Author Topic: 3U Maths Question Thread (Read 1439993 times) Tweet Share

jamonwindeyer · « **Reply #3075 on:** November 21, 2017, 10:43:07 pm »

Quote from: Dragomistress on November 21, 2017, 09:03:34 pm

Sorry for coming back here so quickly with more questions but this will be the last of them

For your first two questions, you are using the equation for chord of contact on your reference sheet. For 29, for example:

$2x=2(y-3)\\2x-2y+6=0\\x-y+3=0$

To find the coordinates where the tangents meet the parabola, you are finding the points of intersection of the parabola and the chord of contact. So you are solving these simultaneously:

$x-y+3=0 (A)\\x^2=4y (B)$

As for 19, the chord of contact equation is general:

$xx_1=2a(y+y_1)$

We know it passes through the point $(0,2a)$, so:

$2a(2a+y_1)=0\\y_1=-2a$

Now the midpoint formula is for the points R and N:

$x=\frac{x_1+0}{2}=\frac{x_1}{2}\\y=\frac{2a-2a}{2}=0$

The equation is therefore just $y=0$, the x-axis

For Q20, find the equations of the tangents using the formula on your reference sheet (in this question, $a=2$):

$y=-2x-8\\y=5x-50$

Then, consider two equations with parameters instead:

$y=tx-at^2\\y=sx-as^2$

Find the POI of these by equating:

$tx-at^2=sx-as^2\\x(t-s)=a(t-s)(t+s)\\x=a(t+s)\\\therefore y=at(t+s)-at^2=ats$

Now we know that the tangents meet at 45 degrees. The gradients of each tangent are just $s$ and $t$, so we can use the angle between straight lines formula:

$\tan{45}=\frac{t-s}{1+st}\\1+st=t-s\\t(1-s)=s+1\\t=\frac{s+1}{1-s}$

Might be little errors

KT Nyunt · « **Reply #3076 on:** November 24, 2017, 05:30:49 pm »

Hello,

I'm having trouble understanding how the answer (see attached) works for this combinations question:

An examination has 10 multiple-choice questions, each with 4 options. In each question, only one option is correct. For each question a student chooses one option at random.
Write an expression for the probability that the student chooses the correct option for exactly 7 questions.

If you could explain this to me that'd be amazing.
Thanks in advance

RuiAce · « **Reply #3077 on:** November 24, 2017, 05:31:28 pm »

Quote from: KT Nyunt on November 24, 2017, 05:30:49 pm

Hello,

I'm having trouble understanding how the answer (see attached) works for this combinations question:

An examination has 10 multiple-choice questions, each with 4 options. In each question, only one option is correct. For each question a student chooses one option at random.
Write an expression for the probability that the student chooses the correct option for exactly 7 questions.

If you could explain this to me that'd be amazing.
Thanks in advance

This is a direct application of the binomial probability formula. Have you covered this topic yet?

KT Nyunt · « **Reply #3078 on:** November 24, 2017, 05:37:49 pm »

Quote from: RuiAce on November 24, 2017, 05:31:28 pm

This is a direct application of the binomial probability formula. Have you covered this topic yet?

Yes but to be honest our teacher brushed over it quite a bit. So I don't quite understand it.

RuiAce · « **Reply #3079 on:** November 24, 2017, 05:49:29 pm »

Quote from: KT Nyunt on November 24, 2017, 05:37:49 pm

Yes but to be honest our teacher brushed over it quite a bit. So I don't quite understand it.

