Edexcel Core 3 - 21st June 2016 AM

I completely messed up this exam. I think I'm gonna get 60 UMS. Is it still possible to get an A overall having 86 average from last year? Also, I'm doing D2 as part of AS further maths, it can count towards my maths if I get a high mark in it, right?

For the 'show that' modulus question [g(x)] (4 part (b), I think), were you just supposed to ignore the modulus sign? It said the root was 'positive' but that just means that x is positive - it didn't mean that 4e^2x - 25 was positive.

I did just ignore the modulus sign completely and ended up with the right thing, but I don't know that what I did was mathematically correct - was I supposed to prove that 4e^2x - 25 was positive for my value of x and therefore the modulus sign didn't affect it? It was only two marks but even if I got both of them, I'm not convinced the question was right... did anyone else think the same thing?

Hey guys. You know the five marker with three parts in the modulus graph . How many mark d o you think the question about the asymotote was ?

what you did was mathematically correct as the root would have been in the non reflected part of the graph which is 4e2x -25 . If it was in the reflected part of the graph you would have had to change it to -4e2x +25

For the 'show that' modulus question [g(x)] (4 part (b), I think), were you just supposed to ignore the modulus sign? It said the root was 'positive' but that just means that x is positive - it didn't mean that 4e^2x - 25 was positive.

I did just ignore the modulus sign completely and ended up with the right thing, but I don't know that what I did was mathematically correct - was I supposed to prove that 4e^2x - 25 was positive for my value of x and therefore the modulus sign didn't affect it? It was only two marks but even if I got both of them, I'm not convinced the question was right... did anyone else think the same thing?

i didnt prove 4e^2x-25>0 i just ignored mod signs, though this gives you the right answer. 43(y-intercept)>21 (asymptote) and 2 (gradient of line) >0 so they intersect when 4e^2x-25>0 (the steep part).

Is it just me that felt really confident going into the exam but feels like they've completely messed up?

True but that's not what the question said nor what I wrote. Part of the graph in quadrant one was reflected, and it just said x was positive. When you come up with a formula for alpha you can't know whether or not you should be using 4e^2x - 25 or 25 - 4e^2x, so surely you should form two different equations (only one of which would have a solution). Did I miss something in the question explicitly saying g(x) is positive?

Does it matter because they gave you the iteraton formula you had to find so you knew what you were aiming for

It mattered to me because I wasted 10 minutes trying to understand why it was valid to just ignore modulus signs. But I finished the whole paper in time, got the iteration formula I was supposed to and now I understand why it's valid, so no harm done.

well, it came up Looooool but you were right i drew arcos by accident instead of arccsin but i think i got the 2nd part right

will i lose a mark for 4(a)(i) if i put (1/2)ln(25/4) instead of ln(5/2)?