Author Topic: 3U Maths Question Thread (Read 1234678 times) Tweet Share

Sine · « **Reply #3135 on:** December 20, 2017, 09:03:16 pm »

Quote from: Dragomistress on December 20, 2017, 08:51:01 pm

May someone show how to do this question since Math in Focus doesn't appear to show this.

integrate the equation this will give you an equation of y in terms of x + c
this constant is unknown but we do know the equation passes through (2,-1) so we sub that in to find c
Once we find c we can express as the equation of the curve with no unknown constants.

Opengangs · « **Reply #3136 on:** December 20, 2017, 10:17:07 pm »

Quote from: Dragomistress on December 20, 2017, 08:51:01 pm

May someone show how to do this question since Math in Focus doesn't appear to show this.

$\frac{dy}{dx} = \frac{x}{(x^2 - 1)^2}$ $u = x^2 - 1$ $du = 2xdx$ $\frac{du}{2} = xdx$ $y = \frac{1}{2}\int \frac{du}{u^2}$ $y = -\frac{1}{2}\left(\frac{1}{u}\right) + C$ $y = -\frac{1}{2}\left(\frac{1}{(x^2 - 1}\right) + C$ $-1 = -\frac{1}{2}\left(\frac{1}{3}\right) + C$ $-1 + \frac{1}{6} = C$ $C = -\frac{5}{6}$
$\therefore y = -\left(\frac{1}{2(x^2 - 1)}\right) - \frac{5}{6}$

Dragomistress · « **Reply #3137 on:** December 22, 2017, 08:55:47 pm »

For integration where there are multiple applications of a method such as use the Simpson's rule x times. Should I memorise the integration formulas for trapezoidal and Simpson's rule or are they usually sane number of uses such as 2 or 3 which allows me to not need to memorise those formulas.

RuiAce · « **Reply #3138 on:** December 22, 2017, 09:12:17 pm »

Quote from: Dragomistress on December 22, 2017, 08:55:47 pm

For integration where there are multiple applications of a method such as use the Simpson's rule x times. Should I memorise the integration formulas for trapezoidal and Simpson's rule or are they usually sane number of uses such as 2 or 3 which allows me to not need to memorise those formulas.

There is usually a sane amount, but it's up to you if you want to memorise the generalised version. It's a lot faster to use, just more effort.

silverplusa · « **Reply #3139 on:** January 02, 2018, 09:53:03 pm »

Hi, just really couldn't get my head around this question

It's from cambridge 3u on circle geometry - alternate segment theorem

RuiAce · « **Reply #3140 on:** January 02, 2018, 10:34:22 pm »

Quote from: silverplusa on January 02, 2018, 09:53:03 pm

Hi, just really couldn't get my head around this question
It's from cambridge 3u on circle geometry - alternate segment theorem

(Image removed from quote.)

$\text{I found a proof but I'm not keen on its elegance factor.}\\ \text{For this proof, draw in the tangent at }S\\ \text{Some reference points have been added in as shown.}$

This type of proof is colloquially known as 'angle chasing'
$\text{Let:}\\ \begin{align*}\angle XSM &= \theta\\ \angle MST &= \alpha\\ \angle TSN&=\beta\\ \angle NSY &= \phi\end{align*}\\ \text{Our objective is to prove that}\\ \boxed{\alpha=\beta}$
_________________________________________________________
$\text{In the big circle, using the tangent drawn at }S,\\ \text{the alternate segment theorem tells us that}\\ \begin{align*}\angle SBT &=\theta\\ \angle SAT &= \phi\end{align*}$
$\text{In the small circle, using the tangent drawn at }S,\\ \text{the alternate segment theorem tells us that}\\ \begin{align*}\angle STM &= \theta\\ \angle STN &= \phi\end{align*}$
$\text{In the small circle, using the tangent given at }T,\\ \text{the alternate segment tells us that}\\ \begin{align*}\angle MTA &= \alpha\\ \angle NTB&=\beta\end{align*}$
_________________________________________________________
$\text{The angle sum of }\triangle SAT\text{ is now}\\ \alpha+(\theta+\alpha)+\phi = 180^\circ$
$\text{The angle sum of }\triangle SBT\text{ is now}\\ \beta + (\phi+\beta) + \theta = 180^\circ$
$\text{Hence by inspection/comparison/simultaneous equations}\\ \text{it follows that }\alpha=\beta$

radnan11 · « **Reply #3141 on:** January 05, 2018, 12:45:03 pm »

