Author Topic: Mathematics Question Thread (Read 1308526 times) Tweet Share

Opengangs · « **Reply #2835 on:** October 19, 2017, 07:00:37 pm »

Quote from: pokemonlv10 on October 19, 2017, 06:38:27 pm

Is there a chance of them asking to integrate absolute values in the exam (Not in MC)? Because i would like to know the algebraic method

We define the absolute value to be x + 1 for x >= - 1 and -(x+1) for x < -1
Perhaps, this helps

asahyoun3 · « **Reply #2836 on:** October 19, 2017, 07:02:20 pm »

Quote from: pokemonlv10 on October 19, 2017, 06:38:27 pm

Is there a chance of them asking to integrate absolute values in the exam (Not in MC)? Because i would like to know the algebraic method

Yes they could, most likely as a question 11 or 12

I have sent the solution to the question you asked, solving it algebraiclly

winstondarmawan · « **Reply #2837 on:** October 19, 2017, 07:10:06 pm »

Quote from: georgiia on October 19, 2017, 05:33:13 pm

I usually store it in the calculator because when you need to find the time later then you sub the calc value in and gove the time to the nearest second, min, hour whatever is asked.
Hope that helps

This is what I usually always do, but in the answers they used the rounded off value.

RuiAce · « **Reply #2838 on:** October 19, 2017, 07:16:25 pm »

Quote from: Mymy409 on October 19, 2017, 05:40:08 pm

A little help with part 2 of this, please. Thanks in advance.

Sorry, the part that's cut off says 'Claire's path' and the corner that's cut off is labelled F and is the farmhouse.

I will get back to this one. This was the dreaded 2001 question that was so hard, apparently the committee got fired for it.

This will take considerable time to explain. I also want to type up a solution for the first part as well, so that it can be added to the compilation in entirety.

Quote from: asahyoun3 on October 19, 2017, 06:19:35 pm

For this question, it is best to solve it graphically, as its more likely to make mistakes doing it algebraically

"Best" is debatable - I've seen students accidentally sketch these wrong. "Most efficient" - yes.

Quote from: pokemonlv10 on October 19, 2017, 06:38:27 pm

Is there a chance of them asking to integrate absolute values in the exam (Not in MC)? Because i would like to know the algebraic method

As Opengangs said, you must take cases.
$|x+1| = \begin{cases}x+1 & \text{if }x\ge -1\\ -(x+1) &\text{if }x < -1\end{cases}$
$\text{So you would consider}\\ \begin{align*}\int_{-3}^2 |x+1|\,dx &= \int_{-3}^{-1} |x+1|\,dx + \int_{-1}^2|x+1|\,dx\\ &= -\int_{-3}^{-1} (x+1)\,dx + \int_{-1}^2 (x+1)\,dx\end{align*}$

Quote from: winstondarmawan on October 19, 2017, 07:10:06 pm

This is what I usually always do, but in the answers they used the rounded off value.

This is my recommendation.

You are safe to use an unrounded value if the question actually told you to compute an approximate value, like the one you've shown. Since they've stated that you must show $k \approx 0.0347 $, you can keep using it if you wish.

If they just ask you to bluntly find the value of k, as opposed to showing an approximation holds, then you should use the exact value.

Note, however, that the one you've shown is the ONLY time I've seen a rounded value get used without problems.

gilliesb18 · « **Reply #2839 on:** October 19, 2017, 07:31:34 pm »

Could someone help me with the following question? Any help whatsoever is appreciated!!
If R is the fixed point(2,3) and P is a movable point(x,y), find the equation of the locus of P if the distance PR is twice the distance from P to the line y=
-1.
Thanks heaps

RuiAce · « **Reply #2840 on:** October 19, 2017, 07:36:12 pm »

Quote from: gilliesb18 on October 19, 2017, 07:31:34 pm

Could someone help me with the following question? Any help whatsoever is appreciated!!
If R is the fixed point(2,3) and P is a movable point(x,y), find the equation of the locus of P if the distance PR is twice the distance from P to the line y=
-1.
Thanks heaps

$\text{Taking }S\text{ to be the point on the line }y=-1\\ \text{the point }S\text{ must have coordinates }(x,-1).$
$\text{Since we require }PR = 2PS\text{ we have}\\ \begin{align*}\sqrt{(x-2)^2+(y-3)^2}&=2\sqrt{(x-x)^2 + (y+1)^2}\\ (x-2)^2 + (y-3)^2 &= 4(y+1)^2\\ (x-2)^2+y^2-6y+9&=4y^2+8y+1\\ (x-2)^2 &= 3y^2+14y-8\end{align*}\\ \text{You can rearrange this however you wish.}$
Aside: It can actually be shown that this will be a hyperbola, but oriented differently to the usual $ y=\frac1x $

hansolo9 · « **Reply #2841 on:** October 19, 2017, 07:39:07 pm »

Need help with this one, thank you

Sukakadonkadonk · « **Reply #2842 on:** October 19, 2017, 07:51:18 pm »

Curious, I've been doing trial 2U past papers from 'good' schools like selective ones and what not, is that a good idea?
Or should I do more past HSC papers?

