Year 13 Maths Help Thread

Announcements Posted on
    Offline

    3
    ReputationRep:
    (Original post by solC)
    Well since we know the points are on the circle |z|=2 the general point on that circle is z=2(cos(\theta)+isin(\theta)). Using the exponential form works too as Physicsmaths said

    Let z=2(cos(\theta)+isin(\theta))
    \Rightarrow w=2(cos\theta+isin\theta) +\frac{4}{2(cos(\theta)+isin(\th  eta))}
    \Rightarrow  w=2(cos\theta+isin\theta) +\frac{2(cos\theta-isin\theta)}{\cos^2\ + \sin^2\theta}
    \Rightarrow  w=4cos\theta
    And the range of this is -4 \leq \theta \leq 4 so K is 4
    Thanks but why did u divide the rhs fraction by cos2+sin2?
    Offline

    2
    ReputationRep:
    (Original post by youreanutter)
    Thanks but why did u divide the rhs fraction by cos2+sin2?
    when you make the denominator real by multiplying top and bottom by cos\theta-isin\theta that's what you get.
    Offline

    0
    ReputationRep:
    (Original post by Ano123)
    I'll volunteer to be a helper. My purpose in this life is to serve...maths to people?
    i've retaken maths 3 times now, i can't do it at all like i've been revising all the time - i dont even do half of my other work (science and other lessons) as im trying to pass maths but i cant do it and i need to pass for the apprenticeship i want to do
    • Thread Starter
    Offline

    2
    ReputationRep:
    Am I the only one who finds that the 'definite integrals involving modulus function questions' in STEP I papers are some of the easiest questions that they can throw at you?
    Offline

    2
    ReputationRep:
    (Original post by savarna)
    i've retaken maths 3 times now, i can't do it at all like i've been revising all the time - i dont even do half of my other work (science and other lessons) as im trying to pass maths but i cant do it and i need to pass for the apprenticeship i want to do
    Is this GCSE or A level?
    Offline

    2
    ReputationRep:
    How were they able to make lamda equal that?

    Name:  image.jpg
Views: 19
Size:  499.0 KB
    Offline

    2
    ReputationRep:
    (Original post by amelienine)
    How were they able to make lamda equal that?

    Name:  image.jpg
Views: 19
Size:  499.0 KB
    ln both sides, leads you to -40lambda = ln(1/2)
    Divide by -40, you get lambda = -ln(1/2)/40
    -ln(1/2) = ln(1/2^-1) = ln(2)
    Offline

    2
    ReputationRep:
    (Original post by amelienine)
    How were they able to make lamda equal that?

    Name:  image.jpg
Views: 19
Size:  499.0 KB

    Take logs of both sides and remember
    ln0.5 = ln(1/2) = ln(2^-1) = -ln2
    Offline

    2
    ReputationRep:
    (Original post by k.russell)
    ln both sides, leads you to -40lambda = ln(1/2)
    Divide by -40, you get lambda = -ln(1/2)/40
    -ln(1/2) = ln(1/2^-1) = ln(2)
    (Original post by solC)
    Take logs of both sides and remember
    ln0.5 = ln(1/2) = ln(2^-1) = -ln2
    Thank you!! (I've just learnt how to multi-reply woohoo)
    Offline

    2
    ReputationRep:
    Name:  image.jpg
Views: 16
Size:  499.5 KB
    Took this from the mark scheme, don't know how T = 93 tho? Put it in my calculator and I'm getting a different answer.
    Offline

    3
    ReputationRep:
    (Original post by amelienine)
    Took this from the mark scheme, don't know how T = 93 tho? Put it in my calculator and I'm getting a different answer.
    That's wrong, it does not go to the second line.
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    That's wrong, it does not go to the second line.
    then how do you find what T is?
    Offline

    3
    ReputationRep:
    (Original post by amelienine)
    then how do you find what T is?
    If you know your exponentials and natural logarithms, you may notice that e^{-\ln(\frac{2}{40})}=e^{\ln(\frac{  40}{2})}=20 and just divide both sides by it for T.

