Hey there! Sign in to join this conversationNew here? Join for free

OCR A2 maths january, C3/C4/S2 question. Watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    C3:
    Hi I'm doing the C3 June 2005 paper. Just have some questions, hoping someone could help me out.
    Heres a link to the paper if anyone wants to help: http://www.school-portal.co.uk/Group...urceID=4777881

    Q8i, The diagram shows part of each of the curves y = e^1/5x and y=(3x+8)^1/3 The curves meet, as shown
    in the diagram, at the point P. The region R, shaded in the diagram, is bounded by the two curves and by the y-axis.
    Show by calculation that the x-coordinate of P lies between 5.2 and 5.3

    So I know this has something to do with proving the sign changes.
    So I put I used (3x+8)^1/3 - e^1/5x. I put in when x = 5.2 and when x = 5.3.
    I got 0.04 and -0.0006. So I did prove the sign changed. However looking in the mark scheme, it has a little table instead saying -0.04 and 0.0006 and then some other values before that. Heres a link to the mark scheme: http://www.school-portal.co.uk/Group...urceID=4777877

    Just wondering where I've gone wrong..



    C4:
    Just finished marking a C4 paper. Relatively simple vector question I'm confused with. Line L1 passes through (2, -3, 1) and (-1, -2, -4).
    Find an equation in the form r = a +tb.
    I thought it would be
    (2,-3,1) and I was always taught to do say if you're finding out the equation for line AB, you would do B - A.
    so I did (-1 -2, -2--3, -4-1) getting (-3, 1, -5).
    But they got the signs the other way round getting (3, -1, 5). Confused!
    Offline

    16
    ReputationRep:
    (Original post by 01234567)
    C3:
    Hi I'm doing the C3 June 2005 paper. Just have some questions, hoping someone could help me out.
    Heres a link to the paper if anyone wants to help: http://www.school-portal.co.uk/Group...urceID=4777881

    Q8i, The diagram shows part of each of the curves y = e^1/5x and y=(3x+8)^1/3 The curves meet, as shown
    in the diagram, at the point P. The region R, shaded in the diagram, is bounded by the two curves and by the y-axis.
    Show by calculation that the x-coordinate of P lies between 5.2 and 5.3

    So I know this has something to do with proving the sign changes.
    So I put I used (3x+8)^1/3 - e^1/5x. I put in when x = 5.2 and when x = 5.3.
    I got 0.04 and -0.0006. So I did prove the sign changed. However looking in the mark scheme, it has a little table instead saying -0.04 and 0.0006 and then some other values before that. Heres a link to the mark scheme: http://www.school-portal.co.uk/Group...urceID=4777877

    Just wondering where I've gone
    Your mark scheme link does not work

    But I would guess they have just subtracted the graphs the other way
    Yours would be fine if this is the case
    Offline

    16
    ReputationRep:
    (Original post by 01234567)

    C4:
    Just finished marking a C4 paper. Relatively simple vector question I'm confused with. Line L1 passes through (2, -3, 1) and (-1, -2, -4).
    Find an equation in the form r = a +tb.
    I thought it would be
    (2,-3,1) and I was always taught to do say if you're finding out the equation for line AB, you would do B - A.
    so I did (-1 -2, -2--3, -4-1) getting (-3, 1, -5).
    But they got the signs the other way round getting (3, -1, 5). Confused!
    You can write it either way

    Similarly you could use B instead of A as the starting point
    • Thread Starter
    Offline

    0
    ReputationRep:
    Hi i have a questions on S2 June 09 I was hoping someone could help me with. Link to the paper: http://www.ocr.org.uk/images/59560-q...atistics-2.pdf

    Q5) In a large region of derelict land, bricks are found scattered in the earth.
    Assume now that the number of bricks in 1 cubic metre of earth can be modelled by the distribution
    Po(3).
    ii) Find the probability that the number of bricks in 4 cubic metres of earth is between 8 and 14
    inclusive.
    iii) Find the size of the largest volume of earth for which the probability that no bricks are found is
    at least 0.4.

    ii) So because its 4 cubic meters not 1, we times Poisson by 4, so now we have Po(12).
    If its inclusive between 8 and 14, that gives us 8 > X < 14, with those signs of inequality don't know how to make it typing. This means we're finding when x is less or equal to 14 and because its poisson, we always do one less than the first value, so is this x is less or equal to 7??

    iii) For this i equated P(x=0) to more or equal to 0.4 ... not really sure where to go now.
    I have the formula of poisson of e^-l times l^0 over 0! but i dont know what to do now..


