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E.g. If we want to simplify \ln{(e^x)}^2 , this actually means that we want to find \ln{(e^x)}\times{\ln{(e^x)}}=x\times{x}=x^2

The power log rule applies. I am not actually pulling (x+y)^2 down, I am pulling (x+y) down and squaring it.

So what I mean is that they actually mean \ln{(e^{x+y})}^2=(\ln{(e^{x+y})})^2

I'm assuming the notation is as such: \ln{(x)}^2=\ln{(x)}\times\ln{(x)}

Answer should be (x+y)^2 , but if the 2 was inside the bracket as such \ln{(e^{2(x+y)})} then the answer would be 2(x+y)

Re: HSC 2015 4U Marathon - Advanced Level Yep I got that for y=-x , I think you meant y=-x but instead you wrote y=x which should instead give f(x^2)=f(x)^2-f(2x)+1 .. (3 posts ago)

Re: HSC 2015 4U Marathon - Advanced Level I tried that as well as plugging in y=-x , but got an equation which I'm not sure how to manipulate.. Not too experienced/familiar with these functional equations.

Re: HSC 2015 4U Marathon - Advanced Level Set x=y=0 to get f(0)=f(0)^2-f(0)+1 \Rightarrow (f(0)-1)^2=0 \Rightarrow f(0)=1 . Not sure how to go from there.. :p I can see that the function defined by f:\mathbb{Q}\rightarrow \mathbb{Q} such that f:=f(x)=1 is a solution, but not sure how to...

I have heard ECON2101 Microeconomics 2 is a relatively easy ASB course if you decent at maths (since it's more mathematically focussed i.e. involves basic Calculus etc.) FINS2624 also seems like an interesting/easier course to do as well, but I don't know how much it differs between Semester 1...

The event A \cap B denotes the event that the sum is 4 or more but is less than 6 i.e the sum can only be 4 or 5. This happens only when you land (1,3), (2,2), (3,1), (1,4) (2,3), (3,2), (4,1). Hence, \mathbb{P}(A\cap{B})=\frac{7}{36} .

Re: UNSW Chit Chat Thread 2015. Haha, yeah tbh they're not the best lecture notes you will see but sometimes they are useful since he uses tutorial questions to explain concepts etc. but not something I would recommend.

2013 Independent Trial Paper (4U) - Q16 (c)

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread We have a function f defined by f:=f(x)=\dfrac{x}{\sqrt{x^2+1}}=\dfrac{1}{\sqrt{1+\frac{1}{x^2}}}\rightarrow{\pm{1}} as x \rightarrow \pm\infty . Check this graphically. (Note that f(x) > 0 when x > 0 and f(x) < 0 when x < 0...

Re: HSC 2015 3U Marathon Re-arranging the equation gives, x^2=(\sqrt{x}-\sqrt{y})^2 \Rightarrow y=(\sqrt{x}\pm{x})^2

Call the UNSW Economics Faculty (or ASB Student Centre), since they will be able to tell you exactly what is needed of you to apply.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Factor out the 'a' term and differentiate via product rule as usual y'(x)=a[(x-b)+(x-a)]=a[2x-(a+b)]

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread So your tangent will be of the form y=mx (since it passes through the origin). Find the derivative of f:=f(x)=x(1-x)^{6} which will be f':=f'(x)=(1-x)^6-6x(1-x)^5=(1-x)^5(1-7x) (what you found) and sub. in the point x = 0 to...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread x^3-3x+2=0 \Rightarrow (x-1)^2(x+2)=0 , so your points of intersection will be (-2,0) and (1,-3) . Also, since x = 1 is a double root, it follows that it the line x + y + 2 = 0 must be tangential to the cubic. Not sure what...

Re: HSC 2015 3U Marathon I get 15.912.. as well, I think the OP of the question made a calculator error.

Re: HSC 2015 3U Marathon Sorry for the bad drawing, but this is how it should look like.