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Tricky Differentiation Question GCSE FURTHER MATHEMATICS

Hello! I do ccea further maths, in northern ireland ...so im in year 12 here, but it's the equivalent of year 11 in England...moving on from the blabber..

I have a maths mock on this thursday and it's everything we have covered in pure maths excluding logs(we havent finished it yet). So i was doing one of the old 'Additional Mathematics papers for pure Maths(PAPER 1) of May/June 2005 by CCEA.

I came across this deadly question, can anyone help me out please? Mock on thursday.

I can't find the mark scheme anywhere on the interweb :smile: and can you tell me whereabouts i went wrong please because i only got up to the differentiation part and got dy/dx= 2ax-5, with b=-1, i don't know what a=...c=4. It's mainly just part (i) that i got stuck on, which prevented me from doing the rest of the question! Please help! I have posted this exact thread on the Maths Exams forum.

I have attached the question http://www.thestudentroom.co.uk/attachment.php?attachmentid=481431

Thank you

RevisionLad
Reply 1
Original post by revisionlad
Hello! I do ccea further maths, in northern ireland ...so im in year 12 here, but it's the equivalent of year 11 in England...moving on from the blabber..

I have a maths mock on this thursday and it's everything we have covered in pure maths excluding logs(we havent finished it yet). So i was doing one of the old 'Additional Mathematics papers for pure Maths(PAPER 1) of May/June 2005 by CCEA.

I came across this deadly question, can anyone help me out please? Mock on thursday.

I can't find the mark scheme anywhere on the interweb :smile: and can you tell me whereabouts i went wrong please because i only got up to the differentiation part and got dy/dx= 2ax-5, with b=-1, i don't know what a=...c=4. It's mainly just part (i) that i got stuck on, which prevented me from doing the rest of the question! Please help! I have posted this exact thread on the Maths Exams forum.

I have attached the question http://www.thestudentroom.co.uk/attachment.php?attachmentid=481431

Thank you

RevisionLad


Really think about what values you're inputting into your derivative. I assume you've gotten to that stage. When the gradient is -5 the curve crosses the y-axis, right? Think about what values will go into your derivative function from here on.
Original post by revisionlad
Hello! I do ccea further maths, in northern ireland ...so im in year 12 here, but it's the equivalent of year 11 in England...moving on from the blabber..

I have a maths mock on this thursday and it's everything we have covered in pure maths excluding logs(we havent finished it yet). So i was doing one of the old 'Additional Mathematics papers for pure Maths(PAPER 1) of May/June 2005 by CCEA.

I came across this deadly question, can anyone help me out please? Mock on thursday.

I can't find the mark scheme anywhere on the interweb :smile: and can you tell me whereabouts i went wrong please because i only got up to the differentiation part and got dy/dx= 2ax-5, with b=-1, i don't know what a=...c=4. It's mainly just part (i) that i got stuck on, which prevented me from doing the rest of the question! Please help! I have posted this exact thread on the Maths Exams forum.

I have attached the question http://www.thestudentroom.co.uk/attachment.php?attachmentid=481431

Thank you

RevisionLad


Hi :smile: This is an interesting question. You've correctly stated that dy/dx is in fact -5 when x = 0. This means that the x value returns 0, and since the form of a quadratic derivative is 2ax+b, then ax = 0 when x is 0, therefore meaning b is -5. So B is -5.
You have also said that when the curve crosses the the x axis the y value is 0. This means that when the curve crosses at (1,0), x = 1 when y = 0.
This can then be solved from here, a(1)^2 - 5*1+c = 0. where c is the y intercept (4).
The rest is basic quadratic maths. If you need further help let me know!!
(edited 8 years ago)
Original post by The-Spartan
Hi :smile: This is an interesting question. You've correctly stated that dy/dx is in fact -5 when x = 0. This means that the x value returns 0, and since the form of a quadratic derivative is 2ax+b, then ax = 0 when x is 0, therefore meaning b is -5. So B is -5.
You have also said that when the curve crosses the the x axis the y value is 0. This means that when the curve crosses at (1,0), x = 1 when y = 0.
This can then be solved from here, a(1)^2 - 5*1+c = 0. where c is the y intercept (4).
The rest is basic quadratic maths. If you need further help let me know!!


Thankyou very much spartan! I realised i was overcomplicating way too much just to find 'a' when i should have subbed (1,0) into my formula..would have got it easily then...i think i got confused earlier and didn't know what to do with (1,0) as i was too focused on getting 'a' any way possible...thanks for ure help

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