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Happy revising.
Collaborative Mathematics
Post your problem here and hopefully someone will get back to you quite quickly. State the question that is giving you trouble and what you’ve tried so far.
Happy revising.
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okay, i cannae get the answers to 3b), 10b) and 12b) on paper 2…. got a D in the actual papers last year… when they werent like past papers but present papers… the worst present ever!! lol… so yeha… i think paper one was fine… cannae mind most of it though… blank out the bad bits me thinks! TA! K
I’m not quite smart enough to get all the answers. But my Enlgish is impecable. Advanced Higher baby! And I just loooove my teacher ‘Christypoo’ i want his babies haha
omg, lorna? is that you? hey….! why his babies…. freakin nutjob! Mr Chrustie! lol… ah dear…
Hey K and L. I think you’re new collaborators here. Welcome. I believe you are not from NBHS like us as I’m not aware of any English teachers going by the name Christepoo. It does worry me that L seems to have a wee crush on her teacher however. I hope you will find this blog useful over the next few weeks. We do have a few rules here. The first is no sweary words, the second is ask for help and the third is try and give help if you can. We are all in this together.
Q3B) If you put the straight line equation from part A equal to the second parabola and then solve what you get, you will find the point of contact.
Q10B) You could try differentiating. Remeber the derivative is also the gradient function so gradient = derivative. You are told the gradient is 1 so f’(x)= 1
Q12b) What a beautiful question. Differentiate. Find your turning points between x=1 and x = 4 (these are the limits of the curve from the diagram). Evaluate the area at you limits and t.p.s and decide which are the greatest and least values.
Hope you find this useful and please comment back in the future.
One other thing. I’m not awfully great at text/youngling speak and would find it easier if you didn’t use it. I had to ask my class earlier to translate some of it for me. I only found out a couple of weeks ago that lol is not a reference to a lollipop
haha that amused me mr stebbing! first time ive smiled all night
HI
CAN I HAVE PDF FILE FOR NOVEMBER -OCTOBER MATH PAPER 2 AND 4 2006
IMAD
Hi Imad. I’m not sure what you are wanting exactly. Can you be more specific please? Thanks
IMAD, has a caps lock problem. I normally ignore people with a caps lock problem.
Thanks for the solution to 12b Craig.
Would I have got the marks if i’d just run x=1 to x=4 ( 0>x
Hi,
How can I open .flp files on Windows XP, Office, PC ?
Robin.
Mr Stebbing,
Thanks for the help on question 12b. I’m a little shady on it though, I can get the right answer by just plugging into the formula all the allowed values of x and selecting minimum and maximum.
But if I differentiate and work through the problem like they did on:
http://www.invergordon.highland.sch.uk/departments/maths/higher/resources/higher2006p2solutions.pdf
(last page), I only get the maximum turning point. I still have to plug in the end-points for the minimum.
My question is, why not just do it the simple way? Or is it because x could be a decimal value, and therefore my method would not be accurate? If then, why doesn’t differentiating give me the minimum stationary points ass well?
A tad confused! Thanks -
James
Okay so I thought the first post didn’t get through. My bad.
im stuck on qu10 on 2006 non calc, any help would be appreciated, duncan says hi. will check l8r for any resaponse big man
It’s a logarithmic scale, so it goes up logarithmically, it has to increase by 4 times as much each time to increase by 1 (write log to the base 4 of 4 then 16 then 64, see what I mean?). Because the graph being plotted on this scale is exponential, it increases by ‘a’ times as much each time, it makes a straight line. They kind of cancel out.
If you take the point the question gives you, it says the y value is 3 but we need the real y value. To get it just make an equation; we know its log to the base 4 of something so:
log to the base 4 (y) = 3
then do the re-organising thing
y = 4 to the power of 3
then substitute into the equation with the x value 6. The number that a equals will be easy to guess because you can’t solve the equation with logs in a non-calc paper.
James.
Hi James. Firstly I don’t mind caps lock problems. Imad maybe wanted to stand out from the crowd. I as a teacher would never stand in his way. Hey ho.
Now to the serious stuff. You wouldn’t get the marks if you plugged values into a calculator. The marks are gained from the differentiation/nature table side of things. If you imagine any function , if you want the maximum and minimum values of the function over a limited domain, these will occur either at the turning points or on the limits of the domain.
Consider the function
If you have a domain of
the minimum value of the function is at the turning point and the maximum value is at x=3
the minimum value will occur at x =3 and the maximum value at x = 7 .
