Higher Mathematics 2004 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.
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.
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.
I can confirm i have completed unit 3. that would be brilliant if you are able to help me on this question.
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.
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???
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
Hey again…
2004 Paper 2 Q1(a) is defeating me at the moment…
Any help would be appreciated!
Hi again Christo
Just remember that you have a couple of formulae for gradient
Using the second formula and by rearranging the formula of the straight line should help you defeat this question.
Mr S.
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.
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
do this for both lines and you should find only one line is perpendicular.
Enjoy
hi please could you help with question 5) from 2004 paper 2? i used differentiation but am now really stuck!
thanks
i cant seem to do question 1 in paper 1 :S
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
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.
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
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