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OCR FSMQ Additional Maths 6th June 2016 Official Thread

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Reply 60
Avatar for kendo9
kendo9
16
Original post by Vanilla Cupcake
Congrats! I did the 2006 paper and found it to be quite a nice one (I got 90 on it :biggrin:)
But I'm pretty inconsistent lately so I'm hoping to do a couple more past papers before the exam to try and keep getting high As
Found out I'm terrible at the algebra manipulation questions so I'll probably be practising some of those in the book and end of chapter tests

Posted from TSR Mobile


Thanks :smile: yeah 2006 was fairly nice, when I completed it I thought Id get around 94/95 but after marking it I found out I made stupid errors, like for one question I didn't write y=x..., I wrote x... :frown: I guess I need to keep doing papers to remove these mistakes :biggrin:
Original post by sosassysofe
For some reason our teacher cant get hold of it...so weird Also, would u say that the 2015 exam was harder than previous years?


I would not say the maths required to do it was much harder than previous years but instead the questions were slightly more obscure. For example on some integration questions you would have to "intepret this answer geometrically" or find the greatest distance between the curves in a bounded region. It was 68/100 for an A and I managed to get one. yay!


Posted from TSR Mobile
If no-one can get hold of the 2015 Paper, I will make a similar calculus question based off of my memory of the question and from unofficial mark schemes. But please note it will not be exact. I may also do the weirder trigonometry question as well :P
Original post by Cryptokyo
If no-one can get hold of the 2015 Paper ...


Unfortunately it is against TSR rules to share the latest password protected paper as schools like to use it as a mock.
Original post by Mr M
Unfortunately it is against TSR rules to share the latest password protected paper as schools like to use it as a mock.


Oh, ok. Sorry. But therefore aren't unofficial mark schemes doing this?
(edited 8 years ago)
Original post by Cryptokyo
Oh, ok. Sorry. But therefore aren't unofficial mark schemes doing this?


No.
Original post by TheOtherSide.
6th June 2016 - OCR FSMQ Additional Maths Exam
I hope that this thread will be helpful for anybody who's taking Additional Maths this year. I've posted up some resources, but feel free to message me about any other resources I could include, as well as using this thread to help each other on different topics in Additional Maths.



Thanks for making this thread OP! I've added it to the Exam Directory Thread which you can find here. Use that thread to find discussions on your other exam papers, and do let me know if you spot any that aren't on the list. Good luck with this exam! :h:
Question about trig. identities.....

For questions such as 'solve tan 3x = −1 for values of x in the interval 0◦ ≤ x ≤ 180◦.'
is there a way to find the values without having to memorise the trig. graph? E.g a set of equations for
values of x in each type of graph?
Original post by Tasha_140
Question about trig. identities.....

For questions such as 'solve tan 3x = −1 for values of x in the interval 0◦ ≤ x ≤ 180◦.'
is there a way to find the values without having to memorise the trig. graph? E.g a set of equations for
values of x in each type of graph?


As far as I know, you'll just be expected to use the trig graphs, but that shouldn't be too hard to do anyway, as long as you remember these:

•

The y-intercept of y = sinx is 0 and for y = cosx, it is 1

•

Each period is 360° long for cos and sine graphs, while it is 180° for tan graphs

•

The tan graph crosses the x axis at 0

•

The sine and cos graphs have maximum and minimum points of 1 and -1, while the tan graph doesn't have maximum or minimum points

Really, as long as you know the general shape of each of the graphs, you should be fine:

Spoiler

Spoiler

Reply 69
Avatar for Zacken
Zacken
22
Original post by Tasha_140
...


CAST diagrams. Watch this to understand what it is, then watch this to see how it's used and examples.
Original post by Zacken
CAST diagrams. Watch this to understand what it is, then watch this to see how it's used and examples.


Thankyou, this was really helpful!
Original post by TheOtherSide.
As far as I know, you'll just be expected to use the trig graphs, but that shouldn't be too hard to do anyway, as long as you remember these:

•

The y-intercept of y = sinx is 0 and for y = cosx, it is 1

•

Each period is 360° long for cos and sine graphs, while it is 180° for tan graphs

•

The tan graph crosses the x axis at 0

•

The sine and cos graphs have maximum and minimum points of 1 and -1, while the tan graph doesn't have maximum or minimum points

Really, as long as you know the general shape of each of the graphs, you should be fine:

Spoiler

Spoiler



Thanks for that! Will try and remember those facts
Anyone have any idea how you would solve:

tan^2θ + 3tanθ - 4 = 0,

for -180° < θ < 180°

I figured I need to solve the equation to end up with 'θ=.....' and then I can use a CAST diagram to find other values of theta,
just unsure how to get up to that stage
Reply 73
Avatar for Zacken
Zacken
22
Original post by Tasha_140

I figured I need to solve the equation to end up with 'θ=.....' and then I can use a CAST diagram to find other values of theta,
just unsure how to get up to that stage


It's a quadratic in tan⁡θ\tan \theta. So perhaps make the substitution u=tan⁡θu = \tan \theta to get u2+3u−4=0u^2 + 3u - 4= 0, solve that and then back-substitute tan⁡θ=two solutions from quadratic\tan \theta = \text{two solutions from quadratic}. Then CAST takes over.

