For the first part of the question, I formed two equations $y =\frac{a\sin(\theta)}{b\cos(\theta)}x$ and $y - b\ sin(\theta) = \frac{b\sin(\theta)}{ea + a\cos(\theta)}(x-a\cos(\theta))$
Solving those obtains $y = \frac{bsin(\theta)}{1 + e\cos(\theta)}$ and $x=\frac{b^2\cos(\theta)}{a + ae\cos(\theta)}

I assumed for the second part that I could just sub both $x$ and $y$ straight into $(x+ea)^2 + y^2 = a^2$ but the fractions I end up manipulating seem overly complex so I was wondering if there was an easier way to go about it

I've manipulated the equation to obtain the required result in (i) and used that to find $$S_3(n) = \frac{1}{4}n^2(n+1)^2$$ and $$S_4(n) = \frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$$ - it doesn't look like I can factorise the quadratic part of $S_4(n)$

Hi!

Since I have never taken a STEP test before, I am wondering how much we are supposed to write on our answer sheets, and in how great detail we have to go. Are the official solutions representative of what we students should write?

And is this the formula booklet we will be given/allowed to use?

When forming my expression for R in part (i), I understand that the vertical part (k) is $5\sqrt{5}t -5t^2$, the north part (j) is $5\sqrt{15}t$ but I don't understand how in the solution they've obtained $50 - 15\sqrt{5}t$ for i.

The horizontal part of the $25\ms^{-1}$ is $25\cos(arctan(\frac{1}{2})) = \frac{2}{\sqrt{5}} * 25$ but multiplying through then by $cos(60)$ obtains $5\sqrt{5}$ not $15\sqrt{5}$

Having seen the solution on the integral math site, my contradictions don't line up with theirs, but I assume that there are multiple ways it could be done.

For part (i) Forming inequalities for all numbers then expanding the brackets you obtain $ab-abc>\frac{4}{27}$ , $ac-abc>\frac{4}{27}$ and $bc-abc>\frac{4}{27}$

If you subtract any one of the inequalities from the other you obtain something of the form $a(b-c)>0$ which without loss of generality, can apply for any permutation of a, b and c.

Hello,

In the preparation section, question 2(vi), I'm not entirely sure why you can simply make the substitution u=x in the final part and keep it equal as I. I maybe thought it was an even function and so there was some symmetry involved, or there was some equating of coefficients going on in the top lines, but they didn't really transfer to the Warm down part... Much appreciate any help.

Damin

For parts (i) and (ii) I've managed to muddle through,
For part (iii)
I assume you start by finding an expression for $E(x)$
So I used something similar to (i) and (ii) to obtain ${dE\over dx} = -2\left({dw\over dx} \right)^2(5cosh(w) - 4sinh(w) - 3)$
Integrating to get $E(x) =\left ( {dw \over dx} \right)^2 + wsinh(w) + cosh(w)$
Setting $5cosh(w) - 4sinh(w) - 3 = 0$ the solution is $w = ln(3)$
So the graph of $E(x)$ is an increasing function for $0 \le x < ln(3)$ and a decreasing function after, with $ln(3)$ as the maximum

I don't know which year the question is from so I can't check my solution, but this is what I've ended up with:

The 6 solutions to the equation are $t = tan(\frac{n\pi}{7})$ where $n \in [2,4,6,8,10,12]$ this is obtained from the binomial expansion of $(cos(\theta) + isin(\theta))^7$ then factorising out $sin(\theta)cos^6(\theta)$
The solution of the equation where $t = tan(\theta)$ are the same as $(cos(\theta) + isin(\theta))^7 = 1$ excluding the solution $\theta = 0$ since we factorised out the $sin(\theta)$ Thus we have the aforementioned solutions.

Hi!

I'm having a lot of trouble understanding the solutions on the internet (e.g. MEI website/TSR) for the second part of i) "Show by means of counter example ...."

I don't really understand where do you get f(x) = 1-x/n from?

I would appreciate it a lot if anyone could explain this part of the question to me

:)

Thanks in advance!!

I'm currently a year 12 looking to do economics or a similar course. I've tried a few step 1 past papers in exam conditions and mostly got low 70s, which is 1 or 2 in most years. Is it worth taking the exam with only this level of ability? I have been told that some universities consider grade 2 or below in step 1 to be a bad result, making the application weaker. It seems unlikely that I will get S. I've been told that I need to decide whether or not to take the exam in three days and I have no idea what to do so advice would be greatly appreciated.