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The Edexcel C3 (14/06/12 - AM) Revision Thread

Reply 580

Arva

14

Original post by JordanS94

I've already done well :L I did it in january but I wanted to secure my A overall for Maths :lol:

i got 90ums for it in january yes I know I'm stupid for retaking :lol:

I think retaking's a good idea if you think it will improve your overall grade!
Nice score in January though, aiming for the 100 this time? :wink:

Reply 581

suncake

14

Original post by Arva

Can you post the full question? Remember, the domain of the inverse = the range of the original, and vice versa.

Question 4 here - http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202011%20-%20QP/6665_01_que_20110616.pdf
:smile:

I did forget that.. but it still seems to not work? dfsjsdfljk

Reply 582

JordanS94

16

Original post by Arva

I think retaking's a good idea if you think it will improve your overall grade!
Nice score in January though, aiming for the 100 this time? :wink:

yeah :L It's because I seriously messed up my C4 exam :lol:

tbh I happy with 91 or higher :biggrin:

Reply 583

emmaaa88

12

Original post by grazie

I'd say it almost certainly does. If there's only 1 mark for c) you won't get any. As you haven't included the definition for f(x) in you're attachment, I can't explain why.

This is the whole question :smile:

thank you!

photo(6).JPG

Reply 584

Arva

14

Original post by suncake

Question 4 here - http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202011%20-%20QP/6665_01_que_20110616.pdf
:smile:

I did forget that.. but it still seems to not work? dfsjsdfljk

It does work, don't worry!
if X is greater than or equal to -1, the range of f(x) can be found out: Y must be less than or equal to 4, since ln(x+2) will always be positive. The range of F(x) = the domain of the inverse.

Reply 585

SamXi

7

Original post by Brand New Eyes

All the normal are on the website apart from Jan 2012 which is in some other thread

sci2.co.uk/maths.html has all the Solomon I think

Thanks man this is what I was looking for!

even tried sci.co.uk

+rep

Reply 586

jjbrown

Can anyone tell me how to do this?

Original post by raheem94

l.

If there is a question regarding the drawing of 2cosec2theta with the range 0<theta<360, why doesn't the graph of 2cosec2theta extend from 0<2theta<720 but remains at 0<theta<360

Reply 588

Arva

14

Original post by jjbrown

Can anyone tell me how to do this?

i.

\log_a x = \frac {\log_b x}{\log_b a}

ii.

\log_a x + \log_a y = \log_a (xy)[br][br]\log_a x - \log_a y = \log_a (\frac{x}{y})

(edited 11 years ago)

Reply 589

SamXi

7

http://www.sci2.co.uk/maths/C3/Solomon/C3L.pdf

Q1. c)

I don't understand how we are meant to know F^-1(x) from the information we have already worked out

Reply 590

raheem94

13

Original post by Extricated

If there is a question regarding the drawing of 2cosec2theta with the range 0<theta<360, why doesn't the graph of 2cosec2theta extend from 0<2theta<720 but remains at 0<theta<360

f(x) = 2 cosec (x)

g(x) = f(2x) = 2cosec (2x)

So g(x) is just a horizontal stretch of scale factor 0.5.

If

0 < 2 \theta < 720

, then to simplify it divide it by 2,

0 < \theta < 360

Remember we label the horizontal axis(x-axis) as

\theta

not

2 \theta

.

(edited 11 years ago)

Reply 591

Dark Lord of Mordor

1

Original post by SamXi

http://www.sci2.co.uk/maths/C3/Solomon/C3L.pdf

Q1. c)

I don't understand how we are meant to know F^-1(x) from the information we have already worked out

ff(x) = x

Therefore f(x) = f^(-1) (x) (The inverse is equal to the original function)

Reply 592

Arva

14

Original post by SamXi

http://www.sci2.co.uk/maths/C3/Solomon/C3L.pdf

Q1. c)

I don't understand how we are meant to know F^-1(x) from the information we have already worked out

If ff(x) returns f(x) to x, then f(x) = f^-1) (x).

Reply 593

Arva

14

Original post by SrijanParmeshwar

ff(x) = x

Therefore f(x) = f^(-1) (x) (The inverse is equal to the original function)

You beat me. :colonhash:

Reply 594

jjbrown

Original post by Arva

i.

\log_a x = \frac {\log_b x}{\log_b a}

ii.

\log_a x + \log_a y = \log_a (xy)[br][br]\log_a x - \log_a y = \log_a (\frac{x}{y})

Thanks!

Reply 595

ninegrandstudent

11

Suppose I'd better subscribe to this now the exams closer!

Reply 596

Arva

14

Original post by jjbrown

Thanks!

No problemo.

Reply 597

Brand New Eyes

2

Original post by Arva

i.

\log_a x = \frac {\log_b x}{\log_b a}

ii.

\log_a x + \log_a y = \log_a (xy)[br][br]\log_a x - \log_a y = \log_a (\frac{x}{y})

with the base (b) in e?

Reply 598

Arva

14

Original post by Brand New Eyes

with the base (b) in e?

Base a would be e for this question.

Reply 599

grazie

2

Original post by emmaaa88

This is the whole question :smile:

thank you!

photo(6).JPG

Wow! Reading the text on that image was tough.

f(x) = e^2x - k

So as x gets very large in the negative direction, e^2x approaches 0, but never reaches it. It approaches the y=0 asymptote. Which means that the range of f is f(x) > -k. If e^2x reached 0 (but it doesn't), the range of f would be f(x)

\geq

-k

The Edexcel C3 (14/06/12 - AM) Revision Thread

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