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

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Lol this guy got banned within like an hour of making his first post. A new record?
Original post by PakINDIA
hOW DO YOU DO THIS, PLS DO IT STEP BY STEP I GET TO (1+SIN0 + COS0)/(1 + COSSIN + SIN + COS) = 1
BTW YOU ARE MEANT TO ONLY WORK FROM ONE SIDE... NO MEETING IN THE MIDDLE APPROACHES PLS BECAUSE THAT'S IMPROPER FOR REASONS I DON'T REALLY UNDERSTAND, JUST IN CASE.
I SEE WHAT THE TRICKY PART IS BUT I DONT GET THE ANSWER.
ANYONE ASAP PLS
Expand to get (Sin^2 x 1 Cos^2 1)/(cos x 1)(sin x 1)=1The top part =2 Therefore 2= (cos x 1)(sin x 1)Then it’s pretty straight forward from here
Original post by Notnek
Here's my first one. There's a very tricky part in this question that you need to watch out for (most A Level students would miss it) but apart from that it's good practice for Level 2 FM.


Solve the equation


sinθcosθ+1+cosθsinθ+1=1\displaystyle \frac{\sin \theta}{\cos \theta + 1} + \frac{\cos \theta}{\sin \theta + 1} = 1


for 0oθ360o0^o \leq \theta \leq 360^o .


Is there a more elegant reason why 180 and 270 don't work from sin x . cos x = 0, besides the fact that when you happen to sub them back into the original equation the numbers don't add up?

How come they're a solution to sin x . cos x = 0 but not to the original if the former came from the latter?
Original post by TheFarmerLad
Is there a more elegant reason why 180 and 270 don't work from sin x . cos x = 0, besides the fact that when you happen to sub them back into the original equation the numbers don't add up?

How come they're a solution to sin x . cos x = 0 but not to the original if the former came from the latter?

Because you multiplied by (cos + 1)(sin + 1) you are introducing extra solutions which are not solutions to the original equation.
Original post by RDKGames
Because you multiplied by (cos + 1)(sin + 1) you are introducing extra solutions which are not solutions to the original equation.

Thanks
Original post by 3317752
Here are some of the general things to show up each year:

Quadratic Nth Term
Simultaneous Equations Using Matrices
Representing Ratios (2 markers)
Algebraic Fractions
At least 1 or 2 3D Trig
Representing a Certain Siny or cos or tan value in terms of another sin or cos or tan value. (i:e if siny = 5/3 whats cosy)
Trigonometry Identities
Solving Trig Identities Questions

I'm a bit late haha but...
atm I have sinxcosx = 0
what do i do now?
\theta=50 \alpha=40 ?
Original post by Notnek
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 P3\displaystyle \mathbf{P}^3 represents a clockwise rotation of 150o150^o about the origin.

Q=(0110)\displaystyle \mathbf{Q} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}

The unit vector (01)\displaystyle \begin{pmatrix}0 \\ 1\end{pmatrix} is transformed by the matrix QP\displaystyle \mathbf{Q}\mathbf{P} to give the vector (cosθcosα)\displaystyle \begin{pmatrix}-\cos \theta \\ \cos \alpha \end{pmatrix}.

Find the value of θ\theta and α\alpha.

\theta=50 \alpha=40 ?
YOUR a donut bruv
Original post by steve Jarvis
YOUR a donut bruv

And you're a bigger donut bruv
p=3q=9???
Original post by Notnek
675x42505x3+2407x263x162(axb)(cx+d)(exc)2675x^4-2505x^3+2407x^2-63x-162\equiv (ax-b)(cx+d)(ex-c)^2

a,b,c,da, b, c, d and ee are different positive integers. Find their values.


Does this include trial and error? I seem to be using a lot of trial and error/ logic and it makes me feel like im missing something. (I haven't expanded the brackets because I assume that is missing the point of the question)
Reply 372
Original post by Random_Student1
Does this include trial and error? I seem to be using a lot of trial and error/ logic and it makes me feel like im missing something. (I haven't expanded the brackets because I assume that is missing the point of the question)

Not worked it through, but the squared factor (ex-c)^2 seems useful and maybe think about how it maps onto squared factors in the quartic coeffs. If you can get that term, then dividing through (if even necessary) its just a quadratic coefficient matching and you already know one coefficient. A small amount of reasoning / trial and error may be necessary, but not a lot.
(edited 10 months ago)
Reply 373
Original post by Random_Student1
Does this include trial and error? I seem to be using a lot of trial and error/ logic and it makes me feel like im missing something. (I haven't expanded the brackets because I assume that is missing the point of the question)

It's been a while since I wrote the question but I'm pretty sure the intention was to be able to do it by just looking at the x^4 and constant coefficients:

bdc^2 = 162
ace^2 = 675

You can think about what c and e must be since when squared they are factors of 162 and 675 which doesn't give too many options.

After that I think there's a nice way to go about this without too much trial and error but at first glance I don't see it. I'll have another look tomorrow when I'm more free.
f(1,2)?
Original post by Notnek

Spoiler


6 years later so i dont expect a reply..... by my calculator doesnt say error when i put 180,270 in. How will i know to exclude these in the exam?
Reply 376
Original post by student1274738
6 years later so i dont expect a reply..... by my calculator doesnt say error when i put 180,270 in. How will i know to exclude these in the exam?

The denominator of the first fraction is equal to 0 when theta equals 180. And you can’t divide by 0. If you’re not getting an error on your calculator then you’ll need to explain what you’re putting into your calculator.
(edited 3 weeks ago)

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