STEP Prep Thread 2016

You've got plenty of time to do it!

Are you year 12?

I wish man, in year 13. But I ain't as good as you Cambridge lot, only doing STEP 1 for Warwick which I have an offer for. But the plan is if I get a high 1 or a S this year in STEP 1- might ditch Warwick take a gap and apply to Cambridge ...

Hello everyone, I wanted to ask a question about STEP I 2005 Q4 last part of b)
It is given that tan3Ф= 3t-t^3/1-3t^2 (where t=tanФ) and that tan3Ф=11/2. Solutions suggest what I intended to do which is make a cubic equation but how am I suppose to find that one of the factors is (2t-1)? I understand the trial and error part but it is usually enough to try -2,-1,0,1,2. Should I attempt to put in 1/2 from now on or is there something I do not see?

If you're this good at STEP after just starting... you're twice as good as me.

I doubt it man, it probably just seems like it because I am only doing the questions that I like the look of, I am a lost man with geometry/vectors and combinations trust me. BTW quick question do you know when I was changing the limits in the second part I actually got -3 and -2 as well. Can i always just choose the positive limits if I ever make another similar substitution ?

As in, you got $u^2 = 4, 9$? Yeah, just take the positive root.

Oh yup thats what I meant , and will it always work ? Because I thought it needed to be an even function for this to work ?

Hello everyone, I wanted to ask a question about STEP I 2005 Q4 last part of b)
It is given that tan3Ф= 3t-t^3/1-3t^2 (where t=tanФ) and that tan3Ф=11/2. Solutions suggest what I intended to do which is make a cubic equation but how am I suppose to find that one of the factors is (2t-1)? I understand the trial and error part but it is usually enough to try -2,-1,0,1,2. Should I attempt to put in 1/2 from now on or is there something I do not see?

This is probably wrong (somebody correct me), but I think that what you're *actually* doing is:

$u = \sqrt{e^x + 1} > 0$ and then squaring both sides, so $u$ is gonna be positive.

Alright this solution took me exactly an hour and like I feel like I have a wishy washy answer so someone please give me a mark and perhaps provide a more rigorous reasoning for what I did ? Thanks again ...

Tbh if what you said is right it would sense BTW not gonna annoy you too much but whenever you have time could you look at ^ , and provide a mark ?

You got the integrals wrong.

It should have been $\displaystyle \int_{\frac{\pi}{2}}^{\alpha} x \sin x - \frac{\pi}{2} \, \mathrm{d}x + \int_{\alpha}^{\pi} \frac{\pi}{2} - x\sin x \, \mathrm{d}x$

Since you can see that $\cosec x \leq \frac{2x}{\pi}$ in the interval $\left[\frac{\pi}{2}, \alpha\right]$ This is the same as saying that $x \sin x \geq \frac{\pi }{2}$ by inverting both sides. (Since if one thing is bigger than the other, then 1/big thing will be smaller than 1/small thing).

Then you can do the same thing in the interval [alpha, pi].

That'll lose you an accuracy mark and 2 method marks.

Same mistake in part(ii), in the interval [0, ln 2] we have that $\displaystyle ||e^x - 1|-1| = 2-e^x$ Since for in the interval [0, ln 2] you're gonna get |e^(some number less than ln 2) - 1| - 1 = |some number less than 2 - 1| - 1 = some number less than 1 - 1 = some number less than 0. And hence negative. So the modulus signs equals effectively negates the function. Although note that e^x - 1 will be positive in that region, so the modulus signs around that simply disappear. If anything needs clarification, please ask.

I'd say you get 13/20.

I am not sure if I quite get you, you know in the interval between pi/2 and alpha, its clear that 2x/pi is greater than cosec x. If 2x/pi corresponds to pi/2 and cosecx corresponds to xsinx, I need to do pi/2-xsinx don't I ? I get the second part of your comment though And, thats beautiful, expect me to throw a lot of solutions at you this coming week

I agree that in that interval you have 2x/pi > cosec x

So 2x/pi > 1/sin x ***************

so pi/(2x) < sin x so pi/2 < x sin x which is the same thing as x sin x > pi/2.

Hence x sin x- pi/2 > 0 in the interval [pi/2, alpha].

Moving from ********* to the second line, we take the reciprocal of both sides and flip the inequalities. If x >y then 1/x < 1/y.

AHH man that actually makes I am getting so careless... BTW could you please offer some guidance on STEP 1 2006 question 2 ? What perplexes me is that if the length of the rope is 4a, which is larger than the largest length possible from a corner of (8^0.5)a which is the diagonal length, why can't the goat just graze the entire field because every part of the barn is accessible? I think I am misunderstanding the question ? Thanks

I remember seeing somewhere that you lost 85% of the marks if you assume the goat is tethered to the corner.

Try letting the goat be tethered a distance x from a corner and then working through the algebra. :-)

Edit: Q3 on that paper is a beauty.

@Zacken
Anyways, that question was quite ugly tbh might attempt it later, but for the moment have done this problem but couldn't really get the last part so help would be nice. So could someone please comment? Preferably a mark out of 20 and feedback ? STEP 1 2006 Q6