OCR Core 3 (Not MEI) - Wednesday 20th January PM

Reply 180

GOAhead

korobeiniki

If you want I can upload the paper? TIA once again for providing us with the solutions!

can you send me the paper please?? :frown:

Reply 181

Kelevra

1

And tan 5? That was a complete mystery for me?

Reply 182

Andylol

14

Mr. Cricket

If you put routex = 0
x = 0

If you put route (2x+3) = 0
x = -1.5

Therefore a shift of -1.5 in x-direction!

Normally a question like this would be really easy, but this one was quite tricky, because it stretch in the y-direction which was odd, and didnt shift by -3, which in a normal translation it would! x

Balls. Yeah =/ Silly mistake no. 1

Keep them coming! :frown:

Reply 183

Ammar21

5

Kelevra

And tan 5? That was a complete mystery for me?

lol i know the only nice number i know for tan is tan 45 and obviously u cant make 5 from 10 and 45 so i thought hmm what if i used tan (15-10)
i typed tan15 into the calculator and then it came out with a very pleasing 2-(3)^.5 or something soo i just used that. but im not too sure whether what i did was right or not - overall a very weird question indeed!

Reply 184

GOAhead

We need the paper please

Reply 185

ilyiylm

yessss upload the paper PLEASE

Reply 186

Mr. Cricket

Andylol

Balls. Yeah =/ Silly mistake no. 1

Keep them coming! :frown:

Lol,
Same here - just didnt think to check it! :/

Reply 187

Kelevra

1

The answer to the last one, I think:

7cos(10 - #) = 3sin(#+10)
7cos10cos# + 7 sin10sin# = 3sin#cos10 + 3sin10cos#

Dividie by cos#

7cos10 + 7 sin10 tan# = 3tan# cos10 + 3sin10

Divide by cos10

7 + 7tan10tan# = 3tan# + 3tan10

subst tan10 for p and factorise

tan#(7p -3) = 3p - 7

tan#=3p-7/7p-3

yeesss? Can anyone confirm this?

Reply 188

mrdude

Kelevra

How? I just guessed with trial and error and got the same! :woo:

I can't fully remember the question but I'll try and explain it.

For the first one you were given that it was converging to 1.973 (or something like that)

Therefore 1.973 = N^1/3(2(1.973) +3)

(I think I have the formula right, I may be wrong but the method still counts)

So 1.973/(2(1.973) +3) = N^1/3

So cubing that left hand side gave you N, which came to 18

The second part, you were given that the 3rd value of x was something like 1.2 and the 4th value was about 1.4 (again, I can't remember the exact numbers).

So then it was the same method as the first part except x,n+1 was the higher value you were given instead of them being the same like in the last part.

Sorry I didn't explain that very well, but I hope it helps!

the exam was nuts

it was !! PAPER pLEZ!!

Reply 191

Kelevra

1

mrdude

I can't fully remember the question but I'll try and explain it.

For the first one you were given that it was converging to 1.973 (or something like that)

Therefore 1.973 = N^1/3(2(1.973) +3)

(I think I have the formula right, I may be wrong but the method still counts)

So 1.973/(2(1.973) +3) = N^1/3

So cubing that left hand side gave you N, which came to 18

The second part, you were given that the 3rd value of x was something like 1.2 and the 4th value was about 1.4 (again, I can't remember the exact numbers).

So then it was the same method as the first part except x,n+1 was the higher value you were given instead of them being the same like in the last part.

Sorry I didn't explain that very well, but I hope it helps!

That sounds right, except the answer was 1.9073 to 4 dp. or something similar. But I did that 3 TIMES... and evey time it didn't come to an integer :no:

Reply 192

Andylol

14

mrdude

You had to find N twice right?

If I remember correctly, I got 18 for the first one and 52 for the second one.

YESSSSSSSS :cool:

Thank god

Reply 193

mrdude

Kelevra

That sounds right, except the answer was 1.9073 to 4 dp. or something similar. But I did that 3 TIMES... and evey time it didn't come to an integer :no:

I never came to an integer exactly, I just got something like 18.00004 and 51.9999, so I just put the answers down as 18 and 52

Reply 194

HitTheLights

Nidhogg_Rider

Well that's cool, I got the first part, too bad I left the next part and forgot to go back to it...

Does anyone know how you could have done the last part of 9?

To do q9iii) i expanded out the 2 addition formulae, then worked out tan10 = p, so sin10 = pcos10. Substituted that in and factorised it.

Came out rather nicely as 7-3p/3-7p or the other way up, im not sure

Reply 195

*MJ*

OP

2

korobeiniki

I can upload the paper if you guys want it?

I'll rep you. :smile:

Do it mehn! :woo:

Reply 196

Mr. Cricket

korobeiniki

If you want I can upload the paper? TIA once again for providing us with the solutions!

Yea - please upload the paper!

Thanks

Reply 197

x01

korobeiniki

If you want I can upload the paper? TIA once again for providing us with the solutions!

Yes can you please upload or PM me please.

Reply 198

*MJ*

OP

2

HitTheLights

To do q9iii) i expanded out the 2 addition formulae, then worked out tan10 = p, so sin10 = pcos10. Substituted that in and factorised it.

Came out rather nicely as 7-3p/3-7p or the other way up, im not sure