Author Topic: Specialist Maths (Exam 2): Discussion, Questions & Potential Solutions (Read 22991 times)

Joseph41 · « **on:** October 09, 2018, 03:25:52 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 09, 2018, 01:19:21 pm »

Excited to see what's in store on Monday when Exam 2 comes out.
As usual, solutions will be posted here, and successively updated to be more thorough (such as CAS screenshots, etc).
Good luck everyone with the weekend's study.

Answers

SECTION A
1.   E
2.   B
3.   D
4.   A
5.   D
6.   C
7.   C
8.   E
9.   D
10.   A
11.   C
12.   A
13.   B
14.   C
15.   E
16.   B
17.   E
18.   D
19.   E
20.   B

AlphaZero · « **Reply #2 on:** November 10, 2018, 11:05:51 am »

VCAA 2018 Specialist Exam 2 Provisional Solutions
Solutions by Daniel Traicos at TWM Publications

Fairly straight forward exam I thought. Some tricky multiple choice questions, but I think VCAA were quite generous with this exam.

Edit 1: Fixed solution to Q17. Fixed typo in Q1d.

Onyx · « **Reply #3 on:** November 12, 2018, 11:51:23 am »

calling it now related rates and vector proofs
And a two tail test for stat

HamConspiracy · « **Reply #4 on:** November 12, 2018, 05:28:00 pm »

Who found the paper fairly easy this year? No real tricky questions except for the top of the fountain one.

AlphaZero · « **Reply #5 on:** November 12, 2018, 05:39:32 pm »

Quote from: crouchie on November 09, 2018, 01:19:21 pm

4. B

I get A for Q4. Can we just check?

jazzycab · « **Reply #6 on:** November 12, 2018, 05:44:07 pm »

MCQ

1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. D
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. B

ERQ1

1a. $\text{Dom}\left(f\left(x\right)\right):\left[-\sqrt{2},\sqrt{2}\right]\\\text{Ran}\left(f\left(x\right)\right):\left[-\pi,\pi\right]$
1b:

1c: $f'\left(x\right)=\frac{4}{\sqrt{2-x^2}}$
1d: $f'\left(x\right)=-\frac{4}{\sqrt{2-x^2}}$
1ei: $x\in\left(-\sqrt{2},0\right)\cup\left(0,\sqrt{2}\right)$
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii:

ERQ2

2a: $\text{Centre: }\left(1,2\right)\\\text{Radius: }2$
2b: $\begin{align*}\left|z+1\right|&=\sqrt{2}\left|z-i\right|\\\sqrt{\left(x+1\right)^2+y^2}&=\sqrt{2\left(x^2+\left(y-1\right)^2\right)}\\\left(x+1\right)^2+y^2&=2\left(x^2+\left(y-1\right)^2\right)\\x^2+2x+1+y^2&=2x^2+2y^2-4y+2\\0&=x^2+y^2-2x-4y+1\\0&=\left(x^2-2x+1\right)+\left(y^2-4y+4-4\right)\\0&=\left(x-1\right)^2+\left(y-2\right)^2-4\\4&=\left(x-1\right)^2+\left(y-2\right)^2\end{align*}$
2c/d:

2e: $\frac{4\pi}{3}-\sqrt{3}$

ERQ3

3a: $\begin{align*}V&=\pi\int_{0}^{h}{x^2}dy\\y&=\frac{\sqrt{4x^2-1}}{2}\\2y&=\sqrt{4x^2-1}\\4y^2&=4x^2-1\\4x^2&=4y^2+1\\x^2&=y^2+\frac{1}{4}\\V&=\pi\int_{0}^{h}{y^2+\frac{1}{4}}dy\\&=\pi\left[\frac{y^3}{3}+\frac{y}{4}\right]_{0}^{h}\\&=\pi\left(\frac{h^3}{3}+\frac{h}{4}\right)\\&=\frac{\pi}{4}\left(\frac{4h^3}{3}+h\right)\end{align*}$
3b: $0.59$
3ci: $\begin{align*}\frac{dV_{in}}{dt}&=\frac{1}{25}\\\frac{dV_{out}}{dt}&=\frac{\sqrt{h}}{20}\\\frac{dV}{dt}&=\frac{1}{25}-\frac{\sqrt{h}}{20}\\&=\frac{4-5\sqrt{h}}{100}\\\frac{dV}{dh}&=\frac{\pi}{4}\left(4h^2+1\right)\\\frac{dh}{dt}&=\frac{4}{\pi\left(4h^2+1\right)}\times\frac{4-5\sqrt{h}}{100}\\&=\frac{4-5\sqrt{h}}{25\pi\left(4h^2+1\right)}\end{align*}$
3cii: $0.0153$
3d: $9.8$
3e: $0.43$
3f: $0.23$

