Cambridge HSC MX1 Textbook Marathon/Q&A (3 Viewers)

davidgoes4wce · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
Unnecessarily expanding feels like a waste of time and does not teach much at all.

They wouldn't ask you for a middle term that way, they will either make it plural (to which therefore yes) or just not ask it.

Alot of the introductory problems with Binomials they want to get you to write the whole expansion out by using Binomial Theorem or Pascal's Triangle. Once you develop better understanding and knowledge, obviously you look at quicker methods.

leehuan · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
In Question 13 (b) they ask for the 2 middle terms eg (a+3b)^5

If they ask for two then use what you said about sixth and seventh term for the higher power. But it's a bit more unlikely to be asked in the exam.

davidgoes4wce said:
Alot of the introductory problems with Binomials they want to get you to write the whole expansion out by using Binomial Theorem or Pascal's Triangle. Once you develop better understanding and knowledge, obviously you look at quicker methods.

Do you really need to do more than 5 questions that make you write out a full binomial expansion to see what's going on?

Plus Pascal's triangle isn't really necessary. It helps to visualise things such as Pascal's identity but other than that so long as you know you can replace the terms with \binom{n}{k} nothing to worry about

seanieg89 · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

On the topic of binomial stuff

$$Prove that: $\sum_{k=0}^n \binom{m+k}{m}=\binom{m+n+1}{m+1}.$

leehuan · Jul 5, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

seanieg89 said:
On the topic of binomial stuff

$$Prove that: $\sum_{k=0}^n \binom{m+k}{m}=\binom{m+n+1}{m+1}.$

I think this is how to do that one

$\text{Compare the coefficient of }x^{m}\text{ in the expansion of }\\ (1+x)^m+(1+x)^{m+1}+\dots+(1+x)^{m+n}=\frac{(1+x)^{m+n+1}}{x}-\frac{1}{x}$

$\text{Note that the coefficient of }x^m\text{ in the RHS is equivalent to the coefficient of }x^{m+1}\text{ in the numerator.}$

Edit: WOAH THERE. Start from (1+x)⁰, not (1+x)^m otherwise the series is wrong

Result is not affected as there is no term in x^m in the expansion of (1+x)^k, where 0 ≤ k ≤ m-1

davidgoes4wce · Jul 10, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Yr 12 Cambridge Ex 10 A Q 23.

I spent more than 30 minutes and still couldn't come to the answer for this one. I had a few attempts at drawing the diagram but still couldn't come up with it.

Q23. A rectangular field is 60 metres long and 30 metres wide. A cow wanders randomly around the field. Find the probability that the cow is:

(a) More than 10 metres from the edge of the field

(b) not more than 10 metres from the corner of the field

Answers are:

$$ (a) $ \frac{2}{9}$

$$ (b) $ \frac{\pi}{18}$

InteGrand · Jul 10, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Yr 12 Cambridge Ex 10 A Q 23.

I spent more than 30 minutes and still couldn't come to the answer for this one. I had a few attempts at drawing the diagram but still couldn't come up with it.

Q23. A rectangular field is 60 metres long and 30 metres wide. A cow wanders randomly around the field. Find the probability that the cow is:

(a) More than 10 metres from the edge of the field

(b) not more than 10 metres from the corner of the field

Answers are:

$$ (a) $ \frac{2}{9}$

$$ (b) $ \frac{\pi}{18}$

$\noindent a) Consider a `concentric' $40$ m by $10$ m rectangle inside the rectangular field (so it has the same centre as the field, its edges are parallel to the field, and it is contained inside the field, so like the image below, but ignore the dimensions in that picture).$

http://s150.photobucket.com/user/jim_thompson5910/media/Algebra dot com/rectangle.png.html

$\noindent The edges of the inner rectangle are 10 m from the edges of the rectangular field. It is easy to see then that if the cow is inside the inner rectangle, it is more than 10 m away from an edge, and if the cow is \emph{outside} the inner rectangle, it must be within 10 m of an edge. So the cow is more than 10 m from an edge if and only if it is inside the inner rectangle. The probability of this happening is just the area of the inner rectangle as a proportion of the area of the whole field, which is $\frac{40\times 10}{60\times 30}= \frac{2}{9}$.$

