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March 29, 2024, 03:18:11 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164277 times)  Share 

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cotangent

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9600 on: January 25, 2020, 02:47:11 pm »
+1
Hello all

I am having trouble finding the range of x over the domain of (pi/2, 3pi/2). I keep getting negative infinity to negative a, but the textbook says negative infinity to positive a?

I think they made an error, it should be -infinity to -a.
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9601 on: January 25, 2020, 03:09:10 pm »
+1
I think they made an error, it should be -infinity to -a.
Same I believe they made a mistake as well.

TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9602 on: January 25, 2020, 06:16:49 pm »
+1
Seems like it. When I started doing the exercises, there were questions that asked for the domain and followed the same method (-negative infinity to a)

MaiSakurajima

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9603 on: January 29, 2020, 02:57:30 am »
0
Hey guys can I get some help on this question Extended response:
1. e. Find the equation of the locus of M when:
i. b =/  1/sqrt(2)
ii b = 1/sqrt(2)

What is this question asking for? Like I am currently stumped.
Thanks
https://imgur.com/a/3VfKonZ
https://imgur.com/a/0z04QjS

uh its not letting me attach images i am sorry
« Last Edit: January 29, 2020, 03:02:01 am by MaiSakurajima »
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Bri MT

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9604 on: January 29, 2020, 07:44:45 am »
0
Hey guys can I get some help on this question Extended response:
1. e. Find the equation of the locus of M when:
i. b =/  1/sqrt(2)
ii b = 1/sqrt(2)

What is this question asking for? Like I am currently stumped.
Thanks
https://imgur.com/a/3VfKonZ
https://imgur.com/a/0z04QjS

uh its not letting me attach images i am sorry

The reason it's not letting you attach the image is because you aren't giving it the right url. Got to the top right corner of the image and lick to open the menu. Then select "get share links" and copy the bbcode over to the forums.

You can test if you have the right link by clicking on the url, if it doesn't show you only that image the url is wrong.


Don't need to apologise for that though, embedding images can be confusing :)

dream chaser

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9605 on: February 03, 2020, 06:39:00 pm »
0
Hi Guys,

I need help with this question. How would you do it?

Find the equation of the tangent at the point (2, 1) of the circle x^2 + y^2 − 4y − 1 = 0

I know the equation is y=m(x-2)+1 from simply subbing (2,1) into y-y1=m(x-x1) formula. I just need to find m now but can't fully solve it.

I subbed y=m(x-2)+1 into x^2+y^2-4y-1=0 and ended up with: x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0. What do I do now?

All help will be much appreciated. Thanks

PS: Sorry, I just realized I accidentally post the message twice from looking at the forum. My apologies for that. I thought it didn't get sent the first time around.

jnlfs2010

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9606 on: February 03, 2020, 07:17:33 pm »
+1
Hi Guys,

I need help with this question. How would you do it?

Find the equation of the tangent at the point (2, 1) of the circle x^2 + y^2 − 4y − 1 = 0

I know the equation is y=m(x-2)+1 from simply subbing (2,1) into y-y1=m(x-x1) formula. I just need to find m now but can't fully solve it.

I subbed y=m(x-2)+1 into x^2+y^2-4y-1=0 and ended up with: x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0. What do I do now?

All help will be much appreciated. Thanks

PS: Sorry, I just realized I accidentally post the message twice from looking at the forum. My apologies for that. I thought it didn't get sent the first time around.

You've got a quadratic. Now get into ax2+bx+c=0 form. Since the line is a tangent, the discriminant of that equation must equal to 0 as there is only one solution. With the value of m, you surely should get the formula of the tangent
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dream chaser

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9607 on: February 03, 2020, 07:24:11 pm »
0
You've got a quadratic. Now get into ax2+bx+c=0 form. Since the line is a tangent, the discriminant of that equation must equal to 0 as there is only one solution. With the value of m, you surely should get the formula of the tangent

But what about the x^2 in front? What am I meant to do with that? Thanks for the reply by the way.

