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February 17, 2019, 10:16:46 am

Author Topic: 3U Maths Question Thread  (Read 422312 times)  Share 

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david.wang28

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Re: 3U Maths Question Thread
« Reply #3900 on: February 05, 2019, 03:43:33 pm »
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You've done nothing wrong. Now cancel out the \( (1+8x)^2\) in the left fraction's denominator and right fraction's denominator (remember - you're multiplying two fractions) and then expand and simplify.
I've got it now; thanks for the help Rui! :)
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Re: 3U Maths Question Thread
« Reply #3901 on: February 05, 2019, 03:50:37 pm »
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Hi,
I need help with the attached question! I need help with Part iii of Q7.
Thanks!

RuiAce

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Re: 3U Maths Question Thread
« Reply #3902 on: February 05, 2019, 04:27:12 pm »
+3
Hi,
I need help with the attached question! I need help with Part iii of Q7.
Thanks!
\[ \text{We know that}\\ \begin{align*} \frac{1}{2\times 2} &< \frac{1}{1\times 2}\\ \frac{1}{3\times 3}&< \frac{1}{2\times 3}\\ \frac{1}{4\times 4} &< \frac{1}{3\times 4}\\ &\vdots\\ \frac{1}{98\times 98} &< \frac{1}{97\times 98}\\ \frac{1}{99\times 99}&< \frac{1}{98\times 99} \end{align*} \]
\[ \text{So upon adding we have}\\ \begin{align*}&\quad \frac{1}{2^2} +\frac{1}{3^2} + \frac{1}{4^2} + \dots + \frac{1}{98^2}+\frac{1}{99^2}\\ & < \frac{1}{1\times 2} + \frac{1}{2\times 3} + \frac{1}{3\times 4} + \dots + \frac{1}{97\times 98} + \frac{1}{98\times 99}\\ &= \frac{99}{100} - \frac{1}{99\times 100} \tag{result from ii)}\\ &< \frac{99}{100} \end{align*}\\ \text{as required.} \]
__________________________________
\[ \text{We also know that}\\ \begin{align*} \frac{1}{2\times 2} &> \frac{1}{2\times 3}\\ \frac{1}{3\times 3}&> \frac{1}{3\times 4}\\ &\vdots\\ \frac{1}{99\times 99} &> \frac{1}{99\times 100} \end{align*} \]
\[ \text{So upon adding we have}\\ \begin{align*}&\quad \frac{1}{2\times 2}+ \frac{1}{3\times 3} + \dots + \frac{1}{99\times 99}\\ &> \frac{1}{2\times 3} + \frac{1}{3\times 4} + \dots + \frac{1}{99\times 100}\\ &+ \frac{99}{100} - \frac{1}{1\times 2} \tag{result from ii)}\\ &= \frac{49}{100} \end{align*}\\ \text{as required.}\]

spnmox

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Re: 3U Maths Question Thread
« Reply #3903 on: February 09, 2019, 10:05:29 pm »
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Someone send help please!

What is the focus if the locus equation I calculated is x^2=16a(y-2a) ?

RuiAce

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Re: 3U Maths Question Thread
« Reply #3904 on: February 09, 2019, 10:43:25 pm »
+1
Someone send help please!

What is the focus if the locus equation I calculated is x^2=16a(y-2a) ?
\[ \text{Recall that }x^2=4ay\\ \text{has focus at }(0,a) \]
\[ \text{Thus the translated parabola }(x-h)^2=4a(y-k)\\ \text{has focus at }(h, a+k) \]
Therefore your parabola has focus at \( (0, 4a + 2a) \), i.e. \( (0, 6a)\)

spnmox

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Re: 3U Maths Question Thread
« Reply #3905 on: February 09, 2019, 11:37:45 pm »
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M (3,9) and N(-2,4) lie on the parabola y=x^2 and P (p,p^2) is a variable point on the parabola. Find the exact perpendicular distance from P to the line MN.

As the numerator is a quadratic, I got a little confused. At what point are you able to remove the absolute value signs in the numerator and why are you able to?

RuiAce

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Re: 3U Maths Question Thread
« Reply #3906 on: February 10, 2019, 08:57:10 am »
+2
M (3,9) and N(-2,4) lie on the parabola y=x^2 and P (p,p^2) is a variable point on the parabola. Find the exact perpendicular distance from P to the line MN.

