Search results

  1. M

    Inverse

    Yeah I think that works
  2. M

    Inverse

    no why do you think that the derivative of inverse f(x) is dx/dy? y=f^{-1}(x) f(y)=x \text{differentiate with respect to x} \frac{dy}{dx} \times f'(y)=1 \frac{dy}{dx}=\frac{1}{f'(y)} \frac{dy}{dx}=\frac{1}{f'(f^{-1}(x))} y=f(x) \implies \frac{dy}{dx}=f'(x) \text{multiplying gradients...
  3. M

    Mathematic Induction Question

    Assume for n=k \tan{\theta}\tan{2\theta}+...+\tan{k\theta}\tan{(k+1)\theta}=\tan{(k+1)\theta}\cot{\theta}-(k+1) To prove for n=k+1 That is \tan{\theta}\tan{2\theta}+...+\tan{(k+1)\theta}\tan{(k+2)\theta}=\tan{(k+2)\theta}\cot{\theta}-(k+1) LHS =...
  4. M

    Question

    bro is this question even correct? like the wording makes no sense to me
  5. M

    Trial Q

    we have \overrightarrow{OD}= \frac{1}{5} (4a+2b) So you need to think about the distance from a / b in terms of \vec{AB}=b-a \overrightarrow{OD}-\frac{2}{5} \overrightarrow{AB} = a \overrightarrow{OD}+\frac{3}{5} \overrightarrow{AB} = b So then its closer to A, idk how they got 1/2. I...
  6. M

    Proofs

    Yeah I think that works fine
  7. M

    do teachers our see raw marks and exam papers?

    I don't think so, definitely not the exam papers. I believe my school asks students what marks they got afterwards, I'm not sure though.
  8. M

    BoS Trials Maths, Physics and Business Studies 2023

    I don't know if this was how they intended but basically my thought process is if it looks like product of roots, just make an equation with roots of tan to get the product of roots. It doesn't really use part i) and is slow so it's not the best solution: \text{let } z=\tan{\frac{m\pi}{2n+1}}...
  9. M

    Challenge integral

    I don't know how you got \int \sqrt{\tan x}+\sqrt{\cot x} d x=\sqrt{2} \sin ^{-1}(|\sin x|-|\cos x|)+C it looks wrong in desmos. I'll list what did to resolve the problem with blob's answer \int \sqrt{\tan x} +\sqrt{\cot x} dx = \int \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}...
  10. M

    Challenge integral

    I agree in part, tbh I dont think they're going to ask a question where you could use this method in the hsc anyway. I also agree that partial fractions leads to a better antiderivative, but for speed its x10 worse and i hate partial fractions algebra.
  11. M

    Challenge integral

    On second thought thinking about a different function f(0) could be anything its just you're unable to tell what it is precisely. So like it could be continuous with a derivative at that point. I'm not so sure about this point though, thinking back to this thread if you want to read it...
  12. M

    Challenge integral

    Ok, I don't really know how this contradicts anything I've said/explained. After simplifying there is no difference between the two methods. This integral is basically the case where there is a pointwise discontinuity if you chose preserve it and not simplify which I wouldn't. Does this mean the...
  13. M

    Challenge integral

    I don't really get what you're trying to say, I pointed out that the original integral is \int u + \frac{1}{u} dx so it doesn't make any sense in the first place that dividing by u on the top and bottom of the fraction would lead to a mistake since \frac{1}{u} is already present (if it was 0...
  14. M

    Challenge integral

    unofiiciallyred12 just made a mistake when evaluating the integral (that's why the blue function is incorrect), the u^2 issue isn't actually an issue at all
  15. M

    Challenge integral

    besides the situation where u=0 is undefined either way since cotx is undefined
  16. M

    Challenge integral

    no I don't think that should matter at all I = 2 \times \frac{1}{\sqrt{2}} \arctan{(\frac{t}{\sqrt{2}})} I = \sqrt{2} \arctan{(\frac{\sqrt{tanx}-\frac{1}{\sqrt{tanx}}}{\sqrt{2}})} I = \sqrt{2} \arctan{(\frac{tanx-1}{\sqrt{2tanx}})} + C You just did it wrong. man I should be doing english
  17. M

    Challenge integral

    Good answer but there's an easier way than doing partial fraction decomposition, so I thought I might share it: I = 2\int \frac{u^2+1}{u^4+1}du I = 2\int \frac{\frac{1}{u^2}+1}{u^2+\frac{1}{u^2}}du The logic is basically trying to find a way to cancel the nominator with a substitution so...
  18. M

    Module C English advanced prediction?

    its better, you can also reference your piece in reflection by name
  19. M

    Module C English advanced prediction?

    It really depends on the quote / excerpt, you probably don't have to write anything verbatim just close enough so that the marker knows your referencing the quote.
  20. M

    Module C English advanced prediction?

    Nah, I think that's a waste. But its better than not addressing the quote / excerpt you should have part of it in there, at least the general idea of the excerpt.
Top