$\text{In the binomial probability formula, what one seeks to do is}\\ \text{to compute the probability of having exactly }\boxed{k\text{ 'successes'}}\\ \text{upon }\boxed{n\text{ trials}}\text{ of a particular event.}$
In our situation, the particular event is the answer we give to one multiple choice question, and the 'success' outcome is getting that multiple choice correct.
$\text{The probability of one particular success is }\boxed{p}\\ \text{(and by consequence, the probability of one particular failure is }q=1-p).$
Here, the success probability is $\frac14$, because we are picking one out of 4. The corresponding failure probability is $ \frac34$.
$\text{The formula asserts that in such an outcome}\\ \text{The probability of exactly }k\text{ successes in }n\text{ trials is given}\\ \boxed{\binom{n}{k}p^k q^{n-k}}\text{ or if you prefer, }\binom{n}{k}p^k(1-p)^{n-k}$
________________________________________________________________________
$\text{The proof of the formula is this.}\\ \text{We first consider the probability of a }\textbf{particular}\text{ way we have }k\text{ successes}\\<br />\text{in }n\text{ distinct trials.}$
In our case, an example would be if we got Q1-7 all correct, and Q8-10 all wrong. Another would be Q1,3,5 all wrong and the rest correct.
$\text{Because we need }k\text{ successes and }n-k\text{ failures in that specific order}\\ \text{the probability of this particular outcome is just }p^k (1-p)^{n-k}$
So for our scenario, we require $ \left(\frac14\right)^3 \left(\frac34\right)^{10-3} $
$\text{But this is just for a }\textbf{particular way}.\\ \text{What we require is the probability that the event happens}\\ \textbf{regardless of HOW it happens}\text{, i.e. we just care that it happens}.$
So what we want is the combined probability of ALL possible ways we get 7 correct and 3 wrong. The above two examples are just two possible ways this can happen; there's obviously more ways it can happen than that.
________________________________________________________________________
$\text{So what we need to do is count the number of }\textbf{ways}\\ \text{we can have }k\text{ successes and }n-k\text{ failures in }n\text{ trials.}\\ \text{That's where the }\binom{n}{k}\text{ comes from.}\\ \text{Let }\mathcal{S}\text{ stand for success, and }\mathcal{F}\text{ stand for failure.}$
$\text{In a straight line arrangement, we literally need to arrange}\\ k\quad \mathcal{S}\text{'s and }n-k\quad \mathcal{F}\text{'s in a row.}\\ \text{From year 11, we know that this will be }\frac{n!}{k!(n-k)!}\\ \text{recalling that we have repeated letters.}$
Note that the straight line arrangement is an equivalent way of analysing the problem. This is because it gets us to the same conclusion - we count the number of orderings in which we can have the failures and successes.
$\text{Which is exactly }\binom{n}{k}\text{ as required}\\ \text{giving us the formula }\boxed{\binom{n}{k}p^k(1-p)^{n-k}}$
From here, they just plug into the formula

KT Nyunt · « **Reply #3080 on:** November 24, 2017, 06:19:14 pm »

Quote from: RuiAce on November 24, 2017, 05:49:29 pm

$\text{In the binomial probability formula, what one seeks to do is}\\ \text{to compute the probability of having exactly }\boxed{k\text{ 'successes'}}\\ \text{upon }\boxed{n\text{ trials}}\text{ of a particular event.}$
In our situation, the particular event is the answer we give to one multiple choice question, and the 'success' outcome is getting that multiple choice correct.
$\text{The probability of one particular success is }\boxed{p}\\ \text{(and by consequence, the probability of one particular failure is }q=1-p).$
Here, the success probability is $\frac14$, because we are picking one out of 4. The corresponding failure probability is $ \frac34$.
$\text{The formula asserts that in such an outcome}\\ \text{The probability of exactly }k\text{ successes in }n\text{ trials is given}\\ \boxed{\binom{n}{k}p^k q^{n-k}}\text{ or if you prefer, }\binom{n}{k}p^k(1-p)^{n-k}$
________________________________________________________________________
$\text{The proof of the formula is this.}\\ \text{We first consider the probability of a }\textbf{particular}\text{ way we have }k\text{ successes}\\<br />\text{in }n\text{ distinct trials.}$
In our case, an example would be if we got Q1-7 all correct, and Q8-10 all wrong. Another would be Q1,3,5 all wrong and the rest correct.
$\text{Because we need }k\text{ successes and }n-k\text{ failures in that specific order}\\ \text{the probability of this particular outcome is just }p^k (1-p)^{n-k}$
So for our scenario, we require $ \left(\frac14\right)^3 \left(\frac34\right)^{10-3} $
$\text{But this is just for a }\textbf{particular way}.\\ \text{What we require is the probability that the event happens}\\ \textbf{regardless of HOW it happens}\text{, i.e. we just care that it happens}.$
So what we want is the combined probability of ALL possible ways we get 7 correct and 3 wrong. The above two examples are just two possible ways this can happen; there's obviously more ways it can happen than that.
________________________________________________________________________
$\text{So what we need to do is count the number of }\textbf{ways}\\ \text{we can have }k\text{ successes and }n-k\text{ failures in }n\text{ trials.}\\ \text{That's where the }\binom{n}{k}\text{ comes from.}\\ \text{Let }\mathcal{S}\text{ stand for success, and }\mathcal{F}\text{ stand for failure.}$
$\text{In a straight line arrangement, we literally need to arrange}\\ k\quad \mathcal{S}\text{'s and }n-k\quad \mathcal{F}\text{'s in a row.}\\ \text{From year 11, we know that this will be }\frac{n!}{k!(n-k)!}\\ \text{recalling that we have repeated letters.}$
Note that the straight line arrangement is an equivalent way of analysing the problem. This is because it gets us to the same conclusion - we count the number of orderings in which we can have the failures and successes.
$\text{Which is exactly }\binom{n}{k}\text{ as required}\\ \text{giving us the formula }\boxed{\binom{n}{k}p^k(1-p)^{n-k}}$
From here, they just plug into the formula

Thank you very much Rui! That clears up a lot of things!