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9608 on: February 05, 2020, 12:45:22 pm »
0
But what about the x^2 in front? What am I meant to do with that? Thanks for the reply by the way.
You can always use calculus. dy/dx of x2+ y2-4y-1 is -x/y-2
Then we can use our coordinates to (2,1) to get dy/dx or gradient is equal to 2. Then find the y intercept of (1)=2(2)+c which gives c=-3. Thus equation of tangent is y=2x-3

Sorry for the late reply

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9609 on: February 05, 2020, 12:58:03 pm »
0
You can always use calculus. dy/dx of x2+ y2-4y-1 is -x/y-2
Then we can use our coordinates to (2,1) to get dy/dx or gradient is equal to 2. Then find the y intercept of (1)=2(2)+c which gives c=-3. Thus equation of tangent is y=2x-3

Sorry for the late reply

Thanks for the reply. This is the solution the book gave. Could someone explain it to me please how the book got it.

For instance, what does it mean about 2 equal roots and why is that the case here?


fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9610 on: February 06, 2020, 11:29:44 am »
+4
Hey there!

When jnlfs2010 said this:
You've got a quadratic. Now get into ax2+bx+c=0 form. Since the line is a tangent, the discriminant of that equation must equal to 0 as there is only one solution. With the value of m, you surely should get the formula of the tangent

They essentially implied expanding and simplifying the quadratic you ended up with, which was \(x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0\), which also happened to be book's solution. They also told you to use the discriminant, and let it equal to zero.

Essentially, when a line cuts the parabola, you have a maximum of two solutions, and a minimum of zero. Equating the line and the parabola will give you the x-values of the points of intersection as roots of a quadratic. If the quadratic is a perfect square, then \(\Delta = 0\), and there is one real root ie. the line is a tangent to the parabola, and does not in fact cut the parabola in two different places like when the discriminant is greater than zero ie. the 'two roots' are equal - there is thus only one real root!

Since we want to find the tangent to the circle at (2, 1), there is only one solution (or if you want to think about it like that, there are two equal solutions!) since there is only one point of contact. Hence, the book (and we!) choose to expand the brackets in this quadratic in x \(x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0\) , then find the discriminant using \(\Delta = b^2 - 4ac\), and let it equal zero, since we only want the one solution. They've got in the book that b2 = 4ac, which is the same as b2 - 4ac equals zero. Since there is only one solution, we expect a perfect square with a quadratic in m, thus finding the gradient of the tangent, then using that to find the equation of the line using point gradient form.

A quicker alternative method if you've learned it is suggested by ^^^111^^^, implicit differentiation. Implicit differentiation is just an extension of the chain rule, ie.

In our case, we have that


Isolating \(\frac{dy}{dx}\) and substituting the coordinates of any point on the given circle will give the gradient of the tangent at that point, just like differentiating any other curve. From here, the steps are the same as the method given in the book.

Hope this clears everything up! :)


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interessant

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9611 on: February 27, 2020, 06:13:04 pm »
0
hey guys, does anyone know how to draw cosec, sec and cot graphs without drawing sin, cos and tan graphs first respectively? i’m taking a really long time to draw these graphs and was wondering if anyone had any tips? thanks :)

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9612 on: February 27, 2020, 07:09:55 pm »
+3
hey guys, does anyone know how to draw cosec, sec and cot graphs without drawing sin, cos and tan graphs first respectively? i’m taking a really long time to draw these graphs and was wondering if anyone had any tips? thanks :)

Yeah, all you have to do is memorise the general shapes of each. You could also memorise equations for where the asymptotes should lie, but I prefer to do it this way (using sec as an example):

Let's say my function is y=2sec(x-pi/3)-1. We know that this function will have asymptotes where cos(x-pi/3)=0. Solve this, you get:

cos(x-pi/3)=0
x-pi/3=pi/2, 2pi-pi/2
x=pi/2+pi/3, 3pi/2+pi/3
=5pi/6, 11pi/6

Graph is periodic in 2pi, so use that to draw in all the asymptotes. Next, all peaks should be halfway between asymptotes, so mark that x-coordinate. To figure out what the y-value is there, simply sub it in to the equation! So, for one of those peaks, it occurs at:

(5pi/6 + 11pi/6)/2
(16pi/6)/2
8pi/6
4pi/3

And the y-coordinate to there is:

y=2/cos(4pi/3 - pi/3)-1
=2/cos(pi) - 1
=2/(-1) - 1
=-3

And you're done! Simply draw the general shapes in and connect the dots. See if you can do the other turning point, now!

