P1 Revision topic - 1:15pm 9th June - Edexcel

chats

but on the mark scheme they dont have the value for f(x) but C

i think it's because they don't give you any marks for just writing f(x) again with the 'C' substituted back in....what they're looking for is that you know how to work out C, if you do that you get the marks
but tbh, it is a bit silly of them not to have written the completed version of f(x)

Reply 163

mockel

oh ok, thats what i thought. i'll never understand edexcel

Reply 164

sorry to be a pain and keep asking questions, but how long would it take any of you to do both review exercise 1 and 2 in the P1 book?

Reply 165

chats

sorry to be a pain and keep asking questions, but how lond would it take any of you to do both review exercise 1 and 2 in the P1 book?

i can't really say, but i'd imagine a pretty damn long time....plus i wouldn't get round to finishing it- i'd get bored of the repetitiveness :biggrin:

why do you ask?

Reply 166

mockel

i can't really say, but i'd imagine a pretty damn long time....plus i wouldn't get round to finishing it- i'd get bored of the repetitiveness :biggrin:

why do you ask?

yeh they do tend to repeat a lot.
i asked because ive been on them for so long and just about half way through. just seeing whether i'm slow, which could effect my time in the exam!

Reply 167

chats

yeh they do tend to repeat a lot.
i asked because ive been on them for so long and just about half way through. just seeing whether i'm slow, which could effect my time in the exam!

i don't think you'll have too much of a problem with time in the exam. i can usually do a paper quite relaxed, and still finish it in 45mins -> 1hr, so you should be alright

Reply 168

Well considering there are 72 exam style questions in the review exercise 2 alone, and that they usually have about 7 on a paper, and a paper is 90 minutes...

I wouldn't be worried if I took quite a while to complete them!

Reply 169

KOH

1-) am sure u know how to find the equation of the line given 2 points. once you do that solve simul. equations to find where it cuts the 2nd line (Q) youll have its coordiantes. Now equate the 1st equation to 0 " thats when any curve cuts X axis when Y=0 now you will have 2 cooridantes P<where it cuts X axis> and Q<where is intersects the 2nd line> now find the distance between 2 points P and Q

|PQ| = [ (Y2-Y1) + (X2-X1) ]^1/2

got it?

SHould that not be:

|PQ| = [ (Y2-Y1)^2 + (X2-X1)^2 ]^1/2

Reply 170

mik1a

Sorry, didn't read this part.

Bascally, you are given two variables (but not explicitly told). These can be likened to x and y. You are given enough constraints to write one in terms of the other. You may need to do some substituting.

Once you have them in the form:
eg.
y = x^2 + x + 1

And you have to "find the value of x for which y is minimum", differentiate and set equal to zero:

dy/dx = 2x + 1 = 0
2x = -1
x = -1/2

So this is the value when y is a minimum. Sorry I'm rushed, gtg, I'll explain in mroe detail later, or maybe someone else will beat me to it. These are typical applications of calculus questions.

i'm trying to get a example but i cant open the pdf files on mrhuges

Reply 171

CHeck the review exercise questions 2, there will be some on there. Usually they come with a diagram.

Reply 172

ive found one. can someone answer all parts, i will be very grateful, and also how would the minimum be found?

Reply 173

chats

ive found one. can someone answer all parts, i will be very grateful, and also how would the minimum be found?

(a) new length = (50-2x)
new width = (40-2x)
height = x

Volume = (50-2x)(40-2x)(x)
V = 4x^3 - 180x^2 + 2000x
V = 4x (x^2 - 45x + 500)

Reply 174

(b) not too sure about this. could be something to do with the new length/width not exceeding 50 / 40, i.e.
2x<40
x<20 ???

(c) V = 4x^3 - 180x^2 + 2000x

dV/dx = 12x^2 - 360x + 2000

dv/dx=0, 12x^2 - 360x + 2000 = 0

3x^2 - 90x + 500 = 0

use quadratic formula to solve for 'x'
remember you can only have the positive value of 'x', or the value of 'x' that will mean that the new length and width remain positive

differentiate again d^2V/dx^2 = 24x - 360
put value of 'x' in to prove it's a maximum (hopefully!)

Edit: actually that last bit comes in (e)

(d) just substitute 'x' in V

Reply 175

thanks for that, i get it....i think.

Reply 176

For (b), the answer should be 0 _< x _< 20

Because x obviously cannot be lower than 0 as it is a cut, and any larger than 20 and you will have a negative width, as the shorter width is 40 and you're cutting x out from either side.

Reply 177

nooneknows

I've just been doing the 'examination style paper' in the back of the Revise P1 book, which IMHO is a bit harder than the actual past papers. Anywho...

Question 8

y = x^2 - x + 1 cuts the y-axis at A and has a stationary point at B

a) Find the coordinates of A and B
b) Explain why y = x^2 - x + 1 has no real roots
c) Divide x^3 + 1 by x^2 - x + 1
d) Hence, or otherwise, find the set of values of k for which the equation: x^3 + 1 = kx + k has three real roots.

I can do the first three parts fine (I just thought I better include them so you guys get the context), however, I am nonplussed on part d) and have no idea how to tackle it. As always, any help would be greatly appreciated.

BTW, the answer is k>3/4

Cheers

Reply 178