ERQ4

4a: $y_{A}=x^2-1\\y_{B}=x+3$
4b: $\begin{align*}t+1&=t^2\\t&=\frac{1\pm\sqrt{5}}{2}\\t&>0\\t&=\frac{1+\sqrt{5}}{2}\\y_{A}\left(\frac{1+\sqrt{5}}{2}\right)&=\frac{3\sqrt{5}+5}{2}\\y_{B}\left(\frac{1+\sqrt{5}}{2}\right)&=\frac{9+\sqrt{5}}{2}\end{align*}$
4c: $\left(2.562,5.562\right)$
4d: $t\in\left[0,\frac{5}{2}\right)$
4e: $t\in\left(91.8,95.8\right)\text{ minutes}\\ 4\text{ minutes total}$

ERQ5

5a:

5bi: $20g\sin{\left(30^\circ\right)}-20v=20a$
5bii: $\begin{align*}g\sin{\left(30^\circ\right)}-v&=a\\\frac{g}{2}-v&=a\\a&=\frac{g-2v}{2}\end{align*}$
5c: $x=-v+\frac{g}{2}\log_{e}{\left(\frac{\frac{g}{2}}{\frac{g}{2}-v}\right)}$
5d: $4.8\text{ ms}^{-1}\text{ down the ramp}$
5ei: $\int_{0}^{4.5}{\frac{2}{g-2v}}dv$
5eii: $2.51$

ERQ6

$6a: H_{0}:\mu=150\\H_{1}\mu<150$
6b: $\sigma_{X}=\frac{3\sqrt{2}}{2}$
6c: $0.0092$
6d: $p<0.05$, therefore, $H_{0}$ should be rejected
6e: $146.52$
6f: $0.24$
6g: $\left(139.5,150.5\right)$

crouchie · « **Reply #7 on:** November 12, 2018, 05:57:58 pm »

Quote from: dantraicos on November 12, 2018, 05:39:32 pm

I get A for Q4. Can we just check?

You're right, my bad, a typo! Fixed.

DinWell · « **Reply #8 on:** November 12, 2018, 06:05:49 pm »

Quote from: jazzycab on November 12, 2018, 05:44:07 pm

MCQ
1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. C
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. B
ERQ1
1a. $\text{Dom}\left(f\left(x\right)\right):\left[-\sqrt{2},\sqrt{2}\right]\\\text{Ran}\left(f\left(x\right)\right):\left[-\pi,\pi\right]$
1b: GRAPH TO COME
1c: $f'\left(x\right)=\frac{4}{\sqrt{2-x^2}}$
1d: $f'\left(x\right)=-\frac{4}{\sqrt{2-x^2}}$
1ei: $x\in\left(-\sqrt{2},0\right)\cup\left(0,\sqrt{2}\right)$
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: GRAPH TO COME
ERQ2
2a: $\text{Centre: }\left(1,2\right)\\\text{Radius: }2$
2b: $\begin{align*}\left|z+1\right|&=\sqrt{2}\left|z-i\right|\\\sqrt{\left(x+1\right)^2+y^2}&=\sqrt{2\left(x^2+\left(y-1\right)^2\right)}\\\left(x+1\right)^2+y^2&=2\left(x^2+\left(y-1\right)^2\right)\\x^2+2x+1+y^2&=2x^2+2y^2-4y+2\\0&=x^2+y^2-2x-4y+1\\0&=\left(x^2-2x+1\right)+\left(y^2-4y+4-4\right)\\0&=\left(x-1\right)^2+\left(y-2\right)^2-4\\4&=\left(x-1\right)^2+\left(y-2\right)^2\end{align*}$
2c/d: GRAPH TO COME
2e: $\frac{2\pi}{3}-\sqrt{3}$
ERQ3
3a: $\begin{align*}V&=\pi\int_{0}^{h}{x^2}dy\\y&=\frac{\sqrt{4x^2-1}}{2}\\2y&=\sqrt{4x^2-1}\\4y^2&=4x^2-1\\4x^2&=4y^2+1\\x^2&=y^2+\frac{1}{4}\\V&=\pi\int_{0}^{h}{y^2+\frac{1}{4}}dy\\&=\pi\left[\frac{y^3}{3}+\frac{y}{4}\right]_{0}^{h}\\&=\pi\left(\frac{h^3}{3}+\frac{h}{4}\right)\\&=\frac{\pi}{4}\left(\frac{4h^3}{3}+h\right)\end{align*}$
3b: $0.59$
3ci: $\begin{align*}\frac{dV_{in}}{dt}&=\frac{1}{25}\\\frac{dV_{out}}{dt}&=\frac{\sqrt{h}}{20}\\\frac{dV}{dt}&=\frac{1}{25}-\frac{\sqrt{h}}{20}\\&=\frac{4-5\sqrt{h}}{100}\\\frac{dV}{dh}&=\frac{\pi}{4}\left(4h^2+1\right)\\\frac{dh}{dt}&=\frac{4}{\pi\left(4h^2+1\right)}\times\frac{4-5\sqrt{h}}{100}\\&=\frac{4-5\sqrt{h}}{25\pi\left(4h^2+1\right)}\end{align*}$
3cii: $0.0153$
3d: $9.8$
3e: $0.43$
3f: $0.23$

