ATAR Notes: Forum

Archived Discussion => Mathematics and Science => 2012 => End-of-year exams => Exam Discussion => Victoria => Mathematical Methods CAS => Topic started by: BubbleWrapMan on November 08, 2012, 06:24:06 pm

Title: My Worked Solutions
Post by: BubbleWrapMan on November 08, 2012, 06:24:06 pm: SECTION 1

Question 1

Answer: C

Period = $\frac{2\pi}{\frac{\pi}{5}}=10$

Question 2

Answer: E

Average rate of change = $\frac{f(3)-f(1)}{3-1}=9$

Question 3

Answer: D

Minimum at x = 1

$f(1)=-9$

$|3-1|=2$ and $|-2-1|=3$ so the left endpoint is the global max, as the x-value is furthest from the minimum

$f(-2)=0$ , so the range is $[-9,0]$

Question 4

Answer: A

$\frac{\mathrm{d}}{\mathrm{d}x}(g(e^{kx}))=\frac{\mathrm{d}}{\mathrm{d}x}(e^{kx})g'(e^{kx})=kg'(e^{kx})e^{kx}$

Question 5

Answer: B

$(x-2)^{2}>0$ when $x\in\mathbb{R}\setminus\left \{2\right \}$ , so the maximal domain of the composite is $x\in\mathbb{R}\setminus\left \{2\right \}$ .

The range is $\mathbb{R}$ since the inner function spans all positive reals.

Question 6

Answer: C

Dilation by a factor of $\frac{1}{2}$ parallel to the x-axis, followed by a translation of $\frac{\pi}{4}$ units in the positive direction of the x-axis is a possible transformation, which was option C.

Question 7

Answer: D

At 2 am, t = 2

At 2 pm, t = 14

So the average temperature is $\frac{1}{14-2}\int_{2}^{14}T(t)\mathrm{d}t=22$

Question 8

Answer: A

The given facts about p and q don't actually tell you anything about the answer, since the intercepts of $f(x)$ do not affect the gradient.

But, we have $a<0$ and $m<0<n$ , which means it is a negative cubic with two distinct stationary points and $x=m$ and $x=n$ , and $n>m$ . Doing a quick sketch of this will show that the gradient is negative for $(-\infty,m)\cup(n,\infty)$ .

Question 9

Answer: D

$\frac{\mathrm{d} y}{\mathrm{d} x}=(-2x)\cdot(\frac{1}{2})\cdot(\sqrt{b-x^{2}})^{-\frac{1}{2}}=-\frac{x}{\sqrt{b-x^{2}}}$

If the normal has a gradient of 3 at x=1, the tangent has a gradient of $-\frac{1}{3}$ (negative reciprocal)

$\therefore -\frac{1}{\sqrt{b-1^{2}}}=-\frac{1}{3}\Rightarrow\sqrt{b-1}=3\Rightarrow b=3^{2}+1=10$

Question 10

Answer: C

Solve $\frac{1}{a}\int_{0}^{a}\sin^{2}(x)\mathrm{d}x=0.4$ with $0\leq a\leq 2\pi$ for $a$ .

Question 11

Answer: E

Inverse normal with p = 0.6, µ = 252 and σ = 12 (as $\Pr(X<x)=1-\Pr(X>x) = 1-0.4 = 0.6$ )

Question 12

Answer: A

Possible combos are WL and LW

She has just won, so $\Pr(WL) = 0.7 \cdot (1 - 0.7) = 0.21$

And $\Pr(LW) = (1 - 0.7) \cdot (1 - 0.6) = 0.12$

Adding these gives 0.33

Question 13

Answer: B

$\Pr(A\cap B)+\Pr(A\cap B')=\Pr(A)=\frac{2}{5}+\frac{3}{7}=\frac{29}{35}$

$\therefore \Pr(B'\mid A)=\frac{\Pr(A\cap B')}{\Pr(A)}=\frac{(\frac{3}{7})}{(\frac{29}{35})}=\frac{15}{29}$

Question 14

Answer: D

$y=\sqrt{x}\Rightarrow x=y^2$

All rectangles are of width 1, so the area is $1\cdot 1^2+1\cdot 2^2+1\cdot 3^2+1\cdot 4^2 =30$

Question 15

Answer: B

The antiderivative of a quadratic is always a cubic so this eliminates A, C and E. $f'(x)=0$ has two distinct solutions so the cubic graph will have two stationary points, which eliminates D.

