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March 29, 2024, 03:41:41 pm

Author Topic: VCE Methods Question Thread!  (Read 4803128 times)  Share 

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Shadowxo

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Re: VCE Methods Question Thread!
« Reply #15075 on: July 18, 2017, 05:47:25 pm »
+3
hi sorry to ask, i was wondering how you would anti-differentiate

-x^3 + 4sin^2x + 4cos^2x

the answer is -1/4x^2 + 4x

i understand the first part of the answer but i dont understand how the circular functions would anti-differentiate into 4x

thank you :)

 

It just uses the identity sin2x+cos2x=1
So -x3+4sin2x+4cos2x = -x3+4(sin2x+cos2x) = -x3+4

Edit: If you haven't already memorised this trig identity, do so. It's the fundamental trig identity and not too difficult to memorise :)
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Lavar Big BBB Balls

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Re: VCE Methods Question Thread!
« Reply #15076 on: July 18, 2017, 06:58:57 pm »
0
Hi,

Could I have some help with these http://imgur.com/a/diNmd

Q8a) I'm abit confused, if the question is asking for a probability function, do we not have to find an actual function? We can just create a table that shows the probabilities of some values of x of the function instead?

Q11c) For my crossed out working, why wouldn't this be correct?

Thanks

VanillaRice

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Re: VCE Methods Question Thread!
« Reply #15077 on: July 18, 2017, 07:23:17 pm »
+3
Hi,

Could I have some help with these http://imgur.com/a/diNmd

Q8a) I'm abit confused, if the question is asking for a probability function, do we not have to find an actual function? We can just create a table that shows the probabilities of some values of x of the function instead?

Q11c) For my crossed out working, why wouldn't this be correct?

Thanks
8a) Not too sure. The answer you gave could be described as a probability distribution, but I'm not sure if this is the same as a probability function?

11c) You can only say Pr(A' n B) = Pr(A') x Pr(B) if A and B are independent. This is not stated in the question, so cannot be assumed.

Hope this helps (for the second part, at least)  :)

EDIT: Actually, for 8a, I believe the answer does indeed describe a probability function. It simply lists all the possible values for Pr(P = p), and their corresponding P values. This is similar to tables where the first row contains your y-values, and the second row contains the corresponding x-values.
« Last Edit: July 18, 2017, 07:27:45 pm by VanillaRice »
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Syndicate

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Re: VCE Methods Question Thread!
« Reply #15078 on: July 19, 2017, 12:32:41 pm »
+5
Hi,

Could I have some help with these http://imgur.com/a/diNmd

Q8a) I'm abit confused, if the question is asking for a probability function, do we not have to find an actual function? We can just create a table that shows the probabilities of some values of x of the function instead?

Q11c) For my crossed out working, why wouldn't this be correct?

Thanks

 11c) you will need to draw a Karnaugh Map.
 A intersection B is calculated from part a) and is equal to 1/24

Karnaugh Map:
      A          A'
B  1/24     5/24   1/4
B' 11/24    7/24  3/4
     1/2       1/2      1

Therefore A' intersection B is equal to 5/24

Pr(A'|B) = Pr(A' intersection B)/ Pr(B) = 5/24/ 1/4 = 5/6
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Lavar Big BBB Balls

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Re: VCE Methods Question Thread!
« Reply #15079 on: July 19, 2017, 07:18:54 pm »
0
Thanks all!

amigos

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Re: VCE Methods Question Thread!
« Reply #15080 on: July 20, 2017, 09:20:08 pm »
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Hey all, can you someone please help me with the following question:

Let g be a differentiable function defined for all positive values of x such that the following three conditions hold:
- g(1) = 0
- The tangent to the graph of g at x = 1 is inclined at 45o to the positive x-axis.
- d/dx g(2x) = g'(2)

a) Determine the value for g'(2).
b) Prove that g(2x) = g(x) + (2).
 
Thanks in advance for all help!  :) :)

Shadowxo

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Re: VCE Methods Question Thread!
« Reply #15081 on: July 21, 2017, 01:20:25 pm »
+5
Hey all, can you someone please help me with the following question:

Let g be a differentiable function defined for all positive values of x such that the following three conditions hold:
- g(1) = 0
- The tangent to the graph of g at x = 1 is inclined at 45o to the positive x-axis.
- d/dx g(2x) = g'(2)

a) Determine the value for g'(2).
b) Prove that g(2x) = g(x) + (2).
 
Thanks in advance for all help!  :) :)


Hi :)
I didn't really understand the third condition so I went searching on the internet, do you mean

So condition 1 is saying when x=1, y=0, ie g(1)=0
Condition 2 is saying the gradient at x=1 is 1 (as an angle of 45º corresponds to a gradient of 1). You can also figure this out by saying m = tan(theta) = tan45º = 1. So g'(1)=1.
Condition 3 is saying the gradient of g(x) is the same as the gradient of g(x) dilated by a factor of 1/2 along the x axis, ie when it's stretched towards the y axis (sorry if this is confusing, this background bit isn't as important)

First you need to find the derivative of g(2x). Using the chain rule:

b)For this, you just need to integrate both sides


Hope this helps :)
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KANYEWEST

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Re: VCE Methods Question Thread!
« Reply #15082 on: July 22, 2017, 01:28:40 pm »
0
Hey guys,
I have my Unit 4 Calculus SAC coming up! Apparently it is going to be more "application" based, meaning it will be worded alot,

If you guys have any tips or advice to give for approaching Calculus question, please do let me know! anything would be appriciated!

thanks in advance!
Kanye West x

Sine

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Re: VCE Methods Question Thread!
« Reply #15083 on: July 22, 2017, 01:42:21 pm »
+3
Hey guys,
I have my Unit 4 Calculus SAC coming up! Apparently it is going to be more "application" based, meaning it will be worded alot,

If you guys have any tips or advice to give for approaching Calculus question, please do let me know! anything would be appriciated!

thanks in advance!
reading the question

and I mean this seriously I think most strong students would say that the majority of marks they have lost have been simply for this reason alone. It is harder than it sounds - to have full concentration for the whole duration of your exam/sac.

