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Expand the RHS of part i Since it is true for all a,b,c,d you can compare coefficients. Imagine on the LHS you have ' +0a + 0b + 0c + 0d'. After expanding the RHS you can compare the apprpriate terms. For example, after expanding, RHS will have a +2ax term. Compare the 'a' terms, RHS = 0a =...

re: HSC Chemistry Marathon Archive I think ive seen questions give K and then asking if the current reaction is at equilibrium and you have to calculate Q and say since Q =/= K then ...

Who wrote that graph question?

re: HSC Chemistry Marathon Archive It's still [products]\[reactants] K is a special case of Q only occurring at equilibrium

Oh sorry :P thx for not putting (2-b, a) in

is there a source for this? That sounds.. interesting

Yeah same I did it at the end. Good thing Carrotsticks didn't put (2-b , a) as a choice

I did that at first too. But if you read it again, PQ is a+ib not just Q lol

You could also think of it like this. "If my curve's values tend to infinity but it's always hugging the y = x line, then square rooting those values will make it go under the y = sqrtx curve" Typing that out made me realize it sounds kind of stupid but its how I remembered it lol

Re: MX2 2015 Integration Marathon = \int \frac{x+k}{\sqrt{x^2 + kx}}dx = \sqrt{x^2 + kx} + k \int \frac{1}{\sqrt{(x+ \frac{k}{2})^2 - \frac{k^2 }{4} } } dx = \sqrt{x^2 + kx} + k \ln \left ( x + \frac{k}{2} + \sqrt{x^2 + kx} \right ) + $ C $ That took so long to type on mobile

98.65

The volumes was interesting did anyone get it out? Did we have to prove that the area was in a ratio? Bc. I just assumed it and it worked

Notice that there are asynptotes at x = 3 and y = 2. For vertical asymptotes (x=3) you need to find restrictions on the DOMAIN of the function. In a fraction the denominator cannot equal zero. And from the graph, x =/= 3 x - 3 =/= 0 So the denominator has to be x-3 which rules out C and D...

Re: MX2 2015 Integration Marathon L'hopital's would show it approaches 0^- but that's not really a proof

Re: MX2 2015 Integration Marathon Were you using by parts?

Re: MX2 2015 Integration Marathon Saw this question online $Show $ \int_0^1 \frac{ (-x\ln x)^n}{n!} dx = (n+1)^{-(n+1)} idk why it doesn't work . Show \int_0^1 (-xlnx)^n / (n!) dx = (n+1)^{-(n+1)}

Re: MX2 2015 Integration Marathon I don't think anyone tried to answer this = \int{ \frac{2x^3 - 1}{x^6 + 2x^3 + 9x^2 + 1}} $ dx $ = \int \frac{2x^{3}-1}{(x^3 + 1)^2 + 9x^{2}} $dx$ = \int \frac{2x - \frac{1}{x^2}}{(x^2 + \frac{1}{x})^2 + 9} $dx$ = \frac{1}{3} \tan ^{-1}\left(...

Label that new curve as g (x) to avoidconfusion and then repeat the steps you did before to find g'(x) (which is the second derivative of the original curve)

What InteGrand said. But if you're able to, you might be able to draw the second derivative from the y = f(x) graph by assessing the concavities at each point. eg. from x = -infinity to x = -2, the concavity is highly negative, until it reaches x = -2 where there's an inflexion point, showing...

If f"(x)<0 then the gradient of y = f'(x) is negative. For the y = f (x), it only tells us that it has a negative concavity (think, a concave down parabola). It only starts at the top for the y=f'(x) curve because for y=f (x), the gradient (and hence f'(x)) is positive. . . . So it starts at...