Question 16

Answer: D

If c = 3 the stationary point at (a,-3), which is 'higher up' than the other stationary point, will 'touch' the x-axis, giving another solution, but if c < 3 there will still be a single solution as this highest stationary point will still lie below the x-axis.

If c = 8, the stationary point at (b,-8), which is 'lower down', will touch the x-axis, so any c > 8 will shift this lowest stationary point above the x-axis. Hence, c < 3 or c > 8 for there to be a single solution.

Question 17

Answer: B

The determinant of the coefficient matrix is $m^2+2m-3$ , which is 0 when m = 1 or m = -3.

However if m = 1 we have

$\begin{bmatrix}1&3\\1&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$

These two lines are clearly the same, so there are infinitely many solutions. Checking the other option:

$\begin{bmatrix}-3&3\\1&-1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\-3\end{bmatrix}$

Which are two parallel but distinct lines, so they have no intersection. Hence the system will have no solutions for m = -3.

Question 18

Answer: A

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{x}$

At x = a, $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{a}$ , so the equation of the tangent is $y=\frac{1}{a}x+c$

Using the point $(a,\ln(a))$ ,

$\ln(a)=\frac{1}{a}a+c\Rightarrow c=\ln(a)-1=\ln(\frac{a}{e})$

So the tangent has equation $y=\frac{1}{a}x+\ln(\frac{a}{e})$

At the point $(b,0)$ ,

$\frac{b}{a}+\ln(\frac{a}{e})=0$

$\Rightarrow \frac{b}{a}=\ln(\frac{e}{a})$

$\Rightarrow b=a\ln(\frac{e}{a})$

Now, $b<0\Rightarrow a\ln(\frac{e}{a})<0\Rightarrow \ln(\frac{e}{a})<0$ (as $a>0$ )

$\Rightarrow \frac{e}{a}<1\Rightarrow a>e$ , so option A is false.

Question 19

Answer: B

$\cos(\pi-\theta)=-\cos(\theta)=-\cos(-\theta)$

Question 20

Answer: E

$\Pr(X=0) = (1-p)^{0}p=p$

$\Pr(X=1)=(1-p)^{1}p=p-p^2$

$\therefore \Pr(X>1)=1-\Pr(X=0)-\Pr(X=1)=1-p-(p-p^{2})=1-2p+p^{2}=(1-p)^{2}$

Question 21

Answer: A

$\frac{\mathrm{d} V}{\mathrm{d} h}=\pi h^2$ and $\frac{\mathrm{d} V}{\mathrm{d} t}=-300$

$\therefore \frac{\mathrm{d} h}{\mathrm{d} t}=\frac{\mathrm{d} h}{\mathrm{d} V}\frac{\mathrm{d} V}{\mathrm{d} t}=\frac{-300}{\pi h^{2}}$

At h = 5, $\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{-300}{25\pi}=\frac{-12}{\pi}$ , so the rate of decrease is $\frac{12}{\pi}$ .

Question 22

Answer: B

(a,b) is in the second quadrant and (c,d) is in the fourth quadrant.

The function has been shifted 2 units in the positive x direction, the absolute value has been taken and then the whole graph has been reflected in the x-axis, in that order.

Both stationary points are translated: (a+2,b) and (c+2,d), retaining their min/max nature

(a+2,b) is already above the x-axis so the modulus has no effect. However, (c+2,d) is part of the section below the x-axis, which is now reflected in the x-axis and so the point is moved above the x-axis to (c+2,-d) and becomes a maximum, so now both points are maxima.

Now the whole graph is reflected in the x-axis, so we have (a+2,b) becoming (a+2,-b) and (c+2,-d) becoming (c+2,d). Both were maxima, so they are now minima.

