Author Topic: Maths Methods (Exam 1): Discussion, Questions & Potential Solutions (Read 47710 times)

Joseph41 · « **on:** October 09, 2018, 03:24:54 pm »

This thread is for all exam-related discussion. Was it easy? Was it hard? What did you get for each question? Feel free to post any and all of your thoughts below.

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crouchie · « **Reply #1 on:** November 07, 2018, 10:30:46 am »

How'd you all find it?

Typed up solutions about to be posted. Stay tuned.

EDIT: Solutions Posted. Please let me know of any suggestions.

EDIT 2: Sorry, formatting stuffed up on first upload. Re-uploaded with fix.

pieholic1 · « **Reply #2 on:** November 07, 2018, 10:32:39 am »

Does anyone have a copy of the exam? Happy to provide worked solutions as I tutor methods

crawlingvines · « **Reply #3 on:** November 07, 2018, 10:34:29 am »

Trainwreck. That's all I have to contribute

jazzycab · « **Reply #4 on:** November 07, 2018, 10:35:44 am »

Question 1a
$\begin{align*}u&=-3x^3+x^2-64&\frac{du}{dx}&=-9x^2+2x\\y&=u^3&\frac{dy}{du}&=3u^2\\\therefore\frac{dy}{dx}&=3u^2\left(-9x^2+2x\right)&&=3\left(-3x^3+x^2-64\right)^2\left(2x-9x^2\right)\end{align*}$

Question 1b
$\begin{align*}u&=e^{x}&u'&=e^{x}\\v&=\cos{\left(x\right)}&v'&=-\sin{\left(x\right)}\\f'\left(x\right)&=\frac{u'v-v'u}{v^2}&&=\frac{e^{x}\cos{\left(x\right)}+e^{x}\sin{\left(x\right)}}{\cos^2{\left(x\right)}}\\\therefore f'\left(\pi\right)&=\frac{e^{\pi}\cos{\left(\pi\right)}+e^{\pi}\sin{\left(\pi\right)}}{\cos^2{\left(\pi\right)}}&&=\frac{-e^{\pi}+0}{\left(-1\right)^2}\\&=-e^{\pi}\end{align*}$

Question 2
$\begin{align*}f\left(x\right)&=\int{\frac{1}{2}-\frac{1}{2x-2}}dx&&=\frac{x}{2}-\frac{1}{2}\log_{e}{\left|x-1\right|}+c,c\in\mathbb{R}\\x&>1&\therefore f\left(x\right)&=\frac{x}{2}-\frac{1}{2}\log_{e}{\left(x-1\right)}+c\\0&=1-\frac{1}{2}\log_{e}{\left(2-1\right)}+c&&=1+c\\\therefore c&=-1&f\left(x\right)&=\frac{x}{2}-\frac{1}{2}\log_{e}{\left(x-1\right)}-1\end{align*}$

Question 3a
$\begin{align*}2\cos{\left(x\right)}&=-1&\cos{\left(x\right)}&=-\frac{1}{2}\\x&=\pi-\frac{\pi}{3},\pi+\frac{\pi}{3}&&=\frac{2\pi}{3},\frac{4\pi}{3}\end{align*}$

Question 3a

Question 4a
$\begin{align*}X&\sim N\left(6,4\right)&\therefore \text{Pr}\left(X>6\right)&=0.5\end{align*}$

Question 4b
$\begin{align*}Z&\sim N\left(0,1\right)&\therefore\text{Pr}\left(X>7\right)&=\text{Pr}\left(Z>\frac{7-6}{\sqrt{4}}\right)\\&=\text{Pr}\left(Z>\frac{1}{2}\right)&&=\text{Pr}\left(Z<-\frac{1}{2}\right)\\\therefore b&=-\frac{1}{2}\end{align*}$

