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March 29, 2024, 12:21:11 am

Author Topic: VCE Methods Question Thread!  (Read 4802379 times)  Share 

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MB_

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Re: VCE Methods Question Thread!
« Reply #18390 on: February 13, 2020, 02:10:10 pm »
+2
hi, i'm having a bit of trouble with this question, especially cii. i got c=1 but im not sure as the answer is c=1 and c>2

i've had a look at the solutions but im still a bit confused

thank you!
The solutions for \(x\) are of the form \(x=\pm\sqrt{c-1}\). When \(c\in{(1,2]},\) there are two real solutions for \(x\) (as \(x\geq-1\)). So \(c=1\) or \(c>2\) gives one real solution.
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MoonChild1234

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Re: VCE Methods Question Thread!
« Reply #18391 on: February 13, 2020, 02:35:31 pm »
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why are there two real solutions for those values? wouldnt a value like 2.54 have 2 real solutions as well?

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Re: VCE Methods Question Thread!
« Reply #18392 on: February 13, 2020, 02:40:50 pm »
+1
hi, i'm having a bit of trouble with this question, especially cii. i got c=1 but im not sure as the answer is c=1 and c>2

i've had a look at the solutions but im still a bit confused

thank you!
The second I saw this, I recalled...oh the 2017 NHT exam! The worst part is, VCAA never provide decent solutions for the NHT exams  :'(

For c)i), you just need to make the RHS = x and solve for c, so it isn't too bad and you get c < 1. As is the case with inverses, they only (in almost all circumstances unless you get into the ridiculous powers...) intersect when x=y, thus making the RHS = x is an acceptable way to solve and successfully get the two marks.

For c)ii), it's a little more difficult. I too got c=1, and upon seeing the solutions, went to my teacher (I did this exam about 3 weeks before the real one. She couldn't explain it either other than trying to do it by inspection! Again, you have to use g(x) = x (as this substitutes for the inverse), work through and you would get to x^2 + 1 - c = 0, and this would get you the c = 1 solution.

However, if you chose not to find the solution by discriminant, and instead found x, you would find x = root(c-1) and -root(c-1). As c had to fit into the domains of both g and g^-1, As the domain of g^-1 is [-1,infinity), and thus x must fit into it, you substitute x = -1 into the equation x < -root(c-1), making it 1 < root (c-1) (as subbing it into the positive would not get any real numbers due to the square root and negative).

thus root(c-1) > 1, and c >2.

Sorry for the awful explanation!

why are there two real solutions for those values? wouldnt a value like 2.54 have 2 real solutions as well?
No, as x > -1 or x = 1. If c = 2.54, the smaller x-value will be less than -1, thus out of the domain of the inverse.
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MB_

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Re: VCE Methods Question Thread!
« Reply #18393 on: February 13, 2020, 02:43:19 pm »
+1
why are there two real solutions for those values? wouldnt a value like 2.54 have 2 real solutions as well?
As \(x\geq -1\), subbing in \(c=2.54\) would give the one solution of \(x=\sqrt{1.54}=1.24\)
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MoonChild1234

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Re: VCE Methods Question Thread!
« Reply #18394 on: February 13, 2020, 05:40:05 pm »
0
The second I saw this, I recalled...oh the 2017 NHT exam! The worst part is, VCAA never provide decent solutions for the NHT exams  :'(

For c)i), you just need to make the RHS = x and solve for c, so it isn't too bad and you get c < 1. As is the case with inverses, they only (in almost all circumstances unless you get into the ridiculous powers...) intersect when x=y, thus making the RHS = x is an acceptable way to solve and successfully get the two marks.

For c)ii), it's a little more difficult. I too got c=1, and upon seeing the solutions, went to my teacher (I did this exam about 3 weeks before the real one. She couldn't explain it either other than trying to do it by inspection! Again, you have to use g(x) = x (as this substitutes for the inverse), work through and you would get to x^2 + 1 - c = 0, and this would get you the c = 1 solution.