SECTION 2

Question 1 (9 marks)

part a. (2 marks)

$A=6480=2\cdot\frac{5x}{2}\cdot x+2\cdot h\cdot x+2\cdot\frac{5x}{2}\cdot h=5x^2+7xh$

$\therefore 7xh=6480-5x^{2} \Rightarrow h=\frac{6480-5x^{2}}{7x}$ , as required.

part b. (2 marks)

We have $V(x)>0$ and $x>0$ , so

$\frac{5x(6480-5x^{2})}{14}>0$

$\Leftrightarrow 1296-x^{2}>0$

$\Leftrightarrow x^2<1296$

$\Leftrightarrow 0<x<36$ (as $x>0$ )

part c. (3 marks)

$V=\frac{5x(6480-5x^{2})}{14}=\frac{16200x}{7}-\frac{25x^{3}}{14}$

$\therefore \frac{\mathrm{d} V}{\mathrm{d} x}=-\frac{75}{14}x^2+\frac{16200}{7}$

part d. (2 marks)

$\frac{\mathrm{d} V}{\mathrm{d} x}=0$ when $x=\sqrt{432}=12\sqrt{3}$ , given $x>0$

When $x=12\sqrt{3}$ , $h=\frac{6480-5(12\sqrt{3})^{2}}{7\cdot12\sqrt{3}}=\frac{120\sqrt{3}}{7}$

Hence the block has maximum volume when $x=12\sqrt{3}$ and $h=\frac{120\sqrt{3}}{7}$

Question 2 (14 marks)

part a. (3 marks)

Hyperbola graph, asymptotes at y = 3 and x = 2, intercepts at $(0,\frac{11}{4})$ and $(\frac{11}{6},0)$ , found by evaluating $f(0)$ and solving $f(x)=0$ , respectively.

part b. i. (1 mark)

$f'(x)=\frac{-2}{(2x-4)^{2}}=\frac{-1}{2(x-2)^{2}}$ (either one)

part b. ii. (1 mark)

$(-\infty,0)$

part b. iii. (1 mark)

The gradient of f is never equal to 0, so f has no stationary points.

part c. (2 marks)

$q=f(p)=\frac{1}{2p-4}+3$ so the tangent passes through $(p,\frac{1}{2p-4}+3)$

The gradient of the tangent is $f'(p)=\frac{-2}{(2p-4)^{2}}$

Let c be the y-intercept of the tangent. Then the equation of the tangent is:

$y=\frac{-2}{(2p-4)^{2}}\cdot x+c$

The tangent passes through $(p,\frac{1}{2p-4}+3)$ , so:

$\frac{1}{2p-4}+3=\frac{-2}{(2p-4)^{2}}\cdot p+c$

$\Rightarrow c=\frac{2p-4}{(2p-4)^{2}}+3+\frac{2p}{(2p-4)^{2}}=\frac{4p-4}{(2p-4)^{2}}+3$

Hence the equation of the tangent is:

$y=\frac{-2}{(2p-4)^{2}}\cdot x+\frac{4p-4}{(2p-4)^{2}}+3$

$\Leftrightarrow y-3=\frac{-2}{(2p-4)^{2}}\cdot x+\frac{4p-4}{(2p-4)^{2}}$

$\Leftrightarrow (2p-4)^{2}(y-3)=-2x+4p-4$ , as required.

(2marks for all this shit wtf?)

part d. (2 marks)

Subbing in $x = -1$ and $y = \frac{7}{2}$ (to find the values of p for which the tangent can pass through this point):

$(2p-4)^{2}(\frac{7}{2}-3)=2+4p-4$

$\Leftrightarrow (4p^{2}-16p+16)(\frac{1}{2})=4p-2$

$\Leftrightarrow 2p^{2}-8p+8-4p+2=0$

$\Leftrightarrow p^{2}-6p+5=0$

$\Leftrightarrow (p-1)(p-5)=0$

$\Leftrightarrow p = 1$ or $p=5$

When $p = 1$ , $q = f(1) = \frac{5}{2}$

And when $p = 5$ , $q = f(5) = \frac{19}{6}$

So the points are $(1,\frac{5}{2})$ and $(5,\frac{19}{6})$

part e. (2 marks)

Let $y=f(x)=\frac{1}{2x-4}+3$ , and let $T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}ax+c\\y+d\end{bmatrix}=\begin{bmatrix}x'\\y'\end{bmatrix}$ , where $x'$ and $y'$ are the images of the ordinates of $f$ (i.e. the ordinates of $g$ ).

