i thought that was quite a damn good exam. Pretty sure I got everything apart from the last part of the cubic question. I was doing it correctly but at the end for some reason my brain went oh the roots are 1/sqr root alphabeta and gamma ffs haha. Anyway hopefully it can still be 100 ums depending on boundaries.
i thought that was quite a damn good exam. Pretty sure I got everything apart from the last part of the cubic question. I was doing it correctly but at the end for some reason my brain went oh the roots are 1/sqr root alphabeta and gamma ffs haha. Anyway hopefully it can still be 100 ums depending on boundaries.
for 4a I didn't notice it saying give it in terms of cosh x so i left it as 2+2cosh2x but in part b i converted this into 4cosh^2x during the integration. Does that mean I'll lose a mark
i ****ed this lads yw for lowering the grade boundaries.
I'm sure you did better than you thought! A lot of people skipped some of the 4 markers (including me), so already everyone's lowered the grade boundaries for this yearπ
I'm sure you did better than you thought! A lot of people skipped some of the 4 markers (including me), so already everyone's lowered the grade boundaries for this yearπ
do u remember what answer(s) you put for question 6 and 7? Like the last parts of them
do u remember what answer(s) you put for question 6 and 7? Like the last parts of them
For question 6 my answer was weird. I tried over and over but couldn't fix it. I got that coshy = +-5/12, but then I couldn't do the inverse of either of those answers soooooooo pffftt I don't know about that. For question 7, my final answer was Ο/3[64sqrt(2)-32].
I'm guessing they'll be pretty similar to last year, if not a couple of marks lower just because some of the 4 markers were quite hard. Last year's grade boundaries were this by the way:
For question 6 my answer was weird. I tried over and over but couldn't fix it. I got that coshy = +-5/12, but then I couldn't do the inverse of either of those answers soooooooo pffftt I don't know about that. For question 7, my final answer was Ο/3[64sqrt(2)-32].
yess i think i put down the same as you for q 7 and yea for q6 you get coshy = +-5/12 but you need to reject the negative solution as coshy>=1
yess i think i put down the same as you for q 7 and yea for q6 you get coshy = +-5/12 but you need to reject the negative solution as coshy>=1 so u just use y=arccosh(5/12) = ln2 i believe.
But I don't understand ughhhhhhhh For question 6, I got that coshy = +-5/12 and rejected the negative (because of the graph). But when I tried to do cosh-1(5/12), I couldn't get it on my calculator or by using the formula... I don't know where I went wrong haha!
But I don't understand ughhhhhhhh For question 6, I got that coshy = +-5/12 and rejected the negative (because of the graph). But when I tried to do cosh-1(5/12), I couldn't get it on my calculator or by using the formula... I don't know where I went wrong haha!
Oh my bad it wouldn't be coshy=5/12 as obviously coshy can't be less than 1. I did get an answer with ln something but i can't remember lol do you remember the question xD?
Oh my bad it wouldn't be coshy=5/12 as obviously coshy can't be less than 1. I did get an answer with ln something but i can't remember lol do you remember the question xD?
Ugh dammit XD I just tried your answer (ln2) it came out with cosh(ln2) = 5/4, if that looks familiar? The first part said you had to prove the differential of cosh-1x, which I got. In the end, you had to find the y-coordinate of the stationary point for y = (a constant) + (a constant)x - cosh-1(3x) or something, but that's all I remember
Ugh dammit XD I just tried your answer (ln2) it came out with cosh(ln2) = 5/4, if that looks familiar? The first part said you had to prove the differential of cosh-1x, which I got. In the end, you had to find the y-coordinate of the stationary point for y = (a constant) + (a constant)x - cosh-1(3x) or something, but that's all I remember
yesss ah i remember oh you're gonna be annoyed when you see this. so you worked out x correctly to be 5/12 but it was cosh-1(3x) not cosh-1(x) so you are actually evaluating cosh-1(3*5/12) =cosh-1(5/4) and since 5/4>1 this will work with the cosh-1(x) formula with ln.