    The -1 in front of ln becomes the exponent when moved inside the logarithm, so we take the reciprocal of the fraction as the result.
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    If you know your exponentials and natural logarithms, you may notice that e^{-\ln(\frac{2}{40})}=e^{\ln(\frac{  40}{2})}=20 and just divide both sides by it for T.

    The -1 in front of ln becomes the exponent when moved inside the logarithm, so we take the reciprocal of the fraction as the result.
    Oh thank you! I'm resitting my C3 this year and I've forgotten so much of it

    And it turns out it wasn't ln(2/40)! It was (ln(2))/40, which gave out a completely different answer... oi I'm hopeless
    Offline

    2
    ReputationRep:
    this question is really ****ing me up, I've tried doing it SO many times using LCM and stuff but can't seem to get the right answer :

    part a) A christmas tree has been decorated with a set of lights thatcontain two circuits. On each individual circuit, all the lights come and go off at the same time, but the two circuits are out of sync.

    Circuit 1: lights ON for 4 seconds, and OFF for 9 seconds.
    Circuit 2: lights ON for 7 seconds and OFF for 2 seconds,

    20 seconds ago, both sets of lights came ON at exactly the same time. In how many seconds from now will both sets of lights go OFF at the same time ?



    part b) consider the same christmas tree as above but with the following circuits:
    Circuit 1: Lights ON for 3 seconds, OFF for 5 seconds
    Circuit 2: lights ON for 2 seconds, OFF for 6 seconds
    Ten seconds ago, both sets of lights went ON at exactly the same time. In how many seconds from now will both sets of lights go ON at the same time?


    answers below, I just don't know how to get to them :
    Spoiler:
    Show
    part a: 23
    part b: they will never go OFF at the same time
    Online

    3
    ReputationRep:
    (Original post by medhelp)
    this question is really ****ing me up, I've tried doing it SO many times using LCM and stuff but can't seem to get the right answer :

    part a) A christmas tree has been decorated with a set of lights thatcontain two circuits. On each individual circuit, all the lights come and go off at the same time, but the two circuits are out of sync.

    Circuit 1: lights ON for 4 seconds, and OFF for 9 seconds.
    Circuit 2: lights ON for 7 seconds and OFF for 2 seconds,

    20 seconds ago, both sets of lights came ON at exactly the same time. In how many seconds from now will both sets of lights go OFF at the same time ?



    part b) consider the same christmas tree as above but with the following circuits:
    Circuit 1: Lights ON for 3 seconds, OFF for 5 seconds
    Circuit 2: lights ON for 2 seconds, OFF for 6 seconds
    Ten seconds ago, both sets of lights went ON at exactly the same time. In how many seconds from now will both sets of lights go ON at the same time?


    answers below, I just don't know how to get to them :
    Spoiler:
    Show
    part a: 23
    part b: they will never go OFF at the same time
    Just say
    'I don't believe in christmas, question is inaccurate qed'


    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    Name:  graph-diff.PNG
Views: 32
Size:  13.0 KB

    0 = x(9-2x^2)^(1/2)
    x = root (9/2)
    = root 9 / root 2
    a = 3 / root 2

    I have a feeling this isn't right, though. ' As a multiple of root 2 '

    could I do 3root2/2 ( by mutlitplying both parts of fraction by root 2 ) or does this not make a difference?
    Online

    3
    ReputationRep:
    (Original post by JaredzzC)
    Name:  graph-diff.PNG
Views: 32
Size:  13.0 KB

    0 = x(9-2x^2)^(1/2)
    x = root (9/2)
    = root 9 / root 2
    a = 3 / root 2

    I have a feeling this isn't right, though. ' As a multiple of root 2 '

    could I do 3root2/2 ( by mutlitplying both parts of fraction by root 2 ) or does this not make a difference?
    Your method seems right, yes they probably want it as 3√2/2
    Offline

    3
    ReputationRep:
    Wut Name:  Screen Shot 2016-10-18 at 18.35.03BST.png
Views: 17
Size:  18.4 KB
    Offline

    3
    ReputationRep:
    (Original post by jamestg)
    Wut
    Take a logarithm of base 3 for both sides and rearrange for \log_3(2)
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: December 7, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
How are you feeling about doing A-levels?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.