The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

This discussion is now closed.

Check out other Related discussions

Edexcel A-Level Core Pure 1 - 3rd June 2019 [Exam Discussion]

Scroll to see replies

Original post by number0
have most people taking further maths this year taken the maths a level last year?

I remember you from the thread :tongue:
Original post by thekidwhogames
I remember you from the thread :tongue:


aha yh, glad my class ended up taking it last year, i couldnt deal with the extra exams
Reply 182
Avatar for mmonk
mmonk
11
Hi guys how many solutions are there to cosh(x)^2=3
I get that coshx = +3^0.5 and -3^0.5
so x = ln(sqrt(3)+sqrt(2)
and x = ln(-sqrt(3) +sqrt(2))
the second value is impossible so i reject that.
In the mark scheme they have ln(sqrt(3)-sqrt(2)) How did they get that??
thanks
Because you have -arcosh(root3+root2) then take the power to (root3+root2)^-1
Original post by mmonk
Hi guys how many solutions are there to cosh(x)^2=3
I get that coshx = +3^0.5 and -3^0.5
so x = ln(sqrt(3)+sqrt(2)
and x = ln(-sqrt(3) +sqrt(2))
the second value is impossible so i reject that.
In the mark scheme they have ln(sqrt(3)-sqrt(2)) How did they get that??
thanks

You should be able to visualise (or sketch) the cosh(x) graph and see that there are two solutions to

cosh⁡x=3\cosh x = \sqrt{3}

The inverse cosh function is created by restricting the domain of the cosh function to x≥0x\geq0 and then inverting it. So just like inverse cos, the output of the inverse cosh function is always positive. cosh(x) is an even function so once you've got one solution from inverse cosh, you can just negate this to find the second solution.

Alternatively you can use the definition of cosh in terms of exponentials and then solve the quadratic which will give you two solutions.
Reply 185
Avatar for mmonk
mmonk
11
Thanks Guys
Reply 186
Avatar for mmonk
mmonk
11
Screen shot hard question.PNG can someone help me please with part b dont understand the markscheme at all
Original post by mmonk
Screen shot hard question.PNG can someone help me please with part b dont understand the markscheme at all

So you can work out the width of the logo in model form right?
Then you'll need to find the scale factor to make it 12m wide. Use the scale factor on the area from (a) to find the area of the logo and from that you should be able to figure out the number of tins of paint :smile:
Reply 188
Avatar for mmonk
mmonk
11
I cant work out the width of logo! there are no axis so cant really find the greatest width
Original post by Lemur14
So you can work out the width of the logo in model form right?
Then you'll need to find the scale factor to make it 12m wide. Use the scale factor on the area from (a) to find the area of the logo and from that you should be able to figure out the number of tins of paint :smile:
Original post by mmonk
I cant work out the width of logo! there are no axis so cant really find the greatest width

Okay, so at which values of theta are the points on the horizontal axis that are furthest away from the origin?
Can someone help me, with how to calculate the value of a. Can do the rest of the question but can’t figure out how to calculate the value of a.image.jpg
You know that as theta increases y increases, therefore at the point when y is 2 theta is pie over 3. Therefore at that point you know x is 1 and theta is pie over 3. Dunno if that makes sense
Original post by satveermann
Can someone help me, with how to calculate the value of a. Can do the rest of the question but can’t figure out how to calculate the value of a.image.jpg
Yep, I understand what you mean. However the bit I’m still a bit puzzled on is why the x-Coordinate is 1.
Does anyone know where the individual grade boundaries for each 2018 paper are (or if they even exist), I'm curious as to how people did on stats/mechanics
can anyone help? part a i get but i'm struggling a bit with part b
Screenshot (9).png333.1KB
Original post by yerselfesteem
can anyone help? part a i get but i'm struggling a bit with part b


Spoiler

Original post by jason0597
Does anyone know where the individual grade boundaries for each 2018 paper are (or if they even exist), I'm curious as to how people did on stats/mechanics


They are found here: https://qualifications.pearson.com/content/dam/pdf/Support/Grade-boundaries/A-level/1806-a-level-notional-component-grade-boundaries1.pdf

Hard to decipher because of all the papers are numbered. Paper 21 is FP1, 22 FP2, 23 FS1, 24 FS2, 25 FM1, 26 FM2, 27 D1, 28 D2

To give a TL;DR: it was 25 for an A in FS1, 23 for an A in FP1, FS2 and D2, and 24 for an A in every other paper.
(edited 4 years ago)
Original post by _gcx
They are found here: https://qualifications.pearson.com/content/dam/pdf/Support/Grade-boundaries/A-level/1806-a-level-notional-component-grade-boundaries1.pdf

Hard to decipher because of all the papers are numbered. Paper 21 is FP1, 22 FP2, 23 FS1, 24 FS2, 25 FM1, 26 FM2, 27 D1, 28 D2

To give a TL;DR: it was 25 for an A in FS1, 23 for an A in FP1, FS2 and D2, and 24 for an A in every other paper.


Thanks, looks like they are here:

52% for an A in statistics/mechanics :teehee:
(edited 4 years ago)
Original post by jason0597
Thanks, looks like they are here:


Ohhh you meant the Single Maths boundaries, thought you meant FM AS since this is the Further Maths thread :tongue:
Original post by _gcx
Ohhh you meant the Single Maths boundaries, thought you meant FM AS since this is the Further Maths thread :tongue:


oops, i thought this was general maths not further maths

Related discussions

Latest

Trending

Trending

Articles for you