However, if you chose not to find the solution by discriminant, and instead found x, you would find x = root(c-1) and -root(c-1). As c had to fit into the domains of both g and g^-1, As the domain of g^-1 is [-1,infinity), and thus x must fit into it, you substitute x = -1 into the equation x < -root(c-1), making it 1 < root (c-1) (as subbing it into the positive would not get any real numbers due to the square root and negative).

thus root(c-1) > 1, and c >2.

Sorry for the awful explanation!
No, as x > -1 or x = 1. If c = 2.54, the smaller x-value will be less than -1, thus out of the domain of the inverse.

thank you so much! this really helped :)

so basically, when unsure, just sub in the end values of the domain?

Poojapriya

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Re: VCE Methods Question Thread!
« Reply #18395 on: February 13, 2020, 09:24:39 pm »
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Hey Guys!

I needed help with a question for Methods 1/2. The question is relating to determining the rule of a quadratic from a graph. Here's the question:

A parabola has the equation y=(ax+b)(x+c). When x=5 ,its graph cuts the x-axis and when y = −10 the graph cuts the y-axis.
a. Show that y= ax^2 +(2−5a)x−10.
b. Express the discriminant in terms of a.
c. If the discriminant is equal to 4, find the equation of the parabola and the coordinates of its other x-intercept.

It would be great if you could attach a picture of the steps used as well!

Cheers,
Prerna

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Re: VCE Methods Question Thread!
« Reply #18396 on: February 13, 2020, 09:50:03 pm »
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Hello, hope everyone is having a wonderful day!

I have a question, if you don't mind. I have tried to solve this problem, but it does not make sense to me. An explanation would be really appreciated!

There is a masonry arch bridge of span 50m. The line AC is the water level and B is the highest point of the bridge, and A is the origin. The biggest height of the bridge is 4.5 metres.  A floating platform 20 m wide is towed under the bridge. What is the greatest height of the deck above water level if the platform is to be towed under the bridge with at least 30 cm horizontal clearance on either side?

The first part is fine, but I'm not sure how we can determine the height of the platform from the horizontal clearance.

Thank you!
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Re: VCE Methods Question Thread!
« Reply #18397 on: February 14, 2020, 12:01:48 am »
+3
Hey there!

Diagram
A diagram is really important to this question, it's enclosed above!

It's also important to notice that the curvature of the underside of the bridge is in fact a parabola. Another important thing to note is that the horizontal distance required between the axis and the parabola is 10.3m ie. a total of 20.6m all the way across.

Once we have this information, we can find the equation of the parabola, and thus find the height of the deck above the water level. Perhaps the easiest form for us to express the parabola in this case is y=a(x-b)(x-c) - we know both the intercepts (given that A is the origin (0,0) and that C is the point (50,0), and the vertex B is (25,4.5)). We also note that the parabola is upside down. From this, we can surmise that for some constant a, y=ax(50-x) (satisfying both A and C). Also, since B also lies on the parabola, we can substitute those points in to find the constant a - ie. 4.5=a*252 to find that the parabola has equation \(\frac{9}{1250}x(50-x)\). From here, we note from before that the platform requires 10.3m of space between the axis and the parabola ie. what is the height of the bridge at (25-10.3)m. We can substitute this into our equation to get that the maximum clearance is f(14.7) - or approximately 3.74m.

Hope this helps :)
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Jimmmy

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Re: VCE Methods Question Thread!
« Reply #18398 on: February 14, 2020, 11:42:21 am »
+1
thank you so much! this really helped :)

so basically, when unsure, just sub in the end values of the domain?
Exactly, and from there you would sub in a couple of numbers at intervals (probably 1.5 and 2), and since 1.5 wouldn't fit but 2 would be on the brink, you would quickly realise c>2 only has one solution (as do numbers greater that 2), so it'd be pretty straightforward from there.

