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December 14, 2019, 01:40:54 pm

### AuthorTopic: Mathematics Question Thread  (Read 623547 times) Tweet Share

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#### Hawraa

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##### Re: Mathematics Question Thread
« Reply #4365 on: October 10, 2019, 10:38:59 pm »
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Hey there!

Remember what the function y=f'(x) actually denotes; it's just the slope of the function y=f(x). If f'(x)>0 over some domain, then f(x) is increasing over that domain. Similarly, if f'(x)<0 over some domain, then f(x) is decreasing over that domain. And if f'(x)=0, we have a stationary point.

Also, since
$\frac{d}{dx}f(x) = f'(x)$
We have
$\int d(f(x)) = \int f'(x) dx \\ \int f'(x) dx = f(x) + C$
ie. the area under the curve of y=f'(x) denotes the function value of f(x) at a specific x value.

From my first point, we see that f'(x)>0 for 0≤x<2, hence it is for the same domain that f(x) is increasing.
From my second point, we see that the maximum signed area under the curve is 4 units. Hence, the maximum value of f(x) is also 4.
From my second point, we note that f(6) will be the total signed area from 0 to 6. Seeing that A1 and A2 cancel, we just note the area between x=4 and x=6 (which is just a rectangle!) to be 6 units squared; hence f(6)=-6 (the area is below the x-axis!).

The graph is rather more difficult.
Key points should be noted on the graph, and can be deduced from the signed area! For example, (0, 0), (2, 4), (4, 0) and (6, -6) should all be on your graph. Since f'(2)=0, there should be a stationary point there. Note that also the gradient of f'(x) is negative at x=2, this indicates that f''(2)<0 and thus we have a maximum at x=2. Also, ensure that the gradient at x=0 looks convincingly like a gradient of 3, and similarly for x=4, make sure the gradient looks convincingly like a gradient of -3. Since between x=4 and x=6 f'(x) is a constant, between x=4 and x=6 f(x) should also be a straight line that has a constant gradient of -3.

If you need an actual picture of the graph, let me know and I'll attach one.

Hope this helps

Thanks a lot 🌷 that actually clarifies the question.

#### fun_jirachi

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##### Re: Mathematics Question Thread
« Reply #4366 on: October 10, 2019, 11:48:59 pm »
+1
And one more question please. This is from the (ATAR Notes topic tests HSC mathematics Edition 1 2017-2019. Page 35 Q9.) I think they accidentally forgot to put the answer for this one. The question is :

A strip of wire 96 cm in length is used to build a square prism. Supposing that the length of the square side is x cm, show that the surface area of the square prism is given by  S= 6x(16 - x). Hence, find the volume of the prism with the maximum surface area.

Thanks guys.

Sorry for missing this one!

The question's wording is actually quite ambiguous! However, if you initially thought that the object was a cube, and the surface area should've been 6x2 (like I did!), you would've noticed there would have been no maximum. Hence, the object is just a square-sided rectangular prism with at least one set of opposite sides being squares.

Let the square sides have length x, and the other edge have length y.
Since the total length of wire is 96cm, we have that 8x+4y=96 ie. y=24-2x.
Now, we have the surface area being 2(x2+2xy).
ie. S=2x(x+2y)
S=2x(x+2(24-2x))
S=2x(48-3x)
S=6x(16-x)

From here, we find that $\frac{dS}{dx} = 96-12x$ ie. there's a maximum when x=8 (you can test for a maximum in whichever way you like) (and therefore also when y=8! basically the object was a cube the whole time, which makes sense.) And thus the volume which maximises the surface area is just 8x8x8, or 512cm3.

Hope this helps
Failing everything, but I'm still Flareon up.

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#### Hawraa

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##### Re: Mathematics Question Thread
« Reply #4367 on: October 11, 2019, 01:01:20 am »
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Sorry for missing this one!

The question's wording is actually quite ambiguous! However, if you initially thought that the object was a cube, and the surface area should've been 6x2 (like I did!), you would've noticed there would have been no maximum. Hence, the object is just a square-sided rectangular prism with at least one set of opposite sides being squares.

Let the square sides have length x, and the other edge have length y.
Since the total length of wire is 96cm, we have that 8x+4y=96 ie. y=24-2x.
Now, we have the surface area being 2(x2+2xy).
ie. S=2x(x+2y)
S=2x(x+2(24-2x))
S=2x(48-3x)
S=6x(16-x)

From here, we find that $\frac{dS}{dx} = 96-12x$ ie. there's a maximum when x=8 (you can test for a maximum in whichever way you like) (and therefore also when y=8! basically the object was a cube the whole time, which makes sense.) And thus the volume which maximises the surface area is just 8x8x8, or 512cm3.

Hope this helps

OMG! This question was actually killing me. But the way you explained it is just awesome! Many thanks 🌷

#### Hawraa

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##### Re: Mathematics Question Thread
« Reply #4368 on: October 11, 2019, 01:34:11 am »
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Hi again,
I'm back with another question 😅. This time it's trig functions.

Q: find the area of the minor segment which subtends an angle of 3/4 pi at the centre of a circle with radius 20m.

Can you please explain how to work it out?

#### Grace0702

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##### Re: Mathematics Question Thread
« Reply #4369 on: October 11, 2019, 10:08:35 am »
+1
Hi again,
I'm back with another question 😅. This time it's trig functions.

Q: find the area of the minor segment which subtends an angle of 3/4 pi at the centre of a circle with radius 20m.

Can you please explain how to work it out?

Hey!

So there is actually a formula to work this out which is:
$A=\frac{1}{\:2}r^2\left(\theta -\sin \left(\theta \right)\right)$

Take a look at the diagram. Basically the formula is finding the area of the sector using:
$A=\frac{1}{\:2}r^2\theta$
and the subtracting the area of the triangle with angle theta using:
$A=\frac{1}{\:2}ab\sin \left(C\right)$
If you sub in the radius and angle theta to the area of a triangle formula you get
$A=\frac{1}{\:2}r^2\sin \left(\theta \right)$

So the Area of sector - Area of triangle = Area of minor segment
$A=\frac{1}{\:2}r^2\left(\theta -\sin \left(\theta \right)\right)$

So then we sub in what we know
$\theta =\frac{3\pi }{4}\:and\:r=20m$
$A=\frac{1}{\:2}\left(20\right)^2\times \left(\frac{3\pi }{4}-\sin \left(\frac{3\pi }{4}\right)\right)$
$200\left(\frac{3\pi }{4}-\sin \left(\frac{3\pi }{4}\right)\right)$
$200\left(\frac{3\pi }{4}-\frac{\sqrt{2}}{2}\right)$
$150\pi -100\sqrt{2}$
$A=\left(150\pi \:\:-100\sqrt{2}\right)m^2$

FYI this formula is not on the formula sheet so you will need to remember it or how to derive it
Hope this helps!
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#### Hawraa

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##### Re: Mathematics Question Thread
« Reply #4370 on: October 11, 2019, 11:14:28 am »
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Hey!

So there is actually a formula to work this out which is:
$A=\frac{1}{\:2}r^2\left(\theta -\sin \left(\theta \right)\right)$

(Image removed from quote.)
Take a look at the diagram. Basically the formula is finding the area of the sector using:
$A=\frac{1}{\:2}r^2\theta$
and the subtracting the area of the triangle with angle theta using:
$A=\frac{1}{\:2}ab\sin \left(C\right)$
If you sub in the radius and angle theta to the area of a triangle formula you get
$A=\frac{1}{\:2}r^2\sin \left(\theta \right)$

So the Area of sector - Area of triangle = Area of minor segment
$A=\frac{1}{\:2}r^2\left(\theta -\sin \left(\theta \right)\right)$

So then we sub in what we know
$\theta =\frac{3\pi }{4}\:and\:r=20m$
$A=\frac{1}{\:2}\left(20\right)^2\times \left(\frac{3\pi }{4}-\sin \left(\frac{3\pi }{4}\right)\right)$
$200\left(\frac{3\pi }{4}-\sin \left(\frac{3\pi }{4}\right)\right)$
$200\left(\frac{3\pi }{4}-\frac{\sqrt{2}}{2}\right)$
$150\pi -100\sqrt{2}$
$A=\left(150\pi \:\:-100\sqrt{2}\right)m^2$

FYI this formula is not on the formula sheet so you will need to remember it or how to derive it
Hope this helps!

That's great, thanks a lot🌹. I actually came across this formula when I was looking for an answer, but I don't remember using it in school, so I thought maybe there is another way to solve the question. . But it makes sense now, especially knowing how to derive it.

#### Estermont

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##### Re: Mathematics Question Thread
« Reply #4371 on: October 17, 2019, 07:54:11 pm »
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I've just started year 12 and I was wondering whether it is worth keeping Advanced Mathematics, I feel that I have the ability to get 80-84 for my Advanced exams and 90-93 if I dropped down to General Mathematics. Is it worthwhile dropping to General maths or should I stick with Advanced? Thanks

#### DrDusk

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##### Re: Mathematics Question Thread
« Reply #4372 on: October 17, 2019, 08:28:49 pm »
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I've just started year 12 and I was wondering whether it is worth keeping Advanced Mathematics, I feel that I have the ability to get 80-84 for my Advanced exams and 90-93 if I dropped down to General Mathematics. Is it worthwhile dropping to General maths or should I stick with Advanced? Thanks
I would honestly advise anyone to stay with advanced. Even though advanced forces you to learn much more advanced concepts, the actual ratio of knowledge/ability vs difficulty is basically the same as Standard Maths. Standard math students know much less Maths than advanced students but to them the exam tests their abilities. In advanced you have much more knowledge and it tests your abilities just as much as Standard does for its students. It's rather unlike the Extension 1 and 2 courses where the difficulty of the exam gets much more than the amount of knowledge/ability that most students will have.
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#### RuiAce

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##### Re: Mathematics Question Thread
« Reply #4373 on: October 18, 2019, 10:45:08 am »
+2
Standard math students know much less Maths than advanced students
Please watch your wording. When you say that it becomes a needlessly derogatory remark to standard maths students instead.

I've just started year 12 and I was wondering whether it is worth keeping Advanced Mathematics, I feel that I have the ability to get 80-84 for my Advanced exams and 90-93 if I dropped down to General Mathematics. Is it worthwhile dropping to General maths or should I stick with Advanced? Thanks
Just some food for thought: If you compare the results in the HSC raw marks database you'll find that the marks in Advanced you're earning also has good chances of attaining a B6. Of course, it may not always be the case - the alignment algorithm does vary from year to year.

Otherwise, perhaps just consider whether or not you'd need Advanced for whatever it is you want to pursue at university. You don't want to fall into the trap of relying on Standard for reasonably maths-intensive disciplines. Also, keep in mind that you may need to relearn a few concepts from Year 11 Standard along the way; note that Year 11 content is examinable in the HSC for mathematics subjects.

But of course, you should also weigh out if Advanced is actually stressing you. (Would be a bit surprised if it did with 80+ raw marks, but can't fault you if it does anyway!)

#### Estermont

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##### Re: Mathematics Question Thread
« Reply #4374 on: October 18, 2019, 06:15:58 pm »
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Thank you both for your informative replies!  I've decided that the advanced course is the most suitable for me.

#### Kombmail

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##### Re: Mathematics Question Thread
« Reply #4375 on: October 19, 2019, 11:17:25 pm »
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Does anyone now how to do this question?
Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

(ii) what is the probability that Pat wins the game on the first or on the second throw?

(iii) fins the probability that pat eventually wins the game
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#### fun_jirachi

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##### Re: Mathematics Question Thread
« Reply #4376 on: October 19, 2019, 11:53:16 pm »
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Does anyone now how to do this question?
Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

(ii) what is the probability that Pat wins the game on the first or on the second throw?

(iii) fins the probability that pat eventually wins the game

Hey there!

This question was previously asked here. There's also a small explanation below that post to have a read as well if you don't understand part ii). If there's any other part you don't understand, have a query and we'll answer it

Hope this helps
Failing everything, but I'm still Flareon up.

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#### boulos

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##### Re: Mathematics Question Thread
« Reply #4377 on: October 20, 2019, 12:22:21 pm »
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Hey, i'm confused with part iv of the attached question, not sure where to start.

#### fun_jirachi

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##### Re: Mathematics Question Thread
« Reply #4378 on: October 20, 2019, 12:56:16 pm »
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Hey there!

Essentially, when we have f(x)-k = 0, we're looking for the solutions as a horizontal line cuts through f(x) (we're just considering where y=f(x) and y=k intersect!). Since you have the sketch of the graph from part ii), you'll notice that above the local maximum, y=k cuts the graph only once, while a similar occurs below the local minimum. Between the local maximum and the local minimum, you'll also notice that the line y=k cuts f(x) three times. Hence, the only values where the line y=k will cut the graph at two distinct places is when k is equal to the y-value of the local max and local min ie. when the horizontal line is tangential to the graph! Hence, k would equal either 7 or -20.

Hope this helps
Failing everything, but I'm still Flareon up.

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#### Hawraa

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##### Re: Mathematics Question Thread
« Reply #4379 on: October 20, 2019, 12:56:53 pm »
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Hi everyone,
Can someone please help me with this probability question.

Gerry has a standard deck of 52 cards in his hands, consisting of 4 different suits, each suit has 13 cards. What is the probability of Gerry picking a queen of spades then picking up any card from the hearts suit?

I know that the P of queen of spades would be 1/52 (which is also correct in the answers given) but for the other part of P of any card of the heart suit, I think it should be 13/51 since he already has one card in his hand but the answers say it's 13/52?  So that's why I'm confused.