BoS Maths Trials 2019 (1 Viewer)

akkjen · Oct 13, 2019

admittedly I just thought to do the same when first looking at it

, maybe because after doing so many binomial questions with n power it just comes to mind. If there was a x there instead of a n, I would have realised straight away...

Paradoxica · Oct 13, 2019

The proof is extremely straightforward. You know n is a positive integer, so you can just expand it and delete all terms after 1+1. QED.

In fact, there were many people who wrote the binomial expansion, and then suddenly gave up (???)

Kingom · Oct 13, 2019

so just marking q11 of the 4u papers I seem to notice a couple of common trends:
-for the first integral many methods and manipulations were used with most of them being fruitless (and someone even told us to research an identity smh)
-for the volumes question, many people got the correct bounds and integral and for some reason switched bounds for the evaluation (don't do this in the hsc please)
-many people thought the locus of R was a straight line and came up with a myriad of justifications including but not limited to "i have perfect 20/20 vision"
-some people left their volumes as negatives which we didn't deduct marks for but should be avoided come hsc (for cylindrical shells remember to absolute)
Also a shoutout must be made to the person who found the approximate root for the cubic in q11 via newtons method (very nicely done albeit a tad overkill for a 2 mark question)

Good luck y'all for HSC (after Friday many of you will revel in never having to do English again, unless ofc you're doing extension).

blyatman · Oct 13, 2019

I remember tutoring a student many many years back, and we were working on a problem where we had just found x = 100. Later in the question, it asks you to calculate what $x_0$ is, and they wrote down $x_0=100_0$ . I was speechless at the sheer ingenuity of it. Gave me a good laugh.

Drdusk · Oct 13, 2019

blyatman said:
I remember tutoring a student many many years back, and we were working on a problem where we had just found x = 100. Later in the question, it asks you to calculate what $x_0$ is, and, I kid you not, they wrote down $x_0=100_0$ . I was speechless at the sheer ingenuity of it.

View attachment 27313

I've seen even worse.

One time in school I was looking at some of the band 2/3 answers for HSC Physics, and I kid you not some kid actually did this.

$V_x^2 = U_x^2 = 2.2^2$

$\therefore x^2 = 2.2$

After that we literally made it into a meme and let's just say our teachers weren't too happy when we started pulling out stuff like that on purpose haha

Paradoxica · Oct 13, 2019

the biggest meme in our school was saying "by inspection" whenever you didn't know the answer

Drdusk · Oct 13, 2019

Paradoxica said:
the biggest meme in our school was saying "by inspection" whenever you didn't know the answer

That and saying 'trivial' when the teacher asked you a question was the 2nd biggest meme at our school.

The biggest one was when our 4u teacher in term 1 made us do detentions if we turned up late to 4u class and funniest thing was everyone always turned up late, so in the small sheet thing where it would say what time you arrived, we would write down stuff like 7:99 AM. After that all the detention stuff stopped lol.
(we got this idea from an actual meme)

TheOnePheeph · Oct 13, 2019

Paradoxica said:
the biggest meme in our school was saying "by inspection" whenever you didn't know the answer

Biggest meme for us was saying let u=x when given an integral to solve by u sub. Also the classic "expand this expression", then redrawing the question with the brackets more spaced out.

Paradoxica · Oct 13, 2019

TheOnePheeph said:
Biggest meme for us was saying let u=x when given an integral to solve by u sub. Also the classic "expand this expression", then redrawing the question with the brackets more spaced out.

or let u = the integrand (in the few instances where it actually legitimately works)

TheOnePheeph · Oct 13, 2019

Paradoxica said:
or let u = the integrand (in the few instances where it actually legitimately works)

Lol well hey, that works with stuff like the integral of the square root of tan x

TheOnePheeph · Oct 13, 2019

Drdusk said:
I've seen even worse.

One time in school I was looking at some of the band 2/3 answers for HSC Physics, and I kid you not some kid actually did this.

$V_x^2 = U_x^2 = 2.2^2$

$\therefore x^2 = 2.2$

After that we literally made it into a meme and let's just say our teachers weren't too happy when we started pulling out stuff like that on purpose haha

How do you view these band 2 answers? Sounds like an untapped meme goldmine lol.