Sorry for lack of nice formatting, did this on phone

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9613 on: February 29, 2020, 11:01:26 am »
0
Need some help! (and hopefully this makes sense ha ha)

So I got this question in a quiz the other day, and I'm having some trouble figuring out where I went wrong and what the correct anser is.

f(x) = sin(2arctan(x/3))
Where the domain of sin(x) is restricted to [-pi/2,pi/2]

The questions say to find the domain and range of the function, so as its a composite function, I did this:

                        Domain.        Range.
sin (x).             [-pi/2,pi/2].   [-1,1]
2arctan(x/3).  R                    (-pi, pi)

And because the ran of g must be a subset of the domain of f, it becomes this: (I think)
                        Domain.        Range.
sin (x).             [-pi/2,pi/2].   [-1,1]
2arctan(x/3).  R                    [-pi/2,pi/2].

Then, therfore the range of f(g(x)) is [-1,1], and the domain is R. Is this correct?? I even sketched out the graph and I didnt get any marks for it because apparently the domain was incorrect? How come CAS software graphed it with the domain being R?

EDIT: Also, one more thing, spec teacher told us that in a composite function, the range of fog is the range of f. Is this correct?? In methods we were ways taught that the domain of fog is the domain of g, and we'd have to find the range based on this rule. (after doing all the restrictions and stuff to make the composite function exist)

Any help would be appreciated  :D
« Last Edit: February 29, 2020, 11:06:12 am by ArtyDreams »

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9614 on: February 29, 2020, 01:42:09 pm »
+2
Need some help! (and hopefully this makes sense ha ha)

So I got this question in a quiz the other day, and I'm having some trouble figuring out where I went wrong and what the correct anser is.

f(x) = sin(2arctan(x/3))
Where the domain of sin(x) is restricted to [-pi/2,pi/2]

The questions say to find the domain and range of the function, so as its a composite function, I did this:

                        Domain.        Range.
sin (x).             [-pi/2,pi/2].   [-1,1]
2arctan(x/3).  R                    (-pi, pi)

And because the ran of g must be a subset of the domain of f, it becomes this: (I think)
                        Domain.        Range.
sin (x).             [-pi/2,pi/2].   [-1,1]
2arctan(x/3).  R                    [-pi/2,pi/2].

Then, therfore the range of f(g(x)) is [-1,1], and the domain is R. Is this correct?? I even sketched out the graph and I didnt get any marks for it because apparently the domain was incorrect? How come CAS software graphed it with the domain being R?

EDIT: Also, one more thing, spec teacher told us that in a composite function, the range of fog is the range of f. Is this correct?? In methods we were ways taught that the domain of fog is the domain of g, and we'd have to find the range based on this rule. (after doing all the restrictions and stuff to make the composite function exist)

Any help would be appreciated  :D

If the domain of sin is restricted to [-pi/2, pi/2], then the domain of sin(2arctan(x/3)) is found by solving -pi/2 ≤ 2arctan(x/3) ≤ pi/2. This gives [-3, 3] for the domain. To find the range, note that the range of 2arctan(x/3), for x in [-3, 3], is [-pi/2, pi/2]. So the range of sin(2arctan(x/3)) is [-1, 1].

If you use the "fog table" method, then if you adjust the range of g to be a subset if the domain of f, you also need to make the appropriate adjustments to the domain of g.

Notice that the domain for sin(2arctan(x/3)), if sin is restricted to [-pi/2, pi/2], can not be R, because if x > 3, then 2arctan(4/3) > pi/2.

Your CAS probably showed the graph as existing for all reals because, if there is no restriction on the domain we have: -pi < 2arctan(x/3) < pi, and the maximal domain of sin is R.

In general it is not true that the range of f(g(x)) = range of f(x). For instance, let f(x) = x^2 and g(x) = sin(x). To find the range of f(g(x)), find the range of f(u), where the domain of u = range of g(x) (allowing that the domain of g(x) may not be its maximal domain).

The maximal domain of f(g(x)) is {x in domain(g) : g(x) is in domain(f)}. So you need to find where the domain of f(x) and the range of g(x) intersect, and then find the subset of g's domain that gets mapped to this intersection.