addisonc121
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This is question seven from OCR January 2008. I'll post a photo below
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nerak99
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It just so happens that I have this question already type up
A curve has equation\frac{xe^{2x}}{x+k} where k is a non-zero constant.
Differentiate x e^{2x} , and show that \frac{{\rm{d}}y}{dx}=\frac{e^{2x}(2x^2+2kx+k)}{(x+k)^2} [5]
Given that the curve has exactly one stationary point, find the value of k, and determine the exact coordinates of the stationary point.
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Zacken
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(Original post by addisonc121)
This is question seven from OCR January 2008. I'll post a photo below
What question?
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the bear
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(Original post by nerak99)
It just so happens that I have this question already type up
A curve has equation\frac{xe^{2x}}{x+k} where k is a non-zero constant.
Differentiate x e^{2x} , and show that \frac{{\rm{d}}y}{dx}=\frac{e^{2x}(2x^2+2kx+k)}{(x+k)^2} [5]
Given that the curve has exactly one stationary point, find the value of k, and determine the exact coordinates of the stationary point.
it looks like you need to use the Quotiont Rule to differentiate the big fraction ?

then remember a fraction can only be zero if the top is zero ?
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nerak99
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The solutions start like this
The numerator of this quotient is a product and so when we require du in differentiating the quotient we will need to have differentiated the product x e^{2x} already.

The differential ofx e^{2x} u(x)=x \,,\;v(x)=e^{2x} is from the product formula f^\prime(x) ={u{\rm{d}}v+v{\rm{d}u}} giving us the differential of the numerator as (x \times2e^{2x})+(e^{2x})=e^{2x}(2x+1)

The rest then flows as follows 

\begin{aligned}f(x)&=\frac{xe^{2x}}{x+k}=\frac{u(x)}{v(x)}\\ 

f^\prime(x)&=\frac{v{\rm{d}}u-u{\rm{d}v}}{v^2}\quad&=\frac{(x+k) \left[ e^{2x}(2x+1)\right]-xe^{2x}.1 }{(x+k)^2 }\\

\end{aligned}

Which will lead you to =\frac{e^{2x}\left (2x^2+2xk+k \right)}{(x+k)^2}
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nerak99
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The last part, which requires a bit of insight, flows from realising that if there is only one solution to the differential being zero then the discriminant of the quadratic in the numerator must be zero in order to have a repeated root. (This is a bit of a favourite with this particular C3)
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nerak99
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(Original post by Zacken)
What question?
If that is a response to me? The heading said Core 3 OCR and the body of the post said Q8 Jan 2008. Which I had already typed up for a different purpose.
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Zacken
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(Original post by nerak99)
If that is a response to me? The heading said Core 3 OCR and the body of the post said Q8 Jan 2008. Which I had already typed up for a different purpose.
Just realised the OP says "This is question seven from OCR January 2008. I'll post a photo below" I skipped over the "question seven" bit for some reason. :lol:
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rainbowcolours1
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These youtube videos should help you out.

https://www.youtube.com/watch?v=23g9...3KaYW&index=14

https://www.youtube.com/watch?v=TAqd...3KaYW&index=15
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nerak99
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(Original post by Zacken)
Just realised the OP says "This is question seven from OCR January 2008. I'll post a photo below" I skipped over the "question seven" bit for some reason. :lol:
I did exactly the same the other day, it was a scrappy post really. A comprehensive reply to a question where the forum had to find the question.
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