STEP 2003 Solutions Thread Watch

toasted-lion
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#61
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#61
That was an impressively quick update on the original post Simon. Would you be able to fix all the dead links to other years' solutions some time maybe?
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SimonM
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#62
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(Original post by toasted-lion)
That was an impressively quick update on the original post Simon. Would you be able to fix all the dead links to other years' solutions some time maybe?
I wasn't aware that there were any dead links, if you would care to show me the threads, I'll do it as soon as I can
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toasted-lion
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#63
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(Original post by SimonM)
I wasn't aware that there were any dead links, if you would care to show me the threads, I'll do it as soon as I can
None of them link here, and very few link to 2002, even though the pages exist. Check a couple, you will see what I mean.
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SimonM
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#64
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(Original post by toasted-lion)
None of them link here, and very few link to 2002, even though the pages exist. Check a couple, you will see what I mean.
I cordially invite you to slap me in the face for misreading what you said and wasting 15 minutes checking every link to a solution on 2002 and 2003 before realising you meant making the solution threads link to one another!
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toasted-lion
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(Original post by SimonM)
I cordially invite you to slap me in the face for misreading what you said and wasting 15 minutes checking every link to a solution on 2002 and 2003 before realising you meant making the solution threads link to one another!
If it makes you feel better, have a preverbial slap in the face. Sorry if my post was ambiguous. Seriously though, much respect for organising these threads, they're invaluable.
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toasted-lion
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(Original post by Adje)
III/1

 \displaystyle \frac{d}{dx} \arcsin \left(\frac{x-1}{x+3}\right)

 \displaystyle = \frac{\sqrt{4}}{(x+3)\sqrt{2+2x}  }}}

 \displaystyle = \frac{2}{\sqrt{2}}\frac{1}{{(x+3  )\sqrt{1+x}}}}}

 \displaystyle \Rightarrow \int \frac{1}{{(x+3)\sqrt{x+1}}}}} \ dx =  \frac{\sqrt{2}}{2}\arcsin \left(\frac{x-1}{x+3}\right) + C
This is what I got too, but it says it has to be valid for x>-1. Isn't the above only valid for x>1?
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Oh I Really Don't Care
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(Original post by toasted-lion)
This is what I got too, but it says it has to be valid for x>-1. Isn't the above only valid for x>1?
 |\frac{x-1}{x+3}| \le 1
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toasted-lion
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#68
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(Original post by DeanK22)
 |\frac{x-1}{x+3}| \le 1
In the question it says "given that x + a > 0 and x + b > 0", but I am not sure whether this means it's not valid if that doesn't apply?
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Oh I Really Don't Care
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#69
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(Original post by toasted-lion)
In the question it says "given that x + a > 0 and x + b > 0", but I am not sure whether this means it's not valid if that doesn't apply?
x >= -1 is required for |(x-1)/(x+3)| <= 1

... however I will look at the question now.

edit; well the conditions are there so you don't have to start using complex numbers.

i.e.

sqrt(a-b) and it is given a > b and sqrt(2x+2) , etc

x is not => -1 but > -1 as if x = -1 you would get division by 0.

I wonder how they spot these things - it is certainly noon trivial approach to guess such functions work.
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toasted-lion
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(Original post by DeanK22)
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x >= -1 is required for |(x-1)/(x+3)| <= 1

... however I will look at the question now.

edit; well the conditions are there so you don't have to start using complex numbers.

i.e.

sqrt(a-b) and it is given a > b and sqrt(2x+2) , etc

x is not => -1 but > -1 as if x = -1 you would get division by 0.

