April 25, 2019, 11:49:05 pm

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#### alexnero7

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« Reply #4005 on: February 12, 2019, 06:02:01 pm »
0
Hey everyone, hope you all well. Can somebody please help me find the primitive function of this attached question? Thanks.

#### Cherre Ho

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« Reply #4006 on: February 12, 2019, 06:03:38 pm »
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Can anyone explain the working out for this? Is there a method that you are supposed to use?
Question: Write this series in sigma notation.
1+1/2+1/4+...+1/512
Also in some examples I have seen like 3+6+12+...+3×2(power n) you are supposed to change the 'n' to a 'k'. Why is that?

#### myopic_owl22

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« Reply #4007 on: February 12, 2019, 06:40:33 pm »
+1
Hey everyone, hope you all well. Can somebody please help me find the primitive function of this attached question? Thanks.

Hi there,
To integrate functions like these, I'd recommend moving the (1/2) bit to the left side of the integral, as it's a constant multiplier. Dealing with the rest...
1. Convert into index notation
$\frac{1}{2}\int (4x-5)^{-3}$
2. Integrate as usual. Treat the bracket as a term - power up by one, divide by new power.
$\frac{1}{2}\cdot \frac{(4x-5)^{-2}}{-2*4}$
We divide through by 4 as well to essentially 'reverse' the effects of the function of a function rule if the answer was to be diffed. General formula:
$\int (ax+b)^{n} = \frac{(ax+b)^{n+1}}{a(n+1)}$
Note that this will only work for functions whose leading power of x is 1. Quadratics, etc. do not follow this rule.

Thankfully, this is about as tricky as 2U integration will get. Hopefully this helps!

Oh and don't forget +c
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#### myopic_owl22

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« Reply #4008 on: February 12, 2019, 07:09:16 pm »
0
Can anyone explain the working out for this? Is there a method that you are supposed to use?
Question: Write this series in sigma notation.
1+1/2+1/4+...+1/512
Also in some examples I have seen like 3+6+12+...+3×2(power n) you are supposed to change the 'n' to a 'k'. Why is that?

Hi there,
Noticing that the terms in your series are all 1 divided by increasing powers of 2 (i.e. 2^0, 2^1, 2^2, up to 2^9), we'd write something like this:
$\sum_{n=0}^{9} \frac{1}{2^{n}}$

The bottom is where n begins (0 in this case for our first term of 1) and the top number is when we finish. You can use trial and error without too much fuss to see what power of 2 that 512 is raised to. Alternatively, you could use logarithms, which is a later topic in the maths course. Here, log2512 = 9. We use n as our only variable as this is a finite series, which will stop when we finish adding 1/512 to our sum.

On the right, is what we're summing over and over again, with different values of n for each time we do it. Hopefully I'm making some sense.

We generally use k if the sum's final term is in general form (i.e. all the terms in this series follow this rule, where k is replaced by the term number). It will either denote some value that we either need to find, or don't need to worry about. It's essentially the math's conventional placeholder, which isn't a major focus of the course anyway, but to see an example of it in action:
$\sum_{n=1}^{k} x^{n} = x^{1} + x^{2} +x^{3} +...x^{k}$
So yeah, it's more focusing on the process of summation rather than the answer. Let me know if I'm not making sense
« Last Edit: February 12, 2019, 08:04:07 pm by myopic_owl22 »
HSC 2018: 2U Maths [98]
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#### Cherre Ho

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« Reply #4009 on: February 15, 2019, 07:40:51 pm »
0
Can someone help with this question? Particularly stuck on how to write the equation for the second bit.
The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms.

#### emilyyyyyyy

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« Reply #4010 on: February 15, 2019, 07:43:55 pm »
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hello can someone please help me how to find the enclosed area between the y-axis and the curve x=y(y-2)??  I'm just not sure what the y-values im meant to be integrating are.
thanks!

#### meerae

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« Reply #4011 on: February 15, 2019, 08:28:33 pm »
+1
hello can someone please help me how to find the enclosed area between the y-axis and the curve x=y(y-2)??  I'm just not sure what the y-values im meant to be integrating are.
thanks!

Hey!
You'd be integrating between y=0 and y=2

Hope this helps!
meerae
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#### jamonwindeyer

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« Reply #4012 on: February 15, 2019, 08:49:51 pm »
+1
Can someone help with this question? Particularly stuck on how to write the equation for the second bit.
The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms.

Hey! So your first equation is:

$S_4=\frac{4}{2}\left(2a+3d\right)\\4a+6d=42$

The next one is tougher, but it is really just the sum of $T_3$ and $T_7$:

$T_3+T_7=46\\a+2d+a+6d=46\\2a+8d=46$

Solving those simultaneously will give you your first term and common difference! And that can be used to find $S_{20}$ - Is that enough to get you rolling?
« Last Edit: February 15, 2019, 09:44:07 pm by jamonwindeyer »

#### jamonwindeyer

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« Reply #4013 on: February 15, 2019, 09:00:38 pm »
+1
hello can someone please help me how to find the enclosed area between the y-axis and the curve x=y(y-2)??  I'm just not sure what the y-values im meant to be integrating are.
thanks!

