Higher Mathematics 2005 Past Paper Help
Published March 16th, 2007 in 5M1, Higher, Past Paper QuestionsPost 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.
Total stuck on question 2, paper one. I thought it seemed really easy at first but now I have no idea of what to do. Please assist immediately lol
Hi Ross. This question is one of those pains in the backside that could lead you up the garden path if you let it. There is no substitution into circle equations or anything like that needed. As the circles are congruent they must have the same radius. This means that P is the mid point of the line AB. Using the formulae list you can get the coordinates of the centres of each circle then use midpoint formula. I think you’ll be able to attempt part b using the same kind of thinking (ie. use distance formula) Hope your holiday is going well. Great weather for revision isn’t it.
Paper 1 Question #8 part c is annoying the heck out of me - I’ve got the max value correct as 9, but the answers state the min value as -35 and all I can muster is -3. The function doesn’t even seem to go that low at all! Where have I gone wrong? (Parts a and b are correct)
Also, question #9 on the same paper has completely flummoxed me - I can get the right answers, but I’m fairly sure it’s the wrong way. What sort of direction should I go in?
Thanks!
it does go that low, just write it into your graphics calculator.
oops
Hi Sam. in answer to #8 c). When you differentiate you’ll get your turning points at x= 0 and x=7/3. Justify these using a nature table. However your minimum value in this case does not lie on one of the turning points. It is on one of your domain limits. For a better explanation of this look at my reply to James on the past paper help for 2006. I hope this helps.
In reply to Gavin it’s a non-calculator paper so you wouldnt be able to do that in the exam !!
Well spotted Martin. Calculators ,who needs them!
would it be possible to receive some help as to how to go about q9 please?
I’ve expanded to: cos^2x -sin^2x
but no clue as to what next. thanks for all you’re other help as well!!!
Hi Pixie.
Use another expansion ie.
Also remember that
square root something at some point is my hint.
cheers I think I was being a bit thick on that 1 and probably this 1 too seeing as it’s only worth 3 marks, q2 has baffled me and i promise this is the last one as any more after now I probably wouldn’t take in anyway, just have to hope nice questions come up tomorrow and I don’t make my stupid mistakes like 1×0 = 1 or something!
he he meant to say q7
hi, i’m really confused about question 7 on paper one, how on earth do you get a=4 from the grap!!!
I have absolutly no idea how to do paper one question 9
cos2x= 7/25
im guessing expand but i got stumped with the fraction!
please help maths exam is a week away ha
Hi Sith
Try using Trig formulae particually [tex] \cos2x = 2\cos^2 x - 1 [tex]
If your having trouble getting some of the solutions and you can’t get hold of a teacher or anyone else- there are very detailed solutions at http://www.sqa.org.uk/sqa/2470.html .
They can be very helpful and there are full solutions for most other subjects and past papers on the SQA site.
Hope I’ve been of some help
Q7 is a tricky little number,
Think of normal log x graphs you’ve come across before –> all of them go through the point (1.0), if the point at y=0 is 5 then the graph has been moved to the right by 4 hense a=4 where f(x) = log b (x-a)
Hi, ive tried doing q(8c) on paper one nad i have looked at the past advice and on the other post, but still cant find the minimum value of ‘f’. i have found that 9 is from the nature table, but i dont understand how you can get -35.
p.s this has been a very useful blog
Ru
i just looked on the link to the sqa website and i have found that you have to do this f(2)=(2x^3)-(7x^2)+9 why would you do this? would you substitute every number of x until you found the lowest?
Ru
i have also got stuck on q(11b) i have used the discriminant to get to (4t^2)-20([t^2]-4)=0 i dont understand how to get from there to t=(5^1/2)
how do u do question 10? ive been tryin for ages but still i am getting it wrong
hey can someone help me with Q.10 on paper 2?