I wonder how they spot these things - it is certainly noon trivial approach to guess such functions work.
So do you think that in this particular question they just want you to do the algebra rather than worry about ranges of validity?
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Oh I Really Don't Care
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#71
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(Original post by toasted-lion)
So do you think that in this particular question they just want you to do the algebra rather than worry about ranges of validity?
Tey would want you to realise that it only works for x > y or x < y, et ; and in the off chance that condition is not given you should say that condition
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toasted-lion
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#72
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STEP III 2003, Q4

Part 2, somebody please put me out of my misery and tell me where I'm going wrong.

 t_1 = \frac{1 - t_0^2}{2t_0} = \frac{1 - tan^2 \frac{7 \pi}{18}}{2tan \frac{7 \pi}{18}} = \frac{1}{tan \frac{7\pi}{9}}



t_2 = \frac{1 - t_1^2}{2t_1} = \frac{1 - \frac{1}{tan^2 \frac{7 \pi}{9}}}{\frac{2}{tan \frac{7 \pi}{9}}}} = \frac{tan^2\frac{7\pi}{9} - 1}{2tan \frac{7\pi}{9}} = \frac{-1}{tan\frac{14\pi}{9}}

Edit: Ok, I am pretty sure the above is not wrong, but I am also pretty sure it is not equivalent to what Tom got as t2.

 \frac{-1}{tan\frac{14\pi}{9}} = - cot \frac{5\pi}{9} = - tan(\frac{\pi}{2} - \frac{5\pi}{9}) = - tan(-\frac{\pi}{18}) = tan\frac{\pi}{18} \not= tan\frac{14\pi}{9}

But my t2 does yield t3 as cot(28pi/9) just the same, which does give the right answer. So I'm quite confused about Tom's post now. Perhaps you could clarify this:

(Original post by tommm)
STEP III 2003, Q4
We can also apply the formula similarly to the cotangent function, so

t_2 = \tan (\frac{14\pi}{9})
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nchen5
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#73
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i need solution to STEP III question 5!
I cant finish the question!
Thanks !
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toasted-lion
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#74
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(Original post by nchen5)
i need solution to STEP III question 5!
I cant finish the question!
Thanks !
I've had a drink now (and maths is a lot harder after a drink) but I'll have a go tomorrow and get back to you.
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SCE1912
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#75
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III Q5.

Spoiler:
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Completing the square on the quadratic
 y = x^2 - 2bx + c = (x-b)^2 + c - b^2
so turning point is  (b, c - b^2)
For two real roots we want the minimum point below the x-axis
so
 c - b^2 &lt; 0 \Rightarrow b^2 &gt; c

For the cubic's turning points set so  \frac{dy}{dx} = 0 to get
 3x^2 - 3b^2 = 0 \Rightarrow x = b \mbox{ or } x = -b
this makes the turning points  (-b, 2b^3 + c) and  (b, c - 2b^3) the former is a max the latter a min.

For three root we want the max above the x-axis and min below the x-axis so

 2b^3 + c &gt; 0 \mbox{ and } c - 2b^3  &lt; 0 \Rightarrow |c| &lt; 2b^3

The same condition holds for the equation  (x-a)^3 - 3b^2(x-a) + c = 0 . The substitution  y = (x-a) transforms it into an equation of the form already considered iff this equation has 3 distinct roots so does the other (just add a)

For the last part we divide thought by 2 to get  x^3 -\frac{9}{2}x^2 + \frac{7}{2}x - \frac{1}{2} = 0. The translation  x = z + \frac{3}{2} gets rid of the cubic term and some binomial expansion later we get  z^3 - \frac{13}{4}z - 2 = 0 or equivalently  (x-\frac{3}{2})^3 - \frac{13}{4}(x-\frac{3}{2}) - 2 = 0

so applying the condition found with c = -2 and b^2 = \frac{13}{12} we have
 2b^3 = \frac{13}{6}\sqrt{\frac{13}{12}} &gt; 2 = |c|
so there are 3 distinct roots.

[That was a lots of latex! If you were doing this question properly a more detailed explanation of why the condition is necessary and sufficient would be good]

EDIT: Thanks to nchen5 for pointing out the i misread the requirement that the roots be positive. This can part is done as follows:

The curve
y = (x-a)^3 - 3b^2(x-a) + c
will have max point (a-b, c + 2b^3) and min (a+b, c - 2b^3) and y-intercept c + 3ab^2 - a^3 .

To ensure the roots are all positive we want the y-intercept below the x-axis c + 3ab^2 - a^3 &lt; 0 and the x-coordinate of the max point greater than zero a &gt; b. This with my previous condition for 3 distinct roots should do.

