Why SDEHS papers so hard MX2 (1 Viewer)

MoonMonster196884 · Dec 4, 2023

Bro, some of the questions in the papers I have seen are harder than Ruse. Almost comparable to BOS Trials, and that's just the first task.

MoonMonster196884 · Dec 4, 2023

do any of u share a similar experience?

carrotsss · Dec 4, 2023

I think I did one at one point and it was pretty easy so idk?

jimmysmith560 · Dec 4, 2023

There are a couple of possible reasons that the exam papers that a school writes for a particular subject (in this case, Mathematics Extension 2) may be difficult. A common reason tends to be that a school wishes to effectively differentiate its students by pushing them to their limits. Assessing students using an exam that would be considered easy or of average difficulty would most likely lead to favourable results from most (if not all) students. This means that students would be ranked very close to the top of the cohort (for example, more than one student could be ranked first) and close to each other in general. On the other hand, a more challenging exam would separate the students who are truly academically capable from those who are not. In such a case, rather than having more than one student place first, only one student could end up placing first, and that would reflect the academic capability of each student more accurately.

Another reason could be that there is simply a teacher(s) at the school who happens to write difficult exams for a particular subject. This means that a change in teachers could impact the difficulty of subsequent exams (either becoming less difficult or more difficult). This would be true especially if the school is not strict in terms of maintaining a certain difficulty when writing its exams.

In any case, it is important to ensure that you are regularly studying and seeking clarification with any aspects of the subject that you may not be fully confident/comfortable with, which will allow you to address areas of improvement as you progress through year 12.

I hope this helps!

MoonMonster196884 · Dec 4, 2023

carrotsss said:
I think I did one at one point and it was pretty easy so idk?

just did the complex numbers part.

carrotsss · Dec 4, 2023

MoonMonster196884 said:
just did the complex numbers part.

which year

MoonMonster196884 · Dec 4, 2023

carrotsss said:
which year

paper not released yet.

carrotsss · Dec 4, 2023

lolcti said:
why have I never heard of this school in my life?

Sydney distance ed high school

asduiashdi · Dec 4, 2023

Could you please share those papers, I would like to try them

I can only find the 2020 one

MoonMonster196884 · Dec 5, 2023

I will give a sketch of a question: find the roots of 3x^2+4x+2i=0 and so it ended up with no solution. More importantly, I left the qn blank, should i get the marks...?

kendricklamarlover101 · Dec 5, 2023

MoonMonster196884 said:
I will give a sketch of a question: find the roots of 3x^2+4x+2i=0 and so it ended up with no solution. More importantly, I left the qn blank, should i get the marks...?

would this not have 2 complex solutions?
$3x^2 +4x+2i = 0$
$x=\frac{-4 \pm \sqrt{16-24i}}{6}$
$\text{Let } \sqrt{16-24i}=a+bi, \text{where a and b are real numbers}$
$\text{From this we have, } a^2 - b^2 = 16 \text{ and, } ab = -12$
$\text{Rearranging the second term as } b=\frac{-12}{a},$
$a^2 - \frac{144}{a^2}=16$
$a^4-16a^2-144=0$
$a^2 = 8+\sqrt{208}, (a^2>0)$
$\text{Using } a^2 - b^2 = 16,$
$b^2 = -8 +\sqrt{208}$
$\therefore \sqrt{16-24i} = \pm\left (\sqrt{8+\sqrt{208}} - i\sqrt{-8 +\sqrt{208}} \right)$
$(\text{Note that b has opposite sign to a as } ab<0)$
$\text{Simplyfing this we have, }$
$\pm\left (2\sqrt{2+\sqrt{13}} - 2i\sqrt{-2 +\sqrt{13}} \right)$
$x=\frac{-4 \pm\left (2\sqrt{2+\sqrt{13}} - 2i\sqrt{-2 +\sqrt{13}} \right)}{6}$
$\therefore x= \frac{-2 \pm\left (\sqrt{2+\sqrt{13}} - i\sqrt{-2 +\sqrt{13}} \right)}{3}$
unless im trippin but im pre sure there is always at least one solution to every polynomial

kendricklamarlover101 · Dec 5, 2023

kendricklamarlover101 said:
unless im trippin but im pre sure there is always at least one solution to every polynomial

wikipedias definition of the fundamental theorem of algebra basically says what i said

