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January 18, 2020, 11:12:34 am

### AuthorTopic: 4U Maths Question Thread  (Read 303661 times) Tweet Share

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#### Aaron Lillis

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« Reply #2355 on: August 07, 2019, 11:27:16 pm »
0

P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.

Find the equation of the locus of M if PQ is a tangent to the parabola y2=4x

Essentially, I determined the locus to be y2=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?

Thank you.

That locus is correct, I think i did that question in a cssa paper. Does the question ask to specify restrictions? I don't see what the restrictions would be as you've eliminated the parameters

#### RuiAce

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« Reply #2356 on: August 08, 2019, 09:33:58 am »
+2

P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.

Find the equation of the locus of M if PQ is a tangent to the parabola y2=4x

Essentially, I determined the locus to be y2=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?

Thank you.
That locus is correct, I think i did that question in a cssa paper. Does the question ask to specify restrictions? I don't see what the restrictions would be as you've eliminated the parameters
My working out and GeoGebra are claiming $y^2 = -\frac{x}{2}$. Note that the line(s) demonstrating that $PQ$ is a tangent to $y^2=4x$ is hidden, but you can put it back in there.
$m_{PQ} = \frac{\frac2q-\frac2p}{2q-2p}=-\frac1{pq}\\ \text{so }PQ\text{ has equation }x+pqy=2(p+q).$
$\text{For this to be tangential to }y^2=4x\text{, upon substitution we'd require}\\ \text{the point of intersection to satisfy}\\ \frac{y^2}{4}+pqy-2(p+q)=0.$
$\text{The tangent will only have one point of intersection with the parabola}\\ \text{and hence this quadratic in }y\text{ has only one solution.}\\ \text{Setting the discriminant equal to 0 gives the constraint}\\ \boxed{p^2q^2+2(p+q) = 0}.$
\text{The coordinates of }M\text{ satisfy}\\ \begin{align*}x&= \frac{2p+2q}{2}=p+q\\ y&= \frac{\frac2p+\frac2q}2 = \frac{q+p}{pq}\end{align*}
$\text{Hence to find the locus of }M\text{, we first see that}\\ y=\frac{x}{pq} \implies pq = \frac{x}{y}.$
$\text{Subbing this and }x=p+q\text{ into the constraint found above gives}\\ \frac{x^2}{y^2}+2x=0 \implies\boxed{ y^2 = -\frac{x}{2}}$
Restrictions on this locus are not so trivially found. First note that the relationship between $p$ and $q$, when treated as a quadratic in $q$, gives $q=\frac{-1\pm \sqrt{1-2p^3}}{p^2}$. The expression under the square root shows that $p < 2^{-1/3}$.

Close examination of the GeoGebra output shows that the restrictions on the locus appear to differ, depending on if the positive or negative root was taken as the correct one. In general, restrictions do apply as the locus will not include every point on its defining equation.

Here, taking the positive root appears to force the condition that $y > 0$, i.e. we only consider the upper branch of the parabola $y^2=-\frac{x}{2}$.

Taking the negative root gives a condition very difficult to describe. There are subcases to consider. When $p < 0$, it can be shown that the locus is restricted to $y <- 2^{-1/3}$. The significance of $-2^{-1/3}$ is that it is the point of intersection between the locus $y^2=-\frac{x}{2}$ and the original hyperbola $xy=4$. On the other hand when $0 < p < 2^{1/3}$, it instead appears as though the locus will always be on the upper branch of the parabola, with a new condition that $x < -2^{-1/3}$ instead.

So overall (combining the two), the locus would be $y^2=-\frac{x}{2}$, subject to the restriction that the parabolic arc from $\left(-2^{5/3}, -2^{1/3}\right)$ to the origin $(0,0)$ is excluded. The exclusion includes the origin, but not the other endpoint.

The main thing making this so awkward is that sometimes $P$ and $Q$ will be in the same quadrant, and other times they won't be. Furthermore, each value of $p$ corresponds to two values of $q$, and it appears as though the algebra works out differently depending on which is taken. For that reason, if you were asked to find the restrictions in an exam, you would be given hints. In general, excluded points on the locus do not need to be found unless the examination has specifically requested for it.  #### diggity

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« Reply #2357 on: August 10, 2019, 03:09:42 pm »
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Hello!

For this question:

"A ball of mass m is projected vertically upwards with speed u. The acceleration acting against the ball is proportional to it speed. Find the time (t) taken to reach the greatest height."

I'm simply confused on how to resolve forces. I resolved them like this: "ma = -mg - kv",  but the answers resolved them as "a = -g - kv"

Why is that the correct resolution? Thanks in advance!
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#### RuiAce

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« Reply #2358 on: August 10, 2019, 03:53:43 pm »
+4
Hello!

