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

18

Original post

by notnek

Can be done without a calculator:

a) Factorise $\ \ \ 2x^2-3x-14$

b) Hence or otherwise solve the equation

$\displaystyle \ \ \ 2(4x^2+9x-2)^2-3(4x^2+9x-2)-14 = 0$

Spoiler

Reply 222

18

Original post

by notnek

I recommend posting a specific question since more people are likely to help. You could ask what topic you need to know to be able to do the question. Threads asking for resources won't get as many views.

Also try not to bump your questions - if you do this then it looks like the thread is answered and fewer people are likely to look at your thread. Leave your question unanswered for a while and you should get replies.

I've learnt all this stuff but I'm not an expert on it anymore so I'm not the best person to help you.

Noted - I just posted a question on it. I'll wait for the replies, thanks for the advice and the support.

Reply 223

9

Original post

by thekidwhogames

Spoiler

it is right. However, I would have said Z = 4x^2 + 9x-2 for avoiding confision.

Reply 224

9

Original post

by notnek

Spoiler

This is why you should always try to simplify (or reduce) into on term in trig, i.e.

sin(Theta)cos(Theta) = 0
(1/2) [2sin(Theta)cos(Theta)] = 0

sin(2 x Theta) = 0

So, 2 x (Theta) = 0 or pi or 2pi

Yes, in sine or cosine, you must be careful when it changes sign (i.e. for sine at pi, 2pi).

However, in trig, the rule of thumb is, because of their periodic nature, take the first values where you find sine, cos or tan values and check by substitution.

Then check for the whole period

i.e. in this case, take 0 and pi, and check for 2pi).

Hope this helps.

Reply 225

9

Original post

by thekidwhogames

Noted - I just posted a question on it. I'll wait for the replies, thanks for the advice and the support.

Hello,

a) Factorise \ \ \ 2x^2-3x-14

b) Hence or otherwise solve the equation

2(4x^2+9x-2)^2-3(4x^2+9x-2)-14 = 0

a) (2x-7)(x+2)

b)
Now try to use a.

So, Let z = (4x^2+9x-2)

2Z^2 - 3Z - 14 = 0

Note the the term in Z is in the form of expression in a.

So, (2Z-7)(Z+2) = 0

So,
2(4x^2+9x-2) - 7 = 0 or (4x^2+9x-2) + 2 = 0.

Reply 226

9

Original post

by Pretish

it was good one lol

This is why you should always try to simplify (or reduce) into on term in trig, i.e.

sin(Theta)cos(Theta) = 0
(1/2) [2sin(Theta)cos(Theta)] = 0

sin(2 x Theta) = 0

So, 2 x (Theta) = 0 or pi or 2pi

Yes, in sine or cosine, you must be careful when it changes sign (i.e. for sine at pi, 2pi).

However, in trig, the rule of thumb is, because of their periodic nature, take the first values where you find sine, cos or tan values and check by substitution.

Then check for the whole period

i.e. in this case, take 0 and pi, and check for 2pi).

Hope this helps.

Reply 227

9

Original post

by Pretish

Spoiler

This is why you should always try to simplify (or reduce) into on term in trig, i.e.

sin(Theta)cos(Theta) = 0
(1/2) [2sin(Theta)cos(Theta)] = 0

sin(2 x Theta) = 0

So, 2 x (Theta) = 0 or pi or 2pi

Yes, in sine or cosine, you must be careful when it changes sign (i.e. for sine at pi, 2pi).

However, in trig, the rule of thumb is, because of their periodic nature, take the first values where you find sine, cos or tan values and check by substitution.

Then check for the whole period

i.e. in this case, take 0 and pi, and check for 2pi).

Hope this helps.

Reply 228

9

Original post

by notnek

Spoiler

This is why you should always try to simplify (or reduce) into on term in trig, i.e.

sin(Theta)cos(Theta) = 0
(1/2) [2sin(Theta)cos(Theta)] = 0

sin(2 x Theta) = 0

So, 2 x (Theta) = 0 or pi or 2pi

Yes, in sine or cosine, you must be careful when it changes sign (i.e. for sine at pi, 2pi).

However, in trig, the rule of thumb is, because of their periodic nature, take the first values where you find sine, cos or tan values and check by substitution.

Then check for the whole period

i.e. in this case, take 0 and pi, and check for 2pi).

Hope this helps.

Reply 229

9

Original post

by thekidwhogames

Find nth term of numerator and denominator so 6n2-n-2/4n2-1. Apply l'hopitals rule to get the limit to infinity to get 3/2.

Yes, that wil work.

But there is series based method.

