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March 29, 2024, 01:06:12 am

Author Topic: QCE Specialist Maths Questions Thread  (Read 25639 times)

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RuiAce

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Re: QCE Specialist Maths Questions Thread
« Reply #45 on: October 12, 2021, 08:11:57 pm »
+2
Does anyone know how (1/2)ln|2x - 2| can simplify to (1/2)ln|x - 1| ??

(its from a question that asks to find the integral of 1/(2x-2))

TIA :)
It does not. The point is that you emphasised how it came from an integral. Remember that indefinite integrals come with the important \(+c\).

Consider the logarithm law \(\log(AB) = \log A + \log B\). Using this, we may obtain
\begin{align*}
\int \frac{1}{2x-2}dx &= \frac12 \ln |2x-2| + c\\
&= \frac12 \ln 2|x-1|+ c\\
&= \frac12 \left(\ln |x-1| + \ln 2 \right)+c\\
&= \frac12 \ln |x-1| + \frac12 \ln 2 + c.
\end{align*}
That \(\frac12\ln 2 + c\) will now just be another constant. So you could replace it with another letter, like say \(b = \frac12 \ln 2 + c\), and obtain
\[ \int \frac{1}{2x-2}dx = \frac{1}{2} \ln |x-1| + b, \]
which reunites both correct answers.

Asides:
- If you are doing a definite integral, it does not matter if you integrate into \(\frac{1}{2} \ln |2x-2|\) or \( \frac{1}{2} \ln |x-1|\). If you are very careful with your log laws, you will be able to obtain the same final answer no matter which antiderivative you use.
- You can also jump directly to the answer of \(\frac{1}{2} \ln |x-1|\) if you first factorise the denominator a little.
\begin{align*}
\int \frac{1}{2x-2}dx &= \int \frac{1}{2(x-1)}dx\\
&= \frac12 \int \frac{1}{x-1}dx\\
&= \frac12 \ln|x-1|+c.
\end{align*}
« Last Edit: October 13, 2021, 09:19:59 pm by RuiAce »

jasmine24

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Re: QCE Specialist Maths Questions Thread
« Reply #46 on: October 13, 2021, 09:12:37 pm »
0
It does not. The point is that you emphasised how it came from an integral. Remember that indefinite integrals come with the important \(+c\).

Consider the logarithm law \(\log(AB) = \log A + \log B\). Using this, we may obtain
\begin{align*}
\int \frac{1}{2x-2}dx &= \frac12 \ln |2x-2 + c|\\
&= \frac12 \ln 2|x-1|+ c\\
&= \frac12 \left(\ln |x-1| + \ln 2 \right)+c\\
&= \frac12 \ln |x-1| + \frac12 \ln 2 + c.
\end{align*}
That \(\frac12\ln 2 + c\) will now just be another constant. So you could replace it with another letter, like say \(b = \frac12 \ln 2 + c\), and obtain
\[ \int \frac{1}{2x-2}dx = \frac{1}{2} \ln |x-1| + b, \]
which reunites both correct answers.

Asides:
- If you are doing a definite integral, it does not matter if you integrate into \(\frac{1}{2} \ln |2x-2|\) or \( \frac{1}{2} \ln |x-1|\). If you are very careful with your log laws, you will be able to obtain the same final answer no matter which antiderivative you use.
- You can also jump directly to the answer of \(\frac{1}{2} \ln |x-1|\) if you first factorise the denominator a little.
\begin{align*}
\int \frac{1}{2x-2}dx &= \int \frac{1}{2(x-1)}dx\\
&= \frac12 \int \frac{1}{x-1}dx\\
&= \frac12 \ln|x-1|+c.
\end{align*}
thank you!

jasmine24

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Re: QCE Specialist Maths Questions Thread
« Reply #47 on: October 17, 2021, 08:34:06 pm »
0
does anyone know how the last line of the solution works? is it some sort of formula?
(Q17 tech active WACE 2020)

RuiAce

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Re: QCE Specialist Maths Questions Thread
« Reply #48 on: October 19, 2021, 06:23:16 pm »
+3
does anyone know how the last line of the solution works? is it some sort of formula?
(Q17 tech active WACE 2020)
The first bit refers to this section screenshotted from the QCE syllabus.

We know that \( \overline{X}\) is approximately normal with mean \(\mu\) and standard deviation \( \frac{\sigma}{\sqrt{n}} \). Therefore \( \frac{\overline{X}-\mu}{\sigma/\sqrt{n}}\) is approximately standard normal. Note that:
a) This is a general rule. In general, for any random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), if we know that \(X\) is normally distributed, then \( \frac{X-\mu}{\sigma}\) is standard normally distributed.
b) It's very similar to the third point where they use \(s\) instead of \(\sigma\). At a high school level, I'd suggest just taking for granted the approximate standard normality works with both \(\sigma\) and with \(s\).

For the second bit, note that in theory that equal sign really should be an approximately equal sign. But let's not worry about that. The point is, first recall that \(\overline{X}\) denotes a sample mean for a larger sample size of \(2n\). And we've calculated that its corresponding standard deviation will therefore be \(9.0192\) (approximately). But also \(\overline{X}\) also has mean \(\mu\) (you should know why this is the case). Therefore here, \( \frac{\overline{X}-\mu}{9.0192} \) will be approximately standard normally distributed. Hence, it is tempting to start by dividing by \(9.0192\), and introducing \(Z\) as a placeholder for a standard normal random variable.
\[ P\left( |\overline{X}-\mu| <10\right) = P\left( \left|\frac{\overline{X}-\mu}{9.0192}\right| <\frac{10}{|9,0192|}\right) =P\left( |Z|<\frac{10}{9.0192} \right) . \]
We still have to remove the absolute value. First recall that the solution to the absolute value inequality \( |x| < a\) is \( -a < x < a\).
\[P\left( |Z|<\frac{10}{9.0192} \right) = P\left( -\frac{10}{9.0192} < Z < \frac{10}{9.0192} \right). \]
But we also know that the density of the standard normal distribution is a bell curve, and in particular symmetric about zero. The symmetry about zero allows you to conclude that \( P\left( -\frac{10}{9.0192} < Z < 0\right)\) and \( P\left( 0 < Z < \frac{10}{9.0192} \right)\) equal one another. Which is where that bit comes from.
\[  P\left( -\frac{10}{9.0192} < Z < \frac{10}{9.0192} \right) = 2  P\left( 0< Z < \frac{10}{9.0192} \right). \]
The final bit is just a calculator plug in. Can be computed by something along the lines of \( \verb|2*(normcdf(10/9.0192,mean=0,sd=1)-normcdf(0,mean=0,sd=1))| \)

Note: Use of symmetry is not required if your graphics calculator can just do \( \verb|normcdf(10/9.0192,mean=0,sd=1)-normcdf(-10/9.0192,mean=0,sd=1)| \) instead.

Studygoals

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Re: QCE Specialist Maths Questions Thread
« Reply #49 on: March 12, 2022, 11:09:40 am »
0
Hi, I need to find a tool to draw diagrams for my PSMT. We are doing it on vectors, and its really hard to draw all the angles and distances and diagrams on word dox. Does anyone know of any good drawing tools?

GreenNinja

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Re: QCE Specialist Maths Questions Thread
« Reply #50 on: March 12, 2022, 11:19:45 am »
+1
Hi, I need to find a tool to draw diagrams for my PSMT. We are doing it on vectors, and its really hard to draw all the angles and distances and diagrams on word dox. Does anyone know of any good drawing tools?

Geogebra is quite easy to use for graphing vectors.