Then ( $y'=y+d\Leftrightarrow y=y'-d$ ) and ( $x'=ax+c\Leftrightarrow x=\frac{x'-c}{a}$ )

The equation for the image of f is then $y'-d=\frac{1}{2(\frac{x'-c}{a})-4}+3\Leftrightarrow y'-d-3=\frac{1}{2(\frac{x'-c}{a})-4}$ , which is equivalent to $y'=\frac{1}{x'}$ .

So we have $y'-d-3=y'$ and $2(\frac{x'-c}{a})-4=x'$

$\Leftrightarrow d=-3$ and $\frac{2}{a}x'-\frac{2c}{a}-4=x'$

$\Leftrightarrow d=-3$ and $\frac{2}{a}=1$ and $-\frac{2c}{a}-4=0$

$\Leftrightarrow d=-3$ and $a=2$ and $c=-2a$

$\Leftrightarrow d=-3$ and $a=2$ and $c=-4$

Question 3 (15 marks)

part a. i. (1 mark)

$X\sim \mathrm{Bi}(n,\frac{1}{4})$

If n = 3, then $\Pr(X=3)=1\cdot(\frac{1}{4})^{3}\cdot(\frac{3}{4})^{0}=\frac{1}{64}$

part a. ii. (2 marks)

$X\sim \mathrm{Bi}(n,\frac{1}{4})$

If n = 20, then $\Pr(X\geq 10)=0.0139$

part a. iii. (1 mark)

$\mathrm{Var}(X)=np(1-p)$

$\Rightarrow \frac{75}{16}=n\cdot\frac{1}{4}\cdot\frac{3}{4}$

$\Rightarrow n=\frac{75}{16}\cdot\frac{16}{3}=25$ , as required.

part b. i. (3 marks)

Let $C_{k}$ stand for a correct answer for question k, and $C'_{k}$ stand for an incorrect answer.

$\Pr(C_{3}C_{4}C_{5})=\Pr(C_{2}C_{3}C_{4}C_{5})+\Pr(C'_{2}C_{3}C_{4}C_{5})$

$=(1-\frac{2}{3})\cdot(\frac{3}{4})^3+\frac{2}{3}\cdot(1-\frac{2}{3})\cdot(\frac{3}{4})^2=\frac{17}{64}$

part b. ii. (2 marks)

Let the transition matrix $T=\begin{bmatrix}\Pr(C_{k}\mid C_{k-1})&\Pr(C_{k}\mid C'_{k-1})\\\Pr(C'_{k}\mid C_{k-1})&\Pr(C'_{k}\mid C'_{k-1})\end{bmatrix}=\begin{bmatrix}\frac{3}{4}&\frac{1}{3}\\\frac{1}{4}&\frac{2}{3}\end{bmatrix}$ , and let the initial state matrix $S_{1}=\begin{bmatrix}0\\1\end{bmatrix}$

Then $S_{k}=T^{k-1}\cdot S_{1}=\begin{bmatrix}\Pr(C_{k})\\ \Pr(C'_{k}) \end{bmatrix}$

$\therefore S_{25}=T^{24}\cdot S_{1}=\begin{bmatrix}\Pr(C_{25})\\ \Pr(C'_{25}) \end{bmatrix}=\begin{bmatrix}0.5714\\ 0.4286 \end{bmatrix}$ to four decimal places.

