Author Topic: 3U Maths Question Thread (Read 1238982 times) Tweet Share

RuiAce · « **Reply #3570 on:** August 04, 2018, 03:20:18 pm »

Quote from: Mate2425 on August 04, 2018, 03:15:50 pm

Thank you so much Rui, i now understand clearly. On a side note could you also please explain what the phase does in the normal displacement of a particle formula because i am struggling to understand where/ why we use it.
Thannnkkkks!

Pretty much whilst the period shifts how often you repeat itself and the amplitude shifts the max velocity, the phase does nothing but shift where your starting point is. This is because all that it does is move the graph left and right (i.e. perform a horizontal translation on the displacement-time graph).

An example of where a phase shift would be useful is this. Assume for simplicity that the height of a tide can be modelled by simple harmonic motion. But you may want to start modelling at 6:00 AM in the morning for convenience. At that point in time, your tide may not necessarily be at low tide, high tide or right in the middle, but rather some random place. So if $t$ is the time-elapsed from 6:00 AM, you insert a phase shift so that your analysis starts at the height it should start at, i.e. the height at 6:00 AM.

Mate2425 · « **Reply #3571 on:** August 05, 2018, 01:08:22 am »

Hi, could someone please help me with solving this question and why they swapped the normal positions in the k +1th formula.
http://educationstandards.nsw.edu.au/wps/wcm/connect/c5cb36cb-4c88-4c84-9295-23767be83c08/maths-ext1-hsc-exam-2012.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-c5cb36cb-4c88-4c84-9295-23767be83c08-lG9Dus7
Q11f ii)

Thankyou !!!!

RuiAce · « **Reply #3572 on:** August 05, 2018, 09:33:14 am »

Quote from: Mate2425 on August 05, 2018, 01:08:22 am

Hi, could someone please help me with solving this question and why they swapped the normal positions in the k +1th formula.
http://educationstandards.nsw.edu.au/wps/wcm/connect/c5cb36cb-4c88-4c84-9295-23767be83c08/maths-ext1-hsc-exam-2012.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-c5cb36cb-4c88-4c84-9295-23767be83c08-lG9Dus7
Q11f ii)

Thankyou !!!!

Not too sure what you mean here. Did you mean that they swapped the positions in the answers or something?

Mate2425 · « **Reply #3573 on:** August 05, 2018, 12:32:17 pm »

Quote from: RuiAce on August 05, 2018, 09:33:14 am

Not too sure what you mean here. Did you mean that they swapped the positions in the answers or something?

Hi Rui, i worked down to 3n - 4k = 0 but unsure how they got to the final statement of 4k -n =0

RuiAce · « **Reply #3574 on:** August 05, 2018, 04:59:27 pm »

Quote from: Mate2425 on August 05, 2018, 12:32:17 pm

Hi Rui, i worked down to 3n - 4k = 0 but unsure how they got to the final statement of 4k -n =0

They probably used $ \binom{n}{k} \left(2x^3\right)^k \left( -\frac{1}{x} \right) ^{n-k} $ instead to get that.

Both conventions will work. This is thanks to the symmetry property of the binomial coefficient. (Quite amazingly the link to the solutions in my compilation is broken so I'm gonna use the other convention of $ \binom{n}{k} \left( 2x^3 \right)^{n-k} \left( -\frac{1}{x} \right)^k $ for this solution, which is presumably what you used, and then update the compilation.)

But yeah, to answer your question before I digress, they used the other convention. The other convention gives $4k - n = 0$ instead of $3n - 4k= 0$
------------------------------------------------------------
$\text{Rewrite }\binom{n}{k} \left( 2x^3 \right)^{n-k} \left( -\frac{1}{x} \right)^k\\ \text{as } \binom{n}{k} 2^{n-k} (-1)^k x^{3(n-k)} x^{-k}.$
$\text{Assume that a constant term, i.e. a term with }x^0\text{ existed.}\\ \text{As correctly stated above (under this convention), this will give }3(n-k) - k = 0\\ \text{i.e. }3n-4k = 0.$
$\text{Now, }n\text{ and }k\text{ are both integers for the sake of the binomial theorem.}\\ \text{If we rearrange to make }k\text{ the subject, }k = \frac{3n}{4}.$
$\text{From here, it is clear that }n\text{ must be a multiple of 4,}\\ \text{for if it weren't a multiple of 4, }k\text{ would not be a whole number.}$

24221 · « **Reply #3575 on:** August 07, 2018, 07:46:43 pm »

Hey, could some please help me with question 11)b)i). Thank You.

RuiAce · « **Reply #3576 on:** August 07, 2018, 11:11:14 pm »

Quote from: 24221 on August 07, 2018, 07:46:43 pm

Hey, could some please help me with question 11)b)i). Thank You.

$\text{The chord of contact has equation}\\ \boxed{xx_0 = 2a(y+y_0)}\text{ from the formula sheet.}$
$\text{Since }T\text{ lies on the axis of the parabola, i.e. }x=0,\\ \text{subbing this in gives }y = -y_0.\\ \text{Therefore }T\text{ is the point }\boxed{T(0, -y_0)}.$
$\text{The tangent to the parabola at the origin }O(0,0)\text{ will just be }y=0,\\ \text{i.e. the }x\text{-axis.}\\ \text{The perpendicular distance of any point from the }x\text{-axis}\\ \text{is essentially determined by just its }y\text{-coordinate.}$
$\text{Since }P\text{ is }(x_0,y_0)\text{, its distance from the }x\text{-axis is }y_0.\\ \text{Since }T\text{ is }(0,-y_0)\text{, its distance from the }x\text{-axis is also }y_0.\\ \text{So clearly they are the same.}$

Mate2425 · « **Reply #3577 on:** August 09, 2018, 12:58:53 pm »

Quote from: RuiAce on August 05, 2018, 04:59:27 pm

They probably used $ \binom{n}{k} \left(2x^3\right)^k \left( -\frac{1}{x} \right) ^{n-k} $ instead to get that.

