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If the dux of your year achieved the same ranks as you last year. You'll probably get an ATAR around 97. But as said before, its all upto you.

I've seen it in textbooks before. Wouldn't be harmful to learn the method if you're doing 4unit though.

Top ones meant to be 1/2*(4+x^2)^2 i think that may alter the outcome. I{1}=8+4x^2+x^4/2+C=4x^2+x^4/2+K I{2}=4x^2+x^4/2+A =>A=K ie A=8+C

Let u=sin(x): du=cos(x) dx I=S cos(x)/(sin(x))^2 dx =S du/u^2 =-1/u +C =-cosec(x)+C

Quite self explanatory if you look at the diagram.

If c=+-1. Delta=0 [you want Delta=0 so there's a kind of double root for log(1+x^2) so tangents only touch once] ie x=-4c/(1+c^2)/2 =-2/2 or +2/2 =+-1 And, y'=+-1 ie at x=+-1 the tangents are mutually perpendicular.

Yeah, sorry about that. Didn't have much time to think about it but your methods what i would have used.

lol, I'm verifying for only a real root. So say z is a real root the inequality holds.

S [0,4] (x+4-4)/sqrt(x+4) dx = S [0,4] sqrt(x+4) dx - S [0,4] 4/sqrt(x+4) dx

i) P'(z)=5z^4+3az^2+b =5[z^4+3az^2/5+b/5] =5[(z^2+3a/10)^2+b/5-9a^2/100] >=5[9a^2/100+b/5-9a^2/100] =5(b/5)=b >0 ie. P'(z)>0 Hence, P(z) is increasing function to infinity since its a poly. Now, P(0)=-c <0. Thus, P(z) has only one real root as it can only cross x-axis once. ii)If...

You can also differentiate arctan(2pif/3)/2pi and you'll get the required answer. At poster above 3unitz, I don't think (sec[x])^2 transformation would do anything.

Best to learn it. It's not very hard and very useful for uni as they won't teach you it during initial coursework.

probs arctan(2pi.f/3)?

Man, get over it. Many people couldn't solve it. Eg: today i momentarily forgot how to do implicit differentiation in my math test but it came back to me. Monsterman may have experienced the same thing but just for a longer time and under the stress of doing well in the first quiz he probs...

Bad luck, there was no inequality questions for us today.

=(1/log(x)) - (1/(log(x))^2)

Well dz/d@=iz dz/z=i d@ ln(z)+C=i@ ln(z)=i@ as C=0 z=e^i@

lol, is this for the upcoming MATH1151 calc test?

dy/dx=((log(x)-1)/(log(x))^2) =1/log(x)-1/(log(x))^2 y=x/log(x) y/x=1/log(x) Thus, dy/dx=(y/x)-(y/x)^2

-512cos(t).e^-sqrt(3)t is the answer i got. But idk if its right because i haven't done anything like this before, might post up a solution later.