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Edexcel FP1 Thread - 20th May, 2016

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Original post by alfmeister
Part a, I tried to find the gradient by doing (4/p - 4/q)/ (4p - 4q) but I am struggling to find it in a simpler form.


4p−4q=4q−4ppq\displaystyle \frac{4}{p} - \frac{4}{q} = \frac{4q - 4p}{pq}

So 4q−4ppq4p−4q=4q−4ppq×14p−4q=−1pq\displaystyle \frac{\frac{4q-4p}{pq}}{4p-4q} = \frac{4q-4p}{pq} \times \frac{1}{4p-4q} = -\frac{1}{pq}
Original post by Zacken
4p−4q=4q−4ppq\displaystyle \frac{4}{p} - \frac{4}{q} = \frac{4q - 4p}{pq}

So 4q−4ppq4p−4q=4q−4ppq×14p−4q=−1pq\displaystyle \frac{\frac{4q-4p}{pq}}{4p-4q} = \frac{4q-4p}{pq} \times \frac{1}{4p-4q} = -\frac{1}{pq}


I got to the second last bit just before you obtained -1/pq, however I am struggling to see how you can cancel out because on top you have 4q - 4p and on the bottom 4p - 4q. I might just be missing something obvious.
Original post by kingaaran
Here: http://goo.gl/s7vHxb

Just put a random function in there (with max degree 3), see what the result is and see if you can show it using the standard formulae :smile:


That's pretty neat thanks :smile: (though I don't know what degree is in this case lol, I'm guessing r and n terms)
Original post by alfmeister
I got to the second last bit just before you obtained -1/pq, however I am struggling to see how you can cancel out because on top you have 4q - 4p and on the bottom 4p - 4q. I might just be missing something obvious.


(4q−4p)=(−1)(4p−4q)(4q-4p) = (-1)(4p-4q)
Original post by yesyesyesno
That's pretty neat thanks :smile: (though I don't know what degree is in this case lol, I'm guessing r and n terms)


It's the highest power of r.

So 3 degree max means don't put something like "sum of r4r^4". Stick to "sum of r3+2r2+⋯r^3 + 2r^2 + \cdots" or whatever, don't make the highest power of r be higher than 3.
Original post by Zacken
(4q−4p)=(−1)(4p−4q)(4q-4p) = (-1)(4p-4q)


Thought it would be something rather obvious, thanks a lot.
Original post by yesyesyesno
That's pretty neat thanks :smile: (though I don't know what degree is in this case lol, I'm guessing r and n terms)


For example, r^3+r^2+r+1 has degree 3.

Just don't start going too crazy and summing terms like r^1000 because there are no standard formulae for it that you need to know :tongue:
Original post by alfmeister
Thought it would be something rather obvious, thanks a lot.


No problem.
Hi, we were taught to memorise that the gradient for a point on a hyberbola is -y/x . Would I be able to just quote this in the exam, and then substitute the points in? Or would I have to differentiate it every time? (I suck at differentiating lol)
Original post by kingaaran
For example, r^3+r^2+r+1 has degree 3.

Just don't start going too crazy and summing terms like r^1000 because there are no standard formulae for it that you need to know :tongue:


cheers lol after all we only have formula for up to r^3
Original post by emilysmith268
Hi, we were taught to memorise that the gradient for a point on a hyberbola is -y/x . Would I be able to just quote this in the exam, and then substitute the points in? Or would I have to differentiate it every time? (I suck at differentiating lol)


You would need to differentiate every time. You'd get no marks for quoting it. Memorising that bit was useless, I'm afraid.
guys you see divisibility tests, do you find the way you do the question a bit random sometimes?

I have a method that after assuming f(k), I see if I get something useful from f(k+1) - f(k) but then I usually end up doing f(k+1) + f(k) and use my calculator to see how useful integer multiples of f(k) expanded are if that makes sense.

I get the proof correct but I want to ask is there a consistent method that works for you guys?
Original post by yesyesyesno
I get the proof correct but I want to ask is there a consistent method that works for you guys?


I just use f(k+1)−f(k)f(k+1) - f(k) and it works every time.
Original post by Zacken
I just use f(k+1)−f(k)f(k+1) - f(k) and it works every time.


yep that's always my first way but after writing like 2 lines of working I imagine an easier way using + integer times f(k) using my calculator :tongue:

btw are you actually sitting this exam? I'm pretty sure I've seen your results somewhere with FP1 in them.
Original post by yesyesyesno
yep that's always my first way but after writing like 2 lines of working I imagine an easier way using + integer times f(k) using my calculator :tongue:

btw are you actually sitting this exam? I'm pretty sure I've seen your results somewhere with FP1 in them.


Bleh, there's not much difference.

Nah - I've already sat C1-4, M1-3, S1-2, FP1 in Jan.
Original post by Zacken
You would need to differentiate every time. You'd get no marks for quoting it. Memorising that bit was useless, I'm afraid.


I was just doing a few questions on this to practice. Would writing y + x(dy/dx) = 0, therefore dy/dx = -y/x show that I've differentiated it?
Original post by emilysmith268
I was just doing a few questions on this to practice. Would writing y + x(dy/dx) = 0, therefore dy/dx = -y/x show that I've differentiated it?


Perfectly so.
Original post by yesyesyesno
guys you see divisibility tests, do you find the way you do the question a bit random sometimes?

I have a method that after assuming f(k), I see if I get something useful from f(k+1) - f(k) but then I usually end up doing f(k+1) + f(k) and use my calculator to see how useful integer multiples of f(k) expanded are if that makes sense.

I get the proof correct but I want to ask is there a consistent method that works for you guys?


Yh I just add or subtract whatever multiple of f(k) makes it easiest
Original post by Zacken
Perfectly so.


thanks so much! i don't know what my teacher was thinking, telling us it was ok to skip that step

just one more thing, for the gradient of parabolas, could you do 2y(dy/dx)=4a, so dy/dx = 4a/2y ?
Original post by emilysmith268
thanks so much! i don't know what my teacher was thinking, telling us it was ok to skip that step

just one more thing, for the gradient of parabolas, could you do 2y(dy/dx)=4a, so dy/dx = 4a/2y ?


Perfect again!

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