Search results

  1. H

    sigma notation

    (ii) We have the line ax + by = c \implies y = (-a/b) x +c/a . So the gradient of the line is -a/b . Now if we look at the vector \binom{-b}{a} treating \vec{i}, \vec{j} as being in the positive x, y directions respectively, then its gradient is rise/run, or a/(-b) = -a/b which is the same...
  2. H

    92 3U HSC Q6c)ii): Binomial Question⌄

    This is probably the most 'condensed' version of what you could do, but in an exam I wouldn't count on being able to see it. The above methods of subbing in to find C are much more reliable. \begin{align*} (1-x)^n &= \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k\ [\text{Binomial Theorem}]\\...
  3. H

    Complex problem solving q

    A more algebraic approach in case the visualisations don't kick in during the exam. \begin{align*} w &= \frac{z_1}{z_2} \\ z_1 &= wz_2 \\ 1 &= \left|\frac{z_1+z_2}{z_1-z_2}\right|\ [\text{Given}]\\ &= \left|\frac{wz_2+z_2}{wz_2-z_2}\right| \\ &=...
  4. H

    complex q

    We continue the ray by substituting into the original locus equation with a free variable that defines the modulus of z. \begin{align*} \text{If}\ \operatorname{arg} z &= \tan^{-1} 1/3 \\ \text{then}\ z &= r(3+i)\ \text{for some}\ r \in \mathbb{R}^{+} \\ \text{Thus,}\ |r(3+i) - (4+3i)| &= 5...
  5. H

    Trig q

    It's essentially two applications of the double angle formulae. We are in quadrant 4, which means that cosine is + and sine is - (as given). So \begin{align*} \sin \theta &= -\frac{5}{12} \\ \cos^2 \theta &= 1- \sin^2 \theta = 1-\frac{25}{144} = \frac{119}{144} \\ \therefore \cos \theta &=...
  6. H

    Trig

    Using the general solution formula is very efficient, but it's not in the current syllabus, which means it's not on the reference sheet, and I fear that under the pressure of an exam one could misremember them resulting in errors. One alternate solution can be found using product to sums (which...
  7. H

    Proofs

    The contrapositive is given by \text{If}\ ax^2+bx+c=0\ \text{has an integer solution in}\ x,\ \text{then at least one of}\ a,b,c\ \text{is even.} Let's prove the contrapositive. We know from the quadratic formula that \begin{align*} x &= \frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ 2ax &= -b\pm...
  8. H

    Proofs 🥲

    4) Let's skip straight to the inductive step. \begin{align*} \text{Assume that}\ \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{k} &< \sqrt{k} \\ \operatorname{LHS}_{k+1} = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{k} + \frac{1}{k+1} &< \sqrt{k} + \frac{1}{k+1}\ \text{by...
  9. H

    Mathematics Extension 1 HSC thoughts

    Q9 was another tricky one (inverse functions). Many people probably paid for it because of the overly simplistic way in which inverse functions are treated in the Year 11 syllabus. The answer is D. Consider y = -x, with inverse function y = -x (same thing). The function intersects with the...
  10. H

    Mathematics Extension 1 HSC thoughts

    I've noticed quite a bit of discussion about Q10 of the multiple choice, the one about the differential equation \frac{dy}{dx} = \sin y + 1 and thought it might be helpful to go through the thought process of solving it not by finding y in terms of x (as you would do a 3 mark question, or...
  11. H

    MX2 HSC 2022 Solutions

    as title
  12. H

    Maths Extension 2 Thoughts and Feelings

    Typeset solutions. Hopefully not too many mistakes : )
  13. H

    Curve Sketching

    Let's say that g(x) = \ln f(x) Suppose the solution to the equation in part (d) is \begin{align*}x=a>0,\ \text{i.e.}\ g(a)=\ln(f(a))= \ln (2a+5)(a+1) = ka+\ln 5\ \ (*)\end{align*} If we look at the secant line running through (0, ln 5) and (a, ka + ln 5), we see graphically that the gradient...
  14. H

    How would I do this?

    \begin{align*} (a)\ \frac{1}{BC} &= \tan 2\alpha = \frac{2\tan\alpha}{1-\tan^2 \alpha} \implies BC = \frac{1-\tan^2 \alpha}{2\tan\alpha}\\(b)\ \frac{1}{BD} &= \sin 2\alpha = 2\sin \alpha \cos \alpha \implies RHS = BD = \frac{1}{2\sin \alpha \cos \alpha} \\ \frac{1}{AC} &= \tan \alpha \implies...
  15. H

    Complex numbers from cambridge

    Call them a,b,c instead for simplicity. A few things of note: z = x+yi,\ x,y \in \mathbb{R} \implies z\overline{z} = (x+yi)(x-yi) = x^2 + y^2 =|z|^2 Also, \text{If } z \text{ lies on the unit circle, then}\ |z| = z\overline{z} = 1 \implies \overline{z} = \frac{1}{z} So...
  16. H

    complex2 q

    Sums of pairs would work but there's an easier way by using sin^2 + cos^2 = 1: \begin{align*} \left( \cos \frac{\pi}{12} + \cos \frac{5\pi}{12} \right)^2 &= \cos ^2 \frac{\pi}{12} + \cos^2 \left(\frac{5\pi}{12} \right) + 2\cos \frac{\pi}{12} \cos \frac{5\pi}{12} \\ &= \cos^2 \frac{\pi}{12} +...
  17. H

    binomial expansion

    \begin{align*} \\&(c)\ \text{By equating the expression for}\ f(x)\ \text{in}\ (b)\ \text{with the expression given at the start, we know that}\\ &\sum_{r=1}^{n} \binom{n}{r} x^{r-1} = \sum_{r=1}^{n} (1+x)^{r-1} \\ &\text{We can now integrate both sides of the expression from} -1\ \text{to}\...
  18. H

    question

    \text{Assume that }\sqrt{4n-2} = p/q,\ \text{where } p,q\in \mathbb{Z}, q\neq 0\ \text{and}\ \operatorname{gcd}(p,q) = 1 \newline \begin{align*} &\implies 4n-2 = p^2/q^2 \\&\implies 2(2n-1)q^2 = p^2 \\&\implies p^2\ \text{is even} \\&\implies p\ \text{is even} \\&\implies p = 2a\ \text{for...
  19. H

    Maths questions help urgent

    (a) \begin{align*} \sin(x)\left\{\sin(x) + \sin(3x) + ... + \sin(2n-1)x \right\} &= \sum_{k=1}^{n} \sin(x) \sin(2k-1)x = \frac{1}{2} \sum_{k=1}^{n} \left[\cos(2x(1-k)) - \cos(2kx) \right]\\&= -\frac{1}{2}\sum_{k=1}^{n} \left[\cos(2kx) -\cos(2(k-1)x) \right]\ \ (\cos x\ \text{is an even...
Top