Both conventions will work. This is thanks to the symmetry property of the binomial coefficient. (Quite amazingly the link to the solutions in my compilation is broken so I'm gonna use the other convention of $ \binom{n}{k} \left( 2x^3 \right)^{n-k} \left( -\frac{1}{x} \right)^k $ for this solution, which is presumably what you used, and then update the compilation.)

But yeah, to answer your question before I digress, they used the other convention. The other convention gives $4k - n = 0$ instead of $3n - 4k= 0$
------------------------------------------------------------
$\text{Rewrite }\binom{n}{k} \left( 2x^3 \right)^{n-k} \left( -\frac{1}{x} \right)^k\\ \text{as } \binom{n}{k} 2^{n-k} (-1)^k x^{3(n-k)} x^{-k}.$
$\text{Assume that a constant term, i.e. a term with }x^0\text{ existed.}\\ \text{As correctly stated above (under this convention), this will give }3(n-k) - k = 0\\ \text{i.e. }3n-4k = 0.$
$\text{Now, }n\text{ and }k\text{ are both integers for the sake of the binomial theorem.}\\ \text{If we rearrange to make }k\text{ the subject, }k = \frac{3n}{4}.$
$\text{From here, it is clear that }n\text{ must be a multiple of 4,}\\ \text{for if it weren't a multiple of 4, }k\text{ would not be a whole number.}$

Thanks Rui for all the help!

Mate2425 · « **Reply #3578 on:** August 09, 2018, 01:03:03 pm »

Hi Rui, for the following function f(x) = x / (4-x^2) there would be two vertical asymptotes at x = +2, -2 and is there also a horizontal asymptote of y = 0?

Thank you!

RuiAce · « **Reply #3579 on:** August 09, 2018, 01:24:42 pm »

Quote from: Mate2425 on August 09, 2018, 01:03:03 pm

Hi Rui, for the following function f(x) = x / (4-x^2) there would be two vertical asymptotes at x = +2, -2 and is there also a horizontal asymptote of y = 0?

Thank you!

Yep that looks correct

Mate2425 · « **Reply #3580 on:** August 09, 2018, 03:18:39 pm »

Agggh, probability.
Could someone please help me understand the method in their approach to this question Q10 MC 2017.
http://educationstandards.nsw.edu.au/wps/wcm/connect/807028e6-87cd-4c19-bf19-9c3c7f6e6be8/2017-hsc-maths-ext-1.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-807028e6-87cd-4c19-bf19-9c3c7f6e6be8-l-jq7Cv
Why is the answer not D?
Thank you.

fun_jirachi · « **Reply #3581 on:** August 09, 2018, 08:47:22 pm »

yo wth i did this question with my mate for fun like 5 months ago to see how hard 3u was, fattest throwback
so anyway, answer attached, soz for trash handwriting

RuiAce · « **Reply #3582 on:** August 09, 2018, 10:20:37 pm »

Quote from: fun_jirachi on August 09, 2018, 08:47:22 pm

yo wth i did this question with my mate for fun like 5 months ago to see how hard 3u was, fattest throwback
so anyway, answer attached, soz for trash handwriting

Lmao it's way easier than that.

Quote from: Mate2425 on August 09, 2018, 03:18:39 pm

Agggh, probability.
Could someone please help me understand the method in their approach to this question Q10 MC 2017.
http://educationstandards.nsw.edu.au/wps/wcm/connect/807028e6-87cd-4c19-bf19-9c3c7f6e6be8/2017-hsc-maths-ext-1.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-807028e6-87cd-4c19-bf19-9c3c7f6e6be8-l-jq7Cv
Why is the answer not D?
Thank you.

$\text{Favourable outcomes: This can literally be counted.}\\ \text{2 diagonal lines;}\\ \text{3 vertical lines;}\\ \text{3 horizontal lines.}\\ \text{Which implies }\boxed{8}\text{ favourable outcomes.}$
$\text{Total outcomes:}\\ \text{We literally just choose 3 squares out of 9 to}\\ \text{put a cross in, which can be done in }\binom93 = \boxed{84}\text{ ways.}$
$\text{So the probability is }\frac{8}{84} = \boxed{\frac{2}{21}}$

Quote from: Mate2425 on August 09, 2018, 03:18:39 pm

Why is the answer not D?
Thank you.

I'm actually not sure at all where you got 8/9 from. That is an insanely huge probability. Intuitively, the answer to this question should be small since it's really hard to get 3 in a line; there's heaps of other possibilities that can happen involving 2 being in a line and 1 floating around elsewhere)

fun_jirachi · « **Reply #3583 on:** August 09, 2018, 10:57:02 pm »

damn wowzers there's an easier way, i didnt think about using combs, nice job!

why do i always overcomplicate things

RuiAce · « **Reply #3584 on:** August 09, 2018, 10:59:45 pm »

Quote from: fun_jirachi on August 09, 2018, 10:57:02 pm

damn wowzers there's an easier way, i didnt think about using combs, nice job!
why do i always overcomplicate things

Believe it or not, for 3U and 4U probability questions I try to go for perms and combs before everything else

May be something to think about and take on board once you actually take the course...

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