Drdusk · Oct 13, 2019

TheOnePheeph said:
How do you view these band 2 answers? Sounds like an untapped meme goldmine lol.

https://arc.nesa.nsw.edu.au/go/hsc/std-packs/

sharky564 · Oct 14, 2019

Paradoxica said:
apparently 5 people think that

$\frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left(-\frac{1}{n^2}\right) n \left( 1+\frac{1}{n} \right)^{n-1}$

We know that by $\frac{\text{d}}{\text{d}n} (f(n))^a = a f'(n) (f(n))^{a-1}$ ,

$\frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left(-\frac{1}{n^2}\right) n \left( 1+\frac{1}{n} \right)^{n-1}$

We also know that by $\frac{\text{d}}{\text{d}n} a^n = a^n \ln(a)$ ,

$\frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left( 1+\frac{1}{n} \right)^{n} \ln \left ( 1 + \frac{1}{n} \right )$

Since we have multiple answers, let's just add them

$\frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left(-\frac{1}{n^2}\right) n \left( 1+\frac{1}{n} \right)^{n-1} + \left( 1+\frac{1}{n} \right)^{n} \ln \left ( 1 + \frac{1}{n} \right )$

so our final answer is

$\frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left( 1+\frac{1}{n} \right )^{n} \left ( \ln \left ( 1 + \frac{1}{n} \right ) - \frac{1}{n+1} \right )$

Find a counterexample to my proof

.

Heresy · Oct 16, 2019

Full marks in 4 Unit Inequalities confirmed

psmao · Oct 17, 2019

When will the results and solutions be released?

marbleblast123 · Oct 20, 2019

Any updates on when the results and solutions be released?

Trebla · Oct 20, 2019

Almost done for Mathematics Extension 2, expect it to be released in the next 2-3 days.

One remark I would like make when marking Q12, was that students need to distinguish between complex numbers, vectors and magnitudes. They are NOT the same thing.

A common response to Q12c)i) was to define the parallelogram on an Argand diagram with point A on the origin, B represented by complex number z, D represented by complex number w and C represented by the complex number z+w.

A number of students who did this then made the following claims:

$AC=z+w$

$BD=z-w$

$AC^2+BD^2 = (z+w)^2+(z-w)^2$

and then proceeded to expand. This is NOT correct because the lengths AC and BD are real numbers.

We say the vector AC (not the length of AC) is represented by the complex number z+w. In other words, z+w simply gives us the information to find the magnitude (through |z+w|) and direction (through arg(z+w)) of the the vector AC.

Lengths should always expressed in terms of the modulus of the complex numbers they represent, so it really should be:

$AC=|z+w|$

$BD=|z-w|$

$AC^2+BD^2 = |z+w|^2+|z-w|^2$

which can then be expanded by breaking the modulus down into the product of a complex number and its conjugate.

TheOnePheeph · Oct 21, 2019

Trebla said:
Almost done for Mathematics Extension 2, expect it to be released in the next 2-3 days.

One remark I would like make when marking Q12, was that students need to distinguish between complex numbers, vectors and magnitudes. They are NOT the same thing.

A common response to Q12c)i) was to define the parallelogram on an Argand diagram with point A on the origin, B represented by complex number z, D represented by complex number w and C represented by the complex number z+w.

A number of students who did this then made the following claims:

$AC=z+w$

$BD=z-w$

$AC^2+BD^2 = (z+w)^2+(z-w)^2$

and then proceeded to expand. This is NOT correct because the lengths AC and BD are real numbers.

We say the vector AC (not the length of AC) is represented by the complex number z+w. In other words, z+w simply gives us the information to find the magnitude (through |z+w|) and direction (through arg(z+w)) of the the vector AC.

Lengths should always expressed in terms of the modulus of the complex numbers they represent, so it really should be:

$AC=|z+w|$

$BD=|z-w|$

$AC^2+BD^2 = |z+w|^2+|z-w|^2$

which can then be expanded by breaking the modulus down into the product of a complex number and its conjugate.

Isn't that question a lot easier to do with cosine rule anyway?

marbleblast123 · Oct 21, 2019

yep very easy with cosine rule

Trebla · Oct 21, 2019

TheOnePheeph said:
Isn't that question a lot easier to do with cosine rule anyway?

Yeah, that was the other common approach.

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