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Archived Discussion => Mathematics and Science => 2012 => End-of-year exams => Exam Discussion => Victoria => Mathematical Methods CAS => Topic started by: Phy124 on November 07, 2012, 04:47:22 pm

Title: 2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: Phy124 on November 07, 2012, 04:47:22 pm: Sorry about the delay, I see some solutions are up already. Too bad I had an exam this morning, not to worry :P

Here are some solutions I typed up for the first methods exam, hope you all did well, if not you've always got exam 2 to make up for it! :D

note: I've added a potential marking scheme that VCAA may use. However, this is not necessarily accurate.

1. a) Differentiate the function using the chain rule:

$\frac{dy}{dx} = 4(x^2-5x)^3(2x-5)$ (1)

Possible methods:

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ where $y = u^4, u = x^2 - 5x$

$\frac{dy}{du} = 4u^3, \frac{du}{dx} = 3u^3, hence: \frac{dy}{dx} = 4(x^2-5x)^3(2x-5)$

$y'(x) = y'(g(x))g'(x)$ where $y(g(x)) = g(x)^4, g(x) = x^2 - 5x$ (Same as above but u = g(x))

b) Differentiate using the quotient rule:

$f'(x) = \frac{\sin(x)\frac{d}{dx}(x) - x \frac{d}{dx}(\sin(x))}{(\sin(x))^2} = \frac{\sin(x) - x\cos(x)}{\sin^2(x)}$ (1)

$f'\left(\frac{\pi}{2} \right) = \frac{\sin\left(\frac{\pi}{2} \right) - \left(\frac{\pi}{2} \right)\cos\left(\frac{\pi}{2} \right)}{\sin^2(\left(\frac{\pi}{2} \right))} = \frac{1 - 0}{1} = 1$ (1)

Using the quotient rule on the formulae sheet: $u = x, v = \sin(x)$

2. $\int \frac{1}{(2x-1)^3} \ dx$

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

$= -\frac{1}{4(2x-1)^2}$ (1)

Using $\int (ax + b)^n \ dx = \frac{(ax+b)^{n+1}}{a(n+1)}, n \neq -1$

3. For inverse swap x and y

$x = 2y^3 + 1$ (1)

$y = \frac{1}{2}\left(x-1\right) = h^{-1}(x)$

$h^{-1}(x) = \left(\frac{1}{2}\left(x-1\right)\right)^{\frac{1}{3}}$ (1)

(Note: You will probably be deducted one mark for not stating "for inverse swap x and y" or similar)

4. a) The mean is given by the sum of the product of $x$ and the $\Pr(x)$

$\mu = \sum x \Pr(x)$ (1)

$= 0(0.2) + 1 (0.2) + 2(0.5) + 3(0.1) = 1.5$ (1)

b) The probability that Daniel receives only one call on each of the days is $\Pr(x=1) \times \Pr(x=1) \times \Pr(x=1)$

$= (\Pr(x=1))^3 = (0.2)^3 = \frac{1}{125}$ (1)

c) Note: This question is quite ambiguous and I'm sure will cause some controversy. However, this is my interpretation of the question, due to it being worth 3 marks.

We want to find the combinations such that Daniel receives four calls over two days. We are told, however ambiguously, that Daniel has already received at least 1 phone call on each of the days.

The combinations possible are:
- 1 call on Monday, 3 calls on Wednesday
- 3 calls on Monday, 1 call on Wednesday
- 2 calls on Monday, 2 calls on Wednesday

The probability of the three combinations is the sum of the 3:

$\Pr(x = 1) \times \Pr(x = 3) + \Pr(x = 3) \times \Pr(x = 1) + \Pr(x = 2) \times \Pr(x = 2) = 0.2 \times 0.1 + 0.1 \times 0.2 + 0.5 \times 0.5 = 0.29$ (1)

The probability that he receives one call on each days is:

$\Pr(x \geq 1) \times \Pr(x \geq 1) = 0.8 \times 0.8 = 0.64$ (1)

Therefore the overall probability is:

$\frac{0.29}{0.64} = \frac{29}{64}$ (1)

5. a)
(http://content.screencast.com/users/phy124/folders/Jing/media/760f9290-3730-40be-8e5e-cae83ad9b704/2012-11-07_1525.png)

Correct graph shape (1)
Correct and labelled intercepts (1)
Correct and labeled endpoints (1)

b) i. Transformations as follows:

$(x,y) \rightarrow (x,-y) \rightarrow (x + 5,-y)$

Therefore the point $(3,2)$ will become $((3)+5,-(2)) = (8,-2)$ (1)

ii. As shown before:

$(x,y) \rightarrow (x + 5,-y)$

$\therefore (x',y') = (x+5),-y)$

$x' =x+5$

$x = x' - 5$

$y' = -y$

$y = -y'$

Sub into equation:

