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September 20, 2018, 01:28:03 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 877178 times)  Share 

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PopcornTime

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9285 on: June 01, 2018, 11:06:22 pm »
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Can someone help me with answering this AMC question?


noregret

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9286 on: June 06, 2018, 07:22:09 pm »
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Hello, I have three questions about Sequences and Series.

1.How do I solve this question:  The sum of the first six terms of a geometric progression is 63 and the ratio of the fourth term to the sum of the second and third terms is -4. Find the first term and common ratio of the progression.

2. How do I prove that the sum n terms of 1^2+(1^2+2^2)^2+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+............. is 1/12(n)(n+1)^2(n+2)?

3. The sum of the first (n+1) terms of an arithmetic series is 5n^2+13n+8. Find the 1st term, the 10 term and the nth term. I get that answer for the 1st term is S(0+1)=5(0)^2+13(0)+8=8, however I do not understand the answer for the 10th and the nth term. If Un=Sn-Sn-1, then Un+1should equal to Sn-1-Sn-2?

TheBigC

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9287 on: June 15, 2018, 12:44:53 pm »
+1
Hello, I have three questions about Sequences and Series.

1.How do I solve this question:  The sum of the first six terms of a geometric progression is 63 and the ratio of the fourth term to the sum of the second and third terms is -4. Find the first term and common ratio of the progression.

2. How do I prove that the sum n terms of 1^2+(1^2+2^2)^2+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+............. is 1/12(n)(n+1)^2(n+2)?

3. The sum of the first (n+1) terms of an arithmetic series is 5n^2+13n+8. Find the 1st term, the 10 term and the nth term. I get that answer for the 1st term is S(0+1)=5(0)^2+13(0)+8=8, however I do not understand the answer for the 10th and the nth term. If Un=Sn-Sn-1, then Un+1should equal to Sn-1-Sn-2?


1.
Therefore, for a geometric series:





Thus,

A simple quadratic is produced:

Thus,
(Common ratio)

Subbing into (1) gives:

Therefore,
(First term)

2. USE MATHEMATICAL INDUCTION OR OTHERWISE.
3.

For first term:

For 10th term:

For the nth term:

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Buoyancy

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9288 on: June 20, 2018, 10:03:09 pm »
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hey guys i need help with finding the area of A2, I can't get my head around it need some explanation.

thanks
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TheBigC

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9289 on: June 21, 2018, 04:43:33 pm »
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hey guys i need help with finding the area of A2, I can't get my head around it need some explanation.

thanks

Refer to attachment
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Guideme

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9290 on: June 30, 2018, 06:04:22 pm »
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Can anyone answer question 4 pls

Thank you in advanced !
:0 :)

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9291 on: July 01, 2018, 08:27:26 pm »
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Can anyone answer question 4 pls

Thank you in advanced !

Formula for Euler's Method is: yn+1 = yn + h.f'(xn,yn)
I can link you videos on youtube if you need to understand how this formula came about.

Answering your question,
x0=0                                          y0=1
x1= 0+0.1=0.1                           y1= 1 + 0.1 x  (13) = 1.1
x2= 0.1+0.1=0.2                        y2= 1.1 + 0.1 x  (1.13) = 1.2331
x3= 0.2+0.1=0..3                       y3= 1.2331 + 0.1 x (1.23313) = 1.4206
x4= 0.3+0.1=0.4                        y4= 1.4206 + 0.1 x  (1.42063) = 1.7073
x5= 0.4+0.1=0.5                        y5= 1.7073 + 0.1 x  (1.70733) = 2.2049

y is approximately 2.2049

You can also use the spreadsheet function in the CAS/Ti-Inspire to make this process a lot faster. Hope it helps!

Mattjbr2

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9292 on: July 04, 2018, 09:56:17 am »
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Could someone please double check for me whether this is correct? How is it possible for c to equal 4?

We know dx/dt=4 when x=0
but x=0 when t=4
hence dx/dt=4 when t=4

So then aren't we supposed to sub in t=4 and dx/dt=4 to get c=44?

Thanks in advance! :)
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RuiAce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9293 on: July 04, 2018, 09:58:01 am »
+1
Could someone please double check for me whether this is correct? How is it possible for c to equal 4?

We know dx/dt=4 when x=0
but x=0 when t=4
hence dx/dt=4 when t=4

So then aren't we supposed to sub in t=4 and dx/dt=4 to get c=44?

Thanks in advance! :)
At first glance, I think that \( \frac{dx}{dt} = 4\) when \( \boxed{x=0}\) has been typo'd. That should've read \( t=0\).

Else, yeah I agree the answers are wrong.
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Mattjbr2

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9294 on: July 04, 2018, 10:03:32 am »
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At first glance, I think that \( \frac{dx}{dt} = 4\) when \( \boxed{x=0}\) has been typo'd. That should've read \( t=0\).

Else, yeah I agree the answers are wrong.

Both the old and the new (corrected) versions of the worked solutions - as well as the back of the textbook - all agree on the same supposedly incorrect answer!  Haha. The back of the textbook has never been wrong as far as i remember.  :-\ That's so strange
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RuiAce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9295 on: July 04, 2018, 10:08:10 am »
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Both the old and the new (corrected) versions of the worked solutions - as well as the back of the textbook - all agree on the same supposedly incorrect answer!  Haha. The back of the textbook has never been wrong as far as i remember.  :-\ That's so strange
If the answers have never been wrong then the question is what's wrong. ;)

(But tbh, yeah, the question is wrong here imo. We know that \(x=0\) when \(t=4\), but that doesn't necessarily mean \(x=0\) ONLY when \(t=4\). What if \(x=0\) when \(t=4, 8\)?

When you consider that, the question suddenly doesn't become so doable.)
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Mattjbr2

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9296 on: July 04, 2018, 10:41:09 am »
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If the answers have never been wrong then the question is what's wrong. ;)

(But tbh, yeah, the question is wrong here imo. We know that \(x=0\) when \(t=4\), but that doesn't necessarily mean \(x=0\) ONLY when \(t=4\). What if \(x=0\) when \(t=4, 8\)?

When you consider that, the question suddenly doesn't become so doable.)

Thanks so much!
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Yertle the Turtle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9297 on: July 12, 2018, 03:10:48 am »
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Could someone explain how to reach an equation for this question, I've been struggling with it for a while.
Thanks :)
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9298 on: July 12, 2018, 07:58:12 am »
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Could someone explain how to reach an equation for this question, I've been struggling with it for a while.
Thanks :)

Rate of population increase is proportional to current population size can be expressed as

where k is some real constant.

Hence, where k' = 1/k.

Then, antidifferentiating with respect to N, we have (from context we can assume N > 0).

At this stage, before solving simultaneously for k' and c, it is convenient to write N in terms of t. This gives



We then use the facts that:
and to solve simultaneously for k' and c in terms of initial population. This gives:

and

And substituting back into our equation for N, we have (with a bit of tidying up):



An alternative (much slicker) method is to begin by recognising that any solution to an equation of the form will be a function of the form Then substitute in known values (t = 0, f(0) = A, and t = 5, f(5) = 2A) to solve for b in terms of A. This gives This agrees with the above approach, because b is the reciprocal of k'.
« Last Edit: July 12, 2018, 04:47:20 pm by S_R_K »

Mattjbr2

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9299 on: July 12, 2018, 03:13:59 pm »
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Hey guys, if we want to give the interval for which f(x)=x^2 is strictly increasing, do we say (0,infinity) or [0,infinity)? In other words, do we include a stationary point? The A+ Methods Exam 1 practice exams book did include the stationary point. I didn't. Do i mark myself wrong, or nah?
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