Another tip would be to be very careful on questions where the answer is used for the remained of the sac. So double check those types of question on the spot, since if you get that wrong the whole rest of your sac will be wrong too and it is up to your teacher's discretion as to how many consequential marks are awarded.

A TART

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Re: VCE Methods Question Thread!
« Reply #15084 on: July 23, 2017, 06:58:19 pm »
0
I'm scratching my head over this question:

Rachel is a keen runner. She is supposed to attend running training five days per week. Rachel finds that if she runs one day, the probability that she will run again the next day is 4/5, and if she does not run one day, the probability that she will not run the next day is 3/4. Suppose that Rachel runs one day:

c)What is the probability that she runs exactly twice in the next three days?

Possible Sample Space:

Run, Run, Not Run--- 4/5 x 4/5 x 1/5 =16/125

Run, Not Run, Run ---- 4/5 x 1/5 x 3/4= 3/25

Not Run, Run, Run-- 1/5 x 1/4 x 4/5 = 1/25

16/125+3/25+1/25= 0.288

The answer my textbook gives is 0.208
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VanillaRice

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Re: VCE Methods Question Thread!
« Reply #15085 on: July 23, 2017, 07:07:30 pm »
+2
I'm scratching my head over this question:

Rachel is a keen runner. She is supposed to attend running training five days per week. Rachel finds that if she runs one day, the probability that she will run again the next day is 4/5, and if she does not run one day, the probability that she will not run the next day is 3/4. Suppose that Rachel runs one day:

c)What is the probability that she runs exactly twice in the next three days?

Possible Sample Space:

Run, Run, Not Run--- 4/5 x 4/5 x 1/5 =16/125

Run, Not Run, Run ---- 4/5 x 1/5 x 3/4= 3/25

Not Run, Run, Run-- 1/5 x 1/4 x 4/5 = 1/25

16/125+3/25+1/25= 0.288

The answer my textbook gives is 0.208
Hi there!
If you look at your second possible event: Run, Not Run, Run ---- 4/5 x 1/5 x 3/4= 3/25
The probability of Rachel running after a day of not running (i.e. Not Run, Run) is 1/4 , so the probability should be 4/5 x 1/5 x 1/4 = 1/25

Therefore, your final answer will be 16/125 + 1/25 + 1/25 = 0.208

Hope this helps  :)
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Gogo14

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Re: VCE Methods Question Thread!
« Reply #15086 on: July 26, 2017, 07:44:55 pm »
0
Having trouble with some prob and understanding: fir discrete
1. What is the significance of Var(x) and why is it =E(x^2)-E(x)^2 and the square root of standard deviation?how are these derived?
2. Why is it that E(ax+b)=E(ax)+E(b)=aE(x)+b?
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RuiAce

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Re: VCE Methods Question Thread!
« Reply #15087 on: July 26, 2017, 07:54:19 pm »
+6
Having trouble with some prob and understanding: fir discrete
1. What is the significance of Var(x) and why is it =E(x^2)-E(x)^2 and the square root of standard deviation?how are these derived?
2. Why is it that E(ax+b)=E(ax)+E(b)=aE(x)+b?


i.e., it is the mean, of the SQUARE of how much you're off the mean of the random variable itself.

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« Last Edit: July 26, 2017, 08:00:48 pm by RuiAce »

koreaboo99

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Re: VCE Methods Question Thread!
« Reply #15088 on: July 27, 2017, 12:59:01 pm »
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I'm having trouble understanding limits and continuity.  How can a limit exist at a certain point if the function itself isn't defined/doesn't exist at that particular point (ie, is discontinuous)?  Like, I don't understand how limits can exist at discontinuous points (I hope I'm making sense)?

Shadowxo

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Re: VCE Methods Question Thread!
« Reply #15089 on: July 27, 2017, 01:15:58 pm »
+3
I'm having trouble understanding limits and continuity.  How can a limit exist at a certain point if the function itself isn't defined/doesn't exist at that particular point (ie, is discontinuous)?  Like, I don't understand how limits can exist at discontinuous points (I hope I'm making sense)?
If you have a limit, you're saying that as you get closer and closer to an x value (eg as x approaches 2), the y value gets closer and closer to a certain value (eg y approaches 4). So, you're saying as x approaches 2, y approaches 4, even if it isn't continuous at x=2 (ie even if the point (2,4) doesn't exist). A limit is approaching that point, not the point itself.
Another example: say you had y=x/x. For any x value except 0, y is 1. But at x=0, y doesn't exist. So we say as x gets closer and closer to 0, y approaches 1 (in this case y is constant) even though (0,1) doesn't exist. So the limit as x approaches 0 for the function x/x is 1.

Hope this helps :) feel free to ask for further clarification
Completed VCE 2016
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Feel free to pm me if you have any questions!