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Hard maths questions for higher maths GCSE

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Original post by TeeEm
then it is officially becomes available to any A level student !!
Definitely it is hard to model ...


do you have a rough time period for when the solution will go up?

As you probably guessed from me asking about the bearing of NE earlier I cant even get the first 3 facts modeled :laugh:
Reply 21
Original post by DylanJ42
do you have a rough time period for when the solution will go up?

As you probably guessed from me asking about the bearing of NE earlier I cant even get the first 3 facts modeled :laugh:


If people are trying I will delay it ...one or two days
Original post by TeeEm
If people are trying I will delay it ...one or two days


would it get more attention if it had its own thread?
Reply 23
Original post by DylanJ42
would it get more attention if it had its own thread?


it would I guess ...
Original post by TeeEm
it would I guess ...


maybe that's an idea, a lot of the really good alevel students probbaly wont see it in a hard gcse questions forum
Original post by TeeEm
it would I guess ...


Can't even draw this. Was planning on using some linear algebra but it's not going too well
Reply 26
Original post by langlitz
Can't even draw this. Was planning on using some linear algebra but it's not going too well


hopefully your face is not like your avatar ...
Original post by TeeEm
hopefully your face is not like your avatar ...


Could you steer us in the right direction, perhaps? I've done a drawing of this, though I'm not sure if that is even how it's supposed to look...

Spoiler

Reply 28
Original post by TheOtherSide.
Could you steer us in the right direction, perhaps? I've done a drawing of this, though I'm not sure if that is even how it's supposed to look...

Spoiler



terrible typo on my behalf (I am notorious) due to copy and paste
I will post a new thread
Original post by TeeEm
terrible typo on my behalf (I am notorious) due to copy and paste
I will post a new thread


I'll probably still have no clue about what's happening...
Reply 30
Original post by TheOtherSide.
I'll probably still have no clue about what's happening...


I changed it here but I will post a new thread
Original post by TeeEm
terrible typo on my behalf (I am notorious) due to copy and paste
I will post a new thread


thank goodness, i thought i was being really silly. i couldn't understand how to sketch it at all.
Reply 32
Original post by DylanJ42
thank goodness, i thought i was being really silly. i couldn't understand how to sketch it at all.


I am sorry
Original post by TeeEm
I am sorry


dont be sorry, i am just relieved that i didnt get wrecked by a gcse question :laugh:
Original post by atsruser
Here's one I've just made up, TeeEm style:

A sphere of radius 1 is placed on top of a sphere of radius 2 so that the line joining their centres is vertical. Find the radius of the base of the smallest right circular cone in which they will fit, with both spheres in contact with the cone.


Spoiler

Find

1(112)(113)(125)(127)(129)(122k+1)\displaystyle \frac{1}{(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{2}{5})(1-\frac{2}{7})(1-\frac{2}{9})\cdots(1-\frac{2}{2k+1})}

where kk is an integer.
Original post by atsruser
Find

1(112)(113)(125)(127)(129)(122k+1)\displaystyle \frac{1}{(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{2}{5})(1-\frac{2}{7})(1-\frac{2}{9})\cdots(1-\frac{2}{2k+1})}

where kk is an integer.


Are you sure that this is GCSE level? I don't even know where to start with this...
Reply 37
Original post by TheOtherSide.
Are you sure that this is GCSE level? I don't even know where to start with this...


Simplify, the denominator is nothing but:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{1}{2} \times \frac{2}{3} \times \frac{3}{5} \times \frac{5}{7} \times \cdots \times \frac{2k-3}{2k-1}\times \frac{2k-1}{2k+1}\end{equation*}



coughs cancel coughs

Edit, solution if you want to check your answer:

Spoiler

(edited 8 years ago)
Original post by Zacken
Simplify, the denominator is nothing but:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{1}{2} \times \frac{2}{3} \times \frac{3}{5} \times \frac{5}{7} \times \cdots \times \frac{2k-3}{2k-1}\times \frac{2k-1}{2k+1}\end{equation*}



coughs cancel coughs


Alright, so those fractions in the first part cancel each other out, so that you get 1/infinity, and that first part is negligible?

And then the (2k - 1) terms cancel each other out so that you get (2k - 3)/(2k + 1)?

I really don't know...

Spoiler

Reply 39
Original post by TheOtherSide.
Alright, so those fractions in the first part cancel each other out, so that you get 1/infinity, and that first part is negligible?

And then the (2k - 1) terms cancel each other out so that you get (2k - 3)/(2k + 1)?

I really don't know...

Spoiler



There's no infinity involved here. If I'm using kk then that's any arbitrary integer that does not imply infinity in anyway.

Everything cancels out, but remember that in the \cdots there's something that cancels the 2k32k-3 as well, so you're only left with...?

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