If you limit the domain to
Questions like 12 are testing whether or not you are able to make an informed judgement on the solution and not whether you can find the answers by trial and error. I hope this has been helpful and please post again.
Robin
In order to open *.flp files you have to download Promethean flipchart viewer or Activstudio student edition. You can find a link to both of these under my post on Flipchart Viewer (There’s a category on the right hand side of this page.) If your school has promethean Interactive whiteboards you can use either program but if it doesn’t you are only allowed to download viewer. I hope this helps.
Hi Hannah.
Thanks to Gav for his help. However I think it would be easier to log both sides of the equation :
=>
leading to
using the log power rule
substitute in the values from the graph and you will have and equation with one unknown (a). Convert this back into exponential form and bob’s yer auntie. I’ll let you try the last bit yourself and if your still unsure post back. Say hi to Duncan for me. Hope ST Andrews is treating him well.
thanks i will try that and get back to you. he was jst at home and wrote that last comment, hence the “big guy” , jst so you dont think i am making personal attacks on you.
thanks
Hi Hananah (can’t you spell your name or the word just for that matter). No offence taken. Lets be honest I am pretty Big at 18 and a half stone. I hope you manage to work that out.
for paper 2 q.10 (b) I’m unsure of why in the answers a=1.29 I got 73.3 which is ever so slightly different.
I’m also kinda stuck on 6(c) and 11
oops sorry 10 part (a) was what I meant to say
Yeah, I got that for 10 part (a) as well! 73.3 seems a much more likely answer for an angle than 1.29 i would say!
Hi Pixie. It’s been a wee while. If you look at the question it is dealing in radians. You’re using degrees. Multiply 73.3 by pi/180 and you should be ok. Setting the calculator in Rad mode from the start would be better though. (Same goes for you Donald)
As for 6 C) If the magnitude of vector PQ is 5units a unit vector has a magnitude of one fifth of this (ie. 1 unit). SO you have to take Vector PQ and divide all components by 5.
And finally Q11.
The amount in the living tree is Ao so the question says A(t) = 0.88 Ao
=>
Taking the natural log (ln) of both sides and using your log rules you should be able to solve the equation and get 1030.9 years which is not 1000 so the calim is false. Post back if you have trouble with the logging.
Mr S.
yup Dave’s having problems with the logging
worked solutions of both 2006 papers here http://www.maths-exam-solutions-scotland.co.uk/ if anyone is still having problems.
Yeah, just noticed that the question asked for radians! However, for Q11 the claim is “that a wheel is made from wood which is over 1000 years old” and so is proved to be true by the solution of 1030.9 years!
I do apologise. Remember to read the question folks. It did indeed claim the tree was over 1000 years old and not 1000 years old as I thought. I wonder if they “watered” the tree, and didn’t chop it down, would it still be standing today?
Thanks for the link paper. Another useful link is this
http://www.invergordon.highland.sch.uk/
If you find the departments tab at the very bottom of the page go to the maths department and you will find solutions for 2005 as well. There are also SQA marking schemes for 2003 -2005 as well but these are less useful to you guys as they don’t always have full solutions. Thanks to James a couple of nights ago for this link.
Who wants solutions though? It’s far more fun to learn the process by collaborating.
Hi stebo and crew, iv been looking at this paper and got stuck on question 5 and 7 … i dont have a clue where to start. Q5 is a question of stationary points and its nature, so i expect nature tables etc but just need help getting there. Q7 is trigonometric formulae possibly but am bewildered about what to do.
it would be good if someone could help if possible thanks.
S.cunning x
oh yeah this is on paper 1. actually can u also explain 9. d
thankyou stebblog! x
Hi
Q5 Stationary points occur when f’(x)=0. So that is what you need to do (using the chain rule)then solve to find values of x. You then use the nature table to test values of f’(x) around each stationary point. This tells you if the curve is increasing or decreasing.
Q7 You need double angle formula - remember sin2x=2sinxcosx. You can then take out a common factor and solve
Q9d You need scalar product cos(theta)=…. it is on the formula sheet. You just need to find u.v and the magnitudes of the 2 vectors then substitute into this formula.
Hope this helps
Q(10) paper 1
Two variable x, and y are connected byt the law y=a^x. THe graph of log4y against x is a straight line passing the origin and point (6,3) find th e value of a
Well, if y=a^x, then
log_4(y)=log_4(a^x),which equals x*log_4(a) by the laws of logarithms. Compare this with the graph, think about gradients and you’ll be there.