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hel

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  • Gender
    Female
  • Exams
    May 2011
  • Country
    United Kingdom
  1. Our class did drama plays so most people chose question 1 (about the masks in drama?) but I chose the one about 'war'. However I interpreted it as ~metaphorical war and wrote more about the concept of conflict within literature. I chose 2 plays Streetcar Named Desire (reality vs illusion, civilized vs animal like, battle of sexes etc) and Amadeus (god vs human, the war of faith in Salieri as a man, genius vs mediocrity etc). I included some reference to Virginia Woolf as I feel like I should have probably wrote about that play instead of Amadeus but I just really loved Amadeus! Anyway I thou
  2. I did the same question for Abnormal psychology. I did it on bulimia, but I feel like I had a lot more to write about sociocultural factors than biological ones. I only wrote about the genetic influence and twin studies for the biological factors meanwhile for the sociocultural ones I used cross cultural studies as well as Social Learning/Social Identity theories and studies. I spoke a bit about gender bias as well but only briefly because I was running out of time. I wish I could have included more about the decrease due to damage in bulimic and their levels of chemicals that stimulate the f
  3. HL maths is impossible. Only like 1% get a 7 so I think to get a high grade you have to be constantly reviewing the learnt material and going an extra mile on independent study. Not only doing the past exam questions but making sure you go over the syllabus thoroughly and: PRACTICE, PRACTICE, PRACTICE. And of course acing the coursework. Although a lot of people ace the coursework and still come out with a 3 or 4 in HL Maths...
  4. hel

    Math SL P1

    There were mixed opinions about P1 in our class. We did have terrible teachers and an awful history of getting 3s in mocks. Some found it hard, other said it was better than what they expected. I personally found it awful. It was a lot harder than what I was expecting, and I left a lot of it unsolved so I doubt I'll get my predicted grade even after P2. Expecting a 4, gutted. Even the simple questions like I think q4 where you had to solve for h'(x) or something with 6cosx etc I couldn't solve.
  5. Psychology was definitely easier than what I was expecting! I hated psychology for two years but now I'm happy I chose it. The exam was quite straight forward :)and I did reliability of re-constructive memory too! A lot of people in my class were caught off guard by the emotion influencing cognitive process surprisingly, for which I used Brown&Kulik, flashbulb memory.
  6. English was such a joy to write... compared to other exams *cough*maths*cough*. I analyzed poem (the more popular option in my IB school) but I mostly interpreted as a work about man/nature rather than about his personal reflection. I did mention it briefly... I wonder what's the ~true interpretation of the poem though.
  7. yup. Sorry about that... From the given diagram, f(x) could be any function that passes through (-2,-8) and has a gradient of 3 at arbitrary x-value greater than 2/3, a. Assuming f(x) passes through the origin and that the gradient at this point is zero (since it looks flat) could probably help but I won't even bother since it says "diagram not drawn to scale". The problem is either really hard or I need to use part (a) to solve it, thus making it 'less harder'. What was (a)? If you want to upload any other images, please upload to imgur.com this time and not imageshack.us... long story ht
  8. hel

    Integration

    that's quite useful, thanks. the graph of f(x)=x^3. You can see the whole question in N10 P1 TZ0, Q10. https://mr-t-tes.wikispaces.com/file/view/Nov+2010+Mathematics+SL+paper+1+QP.pdf
  9. hel

    Integration

    I was having difficulties integrating and answering this question... If anyone could help, I would really appreciate it!
  10. The equation of the tangent at P is y=3 x - 2 . Let T be the region enclosed by the graph of f , the tangent [PR] and the line x= k , between x = −2 and x= k where − 2< k < 1. Given that are of T is 2k+4, show that k satisfies the equation k^4-6k^2+8=0. So I know you have to integrate x^3-3x-2 for the x=k and x=-2 but I struggled to integrate because I got k^4/4-3k^2/2-4k-10=0 I tried equating the integrated equation to 2k+4 but I didn't get the right answer, k^4-6k^2+8=0. (if you want the whole question is Q9.b, Nov 10/TZ0 P1)
  11. The lines (AC) and (BD) intersect at the point P(3,K) Show K=1. AC=(4,2) or (x,y)=(5,2)+s(4,2) BD=(3,-6) or (x,y)=(1,5)+t(3,-6) A(1,0), B(1,5), C(5,2), D(4,-1)
  12. no I don't think it is... thanks anyway! if it helps the mark scheme gives the answers as... x=1/2ln1/9, x=-1/2ln9, x=ln1/3, a=-1/2 and b=9, x=-ln3 (accept a=-1 and b=3) I don't understand why there are so many options? and which ones are right?
  13. Hey guys I need help how to solve this equation... Express your answer in the form alnb 9e^4x-e^2x=0 This is as far as I got 9e^4x=e^2x ln(9e^4x)=ln(e^2x)
  14. hel

    Finding Gradient

    Hi, I was working through an exam problem(May 2009 P1) and I have a question.... Basically the problem is worth 6 marks. You're given a function, f(x)=e^x cosx and need to find the gradient of the normal to the curve of f at x=π (pi). So I used the product rule to differentiate and then substituted π. I simplified the answer which gave -e^π and I left that as an answer. However it says that you should take negative reciprocal -1/f'(π) which results in the gradient being 1/e^π. Why do we need to take the negative reciprocal? Thanks in advance.
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