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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Can't find (d) in your picture: http://community.boredofstudies.org/attachments/13/mathematics-extension-1/32912d1455616894-year-11-mathematics-3-unit-cambridge-question-answer-thread-capture.png

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent We have $4(x^2-4x)^3=0 \Longleftrightarrow x^2 - 4x = 0 \Longleftrightarrow x(x-4)=0$, so $x=0$ or $4$. So you must have an incorrect expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ (we need to multiply the...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Yeah

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread No, into the equation for y, which is given to us.

This is known as a transcendental equation (https://en.wikipedia.org/wiki/Transcendental_equation). We can't find closed-form solutions for your equation in terms of elementary functions.

Re: HSC 2016 MX2 Combinatorics Marathon Number of teams (in braintic's there were four).

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent The tangents being horizontal means that the derivative is 0, so set the derivative equal to 0 and solve. The solutions are the $x$-values where the curve has a horizontal tangent. Plug these $x$-values into the...

Re: HSC 2016 MX2 Combinatorics Marathon $\noindent \textbf{NEW QUESTIONS}$ $\noindent 1) Generalise the result in braintic's previous question.$ $\noindent 2) By considering a `story', come up with a combinatorial identity of your own.$

Re: HSC 2016 MX2 Combinatorics Marathon $\noindent In my opinion, much nicer to prove as a `story proof' (aka double counting proof or combinatorial proof).$ $\noindent Consider a group of $n$ people whom we wish to sort into four labelled teams $A, B, C, D$. On the one hand, this can be done...

Re: HSC 2016 2U Marathon $\noindent As leehuan said, work with $A^2$ rather than $A$, this really simplifies the differentiation required, and $A$ is optimised precisely when $A^2$ is, since $A \geq 0$ and the function $f(x) = x^2$ is increasing on $\left[0, \infty\right)$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent If it's just the constants that are confusing you, you can write $A = a^2$ and $B = b^2$, and then differentiate. Then at the end, replace $A$ and $B$ with $a^2$ and $b^2$.$

Didn't Drsoccerball do it in the HSC Physics exam or something?

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Here's how to do Q.26. Let $\left(u,v\right)$ be an arbitrary point in the plane. Let the cubic be $f(x) = ax^3 + bx^2 + cx +d$, so $f^\prime (x) = 3ax^2 + 2bx + c$. Hence the slope of the tangent to the cubic at...

I think Drsoccerball did do that. He actually did it right though in the end I think.

Re: HSC 2016 4U Marathon - Advanced Level Can't we essentially use the fact that if two at-most-(n-1)th degree polynomials agree at n points, they are identical (*)? So this is the one and only polynomial of at most (n-1)th degree passing through these points, so no different polynomial (of...

Re: HSC 2016 4U Marathon - Advanced Level Correct! (However, I suspect the average HSC 4U student doesn't know about this and would struggle with the Q. without guidance.)

You can use calculus to check your answer, but it's probably best to avoid it as part of your working out (unless you can't see any other way to do it).

$\noindent Expanding the LHS, $LHS = 2 \times 2 - 6\sqrt{10} - \sqrt{10} + 3 \times 5 = 19-7\sqrt{10}$ $\Big{(}$note $\sqrt{2}\times \sqrt{5} = \sqrt{10}$$\Big{)}$. Hence $a=19$, $b=-7$, assuming we are after rational solutions $(a,b)$.$

Obtain LaTeX: https://latex-project.org/ftp.html . Depending on whether you have a Mac, Windows or Linux, go to the appropriate link for that on that page.

An ATAR of 99.95 isn't required for Medicine apart from at USyd (which doesn't require UMAT though).