Turn on thread page Beta
    Offline

    0
    ReputationRep:
    (Original post by Featherflare)
    Yea iv done the unthinkable.. given up. I dont think it can be done with c1- c4 knowledge which is all i have
    Its like trying to solve 5x = 3x.

    I have a question which should be easy but I don't understand it.

    (Original post by C3)
    The function f is defined for positive real values of x by

    f(x) = 12lnx - x^3/2

    Write down the set of values of x for which f(x) is an increasing function of x
    Since its the differentiation chapter I guess you have to find f'(x) and then what?

    Thanks in advance.
    Offline

    0
    ReputationRep:
    (Original post by ntrik)
    Its like trying to solve 5x = 3x.

    I have a question which should be easy but I don't understand it.



    Since its the differentiation chapter I guess you have to find f'(x) and then what?

    Thanks in advance.
    Yes you need to find f'(x) and do f'(x) > 0 as it is an increasing function (ie gradient is more than zero)
    Offline

    0
    ReputationRep:
    f'(x) = 12/x - (3/2)x^0.5
    f'(x) > 0
    12/x - (3/2)x^0.5 > 0
    (3/2)x^0.5 < 12/x
    x^(3/2) < 8 (cross multiply)
    x< 8^(2/3)
    x<4

    sorry i only just saw the x values should be positive so
    0<x<4
    Offline

    0
    ReputationRep:
    Thanks.

    Heres another question I'm having trouble on if anyone wants to have a go.

    (Original post by C3)
    The maximum point on the curve with equation x√sinx, 0<x<180, is the point A. Show that the x-cordinate of point A satisfies the equation 2tanx + x = 0.
    • Thread Starter
    Offline

    2
    ReputationRep:
    Aight..in terms of formulas, we only need to know these:

    cos²x + sin²x = 1
    sec²x = 1+tan²x
    cosec²x = 1 + cot²x

    and

    the R formula for e.g. AcosX + BsinX = c i.e. tanσ = b/a R = √a²+b²

    the double angles can be worked out using the sin(A+B) = sinAcosB + cosAsinB formula, which is given in the formula booklet.
    Offline

    16
    ReputationRep:
    We also need to know the qoutient rule aswell i think. Is there anything else we need to know?
    • Thread Starter
    Offline

    2
    ReputationRep:
    Hmm I never thought of the quotient rule as a formula =o, cus then you have chain rule..etc
    Offline

    18
    ReputationRep:
    (Original post by TheWolf)
    Hmm I never thought of the quotient rule as a formula =o, cus then you have chain rule..etc
    yep

    chain rule: dy/dx = dy/du x du/dx
    product rule: if y = uv => dy/dx = u.dv/dx + v.du/dx
    quotient rule: if y = u/v => dy/dx = [v.du/dx - u.dv/dx]/v²

    dy/dx = 1/[dx/dy]
    Offline

    2
    ReputationRep:
    Can someone please explain 3c from the specimen paper please?
    I get that theta can be equal to 0 and tan theta = -1
    I just don't understand the last step they did, I forgot this stuff
    Thanks
    Offline

    2
    ReputationRep:
    (Original post by TheWolf)
    Hmm I never thought of the quotient rule as a formula =o, cus then you have chain rule..etc
    Quotient rule is given.

    [ although it is a fairly ugly form: loads of g(x) and f(x) ]
    Offline

    2
    ReputationRep:
    (Original post by violet)
    Can someone please explain 3c from the specimen paper please?
    I get that theta can be equal to 0 and tan theta = -1
    I just don't understand the last step they did, I forgot this stuff
    Thanks
    You get: 2sin(x/2)[cos(x/2) + sin(x/2)] = 0

    For this to be true either 2sin(x/2) = 0

    OR cos(x/2) + sin(x/2) = 0

    ==> Two solutions.

    The second give tan(x/2) = -1

    As: sin(x/2) = -cos(x/2)

    => tan(x/2) = -1
    Offline

    1
    ReputationRep:
    (Original post by Featherflare)
    Yea iv done the unthinkable.. given up. I dont think it can be done with c1- c4 knowledge which is all i have
    Ahhh noo u shouldn't have given up...I think I have the answer...

    sin3x=cosx
    sin(2x+x)=cosx
    sin2xcosx+cos2xsinx=cosx
    (2sinxcosx)cosx+(1-2sin²x)sinx=cosx
    2sinxcos2x+sinx-2sin³x=cosx
    2sinx(1-2sin²x)+sinx-2sin³x=cosx
    2sinx-4sin³x+sinx-2sin³x=cosx
    (3sinx-6sin³x)²=(cosx)² <----(square both sides)
    9sin²x-36sin^4 x+36sin^6 x=1-sin²x
    36sin^6 x-36sin^4 x+10sin²x-1=0

    let y=sin²x:
    36y³-36y²+10y-1=0
    gives 3 solutions...2 of which are complex, 1 which is real.