Question 5
$\begin{align*}f\left(x\right):x&>2,&y&>0\\\therefore f^{-1}\left(x\right):x&>0,&y&>2\\x&=\frac{1}{\left(f^{-1}\left(x\right)-2\right)^2}&\frac{1}{x}&=\left(f^{-1}\left(x\right)-2\right)^2\\\pm\sqrt{\frac{1}{x}}&=f^{-1}\left(x\right)-2&f^{-1}\left(x\right)&=2\pm\frac{1}{\sqrt{x}}\\f^{-1}\left(x\right)&>2,&x&>0\\\therefore f^{-1}\left(x\right)&=2+\frac{1}{\sqrt{x}},&x&>0\end{align*}$

Question 6a
$\begin{align*}\text{Pr}\left(\text{Black}\right)&=\frac{1}{2}\times 1+\frac{1}{2}\times\frac{1}{2}&&=\frac{1}{2}+\frac{1}{4}\\&=\frac{2+1}{4}&&=\frac{3}{4}\end{align*}$

Question 6b
$\begin{align*}\text{Pr}\left(\text{Box1 }|\text{Black}\right)&=\frac{\text{Pr}\left(\text{Box 1}\cap\text{Black}\right)}{\text{Pr}\left(\text{Black}\right)}&&=\frac{\frac{1}{2}\times 1}{\frac{3}{4}}\\&=\frac{1}{2}\div\frac{3}{4}&&=\frac{1}{2}\times\frac{4}{3}\\&=\frac{2}{3}\end{align*}$

Question 7a
$\begin{align*}m_1&=2&\therefore m_2&=1\frac{1}{2}\\y_2-0&=m_2\left(x-0\right)&\therefore y_2&=-\frac{1}{2}x\\-\frac{1}{2}x_{P}&=2x_{P}-4&-x_{P}&=4x_{P}-8\\0&=5x_{P}-8&5x_{P}&=8\\\therefore x_{P}&=\frac{8}{5}&y_{P}&=-\frac{1}{2}\times\frac{8}{5}\\y_{P}&=-\frac{4}{5}&\therefore P:&\left(\frac{8}{5},-\frac{4}{5}\right)\end{align*}$

Question 7b
$\begin{align*}OP&=\sqrt{\left(\frac{8}{5}-0\right)^2+\left(-\frac{4}{5}-0\right)^2}&&=\sqrt{\frac{64}{25}+\frac{16}{25}}\\&=\frac{\sqrt{80}}{5}&&=\frac{\sqrt{16\times 5}}{5}\\&=\frac{4\sqrt{5}}{5}\end{align*}$

Question 8a
$\begin{align*}u&=x^2&u'&=2x\\v&=e^{kx}&v'&=ke^{kx}\\f'\left(x\right)&=u'v+v'u&&=2xe^{kx}+kx^2e^{kx}\\&=xe^{kx}\left(2+kx\right)&&=xe^{kx}\left(kx+2\right)\end{align*}$

Question 8b
$\begin{align*}x^2e^{kx}&=xe^{kx}\left(kx+2\right)&0&=x^2e^{kx}-xe^{kx}\left(kx+2\right)\\0&=xe^{kx}\left(x-kx-2\right)&0&=xe^{kx}\left(x\left(1-k\right)-2\right)\\e^{kx}&\ne 0&\therefore x&=0,\frac{2}{1-k}\\\therefore 0&=\frac{2}{1-k}&&\text{or }\frac{2}{1-k}\text{ is undefined}\\\therefore k&=1\end{align*}$

Question 8c
$\begin{align*}A&=\int_{0}^{2}{f\left(x\right)-g\left(x\right)}dx\end{align*}$