How do u do part b, tried 2 different ways got two different answers that were wrong

brightsky · « **Reply #3142 on:** January 05, 2018, 01:14:54 pm »

Let a = arctan(x) and b = arctan(2x).

tan(a+b) = [tan(a) + tan(b)]/[1-tan(a)tan(b)] = (x + 2x)/(1-2x^2) = 3x/(1-2x^2)

We are required to solve the equation a + b = arctan(3).

a + b = arctan(3)
tan(a+b) = 3

3x/(1-2x^2) = 3
x/(1-2x^2) = 1
x = 1-2x^2
2x^2 + x - 1 = 0
(2x - 1)(x + 1) = 0
x = 1/2 or x = -1

Substituting each of these solutions into the original equation, you'd find that only x = 1/2 actually satisfies the equation. Hence, the answer to part b is x = 1/2.

RuiAce · « **Reply #3143 on:** January 05, 2018, 10:44:25 pm »

Quote from: brightsky on January 05, 2018, 01:14:54 pm

Let a = arctan(x) and b = arctan(2x).

Keep in mind that in the HSC, they are not introduced to the notation of "arcsin, arctan" and are only expected to know the sin^-1 and tan^-1 notations.

Method should be correct of course

Lefkiiii6 · « **Reply #3144 on:** January 06, 2018, 11:53:56 am »

Hi I have a question. The number of arrangements of 2n+2 different objects taken n at a time is to the number of arrangements of 2n different objects taken n at a time as 14:5. Find the value of n. Could someone please assist? Thankyou!

RuiAce · « **Reply #3145 on:** January 06, 2018, 11:57:04 am »

Quote from: Lefkiiii6 on January 06, 2018, 11:53:56 am

Hi I have a question. The number of arrangements of 2n+2 different objects taken n at a time is to the number of arrangements of 2n different objects taken n at a time as 14:5. Find the value of n. Could someone please assist? Thankyou!

This question lacks clarity.
- Do we assume a straight-line arrangement?
- Is the question suggesting to select $n$ objects before arranging them?

(Alternatively, if you are unsure, please provide the source of the question and/or the given answer.)

Lefkiiii6 · « **Reply #3146 on:** January 06, 2018, 12:07:45 pm »

Quote from: RuiAce on January 06, 2018, 11:57:04 am

This question lacks clarity.
- Do we assume a straight-line arrangement?
- Is the question suggesting to select $n$ objects before arranging them?

(Alternatively, if you are unsure, please provide the source of the question and/or the given answer.)

I took the question straight from Fitzpatrick 3U and the answer is 3.

RuiAce · « **Reply #3147 on:** January 06, 2018, 12:37:59 pm »

$\text{Backtracking from the given answer,}\\ \text{it would appear as though a straight line arrangement was necessary.}$
If this is the old Fitzpatrick textbook then, well trust it to omit important details.
$\text{There are two ways of starting the question,}\\ \text{but they coincide at the end.}$
$\text{Method 1: This is the standard scenario of picking }n\text{ objects}\\ \text{out of }2n+2\text{, to arrange in a line.}$
An example would be arranging 5 people in a line out of 5.
$\text{By formula, this is just }^{2n+2}P_n.$
$\text{Similarly, if we only have }n\text{ people,}\\ \text{then we have }^{2n}P_n\text{ arrangements instead.}$

Method 2

$\text{Alternatively, we first choose }\textbf{as an unordered selection}\\ n\text{ people out of }2n+2: \binom{2n+2}{2}\text{ arrangements.}$
$\text{After choosing them without any ordering, we assign the ordering manually}\\ \text{in the same way we arrange }n\text{ objects in a straight line.}\\ \text{This can be done in }n!\text{ ways.}$
$\text{So when we deal with }n\text{ out of }2n+2\text{ people, we have }\binom{2n+2}{n} n!\text{ possible outcomes.}$
$\text{Similarly, when we do this with }n\text{ out of only }2n\text{ people},\\ \text{we have }\binom{2n}{n}n!\text{ possible outcomes.}$
Note: $ \binom{N}{K} = \frac{N!}{K!(N-K)!} $

