Author Topic: Mathematics Question Thread (Read 1626096 times) Tweet Share

winstondarmawan · « **Reply #2925 on:** October 21, 2017, 02:29:18 pm »

Hello!
Can someone please explain why for this question:
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22689757_1356165251175661_1010246494_o.png?oh=9c15550ed02a06f9bd76c168e6a75e59&oe=59ECDE84
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22685022_1356184464507073_13089599_n.png?oh=04eab45a00570206fe870ce096865e5c&oe=59EDC439
Why the graph tends towards zero?
TIA.

RuiAce · « **Reply #2926 on:** October 21, 2017, 02:38:08 pm »

Quote from: winstondarmawan on October 21, 2017, 02:29:18 pm

Hello!
Can someone please explain why for this question:
Q: https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22689757_1356165251175661_1010246494_o.png?oh=9c15550ed02a06f9bd76c168e6a75e59&oe=59ECDE84
A: https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/22685022_1356184464507073_13089599_n.png?oh=04eab45a00570206fe870ce096865e5c&oe=59EDC439
Why the graph tends towards zero?
TIA.

Because think about what the derivative is and what happens to the original graph as well.

As the original graph flattens out and tends towards its horizontal asymptote, the gradient of the tangent to the curve is tending towards 0.
This is what the derivative is; the gradient of the tangent. So on the sketch of the derivative, it must tend to 0.

winstondarmawan · « **Reply #2927 on:** October 21, 2017, 02:46:22 pm »

For this question:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22712062_1356191744506345_1426088660_o.png?oh=a041315ba9f538f3e83a8837421517c6&oe=59ECE7F5
Part (iii).
Would it be enough for 2 marks to state that for t>0, v>0 and a>0 (after finding the acceleration).

RuiAce · « **Reply #2928 on:** October 21, 2017, 02:53:20 pm »

Quote from: winstondarmawan on October 21, 2017, 02:46:22 pm

For this question:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22712062_1356191744506345_1426088660_o.png?oh=a041315ba9f538f3e83a8837421517c6&oe=59ECE7F5
Part (iii).
Would it be enough for 2 marks to state that for t>0, v>0 and a>0 (after finding the acceleration).

Yeah.

Although you probably should explain why for v. It's obvious that a>0 but not so obvious that v>0.

RuiAce · « **Reply #2929 on:** October 21, 2017, 03:25:24 pm »

Quote from: sidzeman on October 21, 2017, 12:09:26 pm

^expanding on that, will we have marks deducted for sketching lets say number plane axis freehand rather than with a ruler?

Also, could someone help me with part iv of this question please, as well as part ii from b

$\text{Let the common ratio be }r\\ \text{If 8 is the first term of the series, then we have}\\ \begin{align*}T_2 = 8r &= p\\ T_3 = 8r^2 &= q\\ T_4 = 8r^3 &= 27\end{align*}$
$\text{The last equation solves to give }r=\frac32\\ \text{You can now use that to find }p\text{ and }q$
_________________________
$\text{The probability is just }\frac{13}{52}\times \frac{13}{51}\times \frac{13}{50}\times \frac{13}{49}\\ \text{because there are 13 in each suit, but of course once}\\<br />\text{we've drawn a card, it is gone.}$

RuiAce · « **Reply #2930 on:** October 21, 2017, 03:31:03 pm »

Quote from: K9810 on October 21, 2017, 11:24:31 am

Can you please help me with part iii) and iv)

Guidelines:

The point of inflexion at A (on the velocity graph) isn't too useful. Because there is a local maximum at B, the displacement will increase (at an increasing rate) until it reaches t=4. Then, it will start increasing at a decreasing rate.
(Effectively, for the displacement graph, t=4 will "almost" exhibit the properties of a point of inflexion)

Until you reach t=5, where the particle is at rest. Then, the displacement graph starts turning around (i.e. decreases) until you reach t=6.
(It may be worth mentioning that between the times t=4 and t=6, the graph will appear to be quite parabolic.)

And then at t=6, because the velocity is now constant, the displacement is now linear (straight line). So the displacement decreases in the same way a straight line with gradient m=-5 does.

RuiAce · « **Reply #2931 on:** October 21, 2017, 03:54:42 pm »

Quote from: bsdfjnlkasn on October 21, 2017, 12:22:27 pm

Hey there!

I've got a bunch of questions from the 2001 but i'll stagger them as I post

Thanks so much!

