Maths HL P 1 - aftermath

Duncannon

JamesJB wrote: »

Thanks for the reply :cool: I sure hope so, I mean it is pretty much accurate I surmise. If they have the same slope it is fair to say they are the same tangent, no? I realise I should've found the actual tangent I just literally forgot, I've been doing that in a few tests...Hoping for an A though since I don't want to go down from the mocks, which were far harder than this imo!

if you get the equation of the tangent at the 2 points you get the same line

Duncannon

Deleted User wrote: »

Yeah, as far as I can see because p is squared it will always be greater than 0 (or -1 if you moved that over like me).

answer is correct perfect squares are always positive

Duncannon

Victoria. wrote: »

I stupidly forgot to put down the make and model of my calculator
Was so stressed. Just opened the paper. Does anyone know if it is a problem?

its not a problem

JamesJB

Duncannon wrote: »

if you get the equation of the tangent at the 2 points you get the same line

Thanks, yeah I know - see I just did the slope idea and said they would be the same. Then, since I did that question 3rd, I said 'ah I'll come back and just do the eqn of the line after I've done the others.... BIG mistake. I just hope they'll give me the majority of the marks for doing like, what, more than half of what they wanted?

JamesJB

I was just looking at one answer given here for 4 (c) (iii)

I think I might've got a different answer. I used 3(n/6(n + 1)(2n + 1) + n) (since you bring the summation into the brackets).

Then I did the sum from 1-30 = 3(n/6)(31)(61) + 30 = 28395
and then from 1-11 = 1529

Then I took the smaller away from the bigger and got 26866. I thought I had the right method but it seems I've gone wrong somewhere...Any ideas?

mark.oc

JamesJB wrote: »

I was just looking at one answer given here for 4 (c) (iii)

I think I might've got a different answer. I used 3(n/6(n + 1)(2n + 1) + n) (since you bring the summation into the brackets).

Then I did the sum from 1-30 = 3(n/6)(31)(61) + 30 = 28395
and then from 1-11 = 1529

Then I took the smaller away from the bigger and got 26866. I thought I had the right method but it seems I've gone wrong somewhere...Any ideas?

It's the sum from 1-10 you take away in order to get your answer inclusive of 11.

EuropeanSon

JamesJB wrote: »

I was just looking at one answer given here for 4 (c) (iii)

I think I might've got a different answer. I used 3(n/6(n + 1)(2n + 1) + n) (since you bring the summation into the brackets).

Then I did the sum from 1-30 = 3(n/6)(31)(61) + 30 = 28395
and then from 1-11 = 1529

Then I took the smaller away from the bigger and got 26866. I thought I had the right method but it seems I've gone wrong somewhere...Any ideas?

Should've taken away the sum from 1-10 I think. 11 was included.

EuropeanSon

mark.oc wrote: »

It's the sum from 1-10 you take away in order to get your answer inclusive of 11.

Lol you got there first.

JamesJB

mark.oc wrote: »

It's the sum from 1-10 you take away in order to get your answer inclusive of 11.

AHA. I see now, thanks. It's just a small enough blunder in any case. Quite literally an arithmetic one so what, - a few marks :rolleyes: I think I still have 6 other questions with near (if not almost definitely) full marks. Hoping marking scheme won't somehow screw me over but oh well, I'm happy.

Duncannon

MollyJayne wrote: »

Ok. In reply to everything...

It was a very nice paper, and I think paper two will be the same!

1. c) iii) so annoyed! I got b=-4 and d=6, but somehow I made c=-3-b = c=-3-4=7, when it should have been c=-3+4=1...only realised now! Kept trying to figure out what I'd done wrong... I'd decided to change d to -10 so it would fit the sequence, but it didn't fit the equation =[
The right answers are definitely -4, 1, 6 as they fit the sequence and the equation...

Another controversial issue...
6. b) ii) It's definitely a curve. It says in the question that its a curve. Yup, parametric equations of a CURVE.
I don't know the answer though... I was going to say that its increasing but if its a curve...it curves!

I thought question 8 was lovely! Didn't quite get the last part of c out 'cause I ran out of time (shouldn't have kept checking that question 1!) but should get most of the marks for it...

I think I got atleast 65% in that paper, and I always find paper two easier so I'm really hoping for a B at this stage!

the "curve" was
the line 2x-5y+1=0
the numbers in 1c were-4,1,6

EuropeanSon

http://en.m.wikipedia.org/wiki/Line_(geometry)?wasRedirected=true

In Euclidean geometry, a line is a straight curve

So a curve does NOT have to be "curvey".

Hope that settles things.

Duncannon

zam wrote: »

For alphaxbeta i got ±4
For alpha+beta i got 0 (which i think is wrong!)

alpha x beata is +4 alpha +beata is 2

JamesJB

Anyone use just regular simultaneous eqns for the matrix question? I wasn't in the mood for doing it out the pre/post multiplication way. Got the same answers obviously but I have this idea that they'll somehow try say you had to use matrix methods even if it didn't specify...

Duncannon

JamesJB wrote: »

Anyone use just regular simultaneous eqns for the matrix question? I wasn't in the mood for doing it out the pre/post multiplication way. Got the same answers obviously but I have this idea that they'll somehow try say you had to use matrix methods even if it didn't specify...

they just said find x and y you can do it anyway, matrices, simultaneous equations ,both methods are fine the answers are x = 4, y = 2,

JamesJB

Duncannon wrote: »

they just said find x and y you can do it anyway, matrices, simultaneous equations ,both methods are fine the answers are x = 4, y = 2,

That's what I got anyway :cool: