I made same mistake along with dozens on other questions. I also got same area for underneath curve. Probably loss of around 2 marks, 1 method and 1 accuracy

Reply 125

-Gifted-

Original post by Bealzibub

I made same mistake along with dozens on other questions. I also got same area for underneath curve. Probably loss of around 2 marks, 1 method and 1 accuracy

Yeah probably, 2 marks there.

Reply 126

Cadherin

Original post by -Gifted-

The area underneath the curve was 0.5*5root5, but need to take it away from the rectangle. How many marks dropped ?

2 maybe?

Reply 127

Joefery

Hi,

The area under the curve was 10ln[(3 + root(5))/2] -5root(5)/2. Integrations calculators agree

Reply 128

ComputerMaths97

I lost so many fricking marks.

1 - I couldn't do the c/s thing because I turned both c and s in terms of theta instead of half theta for some retarded reason.

For the area bounded by y=5 and curve, didn't read the question correct so just got area under curve

Didn't bother with the +c for question 1. Not sure why, clearly just forgot.

Only had the positive root for the ln(3+root(5)/2) one

Only drew the graph for positive x values because I thought cosh(x) doesn't exist for negative x.

And here was me hoping for an A :/ Defo missed my cam offer now, 3 huge basic mistakes and a few more mistakes just ruined everything looool. I even got full UMS in S2 near abouts, but still no A* in FM, great :biggrin:

Reply 129

leaf3

Original post by ComputerMaths97

What about your third module for FM though, the other two might bring this one up? I always see you around the maths threads, you're a maths whizz! I hope you prove thoughts here wrong on results day :P

Reply 130

MintyMilk

Where am I meant to find the motivation to study for Physics Unit 5 tomorrow after that? jfc

Reply 131

username1432214

Original post by ComputerMaths97

Nice! You nearly made half as many mistakes as I made. This is not even funny. How many marks for A? How other people find it in your school?

Reply 132

decombatwombat

Original post by ComputerMaths97

So at most 2 marks for the first one, 2 marks for the are bounded, 1 for the +c, 1 for only the positive root, 1 for the graph drawing. So even at the absolute worst you've lost 6-7 marks, That is almost certainly gonna be an A* judging by reactions to the paper.

Reply 133

leaf3

Hi, could someone verify the method for equating bit for the show that "c/s was cot(nθ/2)" when n is an even power for me? Ie which bit was C and which S?Irritatingly hard paper (thanks MEI...)

Reply 134

-Gifted-

Original post by leaf3

Hi, could someone verify the method for equating bit for the show that "c/s was cot(nθ/2)" when n is an even power for me? Ie which bit was C and which S?Irritatingly hard paper (thanks MEI...)

C= (-2j)^n*(sin(x/2))^n*cos(nx/2)
S=(-2j)^n*(sin(x/2))^n*sin(nx/2)

From memory, and dividing gives result. x is theta.

Reply 135

WhiteBison

lol rip.

I couldn't do the binomial, so, probably 2/8 for that question... urgh. Oh and I left my area between the curve and y=5 as some multiple of ln[(3 + root(5))/2] and a string of surds, so... Probably lost at least a mark there for not simplifying. Oh and my 2 cents on the matrices bit: I said M^n's limit was something like (3 4 / 3 4), and (M^-1)^n didn't converge upon a limit; it oscillated between negative and positive infinity. Can't remember the exact language I used though. I reckon I got about 80% raw, so probably a low B. *worlds smallest violin*

Maybe y'all can clarify: can we or can we not swap, say, S2 and M3, between normal Maths and F. Maths? Because I think that I got 72/72 on the latter and if I can make that a F. Maths subject it would probably keep me firmly in the A* A range respectively. I'd like that yes sir.

Reply 136

BriO

That was a harrrd paper. Hardest one I've done. I recon boundaries will be like 58/59 maybe? for A? 66 for A*?

I had no time to simplify at the end and left my answer as 10ln(3+root(5)/2) - a string of (3+root(5))^2 and so on. It all add ups to 0.5*5*root*5 but i left it unsimplified.

For M^-1 i wrote its P^-1*D^-1*P Instead of P*D^-1*P^-1 but proved the result anyway.

Forgot +c for the arctan thing

Got all the way down with last part of q2 except c/s which i did a small mistake on and thus couldnt do.

How many lost marks you guys recon?

Reply 137

ComputerMaths97

Original post by leaf3

I got 91 in M2 last year. So, considering the fact I think I got at least 70/72 in S2 this year, I was hoping an A would be enough for my A* :smile:

Thanks a ton mate, cheers :biggrin:

Original post by Bealzibub

Nice! You nearly made half as many mistakes as I made. This is not even funny. How many marks for A? How other people find it in your school?

Everyone else said they found it okay, but I'm not too bothered by what people say. The fact of the matter is, there's a difference between getting a question wrong, and making a dumb mistake. I made a lot of dumb mistakes, meaning I can't of done that well :smile:

However apparently I only lost 6-7 marks there, which may be an A*, so you might still have an A mate.

Original post by decombatwombat

I reckon the first one will be 3 marks (because you also had to consider the "when n is even" case which would've been worth a mark). The second one yes I agree probably 1. I'm hoping forgetting the negative root is only 1 mark, but I don't know where I was supposed to get the second root from? Was it from the sqrt(x^2 - a^2) or whatever that part was? But 1 or 2 for the negative root in my opinion. The graph drawing I don't know because I have no clue how many marks it was. If it was 2 marks, then yeah -1, if it was 3 marks then I reckon -2 or -1.

To me that seems more like 8 marks lost, and that requires there being no more mistakes in the rest of the paper than I'm yet to notice.

I've already accepted I haven't got an A*, but thanks a ton, maybe I'll still have a high/ mid A.

Reply 138

Aph

the limit as n\rightarrow\infty[\LATEX]
was

\frac{1}{7}*\begin{pmatrix} 3 & 4 \\3 & 4 \end{pmatrix}

right?

(edited 7 years ago)

Reply 139

ComputerMaths97

For the M^n = PD^nP^(-1)
When you multiplied out the matrices, aren't you supposed to do the two on the right first? Might have messed that up badly too.