Dragomistress · « **Reply #3081 on:** November 24, 2017, 07:34:38 pm »

Hello again!

I am rather confused on this question as I do not quite understand how to approach this.

Use mathematics induction to prove the divisibility for all positive integers n, x^n -1 is divisible by x-1.

RuiAce · « **Reply #3082 on:** November 24, 2017, 07:44:29 pm »

Quote from: Dragomistress on November 24, 2017, 07:34:38 pm

Hello again!
I am rather confused on this question as I do not quite understand how to approach this.

Use mathematics induction to prove the divisibility for all positive integers n, x^n -1 is divisible by x-1.

$\text{Of course }x^1-1\text{ is divisible by }x-1\\ \text{Now assume that }x^k-1 = P(x)(x-1)\text{ for some polynomial }P(x)$
Note how we use a polynomial here; because we're talking about polynomial division and not ordinary integer/real number division, the polynomial is more appropriate.
$\text{To prove that }x^{k+1}-1\text{ is divisible by }x-1\text{ now}\\ \text{we're going to have to do a bit of fudging.}\\ \text{Else, we can't use the inductive hypothesis.}$
There are at least two ways you can fudge here. This is one of the two ways.
$\begin{align*}x^{k+1}-1&=x^{k+1}-x^k + x^k-1\\ &= x^{k+1}-x^k + (x-1)P(x)\tag{assumption}\\ &= (x-1)x^k + (x-1)P(x)\tag{ordinary factorising}\\ &= (x-1)\left(x^k + P(x)\right)\end{align*}\\ \text{which is divisible by }x-1\text{ as required.}$

Mate2425 · « **Reply #3083 on:** November 24, 2017, 11:41:56 pm »

Hi could i please get some help with this integration question using substitution:

By Using the substitution; u=2x +7 and u^2 = 2x +7 find.... indefinte integral ×(2x+7)^0.5 dx

Thanks heaps 😃😃

RuiAce · « **Reply #3084 on:** November 25, 2017, 06:45:29 am »

Quote from: Mate2425 on November 24, 2017, 11:41:56 pm

Hi could i please get some help with this integration question using substitution:

By Using the substitution; u=2x +7 and u^2 = 2x +7 find.... indefinte integral ×(2x+7)^0.5 dx

Thanks heaps 😃😃

$\text{For }\boxed{u=2x+7}\implies \boxed{du = 2\,dx}\text{ we have}\\ \begin{align*}\int x\sqrt{2x+7}\,dx &= \int \frac{u=7}2 \sqrt{u}\, \frac{du}{2}\\ &= \frac14\int \left(u^{3/2}-7u^{1/2}\right)du\\&=\frac14\left(\frac{2u^{5/2}}{5}-\frac{14u^{3/2}}{3}\right)+C\\ &= \frac{\sqrt{(2x+7)^5}}{10}+\frac{7\sqrt{(2x+7)^3}}6+C\end{align*}$
______________________________________________
$\text{Assuming W.L.O.G. that }u\ge 0\\ \text{For }\boxed{u^2=2x+7} \implies \boxed{2u\,du = 2\,dx} \implies \boxed{dx = u\,du}\text{ we have}\\ \begin{align*}\int x\sqrt{2x+7}\,dx &= \int \frac{u^2-7}{2} \,u\,u\,du\\ &= \frac12 \int \left(u^4-7u^2\right)du\\ &= \frac12 \left(\frac{u^5}5 - \frac{7u^3}{3}\right)+C\\ &= \frac{\sqrt{(2x+7)^5}}{10}+\frac{7\sqrt{(2x+7)^3}}6+C\end{align*}$
Note: To go from $u^2 = 2x+7$ to $2u\,du = 2\,dx$ one may first rearrange to get $ x = (u^2-7)/2 $ first and then differentiate w.r.t. $u$. Alternatively, just use implicit differentiation.

