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<a https://www.thestudentroom.co.uk/showthread.php?t=7467335'> Year 12s: what will be the first way you research your future options? >>

C1 - I.Y.G.B paper E help

Question 6:


The curve\mathrm{The\ curve} CC has equation\mathrm{has\ equation}


y=4x5 −1 , x≥0y = 4 \sqrt x^5\ - 1\ , \ x\geq 0


Show clearly that\mathrm{Show\ clearly\ that}

4x2d2ydx2 −15 y=k4x^2 \frac{d^2 y}{d x^2}\ -15\ y=k

Where\mathrm{Where} kk is an integer to be found\mathrm{is\ an\ integer\ to\ be\ found}

I've never covered this so I'm unsure what the interaction is when d2ydx2\frac{d^2 y}{d x^2} Is part of the equation. Or even how to approach this.

Anyway I've done the easy part: d2ydx2 = 15x1/2\frac{d^2 y}{d x^2}\ =\ 15x^{1/2}
(edited 7 years ago)
Reply 1
Avatar for Notnek
Notnek
21
Original post by 34908seikj
Question 6:


The curve\mathrm{The\ curve} CC has equation\mathrm{has\ equation}


y=4x5 −1 , x≥0y = 4 \sqrt x^5\ - 1\ , \ x\geq 0


Show clearly that\mathrm{Show\ clearly\ that}

4x2d2ydx2 −15 y=k4x^2 \frac{d^2 y}{d x^2}\ -15\ y=k

Where\mathrm{Where} kk is an integer to be found\mathrm{is\ an\ integer\ to\ be\ found}

I've never covered this so I'm unsure what the interaction is when d2ydx2\frac{d^2 y}{d x^2} Is part of the equation. Or even how to approach this.

Anyway I've done the easy part: d2ydx2 = 15x1/2\frac{d^2 y}{d x^2}\ =\ 15x^{1/2}

There isn't anything that is beyond C1 here.

The question is just asking you to show that the equation is true where k is an integer.

So multiply d2ydx2\frac{d^2 y}{d x^2} by 4x24x^2 then subtract 15y15y and you should end up with an integer once you've simplified.

EDIT : If you've written the question correctly then your d2ydx2\frac{d^2 y}{d x^2} is wrong.
(edited 7 years ago)
Reply 2
Avatar for Zacken
Zacken
22
If that x5\sqrt{x}^5 is meant to mean x52x^{\frac{5}{2}} then your d2ydx2\frac{d^2y}{dx^2} seems correct. Once you multiply that by 4x24x^2 you are left with 60x5/260x^{5/2}.

Now subtract 15y=15(4⋅x5/2−1)=60x52−1515y = 15(4 \cdot x^{5/2} - 1) = 60x^{\frac{5}{2}} - 15 away from that, what integer are you left with?
Reply 3
Avatar for 34908seikj
34908seikj
OP
2
Original post by Zacken
If that x5\sqrt{x}^5 is meant to mean x52x^{\frac{5}{2}} then your d2ydx2\frac{d^2y}{dx^2} seems correct. Once you multiply that by 4x24x^2 you are left with 60x5/260x^{5/2}.

Now subtract 15y=15(4⋅x5/2−1)=60x52−1515y = 15(4 \cdot x^{5/2} - 1) = 60x^{\frac{5}{2}} - 15 away from that, what integer are you left with?


Ah, so It would just be 15, K=15.

Since this is obviously not C1, would this be covered in C2. If not what unit?
Reply 4
Avatar for Zacken
Zacken
22
Original post by 34908seikj
Ah, so It would just be 15, K=15.

Since this is obviously not C1, would this be covered in C2. If not what unit?


Yeah, sounds about right.

idk, looks like standard C1 to me.
Reply 5
Avatar for Notnek
Notnek
21
Original post by 34908seikj
Ah, so It would just be 15, K=15.

Since this is obviously not C1, would this be covered in C2. If not what unit?

Ignore what I said about your d2ydx2\frac{d^2y}{dx^2} being wrong.

This is a C1 question. Maybe the question is a bit different to something that you would see in e.g. an Edexcel paper.

What about it makes you think it's beyond C1?
Reply 6
Avatar for 34908seikj
34908seikj
OP
2
Original post by notnek
Ignore what I said about your d2ydx2\frac{d^2y}{dx^2} being wrong.

This is a C1 question. Maybe the question is a bit different to something that you would see in e.g. an Edexcel paper.

What about it makes you think it's beyond C1?


Maybe it just uses concepts from C1, just on steroids.

I've done every official Edexcel C1 exam paper, and I've never come across a question like this.


The most you do with d2ydx2\frac{d^2y}{dx^2} is literally just differentiating a second time, that's it.
Reply 7
Avatar for math42
math42
20
Original post by 34908seikj
Maybe it just uses concepts from C1, just on steroids.

I've done every official Edexcel C1 exam paper, and I've never come across a question like this.


The most you do with d2ydx2\frac{d^2y}{dx^2} is literally just differentiating a second time, that's it.

It's not on steroids, it is a fairly simple application, just somewhat unusual for Edexcel. It is good to look at IYGB because its slightly offbeat questions like this one are still entirely on the Edexcel syllabus and might be similar to the sort of thing Edexcel sometimes throw in to differentiate (not in the maths way..) at the top level.
Reply 8
Avatar for 34908seikj
34908seikj
OP
2
Original post by 1 8 13 20 42
It's not on steroids, it is a fairly simple application, just somewhat unusual for Edexcel. It is good to look at IYGB because its slightly offbeat questions like this one are still entirely on the Edexcel syllabus and might be similar to the sort of thing Edexcel sometimes throw in to differentiate (not in the maths way..) at the top level.


Yup, completely agree. I've learnt some really neat things from doing I.Y.G.B papers.
Reply 9
Avatar for Zacken
Zacken
22
Original post by 34908seikj
Maybe it just uses concepts from C1, just on steroids.


No, it is a simple and standard concept from C1. This would have been a very easy GCSE question for me.
Reply 10
Avatar for 34908seikj
34908seikj
OP
2
Original post by Zacken
No, it is a simple and standard concept from C1. This would have been a very easy GCSE question for me.


Well that's good for you & I'm sorry I'm not as "good" at maths as you were when you were doing GCSEs/ AS's.


It's clearly not a standard C1 concept, at least on an Edexcel paper, given by the fact I've never seen that type of question before. That said no need to be so aggressive man :/

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