Hence the probability that she answers question 25 correctly is 0.5714, correct to four decimal places.

part c. (2 marks)

$Y\sim \mathrm{Bi}(25,p)$

$\Pr(Y>23)=\Pr(Y=24)+\Pr(Y=25)=25p^{24}(1-p)+p^{25}=p^{24}(25-24p)$

$6\Pr(Y=25)=6p^{25}$

Therefore if $\Pr(Y>23)=6\Pr(Y=25)$ , then $p^{24}(25-24p)=6p^{25}$

$\Rightarrow 25-24p=6p$ (as p > 0)

$\Rightarrow p = \frac{5}{6}$

part d. (4 marks)

$\Pr(W\geq20)=\Pr(Z\geq\frac{20-a}{b})=\Pr(Z\leq\frac{a-20}{b})$

Inverse normal with p = Pr(Y≥18), µ = 0 and σ = 1 gives

$\frac{a-20}{b}=1.698233...\Leftrightarrow b=\frac{a-20}{1.698233...}$ $(1)$

$\Pr(W\geq25)=\Pr(Z\geq\frac{25-a}{b})=\Pr(Z\leq\frac{a-25}{b})$

Inverse normal with p = Pr(Y≥22), µ = 0 and σ = 1 gives

$\frac{a-25}{b}=-0.301369...\Leftrightarrow b=\frac{a-25}{-0.301369...}$ $(2)$

Equating b from (1) and (2) gives $\frac{a-20}{1.698232...}=\frac{a-25}{-0.301369...}\Rightarrow a=24.24643...$

Substituting this value of a into (1), we have $b=\frac{(24.24643...)-20}{1.698233...}=2.50049...$

Hence a = 24.246 and b = 2.500 (or 2.501, seems to vary depending on decimals retained, friggen VCAA)

Question 4 (12 marks)

part a. i. (1 mark)

$\frac{r}{h}=\frac{2}{10}=\frac{1}{5}$ (similar triangles)

$\therefore r = \frac{h}{5}$

part a. ii. (1 mark)

$V=\frac{\pi}{3}r^{2}h=\frac{\pi}{3}\cdot (\frac{h}{5})^{2}\cdot h=\frac{\pi}{3}\cdot\frac{h^{2}}{25}\cdot h=\frac{\pi h^{3}}{75}$ , as required.

part b. (1 mark)

At t = 20,

$h=10+\frac{1}{1600}(8000-24000)=10-\frac{16000}{1600}=10-10=0$ . Since V = 0 when h = 0, the tank is empty when t = 20, as required.

part c. i. (1 mark)

When t = 5, $h=10+\frac{1}{1600}(125-6000)=\frac{405}{64}$

part c. ii. (3 mark)

$\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{3}{1600}(t^{2}-400)$ and $\frac{\mathrm{d} V}{\mathrm{d} h}=\frac{\pi h^{2}}{25}$

$\therefore \frac{\mathrm{d} V}{\mathrm{d} t}=\frac{\mathrm{d} V}{\mathrm{d} h}\cdot \frac{\mathrm{d} h}{\mathrm{d} t}=\frac{3\pi h^{2}(t^{2}-400)}{40000}$

When t = 5, $h=\frac{405}{64}$ , so

$\frac{\mathrm{d} V}{\mathrm{d} t}=\frac{3\pi (\frac{405}{64})^{2}(5^{2}-400)}{40000}\approx -3.5$

Hence the volume is decreasing at a rate of 3.5 cubic metres per minute, correct to one decimal place.

part d. (2 marks)

When h = 2 and 0 ≤ t ≤ 20,

$\frac{1}{1600}(t^{3}-1200t)=2\Rightarrow t=12.2$ minutes, correct to one decimal place.

part e. (2 marks)

When h = 2, $V=\frac{\pi (2)^{3}}{75}=\frac{8\pi}{75}$

$\frac{\mathrm{d} V}{\mathrm{d} t}=0.2=\frac{1}{5}$ , so $V=\frac{t}{5}$ (since V = 0 when t = 0, where t is the time in minutes after the tank is first empty)

So when h = 2, $V=\frac{8\pi}{75}=\frac{t}{5}\Rightarrow t=\frac{8\pi}{15}$ minutes

part f. (1 mark)

He enters the tank when t = 12.2, which is 20 - 12.2 = 7.8 minutes before the tank is first empty.