Keep in mind, that question was the last one of the examination, meaning it was probably designed to differentiate the 40s from the high 40s/50s. Good to know, but far from your typical Methods question. All you can do there is do your best to apply the principles you've learnt!
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Re: VCE Methods Question Thread!
« Reply #18399 on: February 14, 2020, 11:30:14 pm »
+2
Hey Guys!

I needed help with a question for Methods 1/2. The question is relating to determining the rule of a quadratic from a graph. Here's the question:

A parabola has the equation y=(ax+b)(x+c). When x=5 ,its graph cuts the x-axis and when y = −10 the graph cuts the y-axis.
a. Show that y= ax^2 +(2−5a)x−10.
b. Express the discriminant in terms of a.
c. If the discriminant is equal to 4, find the equation of the parabola and the coordinates of its other x-intercept.

It would be great if you could attach a picture of the steps used as well!

Cheers,
Prerna

Hey there! Totally missed this question, sorry about that!

I think for this question, it's best to guide you as opposed to giving you the answer outright with all the steps. If you need further help though, please do ask! :)
a) Showing that the equation is \(y=ax^2+(2-5a)x-10\) is fundamentally using the fact that x=5 is an x-intercept and y=-10 is a y-intercept. ie. we can substitute x=5, y=0 and x=0, y=-10 into the equation \(y=(ax+b)(x+c)\) and leave a as some constant to find the result given in the question.
b) Recall that for some quadratic \(y=ax^2+bx+c\), that the discriminant is \(\Delta = b^2 - 4ac\). Use this fact to substitute in values for a, b and c as given in the equation you've just worked out in part a), expand the brackets and simplify.
c) Here, you're going to be solving another quadratic, but in terms of a. You can use any of the methods you've learned (completing the square, factorisation or the quadratic formula), but remember that the discriminant is equal to 4, not 0 like many other quadratics that you are asked to solve.

Hope this helps :)
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cutiepie30

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Re: VCE Methods Question Thread!
« Reply #18400 on: February 23, 2020, 02:29:41 am »
0
Hello everyone ,

for this question i was wondering how i would do it using x' and y' notation so i could states the sequence of transformations from one graph to the other.

Thanks  :)

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Re: VCE Methods Question Thread!
« Reply #18401 on: February 23, 2020, 09:18:50 am »
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Hi, can anyone help me with this question?

Find the values of p for which the equation px² + 2(p+2)x + p + 7 = 0, has no real solutions.

Thank you

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Re: VCE Methods Question Thread!
« Reply #18402 on: February 23, 2020, 09:48:16 am »
+3
Hi, can anyone help me with this question?

Find the values of p for which the equation px² + 2(p+2)x + p + 7 = 0, has no real solutions.

Thank you

Hi!

For this question, you need to use the discrimnant to find the values of p.

For there to be less than no solutions, the discriminant has to be less than 0.

b^2 - 4ac < 0

And then you substitute the values and solve
i.e. b= 2p+4
a = p
c = 7

Let me know if you need further help in solving the equation!

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Re: VCE Methods Question Thread!
« Reply #18403 on: February 23, 2020, 10:53:11 am »
+6
Hi!

For this question, you need to use the discrimnant to find the values of p.

For there to be less than no solutions, the discriminant has to be less than 0.

b^2 - 4ac < 0

And then you substitute the values and solve
i.e. b= 2p+4
a = p
c = 7

Let me know if you need further help in solving the equation!

Just be careful here, the constant term c is defined as cx0 - here, c would be p + 7. This is otherwise completely correct! :) It's just important to be wary that just because something is a pronumeral, doesn't mean it's not a constant :)

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Re: VCE Methods Question Thread!
« Reply #18404 on: February 23, 2020, 07:41:41 pm »
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A tug-o-war team produces a tension in a rope described by the rule T=290(8t−0.5t2−1.4) units, where t is the number of seconds after commencing the pull.

1. Sketch a graph of T against t, stating the practical domain.
2. What is the greatest tension produced during a ‘heave’?

At the moment i am clueless and don't know where to start so it would be great if someone could help! :)