For this question:

"A ball of mass m is projected vertically upwards with speed u. The acceleration acting against the ball is proportional to it speed. Find the time (t) taken to reach the greatest height."

I'm simply confused on how to resolve forces. I resolved them like this: "ma = -mg - kv",  but the answers resolved them as "a = -g - kv"

Why is that the correct resolution? Thanks in advance!
Considering they said acceleration acting against the ball instead of the "force", $a=-g-kv$ would be the correct answer.  #### blyatman

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« Reply #2359 on: August 10, 2019, 10:15:41 pm »
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Hello!

For this question:

"A ball of mass m is projected vertically upwards with speed u. The acceleration acting against the ball is proportional to it speed. Find the time (t) taken to reach the greatest height."

I'm simply confused on how to resolve forces. I resolved them like this: "ma = -mg - kv",  but the answers resolved them as "a = -g - kv"

Why is that the correct resolution? Thanks in advance!

They're both correct. However, the force balance equation should always be the first line in any of these types of problems. The equations of motion are always derived from the force balance equation. In this case, "ma = -mg - kv", which means that a = -g-(k/m)*v. Note that k/m is still just a proportionality constant, so you could replace k/m with another constant, called k, for simplicity.
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#### RuiAce

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« Reply #2360 on: August 11, 2019, 08:48:17 am »
+1
They're both correct. However, the force balance equation should always be the first line in any of these types of problems. The equations of motion are always derived from the force balance equation. In this case, "ma = -mg - kv", which means that a = -g-(k/m)*v. Note that k/m is still just a proportionality constant, so you could replace k/m with another constant, called k, for simplicity.
Although this makes perfect sense in the real world, from what I have seen they are far more specific about things in the HSC (in what they're looking for). The questions are worded so that students will always go along the exact same line of thinking, with the exact same constant $k$ in mind. (As opposed to swapping it out for another.)

If we want to follow the rule that the force resolution should be the start, if anything they would like you to turn the resistance into $mkv$. And then write $ma = -mg-mkv$.  #### blyatman

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« Reply #2361 on: August 11, 2019, 10:56:35 am »
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Although this makes perfect sense in the real world, from chat I have seen they are far more specific about things in the HSC (in what they're looking for). The questions are worded so that students will always go along the exact same line of thinking, with the exact same constant $k$ in mind. (As opposed to swapping it out for another.)

If we want to follow the rule that the force resolution should be the start, if anything they would like you to turn the resistance into $mkv$. And then write $ma = -mg-mkv$.

Yeh I'm aware of that, but in the question it doesn't state what the propotionality constant is, so it should be perfectly valid to use k/m and then combine that into one final constant. They shouldn't lose marks for it - I remember doing this back in the day and it was fine, according to my teacher as well as my friends teachers.
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#### RuiAce

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« Reply #2362 on: August 18, 2019, 11:26:07 am »
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Yeh I'm aware of that, but in the question it doesn't state what the propotionality constant is, so it should be perfectly valid to use k/m and then combine that into one final constant. They shouldn't lose marks for it - I remember doing this back in the day and it was fine, according to my teacher as well as my friends teachers.
What concerns me about this is that whilst it may have been fine once upon a time, marking guidelines may fluctuate in how strict they are. (For example, stuff like LHopitals rule is instantly penalised now.)

It feels too much like gambling on chance to rely on these handwaves. Although I can respect what teachers have told you, and that they probably still are examiners, this would be something I want to hear from this years examiners first-hand as well. Ive never been comfortable recommending tricks that could potentially be taken as a stretch unless I have first-hand permission granted with it.  #### blyatman

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« Reply #2363 on: August 18, 2019, 01:13:53 pm »
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What concerns me about this is that whilst it may have been fine once upon a time, marking guidelines may fluctuate in how strict they are. (For example, stuff like LHopitals rule is instantly penalised now.)

It feels too much like gambling on chance to rely on these handwaves. Although I can respect what teachers have told you, and that they probably still are examiners, this would be something I want to hear from this years examiners first-hand as well. Ive never been comfortable recommending tricks that could potentially be taken as a stretch unless I have first-hand permission granted with it.
Fair enough. I wasn't even aware that there was ever a time where L'Hopitals rule was acceptable lol.
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#### RuiAce

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« Reply #2364 on: August 18, 2019, 01:27:44 pm »
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Fair enough. I wasn't even aware that there was ever a time where L'Hopitals rule was acceptable lol.
Oddly enough, once upon a time it was ok. But the stance now is strictly against it and I think it's a good call by NESA here.  #### david.wang28

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« Reply #2365 on: August 25, 2019, 08:47:48 pm »
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Hello,
I am stuck on an integration question in the attachment below. I solved it, but I can't seem to find errors in my working out even though the answer required is different. Can anyone please help me out? Thanks, David.
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#### fun_jirachi

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« Reply #2366 on: August 26, 2019, 09:11:15 am »
+1
Hey there!