RuiAce · « **Reply #2843 on:** October 19, 2017, 07:53:34 pm »

Quote from: hansolo9 on October 19, 2017, 07:39:07 pm

Need help with this one, thank you

$\text{Since }\lim_{x\to 0}\frac{\sin x}{x} = 1\\ \text{by using the substitution }x=a\theta\text{ we arrive at}\\ \boxed{\lim_{\theta \to 0}\frac{\sin a\theta}{a\theta} = 1}$
for some constant $ a$.
$\text{Let }a = \frac14\text{. Then, }\boxed{\lim_{\theta \to 0}\frac{\sin \frac\theta4}{\frac\theta4} = 1}\\ \text{So multiplying }\frac{1}{4}\text{ on both the left and the right,}\\ \boxed{\lim_{\theta \to 0}\frac{\sin \frac{\theta}4}{\theta}=\frac14}$

RuiAce · « **Reply #2844 on:** October 19, 2017, 07:55:13 pm »

Quote from: Sukakadonkadonk on October 19, 2017, 07:51:18 pm

Curious, I've been doing trial 2U past papers from 'good' schools like selective ones and what not, is that a good idea?
Or should I do more past HSC papers?

You should be the judge of what papers you are doing.

Personally, however, I never bothered with any trial papers during the lead-up to the HSC and focused exclusively on past HSC papers. I also feel that some "good" selective schools don't have the "best" selective papers either; i felt many of James Ruse's papers were easier than Sydney Grammar's by far.

Sukakadonkadonk · « **Reply #2845 on:** October 19, 2017, 07:59:00 pm »

Quote from: RuiAce on October 19, 2017, 07:55:13 pm

You should be the judge of what papers you are doing.

Personally, however, I never bothered with any trial papers during the lead-up to the HSC and focused exclusively on past HSC papers. I also feel that some "good" selective schools don't have the "best" selective papers either; i felt many of James Ruse's papers were easier than Sydney Grammar's by far.

Thats very true. I will probably focus on HSC past papers then. Or maybe do some Sydney Grammar??

RuiAce · « **Reply #2846 on:** October 19, 2017, 08:24:26 pm »

Quote from: Mymy409 on October 19, 2017, 05:40:08 pm

A little help with part 2 of this, please. Thanks in advance.

Sorry, the part that's cut off says 'Claire's path' and the corner that's cut off is labelled F and is the farmhouse.

$\text{For the distance }FP\text{, using trigonometry in }\triangle FPG\\ \cos\theta = \frac{FG}{FP} \implies \boxed{FP = 250\sec \theta}$
$\text{Using }\text{speed}=\frac{\text{distance}}{\text{time}}\iff \text{time} = \frac{\text{distance}}{\text{speed}}\text{, where Claire's speed is 4 m s}^{-1}\\ \text{the time Claire needs to get from }F\text{ to }P\text{ will be }\frac{125}{2}\sec \theta\text{ seconds.}$
$\text{Using trigonometry again, }\boxed{GP = 250\tan\theta}\\ \text{so we have the distance }DP = 2000 + 250\tan \theta\\ \text{And again using that formula, with the bus's speed of 15 m s}^{-1}\\ \text{the time the bus takes to get from }D\text{ to }P\text{ will be }\frac{400}{3} + \frac{50}{3}\tan\theta\text{ seconds}$
$\text{However, for some 'niceness', we will not use this.}\\ \text{To make the next question more manageable, we convert the times to }\textbf{minutes}\\ \text{The time taken in minutes for Claire is }\boxed{\frac{25}{24}\sec \theta}\\ \text{and the time taken in minutes for the bus is }\boxed{\frac{5}{18}(8+\tan \theta)}$
____________________________________________________________________

Lemma: The derivative of sec

\begin{align*}\frac{d}{dx}\sec x &= \frac{d}{dx} (\cos x)^{-1}\\ &= - (-\sin x)(\cos x)^{-2}\tag{chain rule}\\ &= \sec x \tan x\tag{using some identities}\end{align*}

