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Author Topic: Compilation of solutions to HARD past HSC papers (2U)  (Read 31854 times)  Share 

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RuiAce

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #15 on: October 22, 2017, 10:55:05 am »
+1
Hey I'm confused as to where I've gone wrong for question 16 part (iv) of the 2016 HSC paper. I've attached my working, and I thought I've differentiated correctly by using the product rule but I'm not getting the same result as Jamon in his worked solutions :/

The computation mistake is responsible for your derivative not matching up. But keep in mind that if your working does not logically flow you can still be penalised.



Ironically enough, I actually think Jamon took a convoluted path. I think the best way of doing that question is to just sketch dy/dt against y (which is a parabola), and then just read off that. Alternatively, from prelim, we know that the axis of symmetry (and hence the x-coordinate of the turning point) of \(y=ax^2+bx+c\) is at \(x=-\frac{b}{2a}\), which can also be used to our advantage.
(I forgot if that minus sign should be there or not.)
« Last Edit: October 22, 2017, 11:11:39 am by RuiAce »

sophiegmaher

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #16 on: October 22, 2017, 11:10:30 am »
0

The computation mistake is responsible for your derivative not matching up. But keep in mind that if your working does not logically flow you can still be penalised.



Ironically enough, I actually think Jamon took a convoluted path. I think the best way of doing that question is to just sketch dy/dt against y (which is a parabola), and then just read off that. Alternatively, from prelim, we know that the axis of symmetry (and hence the x-coordinate of the turning point) of \(y=ax^2+bx+c\) is at \(x=-\frac{b}{2a}\), which can also be used to our advantage.
(I forgot if that minus sign should be there or not.)

That makes so much more sense, thank you so much! And yes, the minus sign should be there :)
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Prerna Kumar

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #17 on: October 22, 2017, 06:09:56 pm »
0
Hi, how would you draw a f(x) graph from the f '(x) graph given in a question?
Thanks :)

Mandynguyenn

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #18 on: February 11, 2018, 03:20:49 pm »
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hi,

from a past paper it mentions "finite region",  in the question: "find the area of the finite region bounded by C and L", to confirm does this just want us to find the area enclosed by C and L?

RuiAce

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #19 on: February 12, 2018, 05:44:03 pm »
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hi,

from a past paper it mentions "finite region",  in the question: "find the area of the finite region bounded by C and L", to confirm does this just want us to find the area enclosed by C and L?
Yeah

talitha_h

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #20 on: June 20, 2018, 07:59:00 pm »
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Could someone please explain question 16b)iv) from the 2013 hsc? Thanks
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itssona

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #21 on: June 20, 2018, 09:42:37 pm »
+2
Could someone please explain question 16b)iv) from the 2013 hsc? Thanks

so we want rate of increase of carp to equal rate of decrease of trout.
rate of increase of carp is given by dN/dt which is -e^0.04t x 0.04


for rate of decrease of carp, we differentiate P and make sure to put a negative sign since we want decrease ;)
so P=P0e^0.02t (this is from the standard formula)

where we know P0 is 10 since 10 carp are introduced

and differentiate this so dP/dt = 0.2e^0.02t

rate of decrease of carp is therefore -0.2e^0.02t

now we can equate such that 0.2e^0.02t = 0.04e^0.04t

5e^0.02t=e^0.04t
ln(5)+ln(e^0.02t)=ln(e^0.04t)
ln(5)=0.02t
t=ln(5)/0.02 = 80.5 approx
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RuiAce

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #22 on: October 10, 2018, 01:23:11 pm »
+1


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Now for the hard part. To obtain some intuition behind solving this part we can do some working backwards.
Observe that if we rearrange \( r < \frac{P}{2} \) we obtain \( P > 2r\). We ask ourselves - is there any good reason for that inequality to be true? And the idea is that yes we can because we know from part i) that \( \boxed{P = r(\theta + 2)} \). Note that \( \theta > 0\), because \(\theta\) is the angle that the sector makes.