As the numerator is a quadratic, I got a little confused. At what point are you able to remove the absolute value signs in the numerator and why are you able to?
I don't believe you can?
\[ \text{The line has equation }x - y + 6 = 0\\ \text{so plugging into the perpendicular distance formula gives}\\ d = \frac{|p - p^2 + 6|}{\sqrt{2}} \]
That should be the correct answer. There's nothing we can do to simplify this, because we can't show that \(-p^2 + p + 6\) is a positive definite or negative definite quadratic to exploit relevant properties. If the answer dropped the absolute values without reasonable justification I would consider that a mistake.
« Last Edit: February 10, 2019, 09:03:51 am by RuiAce »

spnmox

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Re: 3U Maths Question Thread
« Reply #3907 on: February 10, 2019, 05:10:22 pm »
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I don't believe you can?
\[ \text{The line has equation }x - y + 6 = 0\\ \text{so plugging into the perpendicular distance formula gives}\\ d = \frac{|p - p^2 + 6|}{\sqrt{2}} \]
That should be the correct answer. There's nothing we can do to simplify this, because we can't show that \(-p^2 + p + 6\) is a positive definite or negative definite quadratic to exploit relevant properties. If the answer dropped the absolute values without reasonable justification I would consider that a mistake.

Ok, thank you!

Jefferson

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Re: 3U Maths Question Thread
« Reply #3908 on: February 12, 2019, 06:19:42 pm »
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Q3 Part (b), Independent of (a)

Could someone help me with why we could make the jump to the positive value of the absolute. Do we need to offer a reason?
Also, is there a better appoarch to this question?
Thanks.

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3909 on: February 12, 2019, 07:01:52 pm »
+1
Q3 Part (b), Independent of (a)

Could someone help me with why we could make the jump to the positive value of the absolute. Do we need to offer a reason?
Also, is there a better appoarch to this question?
Thanks.

Hey!! So the justification is:

- The particle starts at the origin with a velocity of \(v=1\), since \(v=1\) when \(x=0\)
- This means the particle will definitely move to the right first. Assuming this, \(v>0\) always, so the particle is always moving right
- Therefore, \(x\) will never be negative, which guarantees that \(|2x+1|=2x+1\)

Do you need to justify this? Probably not, the absolute values tend to be ignored in every HSC question I've seen requiring a log integral! :)

david.wang28

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Re: 3U Maths Question Thread
« Reply #3910 on: February 12, 2019, 07:38:36 pm »
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Hello,
Can someone please help me with the question in the link below (my working out is below)? Thanks :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3911 on: February 12, 2019, 07:47:49 pm »
+4
Hello,
Can someone please help me with the question in the link below (my working out is below)? Thanks :)
Your working out looks fine. Now just recall that \( \operatorname{cosec}x = \frac{1}{\sin x} \) and \( \cot x = \frac{1}{\tan x}\). This will give you \( \operatorname{cosec}\frac\pi4 = \sqrt{2} \), \( \cot \frac\pi4 = 1\), \( \operatorname{cosec} \frac\pi6 = 2\) and \(\cot \frac\pi6 = \sqrt{3}\).

david.wang28

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Re: 3U Maths Question Thread
« Reply #3912 on: February 12, 2019, 08:00:33 pm »
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Your working out looks fine. Now just recall that \( \operatorname{cosec}x = \frac{1}{\sin x} \) and \( \cot x = \frac{1}{\tan x}\). This will give you \( \operatorname{cosec}\frac\pi4 = \sqrt{2} \), \( \cot \frac\pi4 = 1\), \( \operatorname{cosec} \frac\pi6 = 2\) and \(\cot \frac\pi6 = \sqrt{3}\).
Thank you! :)
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shaynec19

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Re: 3U Maths Question Thread
« Reply #3913 on: February 14, 2019, 02:05:20 pm »
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Could someone please help on how to find the general solutions for this.
If there is multiple ways could you please tell me.



RuiAce

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Re: 3U Maths Question Thread
« Reply #3914 on: February 14, 2019, 08:08:47 pm »
+2
Could someone please help on how to find the general solutions for this.
If there is multiple ways could you please tell me.



You can use an auxiliary angle transformation.
\begin{align*}\sqrt{2}\sin \left(x+ \frac\pi4 \right)&=1\\ \sin \left( x+\frac\pi4 \right)& = \frac{1}{\sqrt2}\\ x+\frac\pi4 &= n\pi + (-1)^n \frac\pi4 \end{align*}
(Using the general solution just for illustrative purposes, because no domain restriction was provided.)