These are the specifics for the questions I need help with:

Finance - ii and iii

Geometry - ii and iii

$\text{Using the same old method we arrive at}\\ B_n = 1000(1.06)^n - 72(1+1.06+\dots +1.06^{n-1})\\ \text{so by considering the geometric sum}\\B_n = 1000(1.06)^n - 72\left(\frac{1.06^n-1}{0.06}\right)$
$\text{From here, we just plug into the calculator.}\\ \begin{align*}B_n&=1000(1.06)^n - \frac{72}{0.06}\left(1.06^n-1\right)\\ &= 1000(1.06)^n - 1200\left(1.06^n - 1\right)\\ &= 1000(1.06)^n - 1200(1.06)^n + 1200\\ &= 1200 - 200(1.06)^n\end{align*}$
____________________________
$\text{After 10 years, we have}\\ B_{10} = 1200-200(1.06)^{10}=841.83046\dots$
You really shouldn't be doing this, but for convenience in typing I will round to 841.83.
$\text{At this point, we }\textbf{reset the counter}.\\ \text{Let }A_n\text{ be the amount in the investment, }\textbf{but starting from 10 years in.}\\ \text{So }A_0 = 841.83$
$\text{Doing the same old thing we now have}\\ \begin{align*}A_1&=841.63(1.06) - 90\\ A_2&=841.63(1.06) - 90(1.06)-90 & \vdots\end{align*}$
$\text{...and eventually }A_n = 841.63(1.06)^n - 90\left(\frac{1.06^n-1}{0.06}\right)$
$\text{We want to solve }A_n = 0.\text{ So}\\ \begin{align*}841.63(1.06)^n - 1500(1.06^n - 1)&=0\\ 658.37(1.06)^n &= 1500\\ n \ln \frac{1500}{658.37}\end{align*}$

~BK~ · « **Reply #2932 on:** October 21, 2017, 04:14:12 pm »

hey RuiAce (or anyone for that matter

)
i would really appreciate some help on the attached question...
i always struggle on these sorta ones and they are worth 6 marks (at least)
TIA

RuiAce · « **Reply #2933 on:** October 21, 2017, 04:32:22 pm »

Quote from: ~BK~ on October 21, 2017, 04:14:12 pm

hey RuiAce (or anyone for that matter )
i would really appreciate some help on the attached question...
i always struggle on these sorta ones and they are worth 6 marks (at least)
TIA

$\text{Since our compounding frequency is monthly}\\ \text{6 percent per annum is converted to 0.5 percent per month.}$
$\text{Follow the usual recursive approach}\\ \begin{align*}A_0 &= 360000\\ A_1 &= A_0(1.005) - M\\ &= 360000(1.005) - M\\ A_2 &= A_1(1.005) - M\\ &= \left[360000(1.005) - M\right](1.005)-M\\ &= 360000(1.005)^2 - M(1+1.005)\end{align*}$
although since this part said "write down" you didn't need to show working if you didn't want to.
______________________________________
$\text{As }A_3 = A_2(1.005) - M\\ \text{it can similarly be shown that }\\ A_3 = 360000(1.005)^3-M(1+1.005+1.005^2)$
$\text{So in general}\\ A_n = 360000(1.005)^n - M(1+1.005+\dots+1.005^{n-1})\\ \text{and so summing the G.P.}\\ A_n = 360000(1.005)^n - M\left(\frac{1.005^n - 1}{0.005}\right)$
$\text{We require }A_{300} = 0\text{, so we end up with}\\ 360000(1.005)^{300} = M\left(\frac{1.005^{300}-1}{0.005}\right)\\ \text{Plugging into the calculator, }M\approx 2319.50$
______________________________________
$\text{Here, we require }A_n < 180000\text{ so:}\\ \begin{align*}360000(1.005)^{n} - M\left(\frac{1.005^{n}-1}{0.005}\right) &<180000\\ 360000(1.005)^n - \frac{2319.50}{0.005}(1.005^n-1) &< 180000\\ 360000(1.005)^n - 463900(1.005^n -1) &< 180000\\ 360000(1.005)^n - 463900(1.005)^n + 463900 &< 180000\\ 283900&<103900(1.005)^n\\ \ln \frac{283900}{103900} &< n \ln 1.005\\ n& > 201.54\end{align*}\\ \text{so 202 months}$

hansolo9 · « **Reply #2934 on:** October 21, 2017, 04:34:16 pm »

Thank you for the previous answers, I understand now.

