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# C1 Graphs

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1. Find the co-ordinates of the stationary points on the curve with equation y=x(x-1)2.

Find the set of real values of k such that the equation x(x-1)2=k2 has exactly one real root.

I really don't know how to do the second part of the question.
2. (Original post by Julii92)
Find the co-ordinates of the stationary points on the curve with equation y=x(x-1)2.

Find the set of real values of k such that the equation x(x-1)2=k2 has exactly one real root.

I really don't know how to do the second part of the question.
Is that really a C1 question? Because I get the set of values to be using a method I just googled, and it is unlikely that it was a C1 method.
3. (Original post by Julii92)
Find the co-ordinates of the stationary points on the curve with equation y=x(x-1)2.

Find the set of real values of k such that the equation x(x-1)2=k2 has exactly one real root.

I really don't know how to do the second part of the question.
Well since must be a square value:

x-1 = x(x-1) OR (x-1)^2 = x

Then solve for x on both (that's the method I can see at first glance).

In fact, scrap that - it doesn't really incorporate k in any way Doesn't make sense to me - are you sure you've written the question correctly?

EDIT: Scrap all of this. Think about the shape of a cubic graph, and where your stationary points are (y-coordinate wise). You need to make sure both are above y = 0 (which you do by adjusting the value of k)
4. (Original post by atti.08)
This might be wrong but don't you bring the k over and then use the b^2-4ac formula, the discriminant?

Wait don't bother, it'is most likely wrong
It's not a quadratic equation so can't use that method
5. (Original post by atti.08)
This might be wrong but don't you bring the k over and then use the b^2-4ac formula, the discriminant?

Wait don't bother, it'is most likely wrong
That only works with quadratics, unfortunately.
6. (Original post by Zii)
Is that really a C1 question? Because I get the set of values to be using a method I just googled, and it is unlikely that it was a C1 method.
That's the right answer - what method was that?
7. math is my enermy
8. (Original post by Julii92)
That's the right answer - what method was that?
Read my post. You need to consider the shape of a cubic graph.

Look at this working here:

Very nasty C1 question!

(Sorry about the messy start, I screwed up the differentiation )
9. (Original post by Julii92)
That's the right answer - what method was that?
The discriminant of a cubic which satisfies is given by (brace yourself)

When the discrimant is strictly less than 0, this corresponds to the cubic having precisely one real root.

So I took your cubic, , expanded the brackets to get , calculated the discrimant and solved for the discriminant . Following through the working gives

which gives the set of values.

Edit: Should have written , not !
10. (Original post by Zii)
The discriminant of a cubic which satisfies is given by (brace yourself)

When the discrimant is strictly less than , this corresponds to the cubic having precisely one real root.

So I took your cubic, , expanded the brackets to get , calculated the discrimant and solved for the discriminant . Following through the working gives

which gives the set of values.
Much better working with what we already know (look at my post above)
11. (Original post by hassi94)
Much better working with what we already know (look at my post above)
I agree but I like my method because it's cool
12. (Original post by Zii)
I agree but I like my method because it's cool
Meh, personally I think it's 'cooler' to use existing knowledge in a clever way rather than getting a complicated formula to solve it but to each their own

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