Question 8d
$\begin{align*}\int{kx^2e^{kx}+2xe^{kx}}dx&=x^2e^{kx}+c,c\in\mathbb{R}&k\int{x^2e^{kx}+\frac{2xe^{kx}}{k}}dx&=x^2e^{kx}+c\\\therefore k\int{f\left(x\right)-g\left(x\right)}&=x^2e^{kx}+c&kA&=\left[x^2e^{kx}\right]_{0}^{2}\\kA&=4e^{2k}-0&k\times\frac{16}{k}&=4e^{2k}\\16&=4e^{2k}&4&=e^{2k}\\\log_{e}{\left(4\right)}&=2k&k&=\frac{1}{2}\log_{e}{\left(4\right)}\\k&=\log_{e}{\left(4\right)^{\frac{1}{2}}}&&=\log_{e}{\left(2\right)}\end{align*}$

Question 9ai
$\begin{align*}\text{If }n\text{ is even or }0&\Rightarrow n+1\text{ is odd}\\\int_{n\pi}^{\left(n+1\right)\pi}{x\sin{\left(x\right)}}dx&=\left[\sin{\left(x\right)}-x\cos{\left(x\right)}\right]_{n\pi}^{\left(n+1\right)\pi}&&=\sin{\left(\pi\left(n+1\right)\right)}-\left(n+1\right)\pi\cos{\left(\pi\left(n+1\right)\right)}-\left(\sin{\left(n\pi\right)}-n\pi\cos{\left(n\pi\right)}\right)\\&=0+\left(n+1\right)\pi-0+n\pi&&=n\pi+\pi+n\pi\\&=2n\pi+\pi&&=\pi\left(2n+1\right)\end{align*}$

Question 9aii
$\begin{align*}\text{If }n\text{ is odd }&\Rightarrow n+1\text{ is even or 0}\\\int_{n\pi}^{\left(n+1\right)\pi}{x\sin{\left(x\right)}}dx&=\left[\sin{\left(x\right)}-x\cos{\left(x\right)}\right]_{n\pi}^{\left(n+1\right)\pi}&&=\sin{\left(\pi\left(n+1\right)\right)}-\left(n+1\right)\pi\cos{\left(\pi\left(n+1\right)\right)}-\left(\sin{\left(n\pi\right)}-n\pi\cos{\left(n\pi\right)}\right)\\&=0-\left(n+1\right)\pi-0+n\pi&&=-n\pi-\pi-n\pi\\&=-2n\pi-\pi&&=-\pi\left(2n+1\right)\end{align*}$

Question 9b
$\begin{align*}u&=x&u'&=1\\v&=\sin{\left(x\right)}&v'&=\cos{\left(x\right)}\\\therefore\frac{dy}{dx}&=\sin{\left(x\right)}+x\cos{\left(x\right)}&\frac{dy}{dx}\Bigg{|}_{x=-\frac{5\pi}{2}}&=-1+0\\\therefore m&=-1&y&=-\left(x+\frac{5\pi}{2}\right)+\frac{5\pi}{2}\\y&=-x\end{align*}$

Question 9c
$\begin{align*}\sin{\left(x+\left(2k+1\right)\pi\right)}&=-\sin{\left(x\right)},k\in\mathbb{Z}&\therefore y\left(x-3\pi\right)&=\left(x-3\pi\right)\sin{\left(x-3\pi\right)}\\&=\left(x-3\pi\right)\left(-\sin{\left(x\right)}\right)&&=\left(3\pi-x\right)\sin{\left(x\right)}\\\therefore a&=3\pi\end{align*}$

Question 9d
$\begin{align*}l_1&=-\left(x-3\pi\right)&l_1\left(0\right)&=3\pi\\l_1&=0&\Rightarrow x&=3\pi\\A&=2\left(\frac{1}{2}\left(3\pi\right)\left(3\pi\right)-\int_{-3\pi}^{-2\pi}{x\sin{\left(x\right)}}dx+\int_{-2\pi}^{-\pi}{x\sin{\left(x\right)}}dx-\int_{-\pi}^{0}{x\sin{\left(x\right)}}dx\right)&&=9\pi^2-2\left(-\pi\left(2\left(-3\right)+1\right)\right)+2\left(\pi\left(2\left(-2\right)+1\right)\right)-2\left(-\pi\left(2\left(-1\right)+1\right)\right)\\&=9\pi^2-10\pi-6\pi-2\pi&&=9\pi^2-18\pi\end{align*}$

