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STEP 2003 Solutions Thread

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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
STEP III 2003, Q4

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

t1=1t022t0=1tan27π182tan7π18=1tan7π9 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}}

Unparseable latex formula:

[br]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.

1tan14π9=cot5π9=tan(π25π9)=tan(π18)=tanπ18tan14π9 \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:

tommm
STEP III 2003, Q4
We can also apply the formula similarly to the cotangent function, so

t2=tan(14π9)t_2 = \tan (\frac{14\pi}{9})
Reply 62
i need solution to STEP III question 5!
I cant finish the question!
Thanks !
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.
Reply 64
III Q5.

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Reply 65
SCE1912
III Q5.

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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
Reply 66
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=(xa)33b2(xa)+cy = (x-a)^3 - 3b^2(x-a) + c
will have max point (ab,c+2b3)(a-b, c + 2b^3) and min (a+b,c2b3)(a+b, c - 2b^3) and y-intercept cc.

To ensure the roots are all positive we want the y-intercept below the x-axis c<0c < 0 and the x-coordinate of the max point greater than zero a>ba > 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
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.
III 5 is basically I 3 2006 with a tincy bit at the end;
DeanK22
III 5 is basically I 3 2006 with a tincy bit at the end;


I thought it seemed familiar! How strange that they repeated a III question on I.
Reply 70
STEP II, Question 12

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Reply 71
STEP II, Question 13

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SimonM
STEP III, Question 6

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I think you've used the wrong trig identity in the second part, it should be sin(a + 1/2)x - sin(a - 1/2)x + .... sin(b - 1 + 1/2)x, which gives the answer as
x(a + b - 1) = pi(2n + 1) or x(b - a) = 2npi
Reply 73
STEP II Question 11

First Part:

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Second Part:

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Reply 74
STEP II, Question 9

Solution One:

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Solution Two:

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Reply 75
Dadeyemi
STEP II 2003 Q5

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I'm fairly sure the answer is

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.
Reply 76
Mark13
I'm fairly sure the answer is

Spoiler

.


Yea I just made some rediculously silly mistakes (e.g. forgetting to multiply by two) I corrected it and get the same answer.
Reply 77
maltodextrin
I think you've used the wrong trig identity in the second part, it should be sin(a + 1/2)x - sin(a - 1/2)x + .... sin(b - 1 + 1/2)x, which gives the answer as
x(a + b - 1) = pi(2n + 1) or x(b - a) = 2npi


I did using 'otherwise' and I got that answer as well.
tommm
STEP III 2003, Q4

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I might be misinterpreting the question but the question asks us to 'show that T = (1 - t^2)/2t and 3t^2 =/ 1'. So this means we should give some justification as to why 3t^2 =/1 right? If so then what would the justification be?

If 3t^2 = 1 then it means that T = (1 + t^2)/4t which I can't see a problem with...
Reply 79
maltodextrin
I might be misinterpreting the question but the question asks us to 'show that T = (1 - t^2)/2t and 3t^2 =/ 1'. So this means we should give some justification as to why 3t^2 =/1 right? If so then what would the justification be?

If 3t^2 = 1 then it means that T = (1 + t^2)/4t which I can't see a problem with...


I think it's because you need T=t to be a repeated root, since its a tangent. Not sure though.

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