Why did we have to restrict the domain of g(x)? I thought that we consider it a different function on it's maximal domain. I interpreted it as, if g(x)=sign(x). Then g(x) = -4 for x<0 and 4 for x>0

mzhao · « **Reply #9 on:** November 12, 2018, 06:10:51 pm »

Quote from: DinWell on November 12, 2018, 06:05:49 pm

Why did we have to restrict the domain of g(x)? I thought that we consider it a different function on it's maximal domain. I interpreted it as, if g(x)=sign(x). Then g(x) = -4 for x<0 and 4 for x>0

Fair point. I think both answers are acceptable, as long as the hybrid function you defined matches the graph you sketched.

iroze007 · « **Reply #10 on:** November 12, 2018, 06:12:36 pm »

Quote from: jazzycab on November 12, 2018, 05:44:07 pm

MCQ
1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. C
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. B
ERQ1
1a. $\text{Dom}\left(f\left(x\right)\right):\left[-\sqrt{2},\sqrt{2}\right]\\\text{Ran}\left(f\left(x\right)\right):\left[-\pi,\pi\right]$
1b: GRAPH TO COME
1c: $f'\left(x\right)=\frac{4}{\sqrt{2-x^2}}$
1d: $f'\left(x\right)=-\frac{4}{\sqrt{2-x^2}}$
1ei: $x\in\left(-\sqrt{2},0\right)\cup\left(0,\sqrt{2}\right)$
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: GRAPH TO COME
ERQ2
2a: $\text{Centre: }\left(1,2\right)\\\text{Radius: }2$
2b: $\begin{align*}\left|z+1\right|&=\sqrt{2}\left|z-i\right|\\\sqrt{\left(x+1\right)^2+y^2}&=\sqrt{2\left(x^2+\left(y-1\right)^2\right)}\\\left(x+1\right)^2+y^2&=2\left(x^2+\left(y-1\right)^2\right)\\x^2+2x+1+y^2&=2x^2+2y^2-4y+2\\0&=x^2+y^2-2x-4y+1\\0&=\left(x^2-2x+1\right)+\left(y^2-4y+4-4\right)\\0&=\left(x-1\right)^2+\left(y-2\right)^2-4\\4&=\left(x-1\right)^2+\left(y-2\right)^2\end{align*}$
2c/d: GRAPH TO COME
2e: $\frac{2\pi}{3}-\sqrt{3}$
ERQ3
3a: $\begin{align*}V&=\pi\int_{0}^{h}{x^2}dy\\y&=\frac{\sqrt{4x^2-1}}{2}\\2y&=\sqrt{4x^2-1}\\4y^2&=4x^2-1\\4x^2&=4y^2+1\\x^2&=y^2+\frac{1}{4}\\V&=\pi\int_{0}^{h}{y^2+\frac{1}{4}}dy\\&=\pi\left[\frac{y^3}{3}+\frac{y}{4}\right]_{0}^{h}\\&=\pi\left(\frac{h^3}{3}+\frac{h}{4}\right)\\&=\frac{\pi}{4}\left(\frac{4h^3}{3}+h\right)\end{align*}$
3b: $0.59$
3ci: $\begin{align*}\frac{dV_{in}}{dt}&=\frac{1}{25}\\\frac{dV_{out}}{dt}&=\frac{\sqrt{h}}{20}\\\frac{dV}{dt}&=\frac{1}{25}-\frac{\sqrt{h}}{20}\\&=\frac{4-5\sqrt{h}}{100}\\\frac{dV}{dh}&=\frac{\pi}{4}\left(4h^2+1\right)\\\frac{dh}{dt}&=\frac{4}{\pi\left(4h^2+1\right)}\times\frac{4-5\sqrt{h}}{100}\\&=\frac{4-5\sqrt{h}}{25\pi\left(4h^2+1\right)}\end{align*}$
3cii: $0.0153$
3d: $9.8$
3e: $0.43$
3f: $0.23$
ERQ4
4a: $y_{A}=x^2-1\\y_{B}=x+3$
4b: $\begin{align*}t+1&=t^2\\t&=\frac{1\pm\sqrt{5}}{2}\\t&>0\\t&=\frac{1+\sqrt{5}}{2}\\y_{A}\left(\frac{1+\sqrt{5}}{2}\right)&=\frac{3\sqrt{5}+5}{2}\\y_{B}\left(\frac{1+\sqrt{5}}{2}\right)&=\frac{9+\sqrt{5}}{2}\end{align*}$
