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2001 paper 2 Question 11b - HSN forum

# 2001 paper 2 Question 11b

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### #1HatingHighers

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Posted 21 March 2005 - 07:43 PM

I'm stuck on the very last question (11b) of the calculator paper from 2001. Anyone kind enough to help?

### #2Floorball Maniac

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Posted 21 March 2005 - 07:59 PM

This is hard! Im not sure myself. I cant do it but this is what I have done so far.

If you dont know, the final answer is 2!

This is what I have done:

y=x+p

If you differentiate y = p + (p-1) - x and work out that gradient, i get
y = p - 1 - 2x

If its a tangent then the gradients are the same.

m = p - 1 - 2x
p = 2x

Some how the x must vanish. Havnt worked out how though.

Hope this is a help. Sorry I have not managed to do it lol. What have you done so far?

### #3HatingHighers

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Posted 21 March 2005 - 08:07 PM

I got it to where you did, lol

I got this:

dy/dx = p - 1 - 2x

m = 1, so

p-1-2x = 1

p = 2x + 2

Then I got stuck...

### #4Floorball Maniac

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Posted 21 March 2005 - 08:12 PM

Oops! I see I made a little mistake.

I have it!
On the diagram, when p is on the y-axis the x is zero. Sub zero in about you get 2!!!!

I hope that is right! Not sure though.

### #5Ally

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Posted 21 March 2005 - 08:23 PM

Equation of curve: y = p + (p -1) - x2

Equation of line: y = x + p

If it's a tangent then both lines are equal to each other.

p + (p -1) - x2 = x + p
p + px - x -x2 = x + p
x2 + 2x - px = 0

Since there is a tangent b2 - 4ac = 0 for equal roots.

x2 + 2x - px = 0
x2 + x(2 - p) = 0

a = 1
b = 2 - p
c = 0

b2 - 4ac = 0
(2 - p)2 - 4(1)(0) = 0
(2 - p)(2 - p) = 0

p=2

Hope that helps...

### #6Floorball Maniac

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Posted 22 March 2005 - 08:37 PM

Thats a big help. Thanks! Even though it wasn't my question I have learned from it.

### #7JiLL

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Posted 14 April 2005 - 05:27 PM

same!!
beast

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