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June 26, 2022, 12:16:38 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 1882295 times)  Share 

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beep boop

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9870 on: February 07, 2022, 07:44:19 pm »
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Can someone please help me w/ 11b and 12aii.

Ty in advance.
« Last Edit: February 07, 2022, 07:46:36 pm by beep boop »
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FlynnGraham

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Re: Specialist 3/4 Question Thread!
« Reply #9871 on: February 27, 2022, 08:50:05 pm »
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how would i do this question:

a graph y = a/(x^2 +2ax +b)  where a,b are a subset of R has range (-infinity, 0) in union with [1/4, infinity) and one of the vertical asymptotes is x=-1
Find a and b

ive attached the answers i get everything up to the blue box, why did they make it equal 1/4, isn't that making an assumption that the maximum of the original will touch the minimum of the reciprical, but thats not always the case for example with y = x^2 + 2x + 3?

Hello anyone, I'm having trouble with this question and the thread on it wasn't very helpful. I'm trying to approach it by finding the values of the denominator function, but I've gotten stuck. So far i have that there's obviously an intercept at (-1,0) and the turning point has a y value of -4 because of the range, but i cant think of anything else. any help would be very appreciated!

Indiantiger

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9872 on: March 03, 2022, 10:04:52 pm »
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Hello there, I suck at proving questions (especially vector proving questions). Do I need to know how to prove vectors for exams?

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9873 on: March 04, 2022, 05:26:15 pm »
+2
Can someone please help me w/ 11b and 12aii.

Ty in advance.

Apologies if this is no longer relevant - missed this the first time.

You know that \(z_3 = \bar{z_2}\) from part i). You can then use the sum and product of roots formulas to figure out the coefficients from the hints given.

For the second question, try expanding out and equating the real and imaginary parts to zero. It's not the hardest to follow from there.

Hello anyone, I'm having trouble with this question and the thread on it wasn't very helpful. I'm trying to approach it by finding the values of the denominator function, but I've gotten stuck. So far i have that there's obviously an intercept at (-1,0) and the turning point has a y value of -4 because of the range, but i cant think of anything else. any help would be very appreciated!

With this approach, you're effectively solving \(x^2+2ax+b = 0 \implies x = \frac{-2a \pm \sqrt{4a^2-4b}}{2} = -a \pm \sqrt{a^2-b}\). You're told that one of the asymptotes is at -1, so \((-1)^2 -2a+b = 0 \implies b = 2a-1\). Substituting, we then have the asymptotes being at \(-a \pm \sqrt{a^2-(2a - 1)} = -a \pm (a-1)\). Clearly, one of the asymptotes is at -1, as expected and the other one is at \(-2a+1\).

I'm not particularly sure why you think the turning point has a y-value of -4 because of the range. The range and the y-values are almost analogous. The turning point must be at 1/4. Since the function is just the reciprocal of a quadratic multiplied by a constant, the extrema of the quadratic must also be extrema of the reciprocal, so what we are effectively looking for is the x-value of the turning point of the quadratic. Recall that for a quadratic of the form \(ax^2+bx+c\) that the turning point is located at \(x = -\frac{b}{2a}\).

Hopefully that's enough to get you a better understanding and make the problem easier to solve.

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KandyZetVo

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9874 on: March 11, 2022, 02:44:23 pm »
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19 .Consider ∣−a∣ = a.
a Find a value for a where ∣−a∣ ≠ a.
b For what values of a does ∣−a∣ = a hold true?
20 When does ∣1 − x∣ = 1 − x and when does ∣1 − x∣ = x − 1?
Hi i have been stuck with this question , wonder if anyone could help ? Thank you

mabajas76

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9875 on: March 11, 2022, 06:53:33 pm »
+1
For a:
The modulus is similar to just saying "How far from zero", so mod(-7) is just 7 from zero, same with mod (7). So a value where the negative is somehow a different magnitude...I'm not too sure, maybe a complex number (imaginary) would work.
b) All real numbers should be the answer.
c) If mod(a-b) has a>b then it is the same as a-b, as it will make a positive number every time. So mod (1-x)=1-x when x<1 and it equals x-1 when x>1 I believe. Sketching both of them on a graph would help.
"Don't give up, and don't put too much effort into things that don't matter"-Albert Einstein, probably.

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9876 on: March 11, 2022, 11:04:00 pm »
+1
Adding on to the answer above:
\[\text{Recall that } |x| = \begin{cases}x & \text{for } x \geq 0  \\ -x & \text{for } x < 0.\end{cases}\]

What can you then deduce about \(|x|\)? You can then use the same intuition for the next part.

b) All real numbers should be the answer.

Might want to be a bit careful here.
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mabajas76

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9877 on: March 13, 2022, 12:44:25 am »
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Adding on to the answer above:
\[\text{Recall that } |x| = \begin{cases}x & \text{for } x \geq 0  \\ -x & \text{for } x < 0.\end{cases}\]

What can you then deduce about \(|x|\)? You can then use the same intuition for the next part.

Might want to be a bit careful here.
My bad, if x<0, then the mod of a negative is positive which can't equal x as it is negative. So all real positive numbers and 0.
"Don't give up, and don't put too much effort into things that don't matter"-Albert Einstein, probably.

beep boop

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9878 on: April 04, 2022, 09:55:00 pm »
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Hi all,

Can someone help me w/ q48 parts d and e? Also q48 part d?

Thanks,

beep boop
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yr 12 stuff :)

beep boop

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9879 on: April 22, 2022, 07:59:49 pm »
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Hi All,

Was wondering if someone could help me with the highlighted q's?
I've found the first and second derivatives for both questions and tried to equate but nothing works.

ty,
beep boop
class of '22
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'22: psych, methods, spesh, chem, eng lang
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yr 12 stuff :)

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9880 on: April 22, 2022, 11:30:47 pm »
+1
Hi All,

Was wondering if someone could help me with the highlighted q's?
I've found the first and second derivatives for both questions and tried to equate but nothing works.

ty,
beep boop

Hello,

I found this website very useful for solving second-order differential equations, so maybe you can check it out:
https://www.mathsisfun.com/calculus/differential-equations-second-order.html

Also, I'm not sure if this is in the study design btw (but I don't know about school SACs, if someone can confirm this that would be great!)