Post by: dskel on August 02, 2019, 01:53:27 pm
Post by: AlphaZero on August 03, 2019, 01:06:30 pm
Not posting full working since I can't be bothered. I'll just provide answers and a few notes for the harder questions. If anyone has any questions, just ask. I did this exam really quickly, so I could've made a few errors. If I have, let me know.
Question 1a
\(\dfrac{dy}{dx}=2e^x+e^{-x}\)
Question 1b
\(f'\left(\dfrac\pi 3\right)=\dfrac{-2\pi}{3}\)
Question 2
\(\displaystyle f(x)=\frac23x^3-\frac34x^{1/3}-\frac53\)
Question 3a
\(\displaystyle \int_2^7 \frac{1}{x+\sqrt3}dx=\log_e(7+\sqrt3)-\log_e(2+\sqrt3)\quad \text{and}\quad \int_2^7 \frac{1}{x-\sqrt3}dx=\log_e(7-\sqrt3)-\log_e(2-\sqrt3)\)
Question 3b
Pretty easy proof. Write the left-hand side on a common denominator. The result follows.
Question 3c
\(\displaystyle \int_2^7 \frac{x}{x^2-3}dx=\frac12\log_e(46)\)
Question 4a
\(\text{domain}(g)=(3,\,\infty)\quad\text{and}\quad \text{range}(g)=\mathbb{R}\)
Question 4b.i
\(y=x-2\)
Question 4b.ii
Use a calculator to check your answer here. Vertical asymptote at \(x=3\). Graph of \(g\) crosses through \((4,\,2)\).
Question 5a
\(x=-1\quad\text{or}\quad x=3\)
Question 5b
\(\displaystyle h^{-1}(x)=\frac12(x+2)^2-\frac32,\quad \text{domain}(h^{-1})=[-2,\,\infty)\)
Question 6a
\(\Pr(\text{event})=\dfrac12\)
Question 6b
\(\text{90% CI}:\ \left(\dfrac{29}{60},\ \dfrac{17}{20}\right)\)
Question 7a
Evaluating \(\displaystyle \int_0^a \!\Big(\sin(\pi x)-\sin(a\pi)\Big)dx\) gives the required result.
Question 7b
\(A\) is strictly increasing from the diagram, so \(A(1)\leq A(a)\leq A\left(\dfrac32\right)\implies \dfrac2\pi\leq A(a)\leq \dfrac1\pi+\dfrac32\).
Question 7c.i
\(A_2=\displaystyle \int_0^{4/3}2\left(\sin(\pi x)+\dfrac{\sqrt{3}}{2}\right)dx=2\int_0^{4/3}\!\left(\sin(\pi x)-\sin\left(\dfrac{4\pi}{3}\right)\right)dx=2\,A\left(\dfrac43\right)\)
Question 7c.ii
\(A_2=\dfrac3\pi +\dfrac{4\sqrt{3}}{3}\)
Question 8a
\(W\sim\text{Bi}\left(50,\ \dfrac16\right)\implies\Pr(W=k)=\displaystyle \binom{50}{k}\left(\dfrac16\right)^k\left(\dfrac56\right)^{50-k}\)
Question 8b
\begin{align*}\frac{\Pr(W=k+1)}{\Pr(W=k)}&=\frac{\displaystyle\binom{50}{k+1}\left(\dfrac16\right)^{k+1}\left(\dfrac56\right)^{49-k}}{\displaystyle\binom{50}{k}\left(\dfrac16\right)^k\left(\dfrac56\right)^{50-k}} \\ &=\frac{50!}{(k+1)!(49-k)!}\times\frac{k!(50-k!)}{50!}\times \frac16\times \frac65\\ &=\frac{50-k}{5(k+1)}\end{align*} \(\text{Note}:\ \ \dfrac{(n+1)!}{n!}=n+1,\ \ \ n\in\mathbb{N}\)
Question 8c
Comparing the result in Q8b to \(1\), we have \[\frac{50-k}{5(k+1)}=1\implies k=7.5,\] and so we try \(k=7\) and \(k=8:\)\[\frac{\Pr(W=8)}{\Pr(W=7)}=\frac{43}{40}\geq 1\implies \Pr(W=7)<\Pr(W=8)\\ \frac{\Pr(W=9)}{\Pr(W=8)}=\frac{42}{45}\leq 1\implies \Pr(W=8)>\Pr(W=9)\] Thus, \(\Pr(W=k)\) is maximum for \(k=8\).
