Level 2 Further Maths - Post some hard questions (Includes unofficial practice paper)

This is the hardest one I've posted so far and requires a very good undrstanding of transformation matrices. Don't worry if you can't do this one.


The transformation matrix $\displaystyle \mathbf{P}^3$ represents a clockwise rotation of $150^o$ about the origin.

$\displaystyle \mathbf{Q} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$

The unit vector $\displaystyle \begin{pmatrix}0 \\ 1\end{pmatrix}$ is transformed by the matrix $\displaystyle \mathbf{Q}\mathbf{P}$ to give the vector $\displaystyle \begin{pmatrix}-\cos \theta \\ \cos \alpha \end{pmatrix}$.

Find the value of $\theta$ and $\alpha$.

Can anyone help me out with this question ?- without using the trial and improvement method - thanks

The nth term of a sequence is n^2 + n
two consecutive terms in the sequence have a difference of 32
work out the two terms

This is the hardest one I've posted so far and requires a very good undrstanding of transformation matrices. Don't worry if you can't do this one.


The transformation matrix $\displaystyle \mathbf{P}^3$ represents a clockwise rotation of $150^o$ about the origin.

$\displaystyle \mathbf{Q} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$

The unit vector $\displaystyle \begin{pmatrix}0 \\ 1\end{pmatrix}$ is transformed by the matrix $\displaystyle \mathbf{Q}\mathbf{P}$ to give the vector $\displaystyle \begin{pmatrix}-\cos \theta \\ \cos \alpha \end{pmatrix}$.

Find the value of $\theta$ and $\alpha$.

If anyone would be interested in more hard questions to be posted let me know. I don't know if there are many level 2 FM students on TSR this year.

Hi... I'd definitely be interested in harder questions like the ones posted on this thread. Also, I posted a written solution to the matrices question so I'd appreciate if you checked it out to verify.

I couldn't really find a clear solution to this question on this thread so I decided it would be best to just write one up myself. This is the way in which I thought the question could most easily be solved but I'm sure there are various other methods. Apologies for the bad handwriting - let me know if anything is unclear or incorrect in the attached image so that I can re-upload a corrected version.

Careful here (highlighted below) since the y coordinate of the transformed unit vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is negative... so you should have $-\sin 50$ instead. Otherwise, your approach seems solid.

The only other comment I would make is that since cosine,sine are have a period of 360 degrees, the solution to $-\cos \theta = -\cos 50$ is firstly $\theta = 50$, as you say, but it can also be $\theta = 50 + 360 = 410$, or $\theta = 50-360 = -310$, and so on... so (50) + (any integer) * 360 is the general solution... but I'm not sure if that's expected at this level.

P.S. For $\cos \alpha = \sin 50$ it would be more elegant to realise that you can apply an identity.. namely $\sin 50 \equiv \cos(90-50)$.

$675x^4-2505x^3+2407x^2-63x-162\equiv (ax-b)(cx+d)(ex-c)^2$

$a, b, c, d$ and $e$ are different positive integers. Find their values.

I'm quite confused with this one, could you provide a hint please? And also I'd appreciate more questions before paper 2 - but I can understand if you aren't able to upload any as you only really have 1 day so no worries if it's not possible

If anyone's looking for some tricky questions to try before paper 2 then check this thread:

https://www.thestudentroom.co.uk/showthread.php?t=4780436

(Don't open the spoiler if you want to work out the answer to the first difficult question on the thread)

(I moved your post to the question thread so it's all in one place).

I'm not really sure what you mean. Is there a solution you are looking at which is confusing you? If there is can you please link to it?

If not can you please post all your working up to the point you get confused?

So we simplify the original question to get sinx * cosx = 0. The solutions to this equation would be 0,90,180,270,360. The points where the sin and cos graph touch the x-axis.
But with the original equation in the question, 180 and 270 aren't solutions because substituting them in would cause a division by zero, which is unsolvable. So does this not mean that the two equations aren't the same since they have different solutions? But we got the equation sinx * cosx = 0 from the original one in the question so I'm not sure what happened in the simplifying process that caused these equations to have different solutions. I hope this makes sense? thanks