$\text{Using method 1, and noting that }^NP_K = \frac{n!}{(n-k)!},\\ \text{we essentially wish to solve }\frac{^{2n+2}P_n}{^{2n}P_n}=\frac{14}{5}$
$\begin{align*}\frac{^{2n+2}P_n}{^{2n}P_n}&=\frac{14}{5}\\ 5 ^{2n+2}P_n &= 14 ^{2n}P_n\\ 5 \times \frac{(2n+2)!}{(n+2)!} &= 14 \times \frac{(2n)!}{n!}\end{align*}$
$\text{Now, observe that}\\ \begin{align*}\frac{(2n)!}{n!}&= \frac{2n (2n-1)(2n-2)\dots (n+1)(n)(n-1)\dots (2)(1)}{n(n-1)\dots(2)(1)}\\ &= 2n(2n-1)\dots(n+1)\end{align*}\\ \text{and similarly,}\\ \frac{(2n+2)!}{(n+2)!}= (2n+2)(2n+1)\dots (n+3)$
$\text{So our equation becomes}\\ 5\times (2n+2)\dots (n+3) = 14\times (2n)\dots(n+1)\\ \text{Cancelling out a lot of terms reduces this down to}\\ 5(2n+2)(2n+1) = 14(n+2)(n+1)$
$\text{This is now just a quadratic equation.}\\ \begin{align*}20n^2+30n+10&= 14n^2+42n+28\\ 6n^2-12n-18&=0\\ 6(n+1)(n-3)&=0\\ n=-1, 3\end{align*}$
$\text{Since }n\ge 0,\text{ our solution is }n=3.$

Lefkiiii6 · « **Reply #3148 on:** January 06, 2018, 04:18:37 pm »

Thanks a lot! I have some other probability questions.

1) On the average, a typist has to correct one word in 800 words. Assuming that a page contains 200 words, find the probability of more than one correction per page. Answer: 0.0625

2) If ten coins are tossed 50 times, in how many cases would we expect the number of head to exceed the number of tails. Answer: 19

3) A manufacture of razor blades find that on average 1 blade in every 20 is faulty. The razor blades are marketed in packets of 5. Out of a batch of 200 packets, how many would be expected to have 2 faulty blades? Answer: 4

These are all from Fitzpatrick.

RuiAce · « **Reply #3149 on:** January 06, 2018, 05:26:46 pm »

Quote from: Lefkiiii6 on January 06, 2018, 04:18:37 pm

Thanks a lot! I have some other probability questions.

1) On the average, a typist has to correct one word in 800 words. Assuming that a page contains 200 words, find the probability of more than one correction per page. Answer: 0.0625

I checked with MLov on this question. He took out the textbook and saw that the answer was 0.0264 which was what both of us were getting. Just double check that you read off the correct answers.
$\text{As for the method, this is just a straightforward application}\\ \text{of binomial probability.}$
As a matter of fact, all of these are related to binomial probability. If you haven't been taught it before, you will not recognise the formula. You may use this post to learn about it.
$\text{For this question, we assume that the probability any ONE word needs correction}\\ \text{is }\frac{1}{800},\text{ and in total we have 200 words.}$
$\text{Seeing as though we are considering the probability of MORE than one correction}\\ \text{the computations are faster when we consider the complement,}\\ \text{i.e. the probability of exactly one or zero corrections.}$
$\text{Using the complement, our desired probability is}\\ 1 - \binom{200}{0}\left(\frac{799}{800}\right) - \binom{200}{1} \left(\frac{1}{800}\right)\left(\frac{799}{800}\right)^{199} = 0.0264$
Now, regarding Q2 and Q3. I managed to get them out but I had to step out of HSC zone. I honestly have no clue at all why the old Fitzpatrick book did that, but they included the concept of expectation which is not taught in the HSC due to the fact we don't cover statistics in our course. If you want to see the solution, I'm happy to provide it anyway, but I will confidently say that it's not actually HSC material.

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