Author Topic: VCE Methods Question Thread! (Read 5853473 times) Tweet Share

Phy124 · « **Reply #1020 on:** October 11, 2012, 07:06:43 pm »

Firstly we know:

$Pr(Z<c)=Pr(Z>-c) = a$

We are asked $Pr(|Z| < c)$

In otherwords $Pr(Z<c) \ and \ Pr(-Z<c)$

Which is the same as $Pr(Z<c) \ and \ Pr(Z>-c)$ (using that rule you learnt early has where you flip the sign when dividing by a negative)

Which can also be written as: $Pr(-c<Z<c)$

$Pr(-c<Z<c) = 1 - \left[Pr(Z<c)+Pr(Z>-c)\right] = 1 - (a+a) = 1 - 2a$

Or you could do:

$Pr(Z>c) = Pr(Z<-c) = 1-a$

$Pr(-c<Z<c) = Pr(Z<c) - Pr(Z<-c) = a - (1 - a) = 2a - 1$

You may find it beneficial to draw up a normal distribution and take c as an arbitrary point and work from there, that's what I did in year 12 when I became stuck.

barydos · « **Reply #1021 on:** October 11, 2012, 07:28:45 pm »

Quote from: rangaaaaaa on October 11, 2012, 07:06:43 pm

Firstly we know:

$Pr(Z<c)=Pr(Z>-c) = a$

We are asked $Pr(|Z| < c)$

In otherwords $Pr(Z<c) \ and \ Pr(-Z<c)$

Which is the same as $Pr(Z<c) \ and \ Pr(Z>-c)$ (using that rule you learnt early has where you flip the sign when dividing by a negative)

Which can also be written as: $Pr(-c<Z<c)$

$Pr(-c<Z<c) = 1 - \left[Pr(Z<c)+Pr(Z>-c)\right] = 1 - (a+a) = 1 - 2a$

Or you could do:

$Pr(Z>c) = Pr(Z<-c) = 1-a$

$Pr(-c<Z<c) = Pr(Z<c) - Pr(Z<-c) = a - (1 - a) = 2a - 1$

You may find it beneficial to draw up a normal distribution and take c as an arbitrary point and work from there, that's what I did in year 12 when I became stuck.

Thanks a lot, you're awesome!

Jenny_2108 · « **Reply #1022 on:** October 11, 2012, 07:34:29 pm »

^ or you can do Pr(-c<Z<c)=1-2(1-a)=2a-1

barydos · « **Reply #1023 on:** October 12, 2012, 11:41:16 pm »

How do you solve for x between [0,2pi] for tan(x) = (sqrt(2))cos(x)?

Phy124 · « **Reply #1024 on:** October 13, 2012, 12:03:11 am »

Quote from: Anonymiza on October 12, 2012, 11:41:16 pm

How do you solve for x between [0,2pi] for tan(x) = (sqrt(2))cos(x)?

Probably an easier method, but how I did it:

$tan(x) = \sqrt{2}cos(x)$

$\frac{sin(x)}{cos(x)}=\sqrt{2}cos(x)$

$sin(x) = \sqrt{2}cos^2(x)$

$sin(x) = \sqrt{2}(1-sin^2(x))$

$sin(x) = \sqrt{2} - \sqrt{2}sin^2(x)$

$\sqrt{2}sin^2(x) + sin(x) - \sqrt{2}=0$

$sin(x) = \frac{-1\pm\sqrt{1^2-4\times\sqrt{2}\times-\sqrt{2}}}{2\times \sqrt{2}}$ <- quadratic formula

$sin(x) = \frac{-1\pm\sqrt{9}}{2\sqrt{2}}$

$sin(x) = \frac{-1\pm 3}{2\sqrt{2}}$

$sin(x) = \frac{2}{2\sqrt{2}} \ or \ \frac{-4}{2\sqrt{2}}$

$sin(x) = \frac{1}{\sqrt{2}} \ or \ \frac{-2}{\sqrt{2}}$

$sin(x) = \frac{1}{\sqrt{2}} \ or \ -\sqrt{2}$

$sin(x) = -\sqrt{2}$ has no solutions

$sin(x) = \frac{1}{\sqrt{2}}$ has solutions $x = \frac{\pi}{4}, \frac{3\pi}{4}$ in the domain.

martin1106 · « **Reply #1025 on:** October 13, 2012, 11:39:16 am »

I subbed different values for n in the matrix and get n=8 when p=0.44..(cannot remember the exact one)

yet, how to solve for the n on the calculator? (vcca 2009 exam2) Given that p≤0.45

thanks

dim_sim · « **Reply #1026 on:** October 13, 2012, 01:02:09 pm »

What's the best way to approach this question?

Find $\frac{dy}{dx}$ if $y=(e^x + e^{-x})^2$

pi · « **Reply #1027 on:** October 13, 2012, 01:07:46 pm »

Expand it out and then diff each term

dim_sim · « **Reply #1028 on:** October 13, 2012, 01:14:19 pm »

Quote from: ρнуѕικѕ ♥ on October 13, 2012, 01:07:46 pm

Expand it out and then diff each term

Oh wow, that was heaps easier than I thought.
I was stuck on it for quite a while hahaha...

Thanks for the quick reply!

soccerboi · « **Reply #1029 on:** October 13, 2012, 05:07:44 pm »

A line with gradient m, m<0, passes through the point (1,2). The value of m for which the area enclosed by the line and the two axes is a minimum is:
A) -6
B) -5
C) -3
D) -3
E) -2

Is there another way of doing this question besides using trial and error?