Note: the above method works because for \(X\sim\text{Bi}(n,\,p)\), where \(0<p<1\), \(\Pr(X=k)\) is strictly increasing for \(k\leq \inf\{\text{mode}(X)\}\) and is strictly decreasing for \(k\geq \sup\{\text{mode}(X)\}\).
Alternatively, since \(0<p\,(=1/6)<1\) and we're told that \(\text{mode}(W)\) is unique, we have \[\text{mode}(W)=\left\lfloor (n+1)p\right\rfloor=\left\lfloor \frac{51}{6}\right\rfloor =8\]
Note: the above method works because for \(X\sim\text{Bi}(n,\,p)\), where \(0<p<1\), \(\Pr(X=k)\) is strictly increasing for \(k\leq \inf\{\text{mode}(X)\}\) and is strictly decreasing for \(k\geq \sup\{\text{mode}(X)\}\).
Alternatively, since \(0<p\,(=1/6)<1\) and we're told that \(\text{mode}(W)\) is unique, we have \[\text{mode}(W)=\left\lfloor (n+1)p\right\rfloor=\left\lfloor \frac{51}{6}\right\rfloor =8\]
Post by: dskel on August 05, 2019, 10:00:08 am
2019 NHT Methods Exam 1: AlphaZero's Answers
Not posting full working since I can't be bothered. I'll just provide answers and a few notes for the harder questions. If anyone has any questions, just ask. I did this exam really quickly, so I could've made a few errors. If I have, let me know.Question 1a\(\dfrac{dy}{dx}=2e^x+e^{-x}\)Question 1b\(f'\left(\dfrac\pi 3\right)=\dfrac{-2\pi}{3}\)Question 2\(\displaystyle f(x)=\frac23x^3-\frac34x^{1/3}-\frac53\)Question 3a\(\displaystyle \int_2^7 \frac{1}{x+\sqrt3}dx=\log_e(7+\sqrt3)-\log_e(2+\sqrt3)\quad \text{and}\quad \int_2^7 \frac{1}{x-\sqrt3}dx=\log_e(7-\sqrt3)-\log_e(2-\sqrt3)\)Question 3bPretty easy proof. Right the left-hand side on a common denominator. The result follows.Question 3c\(\displaystyle \int_2^7 \frac{x}{x^2-3}dx=\frac12\log_e(46)\)Question 4a\(\text{domain}(g)=(3,\,\infty)\quad\text{and}\quad \text{range}(g)=\mathbb{R}\)Question 4b.i\(y=x-2\)Question 4b.iiUse a calculator to check your answer here. Vertical asymptote at \(x=3\). Graph of \(g\) crosses through \((4,\,2)\).Question 5a\(x=-1\quad\text{or}\quad x=3\)Question 5b\(\displaystyle h^{-1}(x)=\frac12(x+2)^2-\frac32,\quad \text{domain}(h^{-1})=[-2,\,\infty)\)Question 6a\(\Pr(\text{event})=\dfrac12\)Question 6b\(\text{90% CI}:\ \left(\dfrac{29}{60},\ \dfrac{17}{20}\right)\)Question 7aEvaluating \(\displaystyle \int_0^a \!\Big(\sin(\pi x)-\sin(a\pi)\Big)dx\) gives the required result.Question 7b\(A\) is strictly increasing from the diagram, so \(A(1)\leq A(a)\leq A\left(\dfrac32\right)\implies \dfrac2\pi\leq A(a)\leq \dfrac1\pi+\dfrac32\).Question 7c.i\(A_2=\displaystyle \int_0^{4/3}2\left(\sin(\pi x)+\dfrac{\sqrt{3}}{2}\right)dx=2\int_0^{4/3}\!