Hence he has $7.8+\frac{8\pi}{15}=9.5$ minutes, correct to one decimal place.

Question 5 (8 marks)

part a. i. (1 mark)

$\int_{-2}^{0}f(x)\mathrm{d}x=1-e^{-2}$

part a. ii. (1 mark)

$A=\int_{-2}^{0}f(x)\mathrm{d}x=1-e^{-2}$

part a. iii. (1 mark)

$A=\int_{-2}^{0}f(x)\mathrm{d}x+\int_{0}^{1}f(x)\mathrm{d}x=1-e^{-2}+e-1=e-e^{-2}$

part b. i. (2 marks)

$\ln(x)=-\ln(a-x)$

$\Rightarrow \ln(x)=\ln(\frac{1}{a-x})$

$\Rightarrow x=\frac{1}{a-x}$

$\Rightarrow ax-x^{2}=1$

$\Rightarrow x^{2}-ax+1=0$

$\Rightarrow x=\frac{a-\sqrt{a^{2}-4}}{2},x=\frac{a+\sqrt{a^{2}-4}}{2}$

part b. ii. (1 mark)

The points are distinct for $a^{2}-4>0\Rightarrow a^{2}>4 \Rightarrow a>2$ (as a > 0)

Hence the set of values is $(2,\infty)$

part c. (2 marks)

The x-coordinate of the midpoint is the midpoint of the x-coordinates, so the x-coordinate of the midpoint of AB is:

$\frac{1}{2}(\frac{a-\sqrt{a^{2}-4}}{2}+\frac{a+\sqrt{a^{2}-4}}{2})=\frac{a}{2}$

If $\frac{a}{2}=\sqrt{2}$ , then $a=2\sqrt{2}$
Title: Re: My Worked Solutions
Post by: sahil26 on November 08, 2012, 06:37:17 pm: Thanks for the answers :)

Only lost 2 at max! Thank the Almighty :P
Title: Re: My Worked Solutions
Post by: rebeckab on November 08, 2012, 06:37:51 pm: awesome, I think I got either 21 or 22 right! :DD Just not sure about 18 - I spent ages working out between two options, and then I think I might have circled the one that was true rather than false :/
Title: Re: My Worked Solutions
Post by: BubbleWrapMan on November 08, 2012, 10:42:42 pm: DONE finally
Title: Re: My Worked Solutions
Post by: pi on November 08, 2012, 10:44:23 pm: Wow, great work!
Title: Re: My Worked Solutions
Post by: Special At Specialist on November 08, 2012, 11:08:03 pm: Shit it looks like I got even more wrong than I was expecting...
Can I get consequential marks for question 5a. iii if I got ii wrong?
For ii, I said area = 2 - integral of f(x) instead of just integral of f(x)
Then I used that area + other integral to get the answer which is wrong only because of part ii.
Title: Re: My Worked Solutions
Post by: BubbleWrapMan on November 08, 2012, 11:09:04 pm: Most likely
Title: Re: My Worked Solutions
Post by: barydos on November 08, 2012, 11:47:10 pm: Great solutions, dude!
Title: Re: My Worked Solutions
Post by: abeybaby on November 09, 2012, 01:29:40 am: Man this would take AGES to type up! I was about to say you have great notation, then I realized you got 50 last year and it goes without saying :)
Title: Re: My Worked Solutions
Post by: WhoTookMyUsername on November 09, 2012, 06:51:42 am: Yes! You have no units for q5 :D hopefully they accept units^2

(Great solutionsa s well !)
Title: Re: My Worked Solutions
Post by: BubbleWrapMan on November 09, 2012, 11:32:48 am: Dis ain't physics, if were an assessor I wouldn't care about units
Title: Re: My Worked Solutions
Post by: #1procrastinator on November 09, 2012, 11:47:52 am: Son of a beach...I assumed x=h for the damn rectangular prism