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$\noindent This is based on the fact that if $a,b,c\in \mathbb{Z}$ and $n\in \mathbb{Z}^{+}$, if $b\equiv c \pmod{n}$, then $ab \equiv ac \pmod{n}$. This is pretty easy to prove from the definition of mod.$

$\noindent Yeah, this is just because $2^6$ is congruent to 1 modulo 9.$

$\noindent For (b), the inequality should be $|z_1 - z_2| \geq \Big{|} |z_1|-|z_2|\Big{|}$ (this is known as the \textit{reverse triangle inequality}). The inequality you typed for (b) can easily have counterexamples found for it (even if we put absolute value signs around the R.H.S., which we...

$\noindent Yeah, we \underline{don't} need \textit{both} of them to be 0 for the equality to hold. The equality will hold even if just one of them is 0. Specifically, if $z_1 = 0$ and $z_2$ is any complex number, then $|z_1+z_2| = |0+z_2| = |z_2|$ and $|z_1| + |z_2| = |0| + |z_2| = 0+|z_2| =...

$\noindent By same direction (for non-zero complex numbers), it means the vectors representing those complex numbers have \textit{exactly} the same orientation, i.e. the complex numbers have the same argument. This is equivalent to saying that one complex number is just a positive multiple of...

Neither of those pairs have equal argument (or a 0 complex number), so there wouldn't be equality.

Re: HSC 2016 3U Marathon I'll show how to do Q.4 (Q.5 isn't too hard, just tedious). $\noindent \underline{QUESTION 4}$ $\noindent The equation of the chord of contact is $xx_1 =2a(y+y_1)$. Since $R$ lies on this line and lies on the directrix $y=-a$, at $R$ we have $x_R x_1 =...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent (b) (i) We have $t=\tan \frac{\theta}{2}$, where $\theta = 225^\circ,\, \frac{\theta}{2}=112\tfrac{1}{2}^{\circ}$. Note that $t<0$ (using ASTC for example), and our equation is $\frac{2t}{1-t^2}=\tan \theta = 1, \...

Well sadly both subjects at HSC level are essentially just watered-down drivel devoid of any real science. In the end, they're similar style subjects, just that the actual content is different (they're still both based on regurgitating rote learnt info in such a way that your response will get...

$\noindent The reciprocal function would have a vertical asymptote at $x=-1$ (in fact, the whole graph gets flipped about the $y$-axis. In other words, that function $f$ satisfies $\frac{1}{f(x)}=f(-x)$ for all $x$ for which this equation makes sense (i.e. no 0 denominators etc.).$

$\noindent It would have a vertical asymptote at $x=1$, since as $x\to 1$, the numerator approaches a non-zero number whilst the denominator approaches 0. This is the same as the graph of the equations you posted in the Original Post.$

Yeah, the equations have the same graph.

$\noindent Based on your query, not sure whether you typed those both correctly (mainly the second one), but yes, the graphs are different. I think you meant to have a different numerator for the second one. Something like $(x+1)(x-1)^2$ so that both equations reduce to $\frac{x+1}{x-1}$ for...

$\noindent Because by definition, the symbol $\sqrt{\cdot}$ refers to the non-negative square root.$

For a), the equality occurs if and only if at least one of the complex numbers is 0 or both complex numbers have the same orientation in their before representation (i.e. equal argument). Edit: Mentioned above (though the k above doesn't need to be strictly positive, it can also be 0, as in...

What did you mean by "No year 11"? You mean no Year 11's are allowed to do that Q.? (And I tried to put the hint in spoiler tags, but it appears that the spoilers are dodgy on this site.)

It's probably Year 12 level haha.

$\noindent Oh yeah, missed his square root part (probably wouldn't have missed it if it were in LaTeX). $\Big{(}$And it was $u=\sqrt{x-1}$, not +1.$\Big{)}$

$\noindent If we take DatAtarLyfe's approach of substituting $u=x-1$, we need to replace all $x$'s in the original equations with $u$. Since $u=x-1$, we have $x=u+1$, so our equation becomes $(u+1)-4\sqrt{u}+2 = 0\Longleftrightarrow u - 4\sqrt{u} + 3 = 0$. We would then solve this for $u$ and...