$\noindent b) We want the probability that the cow is within 10 m from a corner. Draw circles of radius 10 m centred at each corner of the rectangular field, and consider their intersecting regions with the field (it'll be a quadrant at each corner). The cow is within 10 m of a vertex if and only if it is inside one of these quadrants. Note these quadrants are non-overlapping, since the sides of the field are longer than 20 m. Now, the total area of the four quadrants is just the area of a full circle of radius 10 m, which is $\pi\cdot 10^2 = 100\pi \text{ m}^2$. So the desired probability is the ratio of this to the entire field's area, which is $\frac{100\pi}{60\times 30}=\frac{\pi}{18}$.$

davidgoes4wce · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$\noindent a) Consider a `concentric' $40$ m by $10$ m rectangle inside the rectangular field (so it has the same centre as the field, its edges are parallel to the field, and it is contained inside the field, so like the image below, but ignore the dimensions in that picture).$

http://s150.photobucket.com/user/jim_thompson5910/media/Algebra dot com/rectangle.png.html

$\noindent The edges of the inner rectangle are 10 m from the edges of the rectangular field. It is easy to see then that if the cow is inside the inner rectangle, it is more than 10 m away from an edge, and if the cow is \emph{outside} the inner rectangle, it must be within 10 m of an edge. So the cow is more than 10 m from an edge if and only if it is inside the inner rectangle. The probability of this happening is just the area of the inner rectangle as a proportion of the area of the whole field, which is $\frac{40\times 10}{60\times 30}= \frac{2}{9}$.$

$\noindent b) We want the probability that the cow is within 10 m from a corner. Draw circles of radius 10 m centred at each corner of the rectangular field, and consider their intersecting regions with the field (it'll be a quadrant at each corner). The cow is within 10 m of a vertex if and only if it is inside one of these quadrants. Note these quadrants are non-overlapping, since the sides of the field are longer than 20 m. Now, the total area of the four quadrants is just the area of a full circle of radius 10 m, which is $\pi\cdot 10^2 = 100\pi \text{ m}^2$. So the desired probability is the ratio of this to the entire field's area, which is $\frac{100\pi}{60\times 30}=\frac{\pi}{18}$.$

I understand Part (a) better , I must admit I read the question in a different way first up. (Im not ashamed of showing my mistake either). I personally don't think they worded the question too well. I now realise it is 10 metres inwards of the rectangle, not outwards (which was my thinking). This is what I drew up first up:

davidgoes4wce · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

This is the required Diagram for Q23 (a) Ex 10 A

davidgoes4wce · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Edge of the field = Refers to the outer boundary of the rectangle.

Hence, the 10 metres is inwards towards the field.

davidgoes4wce · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Q23 (b) I had less trouble with and this is the diagram I drew out:

davidgoes4wce · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

is there a mistake with Q 16 (c), when asking for 'the distance from each particle after 3 seconds'. For part (c), my thinking is to substitute t=3 into each of the functions. (Shown in the graph in : Red, Green and Blue). I was able to get one of the answers of 12.5 metres but surprisingly the Cambridge textbook only showed 1 answer when I thought there should have been two.

$Cambridge Yr 12 Exercise 3C Q 16$

$(a) \ v_1=6+2t \ v_2=4-2t$

$x_1=\int 6+2t \ dt =6t+t^2+C \ and \ x_2=\int 4-2t \ dt=4t-t^2+C$

$For $ x_1 , $ when t=0 , x=2$

$For $ x_2 , $ when t=0 , x=1$

$x_1=6t+t^2+2 \ and \ x_2=4t-t^2+1$

$D=x_1-x_2=(6t+t^2+2)-(4t-t^2+1)=2t^2+2t+1$

$(b) D=2t^2+2t+1$

$t cannot be a negative value$

$\frac{dD}{dt}=4t+2$

$\frac{dD}{dt}=0 \implies t=\frac{-1}{2}$

$At t=0, it represents the minimum Distance between the two particles, which in this case is 1 metre.$

$(c)$

$Mid Point=\frac{x_1+x_2}{2}=\frac{10t+3}{2} $ (Red Line) $$

$Constant Velocity=5$

$Let t=3$

$$ Distance from Particle 1 to Mid-Point $: (6t+t^2+2)-(\frac{10t+3}{2})=29-16.5=12.5 \ metres$

$$ Distance from Particle 2 to Mid-Point $: \frac{10t+3}{2}-(4t-t^2+1)=16.5-4=12.5 \ metres$

davidgoes4wce · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

I just realised that they are both 12.5 metres , my bad.

I guess my other question would be does the mid-point always have to be equidistant for two quadratics when we add them?

davidgoes4wce · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

When I write questions on the Bored of Studies forum, I understand it better than when I write it by hand on paper. I don't know why , not sure if you guys are finding it the same thing.

davidgoes4wce · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

This was the answer in the back of the book for Q 16 (c).

What threw me off, was there was only 1 answer for the distance for each particle 1 and 2. Its ironic that both answers are 12.5 metres.

seanieg89 · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Why are you surprised that the midpoint of two points is equidistant from these two points?