    Q7iv) I have no idea how they got the new parameters.. anyone?

    8ii) It is required to redesign the test so that the probability of making a Type I error is less than 0.01
    when the sample mean is 77.0 s. Calculate an estimate of the smallest sample size needed, and
    explain why your answer is only an estimate.

    - I'm really not sure what they're asking here.. any pointers? I know that with Type I errors, its the same as the significance level but I'm not really sure how to calculate the smallest sample needed.



    If anyone could help with the questions, I would really really appreciate it!
    • Thread Starter
    Offline

    0
    ReputationRep:
    Just did OCR c4 january 2007,
    The points A and B have position vectors a and b relative to an origin O, where a = 4i + 3j − 2k and
    b = −7i + 5j + 4k.
    ) Use a scalar product to find angle OAB.

    I got the answer as 137 degrees. However the answer is actually 43 degrees. Which is 180 - 137... why do I have to do 180 - the answer.... ?
    • Thread Starter
    Offline

    0
    ReputationRep:
    C3 Jan 2007:

    8 The parametric equations of a curve are x = 2t^2, y = 4t. Two points on the curve are P (2p^2, 4p) and
    Q (2q^2, 4q).
    (i) Show that the gradient of the normal to the curve at P is −p. [2]
    (ii) Show that the gradient of the chord joining the points P and Q is
    2 over p + q
    (iii) The chord PQ is the normal to the curve at P. Show that p^2+ pq + 2 = 0. [2]
    (iv) The normal at the point R (8, 8) meets the curve again at S. The normal at S meets the curve
    again at T. Find the coordinates of T.

    I've managed to do part i, ii and iii.
    No idea how to do part iv.. any pointers? I don't see how point R or S comes into the question.. The mark scheme has equated (8,8) to
    • Thread Starter
    Offline

    0
    ReputationRep:
    Anyone....?
    Offline

    16
    ReputationRep:
    (Original post by 01234567)
    Anyone....?
    Would have been nice if you had acknowledged my help on the earlier questions
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by TenOfThem)
    Would have been nice if you had acknowledged my help on the earlier questions

    Oh sorry - I usually do I was so busy trying to get the paper done.
    Belated thanks.
    Offline

    16
    ReputationRep:
    (Original post by 01234567)
    Just did OCR c4 january 2007,
    The points A and B have position vectors a and b relative to an origin O, where a = 4i + 3j − 2k and
    b = −7i + 5j + 4k.
    ) Use a scalar product to find angle OAB.

    I got the answer as 137 degrees. However the answer is actually 43 degrees. Which is 180 - 137... why do I have to do 180 - the answer.... ?
    Make sure that you use the vectors in the correct direction

    AO and AB
    Offline

    16
    ReputationRep:
    (Original post by 01234567)
    C3 Jan 2007:

    8 The parametric equations of a curve are x = 2t^2, y = 4t. Two points on the curve are P (2p^2, 4p) and
    Q (2q^2, 4q).
    (i) Show that the gradient of the normal to the curve at P is −p. [2]
    (ii) Show that the gradient of the chord joining the points P and Q is
    2 over p + q
    (iii) The chord PQ is the normal to the curve at P. Show that p^2+ pq + 2 = 0. [2]
    (iv) The normal at the point R (8, 8) meets the curve again at S. The normal at S meets the curve
    again at T. Find the coordinates of T.

