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

    0
    ReputationRep:
    Can't believe I couldnt solve question 4 properly!!!
    Mind went completely blank!
    Offline

    15
    ReputationRep:
    (Original post by ActaNonVerba)
    -4sinxcosx + 2cosx = 0, 4sinxcosx = 2cosx, divide by cosx to get 4sinx = 2, then sinx=0.5..
    You haven't disproved what I said. Given that there were three stationary points, you've actually made my point.
    One SP was cos(x)=0. Standard C2 level question...


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by a10)
    but cos x cant be zero the limits of x where greater than zero and less than pi? therefore u can divide by cos?

    Pi/2
    Offline

    14
    ReputationRep:
    Proof there are 3 stationary points for those wondering:

    Name:  IMG_1040.jpg
Views: 248
Size:  504.2 KB
    Name:  IMG_1041.jpg
Views: 134
Size:  499.3 KB
    Offline

    10
    ReputationRep:
    (Original post by a10)
    but cos x cant be zero the limits of x where greater than zero and less than pi? therefore u can divide by cos?
    cosx=0 gives you pi/2 which is within the limits though. The limits are for x, not for cosx
    Offline

    1
    ReputationRep:
    For the balloon question, for inverse proportion I did 1/k multiplied by Sq. root of R, instead of k/Sq.root R. As a result, I got the answers wrong. But I did the correct integration etc. for the differential equation.

    Does anyone know if I lost the whole 9 marks ? (( Or will I get error carried forward, and only lose about 2-3 marks? Thanks, I'm really scared, as I need an a*
    Offline

    0
    ReputationRep:
    (Original post by habs10)
    For the balloon question, for inverse proportion I did 1/k multiplied by Sq. root of R, instead of k/Sq.root R. As a result, I got the answers wrong. But I did the correct integration etc. for the differential equation.

    Does anyone know if I lost the whole 9 marks ? (( Or will I get error carried forward, and only lose about 2-3 marks? Thanks, I'm really scared, as I need an a*
    I'd guess you should get 4/5 if you subbed in their values to give you a k and c
    Offline

    14
    ReputationRep:
    (Original post by tooambitious)
    No, because it could be 0, can you divide by 0?


    Posted from TSR Mobile
    Trueee,.. owned him
    Offline

    0
    ReputationRep:
    (Original post by tooambitious)
    You haven't disproved what I said. Given that there were three stationary points, you've actually made my point.
    One SP was cos(x)=0. Standard C2 level question...


    Posted from TSR Mobile
    thought u were saying you couldn't take out cosx.. but yh, I only calculated 2 anyways, we're was the third?
    Offline

    20
    (Original post by Holz888)
    cosx=0 gives you pi/2 which is within the limits though. The limits are for x, not for cosx
    Oh noooo I dropped that value, I completely forgot about it.

    Basic error right there

    I let myself down SO much in this exam. So many stupid mistakes caused by me just being stupid and losing my head.
    Offline

    15
    ReputationRep:
    (Original post by Namige)
    Trueee,.. owned him
    Don't mean to sound like a prick, just really surprised people are asking this question at C4 level, how did you do C2 :erm:


    Posted from TSR Mobile
    Offline

    14
    ReputationRep:
    (Original post by habs10)
    For the balloon question, for inverse proportion I did 1/k multiplied by Sq. root of R, instead of k/Sq.root R. As a result, I got the answers wrong. But I did the correct integration etc. for the differential equation.

    Does anyone know if I lost the whole 9 marks ? (( Or will I get error carried forward, and only lose about 2-3 marks? Thanks, I'm really scared, as I need an a*
    It doesn't matter, If you work out 1/k, it will still be the same as k if you did it properly. As it's just a constant.

    For example

    n = 1/5
    1/n = 5


    n=5
    1/n=5
    Offline

    0
    ReputationRep:
    (Original post by Benjy100)
    This is effectively a preliminary markscheme until Mr M posts his full definitive solutions. Some of my answers may be wrong. Spoilered incase you don't want to see them

    *A few additions have been made to the markscheme* - namely the cartesian equation (forgot to put it in)

    Spoiler:
    Show
    1. \frac{4}{{x + 2}} - \frac{3}{{x - 1}} + \frac{2}{{{{(x - 1)}^2}}}

    2. \frac{1}{9}{x^9}\ln 3x - \frac{1}{{81}}{x^9} + k

    3. Skew. Solved the first two equations for \lambda  = 3 and \mu  = 8 and then substituted into the third equation for 18=-38 which is clearly untrue, thus the equations are inconsistent and the lines are skew.