\left(\sin(\pi x)-\sin\left(\dfrac{4\pi}{3}\right)\right)dx=2\,A\left(\dfrac43\right)\)Question 7c.ii\(A_2=\dfrac3\pi +\dfrac{4\sqrt{3}}{3}\)Question 8a\(W\sim\text{Bi}\left(50,\ \dfrac16\right)\implies\Pr(W=k)=\displaystyle \binom{50}{k}\left(\dfrac16\right)^k\left(\dfrac56\right)^{50-k}\)Question 8b\begin{align*}\frac{\Pr(W=k+1)}{\Pr(W=k)}&=\frac{\displaystyle\binom{50}{k+1}\left(\dfrac16\right)^{k+1}\left(\dfrac56\right)^{49-k}}{\displaystyle\binom{50}{k}\left(\dfrac16\right)^k\left(\dfrac56\right)^{50-k}} \\ &=\frac{50!}{(k+1)!(49-k)!}\times\frac{k!(50-k!)}{50!}\times \frac16\times \frac65\\ &=\frac{50-k}{5(k+1)}\end{align*} \(\text{Note}:\ \ \dfrac{(n+1)!}{n!}=n+1,\ \ \ n\in\mathbb{N}\)Question 8cComparing the result in Q8b to \(1\), we have \[\frac{50-k}{5(k+1)}=1\implies k=7.5,\] and so we try \(k=7\) and \(k=8:\)\[\frac{\Pr(W=8)}{\Pr(W=7)}=\frac{43}{40}\geq 1\implies \Pr(W=7)<\Pr(W=8)\\ \frac{\Pr(W=9)}{\Pr(W=8)}=\frac{42}{45}\leq 1\implies \Pr(W=8)>\Pr(W=9)\] Thus, \(\Pr(W=k)\) is maximum for \(k=8\).
Note: the above method works because for \(X\sim\text{Bi}(n,\,p)\), where \(0<p<1\), \(\Pr(X=k)\) is strictly increasing for \(k\leq \inf\{\text{mode}(X)\}\) and is strictly decreasing for \(k\geq \sup\{\text{mode}(X)\}\).
Alternatively, since \(0<p\,(=1/6)<1\) and we're told that \(\text{mode}(W)\) is unique, we have \[\text{mode}(W)=\left\lfloor (n+1)p\right\rfloor=\left\lfloor \frac{51}{6}\right\rfloor =8\]
I got the same, just that question 7C) I didn't set a value for a, did you just pick 4/3 or does that come from somewhere? The same thing occurred it equals 2A(a) just thought it was a strange question.
Post by: AlphaZero on August 05, 2019, 02:25:42 pm
I got the same, just that question 7C) I didn't set a value for a, did you just pick 4/3 or does that come from somewhere? The same thing occurred it equals 2A(a) just thought it was a strange question.
You have to find the particular value of \(a\) that gives the area of the region, which is \(\dfrac43\).
The expression \(2\,A(a)\) is just a function of \(a\). It doesn't give the area of the required region.
Post by: bec.hodges on October 14, 2019, 08:19:38 pm
Post by: S_R_K on October 14, 2019, 08:55:29 pm
Hi, I'm currently struggling with 8b) and I was wondering if you could post the full solutions? I've tried so many times and I can't get the final answer :(
Now cancel common factors.
Post by: Matthew_Whelan on October 18, 2019, 07:04:29 pm
Now cancel common factors.
Regarding 8b), how do you go from the first line of working to that? More so the conceptual reasoning behind it, I don't understand how to equate that from the first expression or why you do that if that makes sense.
i.e. the line of working before that is confusing..
Edit: I watched a tutorial and it makes sense now. Are there any ways of practising these types of abstract questions? The VCAA exams prior to 2018 are good practise but not quite at this calibre imo.