This 'disguised' quadratic thing shows up in many guises, be it trigonometrical or exponential, so for example 52x+5x−10=0  ⟺  (5x)2+5x−10=0  ⟺  u2+u−10=05^{2x} + 5^x - 10 = 0 \iff (5^x)^2 + 5^x - 10 = 0 \iff u^2 + u - 10 = 0 where u=5xu = 5^x is a quadratic in 5x5^x. You need to get used to spotting these things and I find that making the explicit substitution u=tan⁡θu = \tan \theta or u=5xu = 5^x is really helpful for most students.
Original post by Zacken
It's a quadratic in tan⁡θ\tan \theta. So perhaps make the substitution u=tan⁡θu = \tan \theta to get u2+3u−4=0u^2 + 3u - 4= 0, solve that and then back-substitute tan⁡θ=two solutions from quadratic\tan \theta = \text{two solutions from quadratic}. Then CAST takes over.

This 'disguised' quadratic thing shows up in many guises, be it trigonometrical or exponential, so for example 52x+5x−10=0  ⟺  (5x)2+5x−10=0  ⟺  u2+u−10=05^{2x} + 5^x - 10 = 0 \iff (5^x)^2 + 5^x - 10 = 0 \iff u^2 + u - 10 = 0 where u=5xu = 5^x is a quadratic in 5x5^x. You need to get used to spotting these things and I find that making the explicit substitution u=tan⁡θu = \tan \theta or u=5xu = 5^x is really helpful for most students.


Thankyou! That makes a lot more sense!
Reply 75
Avatar for Zacken
Zacken
22
Original post by Tasha_140
Thankyou! That makes a lot more sense!


No worries. :-)
Show that the equation '3cos²θ = sinθ + 1' can be written as '3sin²θ + sinθ - 2 = 0'

How do you solve something like this? I know the rule 'sin²θ + cos²θ = 1', but I can't see how this can be used - unless there is another formula I need or I'm not solving it correctly?
Reply 77
Avatar for Zacken
Zacken
22
Original post by Tasha_140
Show that the equation '3cos²θ = sinθ + 1' can be written as '3sin²θ + sinθ - 2 = 0'

How do you solve something like this? I know the rule 'sin²θ + cos²θ = 1', but I can't see how this can be used - unless there is another formula I need or I'm not solving it correctly?


cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1.

Re-arrange this: cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta

So 3cos⁡2θ=3(1−sin⁡2θ)=3−3sin⁡2θ3\cos^2 \theta = 3(1-\sin^2 \theta) = 3 - 3\sin^2 \theta.

Hence: 3cos⁡2θ=sin⁡θ+1  ⟺  3−3sin⁡2θ=sin⁡θ+13\cos^2 \theta = \sin \theta +1 \iff 3 - 3\sin^2 \theta = \sin \theta + 1

Re-arrange this to get it in the required quadratic form and get used to doing cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1-\sin^2 \theta and sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1- \cos^2 \theta because you will be using these identities a lot over your mathematical career.
Original post by Zacken
cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1.

Re-arrange this: cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta

So 3cos⁡2θ=3(1−sin⁡2θ)=3−3sin⁡2θ3\cos^2 \theta = 3(1-\sin^2 \theta) = 3 - 3\sin^2 \theta.

Hence: 3cos⁡2θ=sin⁡θ+1  ⟺  3−3sin⁡2θ=sin⁡θ+13\cos^2 \theta = \sin \theta +1 \iff 3 - 3\sin^2 \theta = \sin \theta + 1

Re-arrange this to get it in the required quadratic form and get used to doing cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1-\sin^2 \theta and sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1- \cos^2 \theta because you will be using these identities a lot over your mathematical career.


Ohh right, got it - thanks again! (:
Just unsure whether I would think to do this if that makes sense? Seeing your explanation it now seems obvious, yet it never came to my head to attempt to solve it that way
Reply 79
Avatar for Zacken
Zacken
22
Original post by Tasha_140
Ohh right, got it - thanks again! (:
Just unsure whether I would think to do this if that makes sense? Seeing your explanation it now seems obvious, yet it never came to my head to attempt to solve it that way


Well, anytime you want to go from something involving cos⁡2θ\cos^2 \theta to something involving sin⁡2θ\sin^2 \theta you'll always need to use that identity. It comes with practice, I suppose. Now that you've seen it, you'll be more aware of it - hopefully.

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