    4. \frac{{dy}}{{dx}} =  - 2\sin 2x + 2\cos x

    Stationary points are (\frac{1}{2}\pi ,1) \left( {\frac{1}{6}\pi ,\frac{3}{2}} \right) \left( {\frac{5}{6}\pi ,\frac{3}{2}} \right)

    5. \frac{1}{{1 - \tan x}} - \frac{1}{{1 + \tan x}} = \frac{{1 + \tan x - (1 - \tan x)}}{{1 - {{\tan }^2}x}} = \frac{{2\tan x}}{{1 - {{\tan }^2}x}} = \tan 2x

    Integral was \frac{1}{4}\ln 3

    6. \ln \left| {1 + \ln x} \right| + \frac{1}{{1 + \ln x}} + k

    7. \left| {AB} \right| = \sqrt {91} \left| {AC} \right| = 3\sqrt 3

    Angle BAC = 171.3 degrees (1 d.p.)

    Show that AD is perpendicular to AB and AC by finding the dot product of AD and AB, and then AD and AC - and showing that both dot products equate to 0.

    I messed up on the volume of the pyramid

    8. \frac{{dr}}{{dt}} = \frac{k}{{\sqrt r }}

    \[r = {(4.86t + 2.7)^{\frac{2}{3}}}\]

    Letting t=0, finding r and substituting gives V = 30.5 cm^3 (3 s.f.)

    9. \[\frac{{dy}}{{dx}} = \frac{{2 - 2{t^{ - 3}}}}{{ - {t^{ - 2}}}} = \frac{{2{t^3} - 2}}{{ - t}} = \frac{{2 - 2{t^3}}}{t}\]

    Solving dy/dx yields t=1 which corresponds to the point (0,3)

    When t>1 x<0 and dy/dx is negative
    When t<1 x>0 and dy/dx is positive

    Hence it is a minimum

    Cartesian equation...

    \[\begin{array}{l}

xt = 1 - t \Rightarrow t(x + 1) = 1 \Rightarrow t = \frac{1}{{x + 1}}\\

y = \frac{2}{{x + 1}} + {(x + 1)^2}

\end{array}\]

    10. Show.

    Let x=0.1 for 0.136

    \[ - \frac{1}{{{x^2}}}(1 + \frac{3}{x} + \frac{6}{{{x^2}}}) =  - \frac{1}{{{x^2}}} - \frac{3}{{{x^3}}} - \frac{6}{{{x^4}}}\]

    Same substitution would be unsuitable as the expansion here is valid for \left| { - \frac{1}{x}} \right| &lt; 1\] which is the same as \left| x \right| &gt; 1\]
    Thankyou!! I forgot to plug x values back in to find y values for question 4.

    Would it be okay to write 1/2ln(root3) ? Did they specify the format?


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by habs10)
    For the balloon question, for inverse proportion I did 1/k multiplied by Sq. root of R, instead of k/Sq.root R. As a result, I got the answers wrong. But I did the correct integration etc. for the differential equation.

    Does anyone know if I lost the whole 9 marks ? (( Or will I get error carried forward, and only lose about 2-3 marks? Thanks, I'm really scared, as I need an a*
    They never specified that it be C/Sq.root R, so surely 1/(K Sq.root R) is still valid as 1/k is just a constant as is C. All this means is that you're final answer will look different to anyone else who did C/Sq.root R, but may still be a correct answer.

    There's still hope!

    However, if you did (1/k) * Sq.root R I would imagine you would lose some marks (such as for M1 at the start and the final answer mark), but I think you would get the rest of the marks for such a big question (i.e. error carried forward).
    Offline

    0
    ReputationRep:
    (Original post by Namige)
    Trueee,.. owned him
    lool calm your tits, I was clearly thinking of something different to you two.. I was just saying you can take it out of the equation to find values for x.. no need for the cheerleading
    Offline

    0
    ReputationRep:
    I forgot to sibstitute x back into u, would that cost me 1 or 3 marks?
    Offline

    15
    ReputationRep:
    (Original post by ActaNonVerba)
    thought u were saying you couldn't take out cosx.. but yh, I only calculated 2 anyways, we're was the third?
    Cos(x) = 0 where x=PI/2
    Sin(x)=1/2 (iirc) where x= 5pi/6 or pi/6


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    (Original post by zmai)
    Thankyou!! I forgot to plug x values back in to find y values for question 4.

    Would it be okay to write 1/2ln(root3) ? Did they specify the format?


    Posted from TSR Mobile
    That's fine they said a ln b.. No indication that b must be an integer
    Offline

    14
    ReputationRep:
    (Original post by zmai)
    Thankyou!! I forgot to plug x values back in to find y values for question 4.

    Would it be okay to write 1/2ln(root3) ? Did they specify the format?


    Posted from TSR Mobile
    Yup, agree with all apart from the volume, which you obviously didn't do lol
    Offline

    2
    ReputationRep:
    (Original post by h2shin)
    I forgot to sibstitute x back into u, would that cost me 1 or 3 marks?
    Mark scheme isnt that harsh. You'll probably lose 1 since it was the last step.
 
 
 
  • 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
    What's your favourite Christmas sweets?
  • 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.