    I've managed to do part i, ii and iii.
    No idea how to do part iv.. any pointers? I don't see how point R or S comes into the question.. The mark scheme has equated (8,8) to
    For R, you know that p=2, so you have the first line
    Solve that with the curve and you have S
    Then you have a new p so that you can find a new line
    Solve that with the curve and you have T
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by TenOfThem)
    For R, you know that p=2, so you have the first line
    Solve that with the curve and you have S
    Then you have a new p so that you can find a new line
    Solve that with the curve and you have T

    Thanks.
    • Thread Starter
    Offline

    0
    ReputationRep:
    The parametric equations of a curve are x = t + 2/t + 1
    , y = 2/t + 3

    Find the cartesian equation of the curve, giving your answer in a form not involving fractions.

    I've done it right up with y = 2/y - 3. I've then put this back into the x equation.
    Got it right into the mark scheme up till 2/y-1 over 2/y-2 = x.

    The mark scheme says eliminating double decker fraction... not sure how it gets the answer: 2x+y=2xy+2
    Offline

    16
    ReputationRep:
    (Original post by 01234567)
    The parametric equations of a curve are x = t + 2/t + 1
    , y = 2/t + 3

    Find the cartesian equation of the curve, giving your answer in a form not involving fractions.

    I've done it right up with y = 2/y - 3. I've then put this back into the x equation.
    Got it right into the mark scheme up till 2/y-1 over 2/y-2 = x.

    The mark scheme says eliminating double decker fraction... not sure how it gets the answer: 2x+y=2xy+2
    Any chance you could use latex as I am not finding the prose easy to read

    if you quote me you can use this as an example to copy

    a + \dfrac{b}{c} - \dfrac{\frac{d}{e+1}}{\frac{f+1}  {g+1}}
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by TenOfThem)
    Any chance you could use latex as I am not finding the prose easy to read

    if you quote me you can use this as an example to copy

    a + \dfrac{b}{c} - \dfrac{\frac{d}{e+1}}{\frac{f+1}  {g+1}}
    Not exactly sure how to use latex but here is a link to the paper: http://www.ocr.org.uk/images/57770-q...hematics-4.pdf

    Q7ii.

    Cheers.
    Offline

    16
    ReputationRep:
    So t=\frac{2}{y}-3

    So x=\dfrac{\frac{2}{y}-1}{\frac{2}{y}-2} = \dfrac{2-y}{2-2y}


    Which gives the answer if you multiply by the denominator
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by TenOfThem)
    So t=\frac{2}{y}-3

    So x=\dfrac{\frac{2}{y}-1}{\frac{2}{y}-2} = \dfrac{2-y}{2-2y}


    Which gives the answer if you multiply by the denominator
    Because you times the y side by y, don't you have to times the x side by y as well, therefore equating it to xy?
    Offline

    16
    ReputationRep:
    (Original post by 01234567)
    Because you times the y side by y, don't you have to times the x side by y as well, therefore equating it to xy?
    I have not multiplied the RHS by y

    Consider

    x = \dfrac{1}{2} = \dfrac{4}{8}

    do you believe that the second fraction is actually 2x?
    • Thread Starter
    Offline

    0
    ReputationRep:
    c3 Jan 2010.

    f(x) = 2 − (√3x + 1)^1/3
    Find the set of values of x for which f(x) = |f(x)|.

    I equated fx to itself and to -fx.. Is this the right way to do the question?
    THe mark scheme has the answer down as x is less or equal to 7.
    From equating fx = fx i didn't get anything, but from equating fx = -fx i got x = 7... how does this then become x is less or equal to 7? is it because you have to look at it from the graph?

    5)ii) Find an expression for the second derivative and hence find the value of
    the second derivative at the stationary point.
    I got the right answer up to a point in the mark scheme.
    To find out the second derivative at the stationary point is this the same as just equating dy/dx = 0?
    From the mark scheme, they appear to have substituted x = 0 to find out the stationary point from the equation... so, is this different for the second derivative?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by TenOfThem)
    I have not multiplied the RHS by y

    Consider

    x = \dfrac{1}{2} = \dfrac{4}{8}

    do you believe that the second fraction is actually 2x?
    Thanks, I've got it now!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Will you be richer or poorer than your parents?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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.