STEP 1 2015 Solutions Thread

Did anyone do any of the statistics questions? I've had a good look through the main step thread and this one and no one's mentioned them that I've seen. I don't remember much but I remember getting 1/2 for two parts of the first statistics question, about the die-rolling. Then the last part came out as a long quadratic in terms of n? I felt like most of that question was wrong, I was a bit rushed.

(iii)

Q7)Someone is going to come along and post a proper solution. These are just placeholders while we have nothing pretty much.

Those aren't right, it should be(Wolfram Alpha agrees btw)

The ranges for a (and the corresponding formulas for M(a)) were:

Q4)

i)

Let's call the coordinates of the rods midpoint $\displaystyle (X,Y)$ and let $\displaystyle \theta$ be the angle between the x axis and the rod.

Since the gradient of the parabola is $\displaystyle \frac{x}{2}$, the rod must have the same gradient if it is tangent. This means that $\displaystyle m=\frac{X}{2}$

But the gradient of the rod is also $\displaystyle \tan(\theta)$ so that means $\displaystyle \frac{X}{2}=\tan(\theta)$.

This gives us $\displaystyle X= 2\tan(\theta)$.

Also as the rod's midpoint lies of the parabola then $\displaystyle Y=\frac{X^2}{4}$.
This implies that $\displaystyle Y=\tan^2(\theta)$.

In order to get one of the coordinates of the midpoint we must travel $\displaystyle (b\cos(\theta), b\sin(\theta))$ down the rod for the bottom end and up the rod for the bottom end.

This gives us the end of the rod the final coordinates:

$\displaystyle (2\tan(\theta) \pm b\cos(\theta), \tan^2(\theta) \pm b\sin(\theta)$.

ii)

First we just find the equation of the rod.
We know that one coordinate is $\displaystyle (0,A)$.

That means that since $\displaystyle y=mx+c$ then:

$\displaystyle \tan^2(\theta)=\tan(\theta) \left ( 2\tan(\theta) \right ) + A$

So $\displaystyle A = -\tan^2(\theta)$.

This means that the area in question is equal to:

$\displaystyle \int_{2\tan(\theta) -b\cos(\theta)}^{2\tan(\theta) + b\cos(\theta)} \tan(\theta)x - \tan^2(\theta) - \frac{x^2}{4} \; dx$

Which is equal to:
$-\frac{1}{4 }\displaystyle \int_{2\tan(\theta) -b\cos(\theta)}^{2\tan(\theta) + b\cos(\theta)} \left ( x - 2\tan(\theta) \right )^2 \; dx$

Evaluating the integral gives:

$\displaystyle (-) \frac{\left ( b\cos(\theta) \right ) ^3}{6}$

I am suddenly a lot less nervous, i was scared I had got 2 wrong as well hahaha, or my misread of -1/3 as 1/3 in question 7 meant my answers were far astray, thanks for settling some nerves.

Correct me if I am wrong, but shouldn't the range for a for $\frac{a^3}{9}$ be $2 \leq a < 3$, so that all the solutions have continuity at the boundaries of the intervals?

That is, so $\frac{3a + 2}{9} = \frac{a^3}{9}$ when $a = 2$, as it is for $\frac{a^3}{9} = 3a - 6$ when $a = 3$?

I may be missing something here, but I did not get the same ranges for a.

Those aren't right, it should be(Wolfram Alpha agrees btw)

You would think they would try to be original in such an important exam rather than giving loads of people the chance to get one or two free problems.

What actually was Q6? I believe I did that question, but I don't have a clue what the question was.

The vectors question.

Is modular arithmetic part of the STEP syllabus? Because anyone who didn't know it would probably find Q1 very hard. There are avenues into the problem without it but not ones that are easy to see or execute for most candidates (and likely those who were good enough to see and do these methods would know modular arithmetic anyway).

Oh yeah of course. That'll be another 5-10marks I can scrape. Cheers
I now understand why nobody has posted a solution yet.
I don't think it is. I thought it was just A-Level Maths spec. + Proof by Induction? And are you talking about STEP II? I didn't take that, but did Q1 and STEP I and wouldn't see how modular arithmetic was useful for that, and found it easy to do without it.

...