$\text{Now, }T_2\text{ really measures the time taken for the bus to reach }P\\ \textbf{but from 8am.}$
$\text{If the arrival time is, say, }t\text{ minutes past 8}\\ \text{then Claire needs to arrive at }P\text{ before }t\text{ minutes past 8 as well}\\ \text{Note, however, that this will directly impact }\textbf{the time Claire must leave the house}.$
(Example: if t = 24, then the bus arrives at 8:24 AM. So Claire must be there by 8:24AM as well.)
$\text{The bus leaves the depot at 8:00 AM}\\ \text{The bus arrives }\frac{5}{18}(8+\tan\theta)\text{ minutes after 8 AM}\\ \text{So Claire must arrive by }\frac{5}{18}(8+\tan\theta)\text{ minutes after 8 AM.}$
$\text{But it takes Claire }\frac{25}{24}\sec \theta\text{ minutes to get there!}\\ \text{Therefore, working backwards, Claire must've left the house}\\ \text{within }\boxed{\frac{5}{18}(8+\tan\theta) - \frac{25}{24}\sec \theta}\text{ minutes past 8 AM}\\ \text{This is the expression we must work with.}$
__________________________________
$\text{Let }T =\frac{5}{18}(8+\tan\theta) - \frac{25}{24}\sec \theta\\ \text{Again, this reflects the time, }\textbf{after}\text{ 8AM, in which Claire can leave.}$
$\text{We want the }\textbf{latest}\text{ time.}\\ \text{This means we want }T\text{ to be }\textbf{as large}\text{ as possible.}\\ \text{It turns out that we have a maxima problem now.}$
__________________________________
$\text{Differentiating, with the aid of the lemma,}\\ \frac{dT}{d\theta} = \frac{5}{18}\sec^2\theta - \frac{25}{24}\sec\theta\tan\theta$
$\text{We set this to 0 to find the max:}\\ \begin{align*}\frac{5}{18}\sec^2\theta - \frac{25}{24}\sec \theta \tan \theta&=0\\ \frac{5}{18}-\frac{25}{24}\sin\theta &= 0\\ \sin \theta &= \frac{4}{15}\\ \theta &= \boxed{\sin^{-1}\frac{4}{15}}\end{align*}$
$\text{We test a bit to each side to ensure this is a max}\\ \text{When }\theta = 0.26, \frac{dT}{d\theta} = +0.01\dots > 0\\ \text{When }\theta = 0.28, \frac{dT}{d\theta} = -0.10\dots < 0\\ \text{So indeed we have a max.}$
__________________________________
$\text{So subbing back in, the maximum value is approximately}\\ T \approx 1.21827553\dots\text{ minutes}\\ \text{Using the degrees-minutes-second button on the calculator, this is approximately}\\ \boxed{\text{1 minute 13 seconds}}$
$\text{Hence the latest time is approximately }\boxed{8:01:13\text{ AM}}$

julia_warren13 · « **Reply #2847 on:** October 19, 2017, 08:25:29 pm »

Could someone please explain how to do part ii) of this question (it's from the 2009 HSC paper)
Thank you

RuiAce · « **Reply #2848 on:** October 19, 2017, 08:29:27 pm »

Quote from: julia_warren13 on October 19, 2017, 08:25:29 pm

Could someone please explain how to do part ii) of this question (it's from the 2009 HSC paper)
Thank you

$\text{Low tide will just be }1-0.7 = 0.3$
This is because if we try to minimise $\sin ax$, the minimum value will be -1. This can be proven via calculus, but is more easily deduced just because the range of $ \sin ax $ is $ -1 \le y \le 1$.
$\text{So we want to solve }h=0.3\\ \text{This could be solved more quickly, but I will take the normal route.}$
$\text{Since we have }0\le t \le 12\text{, our domain of interest is }\boxed{0 \le \frac{\pi t}{6} \le 2\pi}\\ \text{We will solve the equation over this domain.}$ \begin{align*}1+0.7\sin \frac{\pi t}{6}&=0.3\\ \sin \frac{\pi t}{6}&=-0.7\\ \frac{\pi t}{6} &=\frac{3\pi}{2}\\ t&= 9\end{align*}
$\text{So since the initial time is 5 am}\\ \text{9 hours later we will be at 2 pm.}$

Mymy409 · « **Reply #2849 on:** October 19, 2017, 08:31:37 pm »

