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February 17, 2019, 10:16:46 am

### AuthorTopic: 3U Maths Question Thread  (Read 422312 times) Tweet 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 »
0
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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#### 006896

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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 »
0
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 »
0
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 »
0
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 »
0
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 »
0
Hello,
Can someone please help me with the question in the link below (my working out is below)? Thanks
HSC 2019: English Advanced(77) (gonna repeat), Chemistry, Physics, Maths Extension 1(35) (gonna repeat), Maths Extension 2, Business Studies(80) (screw this)

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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!
HSC 2019: English Advanced(77) (gonna repeat), Chemistry, Physics, Maths Extension 1(35) (gonna repeat), Maths Extension 2, Business Studies(80) (screw this)

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#### shaynec19

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##### Re: 3U Maths Question Thread
« Reply #3913 on: February 14, 2019, 02:05:20 pm »
0
Could someone please help on how to find the general solutions for this.
If there is multiple ways could you please tell me.

$sinx+cosx~=1$

#### 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.

$sinx+cosx~=1$
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.)