Shenz0r · « **Reply #1030 on:** October 13, 2012, 05:10:32 pm »

Quote from: martin1106 on October 13, 2012, 11:39:16 am

I subbed different values for n in the matrix and get n=8 when p=0.44..(cannot remember the exact one)

yet, how to solve for the n on the calculator? (vcca 2009 exam2) Given that p≤0.45

thanks

You can't solve for n using the solve function on the CAS, unfortunately. It gives you an error. The only way to do this question is definitely trial and error.

b^3 · « **Reply #1031 on:** October 13, 2012, 05:24:28 pm »

EDIT:..... forgot it was multiple choice... just use technology.....
$\mathrm{solve}(y=m\cdot x+c,c)|x=1\:\mathrm{and}\:y=2$
$\mathrm{solve}(m\cdot x+2-m=0,x)$
$\mathrm{fMin}(\frac{1}{2}(2-m)(\frac{m-2}{m}),m)|m\leq0$

The by hand method anyway....
Firstly find the equation of the line.
$\begin{alignedat}{1}y & =mx+c \\ 2 & =m(1)+c \\ c & =2-m \\ y & =mx+2-m \\ y & =m(x-1)+2 \end{alignedat}$
Find where it intersects the axis.
$\begin{alignedat}{1}0 & =m(x-1)+2 \\ x-1 & =\frac{-2}{m} \\ x & =1-\frac{2}{m} \end{alignedat}$
Since m is negative this means this x-intercept is greater than 1. Now the gradient is negative, so that will mean the area will be from $0$ to $1-\frac{2}{m}$ .
Now we could integrate it, but since its just a triangle, I'm going to do it the easier way.
$\begin{alignedat}{1}A & =\frac{1}{2}bh \\ & =\frac{1}{2}\left(1-\frac{2}{m}\right)\left(2-m\right) \\ & =\frac{1}{2}\left(\frac{m-2}{m}\right)\left(2-m\right) \\ & =-\frac{1}{2m}\left(m-2\right)^{2} \end{alignedat}$
Find where the area is a minimum.
$\begin{alignedat}{1}\frac{dA}{dm} & =\frac{-1}{2m}\left(2\left(m-2\right)\right)+\left(m-2\right)^{2}\left(\frac{1}{2m^{2}}\right)=0 \\ \frac{1}{2m}\left(m-2\right)\left(\frac{m-2}{m}-2\right) & =0 \\ m-2=0 & \mathrm{or\:\frac{m-2}{m}-2=0} \\ m=2(\mathrm{but\: m<0)\:\:\or\:\:} & m-2=2m \\ & -2=m \\ \therefore & m=-2 \end{alignedat}$

barydos · « **Reply #1032 on:** October 13, 2012, 07:28:10 pm »

Quote from: rangaaaaaa on October 13, 2012, 12:03:11 am

Quote from: Anonymiza on October 12, 2012, 11:41:16 pm
How do you solve for x between [0,2pi] for tan(x) = (sqrt(2))cos(x)?
Probably an easier method, but how I did it:

$tan(x) = \sqrt{2}cos(x)$

$\frac{sin(x)}{cos(x)}=\sqrt{2}cos(x)$

$sin(x) = \sqrt{2}cos^2(x)$

$sin(x) = \sqrt{2}(1-sin^2(x))$

$sin(x) = \sqrt{2} - \sqrt{2}sin^2(x)$

$\sqrt{2}sin^2(x) + sin(x) - \sqrt{2}=0$

$sin(x) = \frac{-1\pm\sqrt{1^2-4\times\sqrt{2}\times-\sqrt{2}}}{2\times \sqrt{2}}$ <- quadratic formula

$sin(x) = \frac{-1\pm\sqrt{9}}{2\sqrt{2}}$

$sin(x) = \frac{-1\pm 3}{2\sqrt{2}}$

$sin(x) = \frac{2}{2\sqrt{2}} \ or \ \frac{-4}{2\sqrt{2}}$

$sin(x) = \frac{1}{\sqrt{2}} \ or \ \frac{-2}{\sqrt{2}}$

$sin(x) = \frac{1}{\sqrt{2}} \ or \ -\sqrt{2}$

$sin(x) = -\sqrt{2}$ has no solutions

$sin(x) = \frac{1}{\sqrt{2}}$ has solutions $x = \frac{\pi}{4}, \frac{3\pi}{4}$ in the domain.

Thanks so much

martin1106 · « **Reply #1033 on:** October 13, 2012, 09:11:56 pm »

Quote from: Shenz0r on October 13, 2012, 05:10:32 pm

Quote from: martin1106 on October 13, 2012, 11:39:16 am
I subbed different values for n in the matrix and get n=8 when p=0.44..(cannot remember the exact one)

yet, how to solve for the n on the calculator? (vcca 2009 exam2) Given that p≤0.45

thanks

You can't solve for n using the solve function on the CAS, unfortunately. It gives you an error. The only way to do this question is definitely trial and error.

thanks for answering my question

sin0001 · « **Reply #1034 on:** October 13, 2012, 09:31:17 pm »

There was a question in a VCAA exam in which you had to find the nature of the stationary points. The nature of one of them was 'local max.,' in the solutions, if i only wrote 'maximum,' would it be correct?

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