I also need help for part 2, 3, and 4. Not that good at these types of questions

~BK~ · « **Reply #2935 on:** October 21, 2017, 04:41:03 pm »

Quote from: RuiAce on October 21, 2017, 04:32:22 pm

$\text{Here, we require }A_n < 180000\text{ so:}\\ \begin{align*}360000(1.005)^{n} - M\left(\frac{1.005^{n}-1}{0.005}\right) &<180000\\ 360000(1.005)^n - \frac{2319.50}{0.005}(1.005^n-1) &< 180000\\ 360000(1.005)^n - 463900(1.005^n -1) &< 180000\\ 360000(1.005)^n - 463900(1.005)^n + 463900 &< 180000\\ 283900&<103900(1.005)^n\\ \ln \frac{283900}{103900} &< n \ln 1.005\\ n& > 201.54\end{align*}\\ \text{so 202 months}$

thanks heaps, clears it up, just wish i could think like that during an exam!!

that last section, as in the bit above, you've done differently to succus one and got the same answer.... personally, your method makes way more sense...
tvm btw- can't believe you're sacrificing your time on the w/e!?

gabi.c · « **Reply #2936 on:** October 21, 2017, 04:49:06 pm »

Do we get marked down for giving more answers than the marking criteria asks for? In the past papers I've done, I keep going past the answer (simplifying more, or factorising)

RuiAce · « **Reply #2937 on:** October 21, 2017, 04:50:57 pm »

Quote from: gabi.c on October 21, 2017, 04:49:06 pm

Do we get marked down for giving more answers than the marking criteria asks for? In the past papers I've done, I keep going past the answer (simplifying more, or factorising)

Not unless you write something wrong.

A whole bunch of correctness - ok. Add a mistake in there - not ok.

RuiAce · « **Reply #2938 on:** October 21, 2017, 04:51:28 pm »

Quote from: hansolo9 on October 21, 2017, 04:34:16 pm

Thank you for the previous answers, I understand now.

I also need help for part 2, 3, and 4. Not that good at these types of questions

$\text{Note that }\angle CDE = \theta\\ \textbf{Not}\text{ the other one.}$
This can be proven. $ \angle ADG = \frac\pi2 - \theta$ from the angle sum of $ \triangle ADG$.
$\text{But we also know that }\angle BDC = \frac\pi4\\ \text{This is because }BD\text{ is a diagonal of the square }ABCD.$
$\text{Hence }\angle BDE = \frac\pi4 + \theta\\ \text{From here, it's just a matter of doing trigonometry in }\triangle BDE\\ \sin \angle BDE=\frac{BE}{BD} \implies\boxed{BE=15\sqrt2\sin \left(\frac\pi4+\theta\right)}$
This part was more or less the hardest part because of that tiny triangle down the corner, which can easily throw you off.
_____________________________
$\text{Since }AD = 15\text{ we can use trigonometry in }\triangle GAD\\ \text{to deduce that }\boxed{GA = 15 \cos \theta}$
$\angle FAG = \pi \text{ (straight angle) and }\angle BAD = \frac\pi2 = \frac\pi2\text{ (angle in a square)}\\ \text{so it follows that }\angle FAB = \frac\pi2 - \theta$
Which, I find funny.
$\text{Also, }BA=15\text{. So using trigonometry in }\triangle FAB\\ FA=\cos \left(\frac\pi2-\theta\right)\\ \text{But we can simplify this using the complementary angle identity to get}\\ \boxed{\sin \theta = FA}$
_____________________________
$\text{Clearly }FBEG\text{ is a rectangle (three right angles)}\\ \text{so }BE = FG$
$\text{Using the previous parts, }FG = 15 \sin \theta + 15 \cos \theta\\ \text{So just subbing in we get}\\ \boxed{15\sin\theta + 15\cos\theta = 15\sqrt{2}\sin \left(\theta + \frac\pi4\right)}\\ \text{Dividing by 15, we are done.}$
I feel as though parts iii and iv were easier than ii. This is an example of a question where it helps to skip previous parts if you are stuck, and then come back to them.

RuiAce · « **Reply #2939 on:** October 21, 2017, 04:52:33 pm »

Quote from: ~BK~ on October 21, 2017, 04:41:03 pm

thanks heaps, clears it up, just wish i could think like that during an exam!!
that last section, as in the bit above, you've done differently to succus one and got the same answer.... personally, your method makes way more sense...
tvm btw- can't believe you're sacrificing your time on the w/e!?

There's multiple methods, but personally I find the recursive one makes more sense all the time. The other method is more or less only if you need it.

Yeah, I knew that these 4 days would be the period where I get the most questions

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