Higher Mathematics 2004 Past Paper Help at Stebblog

Higher Mathematics 2004 Past Paper Help

Published by Craig Stebbing March 16th, 2007 in 5M1, Higher and Past Paper Questions.

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.

23 Responses to “Higher Mathematics 2004 Past Paper Help”

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1 Richard Apr 2nd, 2007 at 5:25 pm

Could you help me on the 2004 past paper 1 question 9? I am unsure of how to start it or work through these types of questions.
2 Craig Stebbing Apr 2nd, 2007 at 11:31 pm

Hi Richard. I’ll gladly help you with this type of question. However before I do so could you tell me whether or not you have finished Unit 3 yet. If not, there is a good chance that you are still to be taught about Expontials and Logarithms. If this is the case you don’t have the skills to solve this type of question. Please post back.
3 Richard Apr 3rd, 2007 at 12:09 am

I can confirm i have completed unit 3. that would be brilliant if you are able to help me on this question.
4 Craig Stebbing Apr 10th, 2007 at 5:20 pm

Hi Richard. Sorry I’ve taken so long to get back to you. I’ve been on holiday. I’ve been having trouble with superscript and subscripts within the blog so I’ve had to write your help as a word document. I’ve just posted it as Help for Richard. I hope it makes you feel important, but more importantly I hope it helps you. Please post back for more help if you need it. A small rule we have here is “Ask for help when you need it, give help where you can.” I am sure students from my school will be grateful for your help when you give it.
5 girv May 13th, 2007 at 7:19 pm

oh thats helpful thats the question i was stuck on. also just a wee thing for q10 on the non calc i got 6/10 but i dont undrestand why its negative???
6 Craig Stebbing May 13th, 2007 at 7:36 pm

Hi Girv. I presume you got DEA = 2x + 90.

Expand cos(2x + 90) and you get -sin2x

This is equal to -2sinx cos x

plug in your values for sin x and cos x form triangle DEC and bob’s yer auntie.

Yeeha
7 Christo May 14th, 2007 at 5:54 pm

Hey again…

2004 Paper 2 Q1(a) is defeating me at the moment…

Any help would be appreciated!
8 Craig Stebbing May 14th, 2007 at 6:44 pm

Hi again Christo

Just remember that you have a couple of formulae for gradient

$m= \frac{y_2-y_1}{x_2-x_1}$ and $m = tan \theta$

Using the second formula and by rearranging the formula of the straight line should help you defeat this question.
Mr S.
9 ossie May 14th, 2007 at 7:48 pm

Could you please help me with hugher maths 2004 paper 1 question 1, i really need to know how to do this for tomorrow, please help.
10 Craig Stebbing May 14th, 2007 at 8:10 pm

No probs Ossie

Use simultaneous equations to find the coordinates of B

Then use the gradient formula (see above) to find the gradient of AB (coords of A are given in the paper)

part b

rearrange each equation into the form y = mx + c

get the gradients of each line

if you multiply the gradients of two perpendicular lines together the product is - 1 $(m_1\times m_2=-1)$

do this for both lines and you should find only one line is perpendicular.

Enjoy
11 Helena Dec 26th, 2007 at 5:19 pm

hi please could you help with question 5) from 2004 paper 2? i used differentiation but am now really stuck!
thanks
12 Tazz Jan 25th, 2008 at 7:28 pm

i cant seem to do question 1 in paper 1 :S
13 samjcus May 12th, 2008 at 7:35 pm

Hi Tazz

Ok so its asking for the gradient of AB- your given A= (7,4) and B is where the two given lines cross.
To find B rearrange the equations so they look like similtanious equations then solve them after all thats what similtanious does- it finds where the two equations are equal or where the two lines meet. After finding B use the gradient equation.

Hope this is of some (late) help
14 samjcus May 12th, 2008 at 7:39 pm

Hi Helena

If you’ve differentiated correctly you will have an equation like dy/dx = some quadratic. Remember that dy/dx is just the gradient of the line so plug 12 in as dy/dx and then solve quadratic equation to find a solution- this solution is where m=12.
15 kevin May 19th, 2008 at 3:48 pm

can i have some help with 2004 paper 1 question 11 parts (a) and (b) please i’m not sure how to get the formula for f(x)

thanks kevin
16 Ben May 19th, 2008 at 4:20 pm

Integrate the equation of the curve and you’ll end up with

2x^3 - 6x^2 + c

Which is equal to 4 when f(1)

So:
2(1)^3 - 6(1)^2 + c = 4
2 - 6 + c = 4
c = 8
17 Ryan Smart Dec 7th, 2008 at 3:58 pm

2004, Calculator, Question 9. I can do it. But does anyone think thats its damn hard for 2 marks?! - Smarty
18 Ryan Smart Dec 7th, 2008 at 3:59 pm

part a)…..and 3 marks* (typo)
19 Lucy Feb 1st, 2009 at 7:10 pm

Can anyone help with 2004 paper 2 question 5 part a? I have absolutely no idea…. :S
20 Neo Feb 4th, 2009 at 11:56 pm

Lucy for Q5 part a), the gradient of the tangent to the curve is obtained from the differentiation of that curve y=6x^2 - x^3

Once you have found dy/dx, you can now make this equation equal to 12. i.e. dy/dx = 12. From there you can solve the quadratic equation obatained for x.

Can anyone help with Q11 of 2004 Paper2?

I’ve integrated the parabola equation between limits of y = 2m and 1.5m, but don’t get 2/3 square metres. Get 47/48 instead.
21 Neo Feb 5th, 2009 at 12:39 am

OK solved Q11, made mistake of solving shaded area for limits of y, instead of finding the limits of x. Doh!
22 Neo Feb 5th, 2009 at 12:39 am

Oh, and if this is any help, go here:
http://www.sqa.org.uk/files_ccc/04mi_Maths_H.pdf
23 mmaher Feb 6th, 2009 at 11:38 am

Just checking that I can access the site, having got myself registered on Edubuzz, following Mr. Jones’ AH Maths class this morning.
Mike Maher