MedVision ad

Search results

  1. I

    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Also, the fact that those vectors are equal in length is not "using the fact that triangle is equilateral", since equilateral means three equal sides. So we use the fact that two of the vectors are equal length, and that their...
  2. I

    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $I didn't use the fact that the triangle is equilateral, I said that the equation tells us that the vectors have equal length. This is because the ratio of the two vectors' lengths is 1 (i.e. the modulus of the polar form...
  3. I

    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Geometrically, this equation tells us that the vector joining $z_1$ to $z_2$ is the same length as that joining $z_1$ to $z_3$, and the angle between these two vectors is $\frac{\pi}{3}=60^\circ$.$ $This is because...
  4. I

    Perms and coms

    Lol, no need to be sorry. :)
  5. I

    Perms and coms

    $Let the probability of a tails be $p_T$ and the probability of a heads be $p_H$. We must have $p_H + p_T = 1$, as the probabilities of all the possibilities of a coin toss (heads or tails) must sum to 1. Also, since heads comes up twice as often as tails, we have $p_H = 2p_T$ (i.e. the...
  6. I

    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $There will be exactly three possible points. If we're given three distinct points in the plane that are non-collinear, there will always be exactly three possibilities for a fourth point to form a parallelogram. See this...
  7. I

    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Think of it in terms of vectors. Draw out and label the three points on the Argand diagram to get a feel for the three places $z_3$ could go, and recall that a parallelogram is a quadrilateral where the vectors representing...
  8. I

    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Note that if we have two vertices of an isosceles right-angled triangle in the plane and we specify which vertex is the right angle, we'll have two possible locations for the third vertex. Furthermore, multiplying z1 by i would...
  9. I

    Cambridge Prelim MX1 Textbook Marathon/Q&A

    Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Let $z_1,z_2,z_3 \in \mathbb{C}$. Let $\frac{z_3 - z_1}{z_2 - z_1}=\lambda$ for some real number $\lambda$ ($\lambda \neq 0$ as $z_3 \neq z_1$, since we have three distinct points). Then we have $z_3 - z_1 = \lambda (z_2 -...
  10. I

    HSC 2016 MX2 Integration Marathon (archive)

    Re: MX2 2016 Integration Marathon $For the last integral, we can replace $\sin^2 x$ with $1-\cos^2 x$ and then substitute $u=\cos x$, $\mathrm{d}u = -\sin x \text{ d}x$.$
  11. I

    HSC 2016 MX2 Marathon (archive)

    Re: HSC 2016 4U Marathon To disprove the statement's converse, one only needs a counter-example. What do you mean general case?
  12. I

    HSC 2016 MX2 Marathon (archive)

    Re: HSC 2016 4U Marathon $Convex means ``concave up'' and concave means ``concave down''. You can find formal definitions on the Wikipedia pages for them:$ • https://en.wikipedia.org/wiki/Convex_function • https://en.wikipedia.org/wiki/Concave_function
  13. I

    HSC 2016 MX2 Marathon (archive)

    Re: HSC 2016 4U Marathon Yes, just need to provide a counter-example.
  14. I

    Cambridge Year 12 3U Question - Inverse Trig Functions

    $Using the identity $\tan \left(A-B \right)=\frac{\tan A - \tan B}{1+\tan A \tan B}$ and calling the expression $X$, we have$ $$\begin{align*}\tan X &= \frac{\frac{ax}{1-bx}-\frac{x-b}{a}}{1+\frac{ax}{1-bx}\cdot \frac{x-b}{a}} \\ &= \frac{a^2x - (1-bx)(x-b)}{a(1-bx) + (ax)(x-b)}\quad...
  15. I

    HSC 2016 MX2 Marathon (archive)

    Re: HSC 2016 4U Marathon $I'm guessing in general it's too much to hope to find a \emph{polynomial} equation with those roots or with arbitrary function of the original roots, with our standard HSC 4U methods anyway. Of course, we can do this when the roots are $\sqrt{\alpha}$ or $\alpha^{3}$...
  16. I

    HSC 2016 MX2 Marathon (archive)

    Re: HSC 2016 4U Marathon $For the case of squared roots, even though $x^2$ is not bijective, using $P\left(\sqrt{z} \right)$ and then rearranging etc. works, despite this being just one branch of the inverse relation, which is a \textsl{multivalued function}. The crucial part is the...
  17. I

    HSC 2016 MX2 Marathon (archive)

    Re: HSC 2016 4U Marathon $This is the typical method of finding a polynomial equation with roots $\alpha^2$ etc. when given a polynomial with roots $\alpha$ etc. You'll learn it later in the HSC 4U Polynomials topic when you get to it. The reason it works is that replacing $z$ with $\sqrt{z}$...
  18. I

    HSC 2016 Maths Marathon (archive)

    Re: HSC 2016 2U Marathon You should always write the exact answer unless specified otherwise. Even when rounding the answer because they asked for a rounded value, you should write the answer out to many decimal places and put a "..." at the end of the long numeral, before rounding it, so...
  19. I

    HSC 2016 MX2 Combinatorics Marathon (archive)

    Re: HSC 2016 MX2 Combinatorics Marathon $Here's a question:$ $Let $n,m,r$ be positive integers with $n,m\geq r$.$ $Show that $C(m,r)\cdot P(n,r)=P(m,r)\cdot C(n,r)$, where $C(n,r)$ is ``$n$ choose $r$'' ($^n C_r$) and $P(n,r)$ is $^nP_r$.$ $Bonus: provide a combinatorial interpretation.$
  20. I

    HSC 2016 Maths Marathon (archive)

    Re: HSC 2016 2U Marathon $To write the percentage sign in LaTeX, do a backslash and then the percentage symbol: $\%$.$
Top