Quote from: RuiAce on October 19, 2017, 08:24:26 pm

$\text{For the distance }FP\text{, using trigonometry in }\triangle FPG\\ \cos\theta = \frac{FG}{FP} \implies \boxed{FP = 250\sec \theta}$
$\text{Using }\text{speed}=\frac{\text{distance}}{\text{time}}\iff \text{time} = \frac{\text{distance}}{\text{speed}}\text{, where Claire's speed is 4 m s}^{-1}\\ \text{the time Claire needs to get from }F\text{ to }P\text{ will be }\frac{125}{2}\sec \theta\text{ seconds.}$
$\text{Using trigonometry again, }\boxed{GP = 250\tan\theta}\\ \text{so we have the distance }DP = 2000 + 250\tan \theta\\ \text{And again using that formula, with the bus's speed of 15 m s}^{-1}\\ \text{the time the bus takes to get from }D\text{ to }P\text{ will be }\frac{400}{3} + \frac{50}{3}\tan\theta\text{ seconds}$
$\text{However, for some 'niceness', we will not use this.}\\ \text{To make the next question more manageable, we convert the times to }\textbf{minutes}\\ \text{The time taken in minutes for Claire is }\boxed{\frac{25}{24}\sec \theta}\\ \text{and the time taken in minutes for the bus is }\boxed{\frac{5}{18}(8+\tan \theta)}$
____________________________________________________________________

Lemma: The derivative of sec
\begin{align*}\frac{d}{dx}\sec x &= \frac{d}{dx} (\cos x)^{-1}\\ &= - (-\sin x)(\cos x)^{-2}\tag{chain rule}\\ &= \sec x \tan x\tag{using some identities}\end{align*}
$\text{Now, }T_2\text{ really measures the time taken for the bus to reach }P\\ \textbf{but from 8am.}$
$\text{If the arrival time is, say, }t\text{ minutes past 8}\\ \text{then Claire needs to arrive at }P\text{ before }t\text{ minutes past 8 as well}\\ \text{Note, however, that this will directly impact }\textbf{the time Claire must leave the house}.$
(Example: if t = 24, then the bus arrives at 8:24 AM. So Claire must be there by 8:24AM as well.)
$\text{The bus leaves the depot at 8:00 AM}\\ \text{The bus arrives }\frac{5}{18}(8+\tan\theta)\text{ minutes after 8 AM}\\ \text{So Claire must arrive by }\frac{5}{18}(8+\tan\theta)\text{ minutes after 8 AM.}$
$\text{But it takes Claire }\frac{25}{24}\sec \theta\text{ minutes to get there!}\\ \text{Therefore, working backwards, Claire must've left the house}\\ \text{within }\boxed{\frac{5}{18}(8+\tan\theta) - \frac{25}{24}\sec \theta}\text{ minutes past 8 AM}\\ \text{This is the expression we must work with.}$
__________________________________
$\text{Let }T =\frac{5}{18}(8+\tan\theta) - \frac{25}{24}\sec \theta\\ \text{Again, this reflects the time, }\textbf{after}\text{ 8AM, in which Claire can leave.}$
$\text{We want the }\textbf{latest}\text{ time.}\\ \text{This means we want }T\text{ to be }\textbf{as large}\text{ as possible.}\\ \text{It turns out that we have a maxima problem now.}$
__________________________________
$\text{Differentiating, with the aid of the lemma,}\\ \frac{dT}{d\theta} = \frac{5}{18}\sec^2\theta - \frac{25}{24}\sec\theta\tan\theta$
$\text{We set this to 0 to find the max:}\\ \begin{align*}\frac{5}{18}\sec^2\theta - \frac{25}{24}\sec \theta \tan \theta&=0\\ \frac{5}{18}-\frac{25}{24}\sin\theta &= 0\\ \sin \theta &= \frac{4}{15}\\ \theta &= \boxed{\sin^{-1}\frac{4}{15}}\end{align*}$
$\text{We test a bit to each side to ensure this is a max}\\ \text{When }\theta = 0.26, \frac{dT}{d\theta} = +0.01\dots > 0\\ \text{When }\theta = 0.28, \frac{dT}{d\theta} = -0.10\dots < 0\\ \text{So indeed we have a max.}$
__________________________________
$\text{So subbing back in, the maximum value is approximately}\\ T \approx 1.21827553\dots\text{ minutes}\\ \text{Using the degrees-minutes-second button on the calculator, this is approximately}\\ \boxed{\text{1 minute 13 seconds}}$
$\text{Hence the latest time is approximately }\boxed{8:01:13\text{ AM}}$

Thanks very much, Rui

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