So,

3/3/, 4/3, 7/5, 10/7, 13/9

leaving first term,

Numerator general term = (3n+1), n>= 1

Denominator general term = (2n+1), n>= 1

Apart from 1st term general term is (3n+1) / (2n+1), n>=1

Including first term, general term is (3n+1) / (2n+1), n>=0

(3n+1) / (2n+1) = (3 + (1/n) ) / (2 + (1/n))

As n -> infinity,
(3n+1) / (2n+1) = (3 + (1/n) ) / (2 + (1/n)) = (3 + 0) / (2 + 0) = 3/2

Reply 230

9

Original post

by chinkinator

hey man can u show me your working out? I somehow got p= -2, q=4

Draw y coordinate lines for C,B,A, and
starting from C, draw to the left, the line parallel to x coordinate for C.

Let say, line from c toward left and parallel to x coordinate for C meets Y coordinate line of B on W and Y coordinate line of A on Z

Now, (BW / AZ) = ( CW / CZ) = 20% = (1/5)

Very careful when you calculate BW and AZ, for example, if you are calculating BW =yB -yC, then you have to do for the same for AZ, i.e. yA - yC, or vice versa, i.e. BW = yC - yB and AZ = yC - yA.

(BW / AZ) = (2p-q) / (p - 2q) = 1/5 ==> q = 3p

Similarly for CW = xC - xB and CZ = xC - xA.

(q-4) / (2q + 3p - 2) = 1/5 ====> 3q - 3p = 20 -2 = 18

6p = 18 ===> p =3
q = 3p = 9.

draw , g

Reply 231

18

Original post

by mathcool

Yes, that wil work.

But there is series based method.

So,

3/3/, 4/3, 7/5, 10/7, 13/9

leaving first term,

Numerator general term = (3n+1), n>= 1

Denominator general term = (2n+1), n>= 1

Apart from 1st term general term is (3n+1) / (2n+1), n>=1

Including first term, general term is (3n+1) / (2n+1), n>=0

(3n+1) / (2n+1) = (3 + (1/n) ) / (2 + (1/n))

As n -> infinity,
(3n+1) / (2n+1) = (3 + (1/n) ) / (2 + (1/n)) = (3 + 0) / (2 + 0) = 3/2

Yeah that's also a good method ofndokjg it. Thanks for the help. By the way do you anything about modular arithmetic and using it to get last 2 digits of a paper and using it to prove no invisibility?

Reply 232

OP

21

Later today I will be posting a full 105 mark calculator paper that I have created. Hopefully it should be useful for anyone who has completed all the AQA papers or just wants an extra challenge.

I will post answers but not solutions so it would be great if you could all share solutions to some of the questions and we'll end up with a full mark scheme.

Reply 233

B0redBrioche

12

Original post

by notnek

Later today I will be posting a full 105 mark calculator paper that I have created. Hopefully it should be useful for anyone who has completed all the AQA papers or just wants an extra challenge.

It won't have solutions though so it would be great if you could all share solutions to some of the questions and we'll end up with a full mark scheme.

Okay thanks!

Reply 234

18

Original post

by notnek

Later today I will be posting a full 105 mark calculator paper that I have created. Hopefully it should be useful for anyone who has completed all the AQA papers or just wants an extra challenge.

I will post answers but not solutions so it would be great if you could all share solutions to some of the questions and we'll end up with a full mark scheme.

Alright thanks a lot. That will be good fun :smile:

Reply 235

Loci Pi

16

Original post

by notnek

Later today I will be posting a full 105 mark calculator paper that I have created. Hopefully it should be useful for anyone who has completed all the AQA papers or just wants an extra challenge.

I will post answers but not solutions so it would be great if you could all share solutions to some of the questions and we'll end up with a full mark scheme.

Thanks. That sounds good :smile:

Reply 236

OP

21

Something else has come up but I should be able to post the exam before the evening. And answers will come a bit later.

Reply 237

boi_

12

Original post

by notnek

Later today I will be posting a full 105 mark calculator paper that I have created. Hopefully it should be useful for anyone who has completed all the AQA papers or just wants an extra challenge.

I will post answers but not solutions so it would be great if you could all share solutions to some of the questions and we'll end up with a full mark scheme.

Thanks man, that'd be really helpful! Did you do paper one on thursday, if so, how'd u find it? And what kinda stuff do you think will come up on paper 2?

Reply 238

OP

21

Original post

by chinkinator

Thanks man, that'd be really helpful! Did you do paper one on thursday, if so, how'd u find it? And what kinda stuff do you think will come up on paper 2?

I didn't do the paper. I teach this stuff but I haven't seen the paper yet.

Reply 239