4c: $\left(2.562,5.562\right)$
4d: $t\in\left[0,\frac{5}{2}\right)$
4e: $t\in\left(91.8,95.8\right)\\ 4\text{ minutes}$

you sure about that segment q. I got 4pi/3 - root(3)

fluff · « **Reply #11 on:** November 12, 2018, 06:21:13 pm »

Quote from: iroze007 on November 12, 2018, 06:12:36 pm

you sure about that segment q. I got 4pi/3 - root(3)

I got the same as you!

jazzycab · « **Reply #12 on:** November 12, 2018, 06:23:36 pm »

Quote from: iroze007 on November 12, 2018, 06:12:36 pm

you sure about that segment q. I got 4pi/3 - root(3)

Not certain about any of them. Did them very quickly. Will check when I get a chance.

mzhao · « **Reply #13 on:** November 12, 2018, 06:25:34 pm »

Quote from: jazzycab on November 12, 2018, 05:44:07 pm

MCQ
5c: $x&=-v+\frac{g}{2}\log_{e}{\left(\frac{g}{g-2v}\right)}$ - NOTE: This is NOT the same form as specified in the question. I haven't had time to have a good look, but my feeling (after speaking with a number of other people) is that there was an error in this exam...

Nothing wrong with this question.

Spoiler

$ x=-v+\frac{g}{2}\log_{e} \left( \frac{g}{g-2v} \right) \\ =-v+\frac{g}{2}\log_{e} \left(\frac{\frac{g}{2}}{\frac{g}{2}-v} \right) $

DinWell · « **Reply #14 on:** November 12, 2018, 06:26:59 pm »

Quote from: mzhao on November 12, 2018, 06:25:34 pm

Nothing wrong with this question.

Spoiler
$ x=-v+\frac{g}{2}\log_{e} \left( \frac{g}{g-2v} \right) \\ =-v+\frac{g}{2}\log_{e} \left(\frac{\frac{g}{2}}{\frac{g}{2}-v} \right) $

It's alright if you subbed in g=9.8 right? But still kept it in the correct form.

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Author Topic: Specialist Maths (Exam 2): Discussion, Questions & Potential Solutions (Read 22991 times)

Joseph41

Specialist Maths (Exam 2): Discussion, Questions & Potential Solutions

crouchie

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AlphaZero

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Onyx

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HamConspiracy

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AlphaZero

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jazzycab

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crouchie

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DinWell

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mzhao

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iroze007

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fluff

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jazzycab

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mzhao

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DinWell

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