Year 12s: what will be the first way you research your future options? >>

Edexcel C3,C4 June 2013 Thread

Scroll to see replies

c3- jan 2007 -question 8iia)

given that: y= arcosx

-1 < x ≤ 1 and 0 ≤ y ≤ pi

express arcsin in terms of y

can anyone help with this? really stuck.. thanks

The exam solutions link will show you the answer, but I did it differently.

arccos(x) = y
so cos(y) = x

now, assuming you know that cos(x) = sin(90-x) [and] you know that:

sin(90-y) = x

and hence arccsin(x) = 90-y

Reply 3041

thelion0

Original post by justinawe

Generally, the domain would be everything except for the value that makes the denominator equal 0, and the range would be anything other than 0.

Of course, it can depend on additional information given in the question as well.

Is there not an algebraic method?

Reply 3042

suncake

Original post by überambitious_ox

Find the exact value of x such that

3arctan (x-2) + pi =0

Please can someone explain how this would be solved. Thank you

Hope this is right... :ninja:

Posted from TSR Mobile

Reply 3043

Chris-69

Original post by überambitious_ox

Find the exact value of x such that

3arctan (x-2) + pi =0

Please can someone explain how this would be solved. Thank you

I'll use degree instead of radians for convenience.

3arctan(x-2) + 180 = 0
3arctan(x-2) = -180
acrtan(x-2) = -60
x-2 = tan(-60)
x-2 = -root3
x = 2 - root3

That's what I got anyway, not sure of the actual answer.

Edit: Looks like I've been beaten to it :tongue:

Reply 3044

joereed

Original post by MathsNerd1

It will be due but just remember they are reflections in line y=x so it's not too hard really

Yer I think theyre ok, but not the nicest questions in the world, especially back in 2007! Apart fro that, theres nothing in the content that strikes me as something 'bad'.

Reply 3045

überambitious_ox

Original post by suncake

Hope this is right... :ninja:

Posted from TSR Mobile

Thank you; yes it is :smile:

Reply 3046

Flounder1

Apparently c3 was leaked and the replacement is way harder !!! :frown:

Posted from TSR Mobile

Reply 3047

Chris-69

Original post by Flounder1

Apparently c3 was leaked and the replacement is way harder !!! :frown:

Posted from TSR Mobile

Why would it be harder? Why not just a different paper?

Reply 3048

ApplePearOrange5297

Original post by Flounder1

Apparently c3 was leaked and the replacement is way harder !!! :frown:

Posted from TSR Mobile

I call bs

Reply 3049

BearOfStown

Original post by Flounder1

Apparently c3 was leaked and the replacement is way harder !!! :frown:

Posted from TSR Mobile

Source?

Reply 3050

Chris-69

Original post by ApplePearOrange5297

I call bs

No, the papers were stolent/leaked/lost, it's on the edexcel website. It affected C3, C4 and some others. I don't see why they'd be harder, though. Just a different paper.

Edit: https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-ash3/268231_10200103014917943_777688065_n.jpg

Reply 3051

physicso

Can anyone help me with this quick problem?

The question states

f: ln(x+k)
g: |2x-k|

Evaluate fg(k/4)

I got that fg = ln(2x-k+k) = ln2x = ln2k/4

Simplifies to lnk/2

HOWEVER, the markscheme says that the answer is ln3k/2?

Reply 3052

kingpro88

Hows everyone feeling about tommorow ? :/ did anyone else make loads of silly mistakes in jan that made them miss the A by 1 mark :/ errh retakes

Reply 3053

AverageDoc

Original post by Flounder1

Apparently c3 was leaked and the replacement is way harder !!! :frown:

Posted from TSR Mobile

Yeah i heard about that too. Just hope theyre nice to us. The c4 paper got leaked too :frown:

Original post by physicso

What papers this from ?

Reply 3056

joereed

Original post by physicso

f(k/4) = |-k/2| = k/2 sub this into f(x) and BOSH!

Reply 3057

PoorLoser

Original post by BearOfStown

Source?

http://www.thestudentroom.co.uk/showthread.php?t=2362486

And I'm guessing the source of that is the edexcel website.

Yes, there's a replacement paper but this doesn't mean it will be harder. As edexcel have said, the replacement papers have gone through the same processes as the originals, and are no harder/easier than the original papers.

So there really isn't anything to worry about.

Edit:

And if you don't believe edexcel, and still think they'll give a harder paper, the FP1 replacement paper was done on Monday, and most candidates found it (fairly) easy. So I doubt they'll screw you guys over :wink: