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March 28, 2024, 11:44:15 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164203 times)  Share 

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deStudent

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Re: Specialist 3/4 Question Thread!
« Reply #8565 on: February 19, 2017, 08:50:51 pm »
0
http://m.imgur.com/a/wdSAD

Q5) I proved the question but I just wanted to double check that my working is actually correct since it's different to the answer provided.

Q2) I did this question without much trouble but the thing bugging was that it felt like I wasn't understanding the actual concept. I know that it's concave up when f''(x)>0 but I'm trying to visualise what happens to f(x) when the second derivative becomes positive or negative. Sorry for this broad question, but it feels like I don't actually understand it apart from those basic conditions of when it's concave up or down.

To clarify one thing, when the second derivative is zero at a stationary point (ie the x value where the first derivative is equal to zero), this tells us nothing useful right? This is different to when the value of a certain x which makes the second derivative equal zero, right? Because this would tell us the point of inflection?

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8566 on: February 19, 2017, 09:22:43 pm »
+3
http://m.imgur.com/a/wdSAD

Q5) I proved the question but I just wanted to double check that my working is actually correct since it's different to the answer provided.

Q2) I did this question without much trouble but the thing bugging was that it felt like I wasn't understanding the actual concept. I know that it's concave up when f''(x)>0 but I'm trying to visualise what happens to f(x) when the second derivative becomes positive or negative. Sorry for this broad question, but it feels like I don't actually understand it apart from those basic conditions of when it's concave up or down.

To clarify one thing, when the second derivative is zero at a stationary point (ie the x value where the first derivative is equal to zero), this tells us nothing useful right? This is different to when the value of a certain x which makes the second derivative equal zero, right? Because this would tell us the point of inflection?

For 5, it looks like your working is correct. Your derivative and second derivative seem to be right, it just looks like they used the calc to get a rearranged form of d2y/dx2. I might just avoid doing eqn = 0, as you should solve the equation then say eqn = 0 as required.

2. dy/dx tells us the gradient, or how the y value is changing with respect to the x value. d2y/dx2 tells us how the gradient is changing with respect to x. So, when concaving up, the graph gets steeper and steeper meaning the gradient is increasing, so the d2y/dx2 is positive, as the gradient is becoming greater.
If dy/dx is positive, the y value is increasing. If dy/dx is negative, the y value is decreasing.
If d2y/dx2 is positive, the gradient is increasing. If d2y/dx2 is negative, the gradient is decreasing.
The gradient decreasing doesn't necessarily mean the y value is decreasing.

If f''(x) = 0, that means a point of inflection, but that point of inflection may be only a slight one. If f''(x) and f'(x) both equal zero, that means it's a point of inflection where the point of inflection is also a stationary point, meaning the gradient at that point is 0.

Hope this helps clarify a couple things :)
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Jimmonash1991

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Re: Specialist 3/4 Question Thread!
« Reply #8567 on: February 20, 2017, 10:31:53 pm »
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Gee whizz differential equations are fun!!

Hey Shadow, got another question for you. I have attached it in an image.

I am aware of the following rules:

If -1<a<1 and cosx = a, then the general solutions are x = 2n*pi +/- arccos(a), where n is any real integer

If -1<a<1 and tanx = a, then the general solutions are x = n*pi +/- arctan(a), where n is any real integer

If -1<a<1 and sinx = a, then the general solutions are x = 2n*pi + arcsin(a) or x = (2n+1)*pi - arcsin(a), where n is any real integer,

I can see from part (a) how you go from sinA = a, to write a solution for sinx = -a, as

x = arcsin(-a) = -arcsin(a) = -A

Thus solutions are x = 2*pi - A and x = pi + A

But when it comes to determining solutions when cosx = a, and we are given sinA = a to start, I am not so sure.

I thought you could use the compound angle formulae somehow, or substitute sinA for a such that

x = arcos(sinA),

Then taking the fact A is restricted to 0 < A < 1/2*pi, find the solution that way......but I am not sure as I have not seen this problem before, even in year 12. I was not much of a frontier math student in those days, and I would like to know how to do it.

You are right Shadow, Specialist Maths is a little difficult, but not if you know the basics and have some original thought behind your methods, like the above problem.

Thanks again,

James
« Last Edit: February 20, 2017, 10:34:21 pm by Jimmonash1991 »

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Re: Specialist 3/4 Question Thread!
« Reply #8568 on: February 20, 2017, 11:25:16 pm »
+1
Gee whizz differential equations are fun!!

Hey Shadow, got another question for you. I have attached it in an image.

I am aware of the following rules:

If -1<a<1 and cosx = a, then the general solutions are x = 2n*pi +/- arccos(a), where n is any real integer

If -1<a<1 and tanx = a, then the general solutions are x = n*pi +/- arctan(a), where n is any real integer

If -1<a<1 and sinx = a, then the general solutions are x = 2n*pi + arcsin(a) or x = (2n+1)*pi - arcsin(a), where n is any real integer,

I can see from part (a) how you go from sinA = a, to write a solution for sinx = -a, as

x = arcsin(-a) = -arcsin(a) = -A

Thus solutions are x = 2*pi - A and x = pi + A

But when it comes to determining solutions when cosx = a, and we are given sinA = a to start, I am not so sure.

I thought you could use the compound angle formulae somehow, or substitute sinA for a such that

x = arcos(sinA),

Then taking the fact A is restricted to 0 < A < 1/2*pi, find the solution that way......but I am not sure as I have not seen this problem before, even in year 12. I was not much of a frontier math student in those days, and I would like to know how to do it.

You are right Shadow, Specialist Maths is a little difficult, but not if you know the basics and have some original thought behind your methods, like the above problem.

Thanks again,

James

Your best bet is to rewrite cos x = sin(pi/2 - x) = sin(alpha)
So now you have pi/2 - x = alpha + 2k*pi, pi - alpha + 2k*pi etc

Cosine is two syllables. The name contains 'sine'. The 'co' part means 'complementary', i.e. sine of the complementary angle, which means pi/2 - angle
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Jimmonash1991

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Re: Specialist 3/4 Question Thread!
« Reply #8569 on: February 20, 2017, 11:31:59 pm »
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Hey thanks mate!

I better check those other identities too then relating secant and cosecant, and tangent and cotangent.

Damn it, why didn't I think of this?!!!!

James

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Re: Specialist 3/4 Question Thread!
« Reply #8570 on: February 21, 2017, 03:53:20 pm »
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Hey fellas,

Sorry to be a bother (AGAIN), but I am just curious how one can go about solving a trigonometric inequality on a given domain algebraically?

I know we could use calculus and solve use the derivative, but I would like to do so without calculus
Say we are given the inequality

sinx - cosx > 0 from 1/2*pi <x< pi, which is true given sinx > cosx on this interval

I could use a number line to show where it was positive and then negative, but this would not be enough proof for the textbook I imagine.

I am doing the chapter on reciprocal and inverse functions, but I imagine the same process applies

One example I am troubled with is

cosecx - cotx > 0 on 0 <x< pi

Thanks again,

James

deStudent

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Re: Specialist 3/4 Question Thread!
« Reply #8571 on: February 21, 2017, 08:33:34 pm »
0
http://m.imgur.com/a/G1opw

Q7 of a)c) my working is attached but the answer said it only had one point of inflection which was at (pi,0). I got 3, whose correct?

Q7 of b)part b) the answer didn't specify if it was a local min/max for this one? Looking at graph f(x), there appears to be no turning point but I found f'(x)=0 and f"(x)>0. Wouldn't this then suggest that it is a concaving down turning point?

Q8b) I'm lost here. The answer says its non-negative, I'm not sure how they concluded that because we're only looking at its gradient? It then gives this linear graph representation that seems to contradict what it just said as it falls in to negative y-values?

c) how do I solve this? I thought it'd be zero since f'(a) = 0. Wasn't confident in my answer either.

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Re: Specialist 3/4 Question Thread!
« Reply #8572 on: February 21, 2017, 09:27:56 pm »
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In the Cambridge book, I've noticed I'm struggling with questions that involve proofs using multiple parameters. I understand about 80% of the questions in the book (i.e. pretty much all questions that do not involve proofs and complicated "show that"), but I am wondering if I should be concerned about this. Are these types of questions likely to be on SACS and the exams?

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8573 on: February 21, 2017, 09:38:29 pm »
+1
Hey fellas,

Sorry to be a bother (AGAIN), but I am just curious how one can go about solving a trigonometric inequality on a given domain algebraically?

I know we could use calculus and solve use the derivative, but I would like to do so without calculus
Say we are given the inequality

sinx - cosx > 0 from 1/2*pi <x< pi, which is true given sinx > cosx on this interval

I could use a number line to show where it was positive and then negative, but this would not be enough proof for the textbook I imagine.

I am doing the chapter on reciprocal and inverse functions, but I imagine the same process applies

One example I am troubled with is

cosecx - cotx > 0 on 0 <x< pi

Thanks again,

James


So for your example


So true for all x in this domain
« Last Edit: February 21, 2017, 11:49:22 pm by Shadowxo »
Completed VCE 2016
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8574 on: February 21, 2017, 11:09:35 pm »
+3
http://m.imgur.com/a/G1opw

Q7 of a)c) my working is attached but the answer said it only had one point of inflection which was at (pi,0). I got 3, whose correct?

Q7 of b)part b) the answer didn't specify if it was a local min/max for this one? Looking at graph f(x), there appears to be no turning point but I found f'(x)=0 and f"(x)>0. Wouldn't this then suggest that it is a concaving down turning point?

Q8b) I'm lost here. The answer says its non-negative, I'm not sure how they concluded that because we're only looking at its gradient? It then gives this linear graph representation that seems to contradict what it just said as it falls in to negative y-values?

c) how do I solve this? I thought it'd be zero since f'(a) = 0. Wasn't confident in my answer either.

7.a)c), there are 3 points of inflection but (π,0) is the minimum, as your working shows, meaning at (π,0) the gradient is a minimum (what the question asks for)
7.b)b)Looking at the graph you can see there is a turning point at x=-1, a minimum, which is also found when f'(x)=0. As f''(x) > 0 at this point, this means the gradient is increasing meaning it is a minimum.
8b)So if it's a local minimum at x=a, it means before x=a (ie when x=a-h) the gradient of f(x) is negative and after x=a (ie when x=a+h) the gradient is positive. We're looking at the gradient of y=f'(x), ie what f''(x) is. Since h is a small value, I believe they intend it such that it's close to zero meaning there's no point of inflection (as I believe a non-SP point of inflection could affect the "positive" statement). So as the gradient is always increasing, the gradient of f'(x) must be positive. This can be drawn as a graph of f'(x) and the y value continuously going up, similar to a linear graph.
Their linear graph shows y=f'(x) and since it's going up, the gradient of that graph is positive.
8.c) The gradient of f(x) is (presumably) always increasing, meaning f''(x) will always be positive. f''(x) is the same as the gradient of f'(x).

Hope this helps, a bit tired so apologies if I'm not too concise with my answer
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Jimmonash1991

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Re: Specialist 3/4 Question Thread!
« Reply #8575 on: February 21, 2017, 11:10:17 pm »
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Thanks Shadow,

Can I ask you where the "1" went when you divided through by cosx to get tanx?

See how I saw it was:

sinx - cosx > 0 on 1/2*pi <x< pi

sinx > cosx

Now you said cosx was negative, which makes sense as it is under the x-axis passing through at 1/2*pi

tanx < 1 <----- notice the 1 instead of the 0?

Now indeed as tanx approaches 1/2*pi (its asymptote) from the right, it approaches negative infinity, and further as it approaches pi, it approaches zero as it has an x-intercept there right? So the inequality holds......

I just thought there was more algebra to it that is all. The last one is self explanatory too so cheers!

