I think there's a stray $a$ in the middle of the solution to Q3(ii) of assignment 19 (page 4 of the hints).

It currently reads: \[\frac{a+7}{b+7}=\frac{a^2+51}{b^2+51}a,.\]

I loved the last part of that question - it took me ages to find the common factor.

Hi all - hope everyone had a great Christmas and New Year. Assignment 19 has just been released, and we will continue to release Foundation modules on Mondays. We will also publish some more STEP II modules in January.

Please keep using the forum to ask any questions (or to point out any mistakes, for which we are always grateful!)

I need hep in 87 step 2 q3

Thing is, I don't even know where to start.

Can anyone give hints as to how I should do this question.

Thanks

Hi,

I'm stuck on the last part of the step question where they ask you to show that c^2 is between 1 and 0, but one of the equations they gave us shows that c^2 is negative. I read the hints where they told us to use the first equation given which involves c^2. However, it is that same equation which shows that c^2 is negative.

Help pleas

I'm stuck on the second Step II calculus question.

Making the substitution of $x=\pi-t$, the integral becomes \[\int^0_\pi(\pi-t)f(\sin(\pi-t))\,(-dt)\] which tidies up to be \[\int^\pi_0(\pi-t)f(\sin t)\,dt\] I then split the integral into two parts: \[\pi\int^\pi_0f(\sin t)\,dt-\int^\pi_0tf(\sin t)\,dt\] and then subbing $x$ back in: \[\pi\int^\pi_0f(\sin x)\,dx-\int^\pi_0(\pi-x)f(\sin x)\,dx\] which obviously just cancels out the original substitution.

As it's Christmas, you can take a short break from STEP preparation today! Don't worry though, if you are desperate to do some maths there will still be people checking the forum to help you out.

What do you think of this way of doing 4(ii) of assignment 17?

We are looking to show that $9\vert(2^n+5^n+56)$. So I looked at $2^n$ and $5^n \pmod{9}$ and saw that they repeated every 6th value of $n$ (in a reverse pattern, interestingly). For $n=0,1,2,\ldots$, $2^n\equiv1,2,4,-1,-2,-4,1,\ldots \pmod{9}$ and $5^n\equiv1,-4,-2,-1,4,2,1\ldots \pmod{9}$.

Sorry to be nitpicky but in the partial solutions to Assignment 16 q2 ii shouldn't we rather find 3sinA - 4sin^3 A ?
Also for q3 i might it not be easier to use cos^2 A + sin^2 A = 1 on cos15 to find sin15 rather than using the sin2A formula?

How much time do you think we should spend on doing a step question before giving up?

How about if your doing step questions for the first time?

Hi all,

Assignment 18 is now up. There will be no Assignment published next week, so have a good Christmas break.

If you really want to do some more maths, and have done all the previous Foundation modules, why not have a look at the STEP II modules.

Merry Christmas!