MoonMonster196884 · Dec 5, 2023

kendricklamarlover101 said:
would this not have 2 complex solutions?
$3x^2 +4x+2i = 0$
$x=\frac{-4 \pm \sqrt{16-24i}}{6}$
$\text{Let } \sqrt{16-24i}=a+bi, \text{where a and b are real numbers}$
$\text{From this we have, } a^2 - b^2 = 16 \text{ and, } ab = -12$
$\text{Rearranging the second term as } b=\frac{-12}{a},$
$a^2 - \frac{144}{a^2}=16$
$a^4-16a^2-144=0$
$a^2 = 8+\sqrt{208}, (a^2>0)$
$\text{Using } a^2 - b^2 = 16,$
$b^2 = -8 +\sqrt{208}$
$\therefore \sqrt{16-24i} = \pm\left (\sqrt{8+\sqrt{208}} - i\sqrt{-8 +\sqrt{208}} \right)$
$(\text{Note that b has opposite sign to a as } ab<0)$
$\text{Simplyfing this we have, }$
$\pm\left (2\sqrt{2+\sqrt{13}} - 2i\sqrt{-2 +\sqrt{13}} \right)$
$x=\frac{-4 \pm\left (2\sqrt{2+\sqrt{13}} - 2i\sqrt{-2 +\sqrt{13}} \right)}{6}$
$\therefore x= \frac{-2 \pm\left (\sqrt{2+\sqrt{13}} - i\sqrt{-2 +\sqrt{13}} \right)}{3}$
unless im trippin but im pre sure there is always at least one solution to every polynomial

nah i doubt this...

C_master · Dec 5, 2023

kendricklamarlover101 said:
would this not have 2 complex solutions?
$3x^2 +4x+2i = 0$
$x=\frac{-4 \pm \sqrt{16-24i}}{6}$
$\text{Let } \sqrt{16-24i}=a+bi, \text{where a and b are real numbers}$
$\text{From this we have, } a^2 - b^2 = 16 \text{ and, } ab = -12$
$\text{Rearranging the second term as } b=\frac{-12}{a},$
$a^2 - \frac{144}{a^2}=16$
$a^4-16a^2-144=0$
$a^2 = 8+\sqrt{208}, (a^2>0)$
$\text{Using } a^2 - b^2 = 16,$
$b^2 = -8 +\sqrt{208}$
$\therefore \sqrt{16-24i} = \pm\left (\sqrt{8+\sqrt{208}} - i\sqrt{-8 +\sqrt{208}} \right)$
$(\text{Note that b has opposite sign to a as } ab<0)$
$\text{Simplyfing this we have, }$
$\pm\left (2\sqrt{2+\sqrt{13}} - 2i\sqrt{-2 +\sqrt{13}} \right)$
$x=\frac{-4 \pm\left (2\sqrt{2+\sqrt{13}} - 2i\sqrt{-2 +\sqrt{13}} \right)}{6}$
$\therefore x= \frac{-2 \pm\left (\sqrt{2+\sqrt{13}} - i\sqrt{-2 +\sqrt{13}} \right)}{3}$
unless im trippin but im pre sure there is always at least one solution to every polynomial

this mx2?

carrotsss · Dec 5, 2023

C_master said:
this mx2?

this post is in the mx2 forum

C_master · Dec 5, 2023

carrotsss said:
this post is in the mx2 forum

oh f Im blind

MoonMonster196884 · Dec 6, 2023

MoonMonster196884 · Dec 8, 2023

From what I saw, ig the highest mark was 33.3%. Is that believable?

carrotsss · Dec 8, 2023

MoonMonster196884 said:
From what I saw, ig the highest mark was 33.3%. Is that believable?

no way what

MoonMonster196884 · Dec 8, 2023

carrotsss said:
no way what

its like bos trials lvl difficulty

Why SDEHS papers so hard MX2 (1 Viewer)

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