Similarly, if we rearrange \( \frac{P}{2(\pi + 1)} < r \) we obtain \( P < r(2\pi + 2) \). And again, there is a good reason for this. It essentially relies on the fact that if \( \theta > 2\pi\), we get something nonsense. That nonsense, is essentially an angle that can't be the angle in the sector we are given. Because how can it make sense to go beyond a full revolution and still have a sector? (In fact, if you go beyond a full revolution, you get the entire circle plus a little bit more.)

With some intuition built up, we prepare to answer the whole thing.



« Last Edit: October 10, 2018, 01:27:55 pm by RuiAce »

emilijab

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #23 on: October 23, 2018, 05:25:46 pm »
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Hiii,

I was having difficulty with Q9 B from the 2010 HSC paper, parts ii and iii. Derivative graphs in general really confuse me and i was wondering, in this question, how do i find the coordinates of the points without an equation to sub the x value into? The sample answers are of no help.

If you could provide some general advice on derivative graphs i'd greatly appreciate it :)
Thanks!!

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #24 on: October 23, 2018, 05:39:08 pm »
0
Remember that to get y=f'(x) from y=f(x) you differentiate the function. In the same way, you integrate y=f'(x) to get y=f(x). What this essentially means is you dont actually care what the graph y=f'(x) actually is, you care a hell of a lot more about the area under the graph, which for the most part in the question is given to you or is easily calculated (for 4<=x<=6).

So with b)ii), you can see that the definite integral of f'(x) has a maximum value of 4, remembering that area above the x-axis is positive and that area under the x-axis is negative. So basically you're looking for when the area on top of the graph is greatest, and its value when it is the greatest. Fortunately for you, it's given in the question :)
A similar thing happens with part iii), the rectangle next to A2 can be calculated to be 6 units squared (3x2), and since A1 and A2 cancel, you get -6.
Because you know that at x=6, f(x)=-6 (from part iii) and that f(x) has a maximum at x=2, and f(x)=4, and that f(x)=0 at x=4 (from the total signed area of f'(x) between 0 and 4), you can accurately draw the graph given in part iv, which is a parabola :)

I guess general tip here is to notice that when there's no equation, theres usually some other way to solve the question that's right in front of you that most times you won't even notice. Sometimes it doesnt have to be as easy as bringing out your integrals and your dxs. When you're given the derivative graph, I guess look for the area under the graph, and not at the graph itself, because more likely than not you're gonna find f(x) anyway!

hope this helps :)
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emilijab

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Re: Compilation of solutions to HARD past HSC papers (2U)
« Reply #25 on: October 23, 2018, 05:48:14 pm »
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Remember that to get y=f'(x) from y=f(x) you differentiate the function. In the same way, you integrate y=f'(x) to get y=f(x). What this essentially means is you dont actually care what the graph y=f'(x) actually is, you care a hell of a lot more about the area under the graph, which for the most part in the question is given to you or is easily calculated (for 4<=x<=6).

So with b)ii), you can see that the definite integral of f'(x) has a maximum value of 4, remembering that area above the x-axis is positive and that area under the x-axis is negative. So basically you're looking for when the area on top of the graph is greatest, and its value when it is the greatest. Fortunately for you, it's given in the question :)
A similar thing happens with part iii), the rectangle next to A2 can be calculated to be 6 units squared (3x2), and since A1 and A2 cancel, you get -6.
Because you know that at x=6, f(x)=-6 (from part iii) and that f(x) has a maximum at x=2, and f(x)=4, and that f(x)=0 at x=4 (from the total signed area of f'(x) between 0 and 4), you can accurately draw the graph given in part iv, which is a parabola :)

I guess general tip here is to notice that when there's no equation, theres usually some other way to solve the question that's right in front of you that most times you won't even notice. Sometimes it doesnt have to be as easy as bringing out your integrals and your dxs. When you're given the derivative graph, I guess look for the area under the graph, and not at the graph itself, because more likely than not you're gonna find f(x) anyway!

hope this helps :)

ahhhh okay, I didn't realise that the area could indicate the value of y like that. Thanks heaps!! :D Glad I came across that sort of question because I wouldn't have had a clue how to do it in the exam.