve and took that away. I hope i was rite!

Crumbs

Howdy. Not a student but I looked over the HL paper - seemed very reasonable. Couple of nice questions. (By nice I mean interesting! ) Here are the answers I got out. Apologies if there are any errors or typos - I did it pretty quickly.

Q.1
(a) x/(x+y)

(b)
(i) k = -3
(ii) 1, -k-2
(iii) k < -2

(c) Use long division twice and then a bit of substitution.


Q.2
(a) x=1, y=-3, z=4

(b)
(i) 2/3
(ii) 6x^2 - 4x + 1 = 0

(c)
(i) End with (x-1)^2 >= 0
(ii) End with [x + (a-3)]^2 >= 0


Q.3
(a) 2x2 zero matrix

(b) z^5 = 4-4i, z^9 = -16+16i

(c)
(i) 4+i, -4-i
(ii) 2-i, 1+3i (use the -b formula and your result from part (i) will come into it)


Q.4
(a) easily shown

(b)
(i) u2 = 5/2, u3 = 5/3, u4 = 5/4
(ii) un = 5/n
(iii) induction

(c)
(i) un = Sn - Sn-1 = (2n-1)ln3 [showing that un+1 - un = 2ln3 (a constant) proves it's arithmetic]
(ii) 18 terms


Q.5
(a) -3, -2, -1, 0, 1

(b)
(i) ur+1 = (9Cr).(-1)^r.(2)^9-r.(x)^9-3r
(ii) -5376

(c)
(i) Sn = x(1-x^n)/(1-x)^2 - n.x^n+1/1-x (probably possible to simplify further)
(ii) s(infinity) = x/(1-x)^2


Q.6
(a) 4x/(x^2 + 1)^2

(b)
(i) by first principles = -1/x^2
(ii) x + 4y - 4 = 0

(c)
(i) f'(x) = 2/4+x^2, g'(x) = -2/4+x^2
(ii) f'(x)+g'(x) = 0 => f(x)+g(x) is constant
(iii) f(x)+g(x) = pi/2


Q.7
(a) 6/5

(b)
(i) 3x - 2y - 14 = 0
(ii) easily shown

(c)
(i) f'(x)=0 => x = 1/3, f(1/3) = -1, f''(x)<0
(ii) max value of f(x)=-1 => doesn't cross x-axis


Q.8
(a)
(i) x^4/4 + c
(ii) -1/2x^2 + c

(b)
(i) 98/3
(ii) f(x) = 6x + cos(x) - 1/2

(c) 20/3

Donald-Duck

Gangsta wrote:

I think that paper was harder than last years?? But saying that i think I got about 97/98% got everything out except 6 c (iii)

I think your the only person I've heard say that

colm-ccfc84

Gangsta wrote:

I think that paper was harder than last years?? But saying that i think I got about 97/98% got everything out except 6 c (iii)

Wow, well done. Was the answer to that c iii 90 degrees? That is what I wrote down.

mathew

Not sure if this is right.. but heres a solution to 6c

i) f(x)=tan-1(x/2)
f'(x)=2/(4+x^2) ... (tables book)

g(x)=tan-1(2/x) u=2/x du/dx= -2/x^2
g'(x)=-2/(4+x^2)

ii) f'(x)+g'(x) = 0 therefore f(x)+g(x) = constant

iii) tanf(x)=x/2 tang(x)=2/x

from tables... tan(A+B)-tan(A+B)tanAtanB = tanA+tanB

tan(f(x)+g(x))-tan(f(x)-g(x))tanf(x)tan(gx) = tanf(x) + tang(x)
tan(f(x)+g(x))-tan(f(x)-g(x))(x/2)(2/x) = tanf(x) +tang(x)
0 = tanf(x) + tang(x)

therefore constant = (pi)/2

Think its right.. cant be sure tho


edit: seems i was beaten to it ^^^

analyse this

Oh my god I just realised something. In the matrices question where we had to determine A^2-2A I wrote down zero instead of a 2X2 matrix of zeros!!! AAAgggghhhhh!!

Jonathan

Yea the answer to HL 6 c(iii) was 90 degrees. I proved it using triangles. The tan^-1 's are opposite angles in a right angled triangle.

90 + tan^-1 (2/x) + tan^-1 (x/2) = 180

justbringit89

IS this AN OL or HL Thread????

ZorbaTehZ

Crumbs wrote:

(Snip)(ii) 18 terms

Did you not get n =17.5?
Therefore 17 terms of the sequence are less than that number. The 18th term is greater.

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analyse this wrote:

Oh my god I just realised something. In the matrices question where we had to determine A^2-2A I wrote down zero instead of a 2X2 matrix of zeros!!! AAAgggghhhhh!!

i came extremely close to doing that

analyse this

How many marks will I lose? Cant believe I got an (a) part wrong!! The shame:o

Jonathan

analyse this wrote:

Oh my god I just realised something. In the matrices question where we had to determine A^2-2A I wrote down zero instead of a 2X2 matrix of zeros!!! AAAgggghhhhh!!

I feel your pain. I did it too.