Paradoxica · Jul 11, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

seanieg89 said:
Why are you surprised that the midpoint of two points is equidistant from these two points?

brutal

davidgoes4wce · Jul 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

I don't want to Google Search the answer , want to put this up in an open forum to discuss the way to get the answer.

I spent around 15 minutes on this question.

$Year 12 Cambridge Exercise 3C Q18$

I tried to bring in some Physics Equations of Motion with my diagram below (sorry it looks like dog shit but meh)

InteGrand · Jul 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
I don't want to Google Search the answer , want to put this up in an open forum to discuss the way to get the answer.

I spent around 15 minutes on this question.

$Year 12 Cambridge Exercise 3C Q18$

I tried to bring in some Physics Equations of Motion with my diagram below (sorry it looks like dog shit but meh)

$\noindent (a) Note the motions are just in the vertical direction, so the trajectories aren't actually parabolas of course (as they're just either dropped or thrown straight up). Let $y=0$ be the ground level. The stone's equation for displacement as a function of time $t$ is $y_S (t) = Vt - \frac{1}{2}gt^2$ and the ball's equation for displacement is $y_{B} (t) = h -\frac{1}{2}gt^2$, where $g=10,h = 180$ for our purposes. So for a collision, we need $Vt - \frac{1}{2}gt^2 = h-\frac{1}{2}gt^2 \Longleftrightarrow Vt = h \Longleftrightarrow t = \frac{h}{V}$.$

$\noindent So if a collision occurs, it does so at $t = \frac{h}{V}$. The collision actually occurs (in the air) iff the displacement at this time is a positive value. The displacement at this time is $h-\frac{1}{2}g\left( \frac{h}{V}\right)^2 = h - \frac{1}{2}g \times \frac{h^2}{V^2}$. Solving $h - \frac{1}{2}g \times \frac{h^2}{V^2} > 0$ yields $V > \sqrt{\frac{1}{2}gh} = \sqrt{900}=30$. So there is a collision iff $V >\sqrt{\frac{1}{2}gh}=30 $.$

$\noindent (b) To get the $V$ for which we have collision halfway up the cliff, we want the $V$ such that $h - \frac{1}{2}g \times \frac{h^2}{V^2}$ (the displacement at the time of collision) is $\frac{1}{2}h$. So just solve $h - \frac{1}{2}g \times \frac{h^2}{V^2} = \frac{1}{2}h$ for $V$. To get the corresponding time, once we have found this $V$, it is $\frac{h}{V}$.$

davidgoes4wce · Jul 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$\noindent (a) Note the motions are just in the vertical direction, so the trajectories aren't actually parabolas of course (as they're just either dropped or thrown straight up). Let $y=0$ be the ground level. The stone's equation for displacement as a function of time $t$ is $y_S (t) = Vt - \frac{1}{2}gt^2$ and the ball's equation for displacement is $y_{B} (t) = h -\frac{1}{2}gt^2$, where $g=10,h = 180$ for our purposes. So for a collision, we need $Vt - \frac{1}{2}gt^2 = h-\frac{1}{2}gt^2 \Longleftrightarrow Vt = h \Longleftrightarrow t = \frac{h}{V}$.$

$\noindent So if a collision occurs, it does so at $t = \frac{h}{V}$. The collision actually occurs (in the air) iff the displacement at this time is a positive value. The displacement at this time is $h-\frac{1}{2}g\left( \frac{h}{V}\right)^2 = h - \frac{1}{2}g \times \frac{h^2}{V^2}$. Solving $h - \frac{1}{2}g \times \frac{h^2}{V^2} > 0$ yields $V > \sqrt{\frac{1}{2}gh} = \sqrt{900}=30$. So there is a collision iff $V >\sqrt{\frac{1}{2}gh}=30 $.$

$\noindent (b) To get the $V$ for which we have collision halfway up the cliff, we want the $V$ such that $h - \frac{1}{2}g \times \frac{h^2}{V^2}$ (the displacement at the time of collision) is $\frac{1}{2}h$. So just solve $h - \frac{1}{2}g \times \frac{h^2}{V^2} = \frac{1}{2}h$ for $V$. To get the corresponding time, once we have found this $V$, it is $\frac{h}{V}$.$

Am I right in saying that the stone fired path is x=vt

$180= V \times t ? $ In this case velocity is directly proportional to time? (I made the assumption in my working out that the initial velocity was $ 0 ms^{-1})$

InteGrand · Jul 18, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

davidgoes4wce said:
Am I right in saying that the stone fired path is x=vt

$180= V \times t$ ? In this case velocity is directly proportional to time?

There is no motion in the horizontal direction.

Cambridge HSC MX1 Textbook Marathon/Q&A (3 Viewers)

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