Subject to minor computational errors - lack of sleep

Mate2425 · « **Reply #3085 on:** November 26, 2017, 10:01:07 pm »

Quote from: RuiAce on November 25, 2017, 06:45:29 am

$\text{For }\boxed{u=2x+7}\implies \boxed{du = 2\,dx}\text{ we have}\\ \begin{align*}\int x\sqrt{2x+7}\,dx &= \int \frac{u=7}2 \sqrt{u}\, \frac{du}{2}\\ &= \frac14\int \left(u^{3/2}-7u^{1/2}\right)du\\&=\frac14\left(\frac{2u^{5/2}}{5}-\frac{14u^{3/2}}{3}\right)+C\\ &= \frac{\sqrt{(2x+7)^5}}{10}+\frac{7\sqrt{(2x+7)^3}}6+C\end{align*}$
______________________________________________
$\text{Assuming W.L.O.G. that }u\ge 0\\ \text{For }\boxed{u^2=2x+7} \implies \boxed{2u\,du = 2\,dx} \implies \boxed{dx = u\,du}\text{ we have}\\ \begin{align*}\int x\sqrt{2x+7}\,dx &= \int \frac{u^2-7}{2} \,u\,u\,du\\ &= \frac12 \int \left(u^4-7u^2\right)du\\ &= \frac12 \left(\frac{u^5}5 - \frac{7u^3}{3}\right)+C\\ &= \frac{\sqrt{(2x+7)^5}}{10}+\frac{7\sqrt{(2x+7)^3}}6+C\end{align*}$
Note: To go from $u^2 = 2x+7$ to $2u\,du = 2\,dx$ one may first rearrange to get $ x = (u^2-7)/2 $ first and then differentiate w.r.t. $u$. Alternatively, just use implicit differentiation.

Subject to minor computational errors - lack of sleep

Thanks RuiAce :-)

3.14159265359 · « **Reply #3086 on:** November 27, 2017, 02:56:45 pm »

Hello! Can someone try this please.

3^n < (n+1)! For n>/= 4

Thank you

RuiAce · « **Reply #3087 on:** November 28, 2017, 02:21:14 am »

Quote from: Mimi5601 on November 27, 2017, 02:56:45 pm

Hello! Can someone try this please.

3^n < (n+1)! For n>/= 4

Thank you

Two questions.

1. Are you asking for an induction? Because this cannot be proved algebraically
2. Did you just mean >=? The >/= symbol is taken to mean NOT greater than or equal to

itssona · « **Reply #3088 on:** November 28, 2017, 05:25:51 pm »

heey ;d
some parametrics ones
a) prove that tangents at the end points of a focal chord meet at 90 degrees on the directrix
b)P(2at,at^2) is a point on the parabola x^2=4ay. If the tangent at P cuts the x axis at A and the y axis at B, find the ratio in which P divides AB

thank youu

RuiAce · « **Reply #3089 on:** November 29, 2017, 09:49:55 am »

Quote from: sssona09 on November 28, 2017, 05:25:51 pm

heey ;d
some parametrics ones
a) prove that tangents at the end points of a focal chord meet at 90 degrees on the directrix
b)P(2at,at^2) is a point on the parabola x^2=4ay. If the tangent at P cuts the x axis at A and the y axis at B, find the ratio in which P divides AB

thank youu

$\text{Assume that }P\text{ is the point }(2ap,ap^2)\text{ in the usual sense.}\\ \text{From the reference sheet, the tangent has equation }y=px-ap^2$
$\text{Subbing }x=0\text{ and }y=0\text{ to find out intercepts, we see that}\\ A\text{ is }(ap.0)\text{ and }B\text{ is }(0,-ap^2)$
$\text{Suppose }P\text{ divides }AB\text{ in the ratio }m:n\\ \text{where }m\text{ and }n\text{ are real numbers.}\\ \text{Then, using the ratio division formula, we have the equations}\\\begin{align*}2ap&=\frac{n ap}{m+n}\tag{1}\\ ap^2&= \frac{-map^2}{m+n}\tag{2}\end{align*}$
$(2)\div (1)\text{ gives}\\ \begin{align*}\frac{ap^2}{2ap}&=-\frac{map^2}{nap}\\ \frac{m}{n}&=-\frac{1}{2}\end{align*}$
$\text{Hence }m:n = -1:2$
a) is pretty common: You're just trying to use the formula $ m_1m_2 = -1 $ for perpendicular lines. Have a go at that and come back if you're stuck

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