Hey! As meerae says, the values are $y=0$ and $y=2$. This is because these are the two points where the curve cuts the y-axis (sideways parabola!), so this defines the start and stop point for the area. The graph is here:

Also notice the area is to the left of the y-axis, meaning you'll need to compensate for the negative as well! Hope this helps

#### goodluck

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« Reply #4014 on: February 15, 2019, 09:22:06 pm »
+1
Hey how would I solve this? (it's absolute values if the writing doesn't make it clear!)

|(x-1)/(x+1)| <1

#### Cherre Ho

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« Reply #4015 on: February 15, 2019, 09:26:52 pm »
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Hey! So your first equation is:

$S_4=\frac{4}{2}\left(2a+3d\right)\\4a+6d=42$

The next one is tougher, but it is really just the sum of $T_3$ and $T_7$:

$T_4+T_7=46\\a+3d+a+6d=46\\2a+9d=46$

Solving those simultaneously will give you your first term and common difference! And that can be used to find $S_{20}$ - Is that enough to get you rolling?
Ah yep thank you so much!
Though I think you wrote T4 instead of T3 by accident??

#### jamonwindeyer

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« Reply #4016 on: February 15, 2019, 09:44:30 pm »
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Ah yep thank you so much!
Though I think you wrote T4 instead of T3 by accident??

I did indeed, sorry!! Just tidied it up but the general idea is the same

#### RuiAce

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« Reply #4017 on: February 15, 2019, 09:50:39 pm »
+3
Hey how would I solve this? (it's absolute values if the writing doesn't make it clear!)

|(x-1)/(x+1)| <1
$\text{The solution is equivalent to that of}\\ |x-1| < |x+1|$
(However, note that $x \neq -1$ because otherwise in the original equation we have dividing by 0 issues.)
Recall
The definition of the absolute value states that
$|x| = \begin{cases} x & \text{if }x\geq 0\\ -x & \text{if }x < 0\end{cases}$
Therefore we have
\begin{align*} |x-1| &= \begin{cases} x-1 & \text{if }x-1 \geq 0\\ -(x-1) & \text{if }x-1 < 0\end{cases} \\ &= \begin{cases} x-1 & \text{if }x \geq 1\\ -x+1 &\text{if }x < 1\end{cases} \end{align*}
And similarly
$|x+1| = \begin{cases} x+1 &\text{ if }x \geq -1\\ -x-1 &\text{if }x < -1\end{cases}$
$\text{For this inequality, noting the above}\\ \text{we can accordingly split the following cases.}$
\text{Case 1: } x < -1.\\ \text{For this case the inequality becomes}\\ \begin{align*} -x+1 &< -x-1\\ 1 &< -1\end{align*}\\ \text{This obviously has no solution}\\ \text{so no values of }x\text{ in this case belong in the solution.}
\text{Case 2: }-1\leq x < 1.\\ \text{For this case the inequality becomes}\\ \begin{align*}-x+1 &< x+1\\ 0 &< 2x\\ x &> 0 \end{align*}
$\text{So overlapping }x > 0\text{ with }-1\leq x < 1\\ \text{we see that }\boxed{0 < x < 1}\text{ is a part of the solution.}$
\text{Case 3: }x\geq 1.\\ \text{For this case the inequality becomes}\\ \begin{align*}x-1 &< x+1\\ -1 &< 1 \end{align*}
$\text{This is obviously true for every value of }x.\\ \text{So all values of }x\text{ in this case, i.e. }\\\boxed{ x\geq 1}\text{, is a part of the solution.}$
$\text{Combining all parts, our solution is thus}\\ \boxed{x > 0}.$

#### goodluck

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« Reply #4018 on: February 16, 2019, 06:11:59 pm »
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Hey, thank you so much for your help, what's the maximal domain and range of y= (1-2sinx)^1/2?  I think the max range is 0<y< sqrt3 but was confused on how you express the domain

#### RuiAce

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« Reply #4019 on: February 16, 2019, 06:41:31 pm »
+1
The groundwork
$\text{Essentially we are solving }1-2\sin x \geq 0\\ \text{which rearranges to }\sin x \leq \frac12.$
$\text{First note that the solutions to }\sin x = \frac12 \text{ for }0\leq x \leq 2\pi\\ \text{are }x=\frac\pi6\text{ or } \frac{5\pi}6.$
$\text{Following the periodicity of the trigonometric functions, the other solutions will be at}\\ x = 2k\pi+\frac\pi6\text{ or }x=2k\pi + \frac{5\pi}{6}\\ \text{where }k\text{ is an integer.}$
_____________________________
$\text{Now for the inequality, we can sketch the curve }y=\sin x\\ \text{along with the line }y=\frac12.\\ \text{The above computations track down each point of intersection for us.}$
$\text{As we are interested in }\sin x \leq \frac12,\\ \text{we're interested in the }x\text{ values for which}\\ y=\sin x\text{ lies }\textbf{below}\text{ the line }y=\frac12.$
$\text{We see that we'll have }-\frac{7\pi}{6} \leq x \leq \frac\pi6, \, \frac{5\pi}{6}\leq x \leq \frac{13\pi}6\\ \text{and so on.}\\ \text{All of these intervals will form our answer.}$
So expressing it in a nice way is reasonably hard, because we have to express an infinite number of intervals. But we can always work around this by just setting $k$ to be an arbitrary integer. One possible answer is $-\frac{7\pi}{6}+2k\pi \leq x \leq \frac\pi6+2k\pi$, where $k$ is an integer. Having this arbitrary integer placeholder lets us get around this issue, because $k$ can then be substituted for every integer possible.

Note that I purposely avoided "general solutions" as this was posted in the 2U section.

Edit: fixed typos
« Last Edit: February 16, 2019, 07:14:37 pm by RuiAce »