Firstly, I think the reason you're not getting to the correct solution is because your dv is incorrect!! Remember that adding one to the power, dividing by the new power and the derivative doesn't actually work unless you have a linear function (which is what I think you did there.)

An alternative solution is below:

If there's anything you don't understand, please ask further Hope this helps Failing everything, but I'm still Flareon up.

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#### david.wang28

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« Reply #2367 on: August 30, 2019, 06:00:23 pm »
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Hey there!

Firstly, I think the reason you're not getting to the correct solution is because your dv is incorrect!! Remember that adding one to the power, dividing by the new power and the derivative doesn't actually work unless you have a linear function (which is what I think you did there.)

An alternative solution is below:

If there's anything you don't understand, please ask further Hope this helps Thank you so much! Sorry for the late reply; I had trials.
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#### classof2019

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« Reply #2368 on: September 13, 2019, 07:06:38 pm »
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The numbers 5,6,7,8 and 9 are chosen without replacement and arranged to form a five-digit number. Find the probability that the number formed has the three odd digits in increasing numerical order.

This is what I've attempted to reason (though it's wrong).

Since the 5, 7, and 9 are increasing numerical order, let them be fixed. We're essentially arranging the 6 and 8 around them. Let crosses denote possible positions for 6 and 8.

I ended up with: XX5XX7XX9XX

I then said that the number of arrangements was 8P2 as we're choosing two X's then arranging them around the 5, 7 and 9 - then divide by the total arrangements (5!) to get the answer. I've used similar logic in other questions and have gotten them right, however in this case this is wrong.

Just wondering, why am I wrong/ how do I go about getting the right answer?

Thank you.
« Last Edit: September 13, 2019, 07:39:15 pm by classof2019 »

#### RuiAce

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« Reply #2369 on: September 13, 2019, 08:39:28 pm »
+1

The numbers 5,6,7,8 and 9 are chosen without replacement and arranged to form a five-digit number. Find the probability that the number formed has the three odd digits in increasing numerical order.

This is what I've attempted to reason (though it's wrong).

Since the 5, 7, and 9 are increasing numerical order, let them be fixed. We're essentially arranging the 6 and 8 around them. Let crosses denote possible positions for 6 and 8.

I ended up with: XX5XX7XX9XX

I then said that the number of arrangements was 8P2 as we're choosing two X's then arranging them around the 5, 7 and 9 - then divide by the total arrangements (5!) to get the answer. I've used similar logic in other questions and have gotten them right, however in this case this is wrong.

Just wondering, why am I wrong/ how do I go about getting the right answer?

Thank you.
The mistake was in how you wrote it: xx5xx7xx9xx.

Consider the following. On one hand, we can do 6x58x7xx9xx. This would give us the number 65879.

On another hand, we can do x65x87xx9xx. This is a different arrangement in how you did 8P2, but observe that it also gives 65879! Hence the number 65879 has been double counted.

The flaw was that you focused too much on the possibility of something like 68579, which actually escapes any double counting. But it only does so because the even numbers got grouped between two odd numbers. It doesnt work when the even numbers are split.

Instead, a better option would be the following. First arrange the 6 by picking one of the 4 crosses in: x5x7x9x.

Then arrange the 8. Note that now that the 6 has been arranged, wed have something like x6x5x7x9x. Therefore we pick one out of 5 crosses for the 8.

In total, we therefore have $4\times5=20$ favourable arrangements, giving a probability of $\frac{20}{120}=6$.

Note: there is also a clever approach here. Recall the problem of arranging the letters ABCDEFG, if say the E must come before the F. But you can match each arrangement with E before F with another one with F before E. (For example, match BCDEFGA with BCDFEGA.) So exactly half of the arrangements have E before F, I.e. $\frac{7!}{2}$ arrangements.

Here, the idea is that we can group any configuration of the odd numbers among each other. Take for example 65879. We can match it with:
- 65897
- 67859
- 67895
- 69875
- 69857

So each number with 5,7,9 in order can be matched with exactly five other numbers, without changing the position of the 6 and the 8, but worn 5,7,9 not in order. Hence only one sixth of arrangements have 5,7,9 in order.

Nevertheless, that was a nice idea with the stars and bars like idea. Just executed with a flaw!
« Last Edit: September 13, 2019, 08:44:50 pm by RuiAce »  