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Re: MX2 2015 Integration Marathon Two more questions \int_0^1 2^{log_e(x)} dx \int_0^1 \log^{log_ex}(2) dx

Re: MX2 2015 Integration Marathon $ I $ = \int_0^{\frac{\pi}{2}} \frac{cosx - sinx}{cosx+sinx} dx = [\ln(cosx+sinx)]_0^\frac{\pi}{2} = 0 ----------------- $ New question $ \int_0^{\frac{\pi}{2}} \frac{1}{1+tan^{\pi}(x)} dx

$ Ball A $: v = 15, \theta = 60 x'' = 0 . y'' = -10 x' = vcos\theta = 15cos60 . y' = -10t + vsin\theta = -10t + 15sin60 x = 15tcos60 . y = -5t^2 + 15tsin60 $ Let y = 0, for time at which ball A lands $ \therefore t = 3sin60 $ Hence, $ x = 15(3sin60)(cos60) x =...

Re: HSC 2015 4U Marathon $ Assume 4n + 2 = (a-b)(a+b) is true $ $ The factors of 4n+2 differ by 2b (where b is an integer) $ \therefore $ If a is even, b is odd, then a+b is odd and a-b is odd $ $ If a is odd, b is even, then a+b is odd and a-b is odd$ $ If a is even, b is even...

You could use inequalities, which is pretty fast v^2=4x-x^2 v^2 = - ((x-2)^2 - 4) v^2 = 4 - (x-2)^2 (x-2)^2 \geq 0 -(x-2)^2 \leq 0 4 - (x-2)^2 \leq 4 v^2 \leq 4 -2 \leq v \leq 2 $ Hence, maximum v = 4

Re: HSC 2015 4U Marathon wouldn't it be in terms of z? how would you integrate with respect to theta?

It's the same as when you have 3x + x = 4x. This case you have 3log(x)y + log(x)y. You add 3 "lots" of log(x)y with 1 "lot" of log(x)y to give 4 "lots" of log(x)y, so 3log(x)y + log(x)y = 4log(x)y

Re: HSC 2015 4U Marathon Dunno how to use those complex identities but you could use the expansions: cos(a-b) = cosacosb+sinasinb cos(a+b) = cosacosb-sinasinb $ Subtracting, cos(a-b) - cos(a+b) = 2sinasinb \frac{cos(a-b) - cos(a+b)}{2} = sinasinb $ Let a = n\theta, b = 2\theta...

how would they determine the degree of sophistication?

Re: MX2 2015 Integration Marathon let u = sin(lnx) + cos(lnx) dv = x^{2} dx du = \frac{dx}{x}(cos(lnx) - sin(lnx)) v = \frac{x^3}{3} $ I = $ \frac{x^3}{3}(sin(lnx) + cos(lnx)) + \frac{1}{3} \int x^2(sin(lnx) - cos(lnx)) dx $Let new integral = J $ J: let u = sin(lnx) -...

Re: MX2 2015 Integration Marathon $ From the identity, $ cot(a+b) = \frac{cotacotb - 1}{cota+cotb} cot^{-1}a - cot^{-1}b = cot^{-1}(\frac{ab - 1}{a+b}) $ I = $ \int cot^{-1}(x(x+1) - 1) dx $ I = $ \int cot^{-1}(\frac{x(x+1) - 1}{(x+1) - (x)}) dx $ I = $ \int cot^{-1}(x+1) -...

Re: MX2 2015 Integration Marathon = sqrt(sin^2 x + cos^2 x + 2sinxcosx) dx = sqrt((sinx + cosx)^2) dx = |sinx+cosx|

Re: HSC 2015 4U Marathon - Advanced Level that seems familiar to a question in the maths competition by AMT last year. couldnt do it lol

My teacher told me that shaking it would cause the molecules to collide, and produce a big CO2 "bubble", while it breaks surface tension allowing it to escape. Also shaking it would slightly increase temperature, shifting it to CO2 (g)

Re: MX2 2015 Integration Marathon It doesn't, collect the inside of your log function and use log rules to take out the division. you get something like + 1/k ln(e^kx) which simplifies to be x

Re: HSC 2015 4U Marathon $ Sketching the curve, y = 1/x $ $ Upper rectangle $ > \int_{1}^{\sqrt{x}} \frac{dx}{x} > 0 0< \int_1^{\sqrt{x}} \frac{dx}{x} < \sqrt{x} - 1 0 < ln(\sqrt{x}) < \sqrt{x} - 1 0 < ln(x) < 2(\sqrt{x} - 1) $ Multiplying by x $ 0 < xln(x) < 2(x\sqrt{x} -...

Re: MX2 2015 Integration Marathon I don't see why it should be equal in my second last line, I used substituion of t = tan2x, where x = 0 and x = pi/2

\frac{dv}{dt} = -10-2v \frac{dv}{5+v} = -2dt log_e_(5+v) = -2t + C $ At t = 0, v = 0 C = log_e_5 log_e_(5+v) = -2t + log_e_5 2t = log_e_(\frac{5}{5+v}) e^{-2t} = \frac{5+v}{5} 5 + v = 5e^{-2t} v = 5(e^{-2t} - 1) \frac{dx}{dt} = 5e^{-2t}- 1 dx = 5e^{-2t} - 1 dt...

Re: MX2 2015 Integration Marathon \int_0^{\pi/2}{\frac{dx}{sin^4x +cos^4x} \int_0^{\pi/2}{\frac{dx}{(sin^{2}x + cos^{2}x)^{2} - 2cos^{2}xsin^{2}x} \int_0^{\pi/2}{\frac{dx}{1 - \frac{1}{2}sin^{2}2x} Let t = tan2x \therefore \frac{t^2}{1+t^{2}} = sin^{2}2x \int_0^1...

If x and y are acute angles, it must lie in the first quadrant, hence sinx, siny, cosx, cosy must be positive $ Given $ sinx = \frac{2}{3} $ Draw up a right-angled triangle, with an angle of 'x' and opposite side being 2 and hypotenuse being 3. $ $ the sine of this angle equals 2/3 and...