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analyse this wrote:

How many marks will I lose? Cant believe I got an (a) part wrong!! The shame:o

if it makes you feel better i made a mess of 1 a

chas_88

analyse this wrote:

you got 18??!!! i got 20/3 and so did everyone else in my class!! damn i was sure I had got that write:(

20/3 is right, everyone in my class got it.

genericgoon

I think it would only be worth 2/3 marks lost

JC 2K3

Q1:
Easy.

Q2:
Had inequalities I didn't like the look of so I avoided it.

Q3:
Easy. Exact same part c(with different numbers) came up in like 1994.

Q4:
Pretty straightforward. I forgot to prove the series was arithmetic in part (c), damn...

Q5:
Easy.

Q6:
Can't believe I didn't get (c)(iii). I'm so annoyed I didn't realise that as it was a constant you could put any number in in place of x and work it out. As for part (ii), I proved it like:
f'(x) = -g'(x)
integ(f'(x)) = -integ(g'(x))
f(x) = -g(x) + c
f(x) + g(x) = c
c being a constant(constant of integration)

Q7:
Easy, except for the last part I just put y=0 and it worked out like e^3x = 3x, which I said could not be true.....

Q8:
Argh. Piece of piss, but I fúcked up the end by forgetting about the small extra bit of space enclosed by the line at the bottom and hence integrated from 0 to 5....

I'm fairly sure I haven't dropped more than 10-20 marks here, so still on course for the A1

Crumbs

ZorbaTehZ wrote:

Did you not get n =17.5?
Therefore 17 terms of the sequence are less than that number. The 18th term is greater.

For the first part, did you get un = (2n-1)loge3 ?

That gives n < 18.5, so 18 terms.

CosaFada

so was d answer to tha matrice in part a oo??I only got the top 2 zeros!tawt it was a bit odd!! oo

rjt

Crumbs wrote:

(ii) max value of f(x)=-1 => doesn't cross x-axis

This is what I did, but afterwards began to think it wasn't rigorous - after all, this is just a local max point, not an absolute maximum for the function (eg. a cubic equation may have a local max, yet extend on to infinity). Would you not need to show that this is not the case with this function too? I'd love if this wasn't the case - it'd give me full marks on this question, but I'm skeptical.

Overall, I was happy enough with the paper. I made some stupid mistakes though, eg. getting 4(b)(i) wrong, one of the easiest things in the paper (but if I'm lucky I'll get a misreading, and given that my (ii) and (iii) were correct apart from the same mistake, I should get a lot of the marks)!

I answered every question, although messed up 8(c). Still, I think I have full marks in 1,2,5(assuming my alternate solution to 5(c)(ii) that doesn't really use 5(c)(i) is acceptable; got the right answer though) and 6. And so apart from not giving a matrix full of 0's for 3(a)(i) and the mistakes I've mentioned in 7(c)(ii) and 8(c), I think I've got an A1 from that paper.

cocoa

Crumbs wrote:

*snip*
(b)
(i) k = -3
(ii) 2, -k-1
(iii) k < -1

answers to ii and iii are incorrect:
(ii) 1, -k-2
(iii)k<-2

apart from that I agree with all your answers. I did every question and only messed up 8(c),7(b)(i) and said plain zero for the matrices question. Anyone know how many marks the matrices mistake will lose me? ah well, looks like the high nineties for me

JC2K3 wrote:

Q8:
Argh. Piece of piss, but I fúcked up the end by forgetting about the small extra bit of space enclosed by the line at the bottom and hence integrated from 0 to 5....

rofl, I made almost the same mistake (only slightly less) I integrated from 1-5 when i should have integrated from 1-3... oh well

rjt

cocoa wrote:

answers to ii and iii are incorrect:
(ii) 1, -k-2
(iii)k<-2

apart from that I agree with all your answers. I did every question and only messed up 8(c),7(b)(i) and said plain zero for the matrices question. Anyone know how many marks the matrices mistake will lose me? ah well, looks like the high nineties for me

I would say a blunder. Actually, the O matrix = a matrix full of 0s according to the text books, so maybe if your 0 was shaped like a O they'll let you away with it :P. But I seriously doubt any more than 3 marks - the majority of the marks will be going for the squaring and multiplying I'd imagine.

JC 2K3

was 1(b)(iii) not k>-3?

since (i) was b^2 - 4ac = 0 and the answer was k=-3, isn't (iii) just b^2 - 4ac > 0

....

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rjt wrote:

I would say a blunder. Actually, the O matrix = a matrix full of 0s according to the text books, so maybe if your 0 was shaped like a O they'll let you away with it :P. But I seriously doubt any more than 3 marks - the majority of the marks will be going for the squaring and multiplying I'd imagine.

how many marks would that be worth anyway?
5/10?

genericgoon

For ppl wanting the marking scheme its simple A parts are worth 10, and B and C parts are worth 20 each. How these are subdivided in multi-part question is depndant on the number of parts and the difficulty of the individual parts.

cocoa

JC 2K3 wrote:

was 1(b)(iii) not k>-3?

since (i) was b^2 - 4ac = 0 and the answer was k=-3, isn't (iii) just b^2 - 4ac > 0

....

hmm, i would think that, but no.... personally i used alpha and beta for part i... actually i think the reason why it doesn't work is because only one root involves k, but i could be wrong...