    The real one is y=0.6282

    so...

    sin²x=0.6282
    sinx= √0.6282
    x = 52.4°



    You'd be extremely unlucky to get a question like this,especially because complex numbers aren't in C3(at least I don't think so), took me a long time to solve and got lots of help from a further mathmo.. :rolleyes:
    Offline

    2
    ReputationRep:
    (Original post by nas7232)
    We also need to know the qoutient rule aswell i think. Is there anything else we need to know?
    Only sin = (diff) => cos

    and cos = (diff) => -sin

    [ the only two not given in the booklet ]
    Offline

    2
    ReputationRep:
    (Original post by samdavyson)
    You get: 2sin(x/2)[cos(x/2) + sin(x/2)] = 0

    For this to be true either 2sin(x/2) = 0

    OR cos(x/2) + sin(x/2) = 0

    ==> Two solutions.

    The second give tan(x/2) = -1

    As: sin(x/2) = -cos(x/2)

    => tan(x/2) = -1
    Yep, I get that far, but I dunno how to get 3/2 pi

    I get theta is 90 which is 1/2 pi :confused:
    Offline

    2
    ReputationRep:
    (Original post by violet)
    Yep, I get that far, but I dunno how to get 3/2 pi

    I get theta is 90 which is 1/2 pi :confused:
    tan-1(-1) = -pi/4 = θ/2

    But the interval is 0 < θ < 2pi ==> 0 < θ/2 < pi

    So you do a CAST diagram or look at the graph (or whatever your method is) and as tan is negative it must be in the second or fourth quadrants.

    But the limits restrict it to the second quadrant for θ/2.

    => Adding pi to the -pi/4 gives the value for θ/2

    => 3pi/4

    So θ = 3pi/2

    This isn't very clear but you should have some method for dealing with stuff like this from chapter 10 of C2.
    Offline

    0
    ReputationRep:
    Aside from

    (Original post by C3)
    The maximum point on the curve with equation x√sinx, 0<x<180, is the point A. Show that the x-cordinate of point A satisfies the equation 2tanx + x = 0.
    Does anyone know how to solve:

    cos2x + sin2x = 0

    :confused:
    Offline

    15
    ReputationRep:
    (Original post by ntrik)
    Aside from



    Does anyone know how to solve:

    cos2x + sin2x = 0

    :confused:
    Divide by cos2x to get tan2x = -1 and then solve
    Offline

    2
    ReputationRep:
    (Original post by samdavyson)
    tan-1(-1) = -pi/4 = θ/2

    But the interval is 0 < θ < 2pi ==> 0 < θ/2 < pi

    So you do a CAST diagram or look at the graph (or whatever your method is) and as tan is negative it must be in the second or fourth quadrants.

    But the limits restrict it to the second quadrant for θ/2.

    => Adding pi to the -pi/4 gives the value for θ/2

    => 3pi/4

    So θ = 3pi/2

    This isn't very clear but you should have some method for dealing with stuff like this from chapter 10 of C2.

    Oh, yep, I see.. Whoops!
    Thanks.
    Offline

    0
    ReputationRep:
    (Original post by RichE)
    Divide by cos2x to get tan2x = -1 and then solve
    tan2x = -1
    2x = -45
    x = -90

    if x = -90

    then cos(-180) + sin(-180) = 0
    -1 + 0 = 0

    :confused:

    The answers are 67.5 and 157.5 apparently
    Offline

    16
    ReputationRep:
    (Original post by ntrik)
    tan2x = -1
    2x = -45
    x = -90

    if x = -90

    then cos(-180) + sin(-180) = 0
    -1 + 0 = 0

    :confused:

    The answers are 67.5 and 157.5 apparently
    ]

    There is your mistake.
 
 
 
Turn on thread page Beta
Updated: June 20, 2005
The home of Results and Clearing

2,804

people online now

1,567,000

students helped last year

University open days

  1. SAE Institute
    Animation, Audio, Film, Games, Music, Business, Web Further education
    Thu, 16 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  3. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
Poll
Do you want your parents to be with you when you collect your A-level results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

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

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

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