Bri MT · « **Reply #5 on:** November 07, 2018, 10:36:16 am »

Quote from: pieholic1 on November 07, 2018, 10:32:39 am

Does anyone have a copy of the exam? Happy to provide worked solutions as I tutor methods

Although it's a nice sentiment to want to help, we can't allow people to uplaod or request copyright materials including the VCE exams. See this post for more information.

erinations · « **Reply #6 on:** November 07, 2018, 10:41:46 am »

that was a horrible exam.

marauder52 · « **Reply #7 on:** November 07, 2018, 10:46:59 am »

What would I be looking at with
~90% SACs top 20 ranks at a selective school
34/40 Exam 1
70/80 Exam 2

Bell9565 · « **Reply #8 on:** November 07, 2018, 10:49:03 am »

missed the last page, spent to long before I realised I had the symmetry on the (N+1)pi question the wrong way round ....

not.yet.a.nerd · « **Reply #9 on:** November 07, 2018, 10:52:42 am »

The first half was suspiciously easy... The second half was like a punch in the face. I guess the last few questions will really be those deciding marks for the top scores :/

Also did you guys simplify the first question? Worth one mark but I think I over-simplified. RIP $:-\$

juicethelemon · « **Reply #10 on:** November 07, 2018, 10:55:32 am »

Quote from: not.yet.a.nerd on November 07, 2018, 10:52:42 am

The first half was suspiciously easy... The second half was like a punch in the face. I guess the last few questions will really be those deciding marks for the top scores :/

Also did you guys simplify the first question? Worth one mark but I think I over-simplified. RIP $:-\$

wow literally thought I was the only one. It was going great the whole way through and then bam last two questions slaughtered me

yebavin · « **Reply #11 on:** November 07, 2018, 10:58:06 am »

What do you think B cutoff will be?

technodisney · « **Reply #12 on:** November 07, 2018, 11:03:22 am »

Quote from: not.yet.a.nerd on November 07, 2018, 10:52:42 am

The first half was suspiciously easy... The second half was like a punch in the face. I guess the last few questions will really be those deciding marks for the top scores :/

That just about sums it up lol. I got close to half of it done in 15 minutes, I was thinking it was way to easy, then I got a surprise when I actually had to think through the questions at the end, how the heck do we do all of that stupid integration at the end. I could have done it but it would be way too over complicated.

Nason · « **Reply #13 on:** November 07, 2018, 11:06:05 am »

I messed up the last question with such a stupid mistake. Pretty sure the answers meant to be 9(pi)^2 -18(pi) but I didn't square the first pi so I got -9pi. Other than that I think I got everything right, aside from the length of OP, somehow I messed that one up and got 4sqrt(10)/5 when it should be 4sqrt(5)/5

Azzzz · « **Reply #14 on:** November 07, 2018, 11:06:38 am »

Quote from: technodisney on November 07, 2018, 11:03:22 am

That just about sums it up lol. I got close to half of it done in 15 minutes, I was thinking it was way to easy, then I got a surprise when I actually had to think through the questions at the end, how the heck do we do all of that stupid integration at the end. I could have done it but it would be way too over complicated.

All that integration at the end made me want to walk out, that was probably the hardest exam I’ve done for methods I’ll be happy with a 37 at this point.

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Author Topic: Maths Methods (Exam 1): Discussion, Questions & Potential Solutions (Read 47710 times)

Joseph41

Maths Methods (Exam 1): Discussion, Questions & Potential Solutions

crouchie

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pieholic1

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crawlingvines

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jazzycab

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Bri MT

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erinations

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marauder52

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Bell9565

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not.yet.a.nerd

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juicethelemon

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yebavin

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technodisney

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