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April 29, 2024, 05:07:29 pm

Author Topic: 4U Maths Question Thread  (Read 665370 times)  Share 

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RuiAce

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Re: 4U Maths Question Thread
« Reply #1065 on: April 07, 2017, 07:58:20 pm »
0
I think it should be sufficient to differentiate and sub as long as you justify it (i.e tangent only intersects the ellipse once)
There's no ellipse here. That's a 5th power on y there.

I'm not sure if there was a typo in the original question or not. Because if the 5 is replaced by two we have \(x^2+2xy+y^2=4\) which is apparently a pair of parallel lines



Edit: Actually it does work, but we can't use any conic theorems because of the \(y^5\)
« Last Edit: April 07, 2017, 08:04:08 pm by RuiAce »

RuiAce

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Re: 4U Maths Question Thread
« Reply #1066 on: April 07, 2017, 08:15:18 pm »
+1



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« Last Edit: April 08, 2017, 09:26:09 am by RuiAce »

hanaacdr

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Re: 4U Maths Question Thread
« Reply #1067 on: April 08, 2017, 10:32:35 am »
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Hi
Could i get some help on question 3b

thank you

RuiAce

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Re: 4U Maths Question Thread
« Reply #1068 on: April 08, 2017, 07:28:29 pm »
+1
Hi
Could i get some help on question 3b

thank you







hanaacdr

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Re: 4U Maths Question Thread
« Reply #1069 on: April 10, 2017, 12:03:07 pm »
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Hi,
could i get some help on this question please

jakesilove

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Re: 4U Maths Question Thread
« Reply #1070 on: April 10, 2017, 02:30:35 pm »
+2
Hi,
could i get some help on this question please

Hey! Firstly, since all the coefficients are real, we know that either ONE of the roots is real, or THREE of the roots are real (as imaginary roots will come in complex conjugates).

The easiest way to answer the first half is by differentiating the function, it proving that it is non-decreasing. So,



For a>0, this derivative is ALWAYS positive. Thus, there will only be one real root, as the graph will never 'turn around'. It's helpful to sketch an example of a non-decreasing function (ie. gradient is always positive), to prove to the marker that you know why this fact results in one real root.

Now, if two roots are equal, we can right them as



Using sum and product rules.



and



We know from above that



so




Now, subbing alpha into the original equation, we get





Not sure where we are going with this; let's keep going?
 
We are trying to prove that



Subbing it what we've found above;



Which looks like it equals zero! Bit of a round about way of getting there, but hey, it works!
« Last Edit: April 10, 2017, 02:46:53 pm by jakesilove »
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ellipse

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Re: 4U Maths Question Thread
« Reply #1071 on: April 10, 2017, 04:08:46 pm »
+2
Hi,
could i get some help on this question please

Here's an alternate method for part ii
99.55
unsw engo/sci

mx2-99
mx1-98
phys-94
chem-94
eng-88 (rip me)

jakesilove

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Re: 4U Maths Question Thread
« Reply #1072 on: April 10, 2017, 04:10:06 pm »
+2
Here's an alternate method for part ii

That's a heaps better method. Cheers for posting it up!
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ellipse

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Re: 4U Maths Question Thread
« Reply #1073 on: April 10, 2017, 04:20:06 pm »
+2
That's a heaps better method. Cheers for posting it up!

No problem :)
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hanaacdr

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Re: 4U Maths Question Thread
« Reply #1074 on: April 10, 2017, 07:04:09 pm »
0
Hey! Firstly, since all the coefficients are real, we know that either ONE of the roots is real, or THREE of the roots are real (as imaginary roots will come in complex conjugates).

The easiest way to answer the first half is by differentiating the function, it proving that it is non-decreasing. So,



For a>0, this derivative is ALWAYS positive. Thus, there will only be one real root, as the graph will never 'turn around'. It's helpful to sketch an example of a non-decreasing function (ie. gradient is always positive), to prove to the marker that you know why this fact results in one real root.

Now, if two roots are equal, we can right them as



Using sum and product rules.



and



We know from above that



so




Now, subbing alpha into the original equation, we get





Not sure where we are going with this; let's keep going?
 
We are trying to prove that



Subbing it what we've found above;



Which looks like it equals zero! Bit of a round about way of getting there, but hey, it works!


THANK YOUU!

beau77bro

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Re: 4U Maths Question Thread
« Reply #1075 on: April 13, 2017, 09:46:03 am »
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Hey guys could someone help me with 3b and c, I can't quite work it out algebraically

RuiAce

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Re: 4U Maths Question Thread
« Reply #1076 on: April 13, 2017, 10:21:04 am »
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(Image removed from quote.)

Hey guys could someone help me with 3b and c, I can't quite work it out algebraically

___________

Side note: It may be worth mentioning that the given questions are the inverse hyperbolic sine and tangent functions, i.e. \( \sinh^{-1}x\), \( \tanh^{-1}x\). What we are trying to find are the exponential forms of the original hyperbolic functions \(\sinh x\) and \(\tanh x\)
« Last Edit: April 13, 2017, 10:23:26 am by RuiAce »

beau77bro

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Re: 4U Maths Question Thread
« Reply #1077 on: April 13, 2017, 12:00:02 pm »
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thanks rui, i didnt even see that you could just cancel out the y squared, rookie mistake sorry ahhaha

beau77bro

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Re: 4U Maths Question Thread
« Reply #1078 on: April 13, 2017, 12:02:47 pm »
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I hope this question is better, not sure how to do 8a, the rest is fine but just abit confused on whether its algebraic or word and where to go? Sub in value less than 0 and show it doesn't work because it x^2 +3x/2 has to be greater than 0?

RuiAce

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Re: 4U Maths Question Thread
« Reply #1079 on: April 13, 2017, 03:28:22 pm »
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(Image removed from quote.)

I hope this question is better, not sure how to do 8a, the rest is fine but just abit confused on whether its algebraic or word and where to go? Sub in value less than 0 and show it doesn't work because it x^2 +3x/2 has to be greater than 0?
The natural domain of the first is x>0 but for the second it's x>0 AND x<-3/2