I have another teaser for you, which I have been brooding over. It involves a pentagon inscribed in a semi circle, and I have to show that the diametr of the circle from A to E in the diagram attached, is p = sin(4x)/sin(x).

Now I can see that the length AB is 1, and that the vector AO is the radius of the circle. By drawing a vector bisecting the angle from O to a point M, which is halfway between A and B, we get two right angled triangles. I have attached a sketch of what I mean.

You can do this to the four triangles in the semi circle, to get eight right angled triangles, but how you get the ratio I have little clue. ..

If you don't mind explaining the geometry a little bit, because my geometry was NEVER good AT ALL  ::)

James

« Last Edit: February 21, 2017, 11:15:44 pm by Jimmonash1991 »

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8576 on: February 21, 2017, 11:13:28 pm »
+1
In the Cambridge book, I've noticed I'm struggling with questions that involve proofs using multiple parameters. I understand about 80% of the questions in the book (i.e. pretty much all questions that do not involve proofs and complicated "show that"), but I am wondering if I should be concerned about this. Are these types of questions likely to be on SACS and the exams?

Could you provide me with an example? :)
I'd try to get your head around the questions/concepts you're struggling with. Depending on the questions you're referring to, they may be on SACs and exams but if you're having any difficulty you can always post them here!
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8577 on: February 21, 2017, 11:15:09 pm »
+1
Thanks Shadow,

Can I ask you where the "1" went when you divided through by cosx to get tanx?

See how I saw it was:

sinx - cosx > 0 on 1/2*pi <x< pi

sinx > cosx

Now you said cosx was negative, which makes sense as it is under the x-axis passing through at 1/2*pi

tanx < 1 <----- notice the 1 instead of the 0?

Now indeed as tanx approaches 1/2*pi (its asymptote) from the right, it approaches negative infinity, and further as it approaches pi, it approaches zero as it has an x-intercept there right? So the inequality holds......

I just thought there was more algebra to it that is all. The last one is self explanatory too so cheers!

I have another teaser for you, which I have been brooding over. It involves a pentagon inscribed in a semi circle, and I have to show that the diametr of the circle from A to E in the diagram attached, is p = sin(4x)/sin(x).

(Image removed from quote.)

Now I can see that the length AB is 1, and that the vector AO is the radius of the circle. By drawing a vector bisecting the angle from O to a point M, which is halfway between A and B, we get two right angled triangles. I have attached a sketch of what I mean.

(Image removed from quote.)

You can do this to the four triangles in the semi circle, to get eight right angled triangles, but how you get the ratio I have little clue. ..

If you don't mind explaining the geometry a little bit, because my geometry was NEVER good AT ALL  ::)

James



Oops my mistake, serves me right for answering questions half asleep, I'll change it right away ;)

Also even though tan(x) is undefined at π/2, at π/2 the inequality holds (as cos(x)=0, tan(x) is undefined but that's from our simplification)
And I'll see if I can help you with that question tomorrow :) going to rest up now. Might have a bit of a late answer as I'll be a bit busy
« Last Edit: February 21, 2017, 11:39:57 pm by Shadowxo »
Completed VCE 2016
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Jimmonash1991

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Re: Specialist 3/4 Question Thread!
« Reply #8578 on: February 21, 2017, 11:33:44 pm »
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God damn it shadow!

I am now relinquishing you of your moderator rights. I am now the captain of this vessel.

Ha ha!!  ;D 8)

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8579 on: February 21, 2017, 11:42:46 pm »
+1
God damn it shadow!

I am now relinquishing you of your moderator rights. I am now the captain of this vessel.

Ha ha!!  ;D 8)

I'm not a mod, aha there IS no captain! >:-)
But there is a very cool trendsetter  8)
Completed VCE 2016
2015: Biology
2016: Methods | Physics | Chemistry | Specialist Maths | Literature
ATAR : 97.90
2017: BSci (Maths and Engineering) at MelbUni
Feel free to pm me if you have any questions!