More formally, to prove sufficiency consider sign changes and apply Intermediate value theorem (i.e sign change rule); to prove necessity use sign change / IVT to show that if one of the conditions is not met then there'll be a negative root

In the specific case given these extra conditions can be verified

Edit: Cheers to GHOST-5 for pointing out a careless slip of mine. It has been corrected

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nchen5
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(Original post by SCE1912)
III Q5.

Spoiler:
Show

Completing the square on the quadratic
 y = x^2 - 2bx + c = (x-b)^2 + c - b^2
so turning point is  (b, c - b^2)
For two real roots we want the minimum point below the x-axis
so
 c - b^2 &lt; 0 \Rightarrow b^2 &gt; c

For the cubic's turning points set so  \frac{dy}{dx} = 0 to get
 3x^2 - 3b^2 = 0 \Rightarrow x = b \mbox{ or } x = -b
this makes the turning points  (-b, 2b^3 + c) and  (b, c - 2b^3) the former is a max the latter a min.

For three root we want the max above the x-axis and min below the x-axis so

 2b^3 + c &gt; 0 \mbox{ and } c - 2b^3  &lt; 0 \Rightarrow |c| &lt; 2b^3

The same condition holds for the equation  (x-a)^3 - 3b^2(x-a) + c = 0 . The substitution  y = (x-a) transforms it into an equation of the form already considered iff this equation has 3 distinct roots so does the other (just add a)

For the last part we divide thought by 2 to get  x^3 -\frac{9}{2}x^2 + \frac{7}{2}x - \frac{1}{2} = 0. The translation  x = z + \frac{3}{2} gets rid of the cubic term and some binomial expansion later we get  z^3 - \frac{13}{4}z - 2 = 0 or equivalently  (x-\frac{3}{2})^3 - \frac{13}{4}(x-\frac{3}{2}) - 2 = 0

so applying the condition found with c = -2 and b^2 = \frac{13}{12} we have
 2b^3 = \frac{13}{6}\sqrt{\frac{13}{12}} &gt; 2 = |c|
so there are 3 distinct roots.

[That was a lots of latex! If you were doing this question properly a more detailed explanation of why the condition is necessary and sufficient would be good]

Thanks for the attempt but i am afraid u have misread the question quite a bit.
I can manage eveything u have done but the question is asking for the necessary and sufficient condition for the cubic equation to have 3 distinct positive roots. and thats where i got stuck. (such as how u might compare the smallest root with a? )
I
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SCE1912
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#77
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(Original post by nchen5)
Thanks for the attempt but i am afraid u have misread the question quite a bit.
I can manage eveything u have done but the question is asking for the necessary and sufficient condition for the cubic equation to have 3 distinct positive roots. and thats where i got stuck. (such as how u might compare the smallest root with a? )
I
Your right. But I think only a slight alteration is required.

The curve
y = (x-a)^3 - 3b^2(x-a) + c
will have max point (a-b, c + 2b^3) and min (a+b, c - 2b^3) and y-intercept c.

To ensure the roots are all positive we want the y-intercept below the x-axis c &lt; 0 and the x-coordinate of the max point greater than zero a &gt; b. This with my previous condition for 3 distinct roots should do.

More formally, to prove sufficiency consider sign changes and apply Intermediate value theorem (i.e sign change rule); to prove necessity use sign change / IVT to show that if one of the conditions is not meet then there'll be a negative root
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Unbounded
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#78
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(Original post by SCE1912)
To ensure the roots are all positive we want the y-intercept below the x-axis c &lt; 0
I imagine you mean  3ab^2 + c - a^3 &lt; 0 in this case :o:
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toasted-lion
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Yes, in words, you require:

1. y-intercept negative
2. turning points different sign of y
3. turning points both have positive x values

A sketch might help to explain? Worth stating that the graph goes from negative infinity to positive infinity and make it clear how you get the conditions I think.
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Oh I Really Don't Care
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#80
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III 5 is basically I 3 2006 with a tincy bit at the end;
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