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definitely easier imo

Yes it's pretty hard considering my teachers couldn't get it.

You are right, the reason why is because whoever made this paper forgot to copy one of the conditions, b_1 \leq b_2 \leq \cdots \leq b_n. This should fix the counter example you have I believe. This makes the reasoning they give in ii) valid since A_r \leq B_r actually holds true regardless of...

Yeah I think that works

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...

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 =...

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

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...

Yeah I think that works fine

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.

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}}...

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}}...

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.

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...

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...

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...

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

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

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

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...