$-y' = -|(x' - 5)-3| + 2$

$y' = |x' -8| - 3$

$y = |x -8| - 3$

Alternatively:

$g(x) = -f(x) = |x - 3| - 2$ (1)

$h(x) = g(x-5) = |(x-5) - 3| - 2 =|x-8| - 2$ (1)

6. a) $\cos(x) = a\sin(x)$

$\frac{\cos(x)}{\cos(x)} = a\frac{\sin(x)}{\cos(x)}$

$a\tan(x) = 1$

$\tan(x) = \frac{1}{a}$ (1)

$x = \frac{\pi}{3}$ is a solution which occurs when $\tan(x) = \sqrt{3}$

$\therefore \frac{1}{a}= \sqrt{3}$

$\therefore a = \frac{1}{\sqrt{3}}$ (1)

b) $\tan(x) = \sqrt{3}$

$x = \frac{\pi}{3}, \frac{4\pi}{3}$

$\therefore x = \frac{4\pi}{3}$ is the other x-coordinate for intersection. (1)

7. $2\log_e(x+2) - \log_e(x) = \log_e(2x+1)$ , $x>0$

Using logarithmic laws:

$\log_e((x+2)^2) - \log_e(x) = \log_e(2x+1)$

$\log_e\left(\frac{(x+2)^2}{x}\right) = \log_e(2x+1)$

$\log_e\left(\frac{(x+2)^2}{x}\right) = \log_e(2x+1)$ (1)

Taking the exponential of each side:

$\frac{(x+2)^2}{x} = 2x+1$ (1)

$(x+2)^2 = 2x^2 + x$

$x^2 + 4x + 4 = 2x^2 + x$

$x^2 - 3x - 4 = 0$

$(x+1)(x-4)$

$x=-1,4$

However $x>0$

$x = 4$ (1)

8.a) $\Pr(X<106) = q$

$\Pr(94<X<100) = \Pr(100<X<106)$

$\Pr(X<100) = \frac{1}{2}$ (1)

$\therefore \Pr(100<X<106) = q - \frac{1}{2}$

$\therefore \Pr(94<X<100) = q - \frac{1}{2} = \frac{1}{2}\left(2q - 1\right)$ (1)

b) $\int_{-\infty}^b f(x) \ dx = \int_{0}^b \frac{x+1}{12} \ dx = \frac{1}{12} \int_0^b (x+1) dx = \frac{5}{8}$ (1)

$\frac{1}{12} \int_0^b (x+1) dx = \frac{1}{12}\left[\frac{x^2}{2}+x\right]^b_0 = \frac{1}{12}\left[\left[\frac{(b)^2}{2}+(b)\right] - \left[\frac{(0)^2}{2}+(0)\right]\right] = \frac{1}{12}\left(\frac{b^2}{2} +b\right)$

$\frac{1}{12}\left(\frac{b^2}{2}+b \right) = \frac{5}{8}$ (1)

$\frac{b^2}{24}+\frac{b}{12}-\frac{5}{8}=0$

$b^2 + 2b - 15 = 0$

$(b-3)(b+5)$

$b=-5,3$

However $0 \leq b \leq 4$

$\therefore b = 3$ (1)

9. a) Use product rule given on formula sheet:

$u =x, v = \sin(x)$ (or the other way around)

$f'(x) = \sin(x) + x\cos(x)$ (1)

b) $\frac{d}{dx}(x\sin(x)) = \sin(x) + x\cos(x)$

$\therefore x\cos(x) = \frac{d}{dx}(x\sin(x)) - \sin(x)$

$\therefore \int x\cos(x) \dx = x\sin(x) - \int \sin(x) \ dx$

$\therefore \int x\cos(x) \dx = x\sin(x) + \cos(x) \ dx$ (1)

$\int^{\frac{\pi}{2}}_{\frac{\pi}{6}} = [x\sin(x) +\cos(x)]^{\frac{\pi}{2}}_{\frac{\pi}{6}} = <br /><br />\left[\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right) + c\os\left(\frac{\pi}{2}\right)\right] - \left[\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{6}\right)\right]$

$= <br /><br />\left[\left(\frac{\pi}{2}\right)(1) + 0\right] - \left[\left(\frac{\pi}{6}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\right]$ (1)

$= \frac{\pi}{2}-\frac{\pi}{12}-\frac{\sqrt{3}}{2} = \frac{5\pi}{12}-\frac{\sqrt{3}}{2}$ (1)

10. a) i. $f(x) = e^{-mx} + 3x$

Differentiate and solve to equal zero for stationary point:

$f'(x) = -me^{-mx} + 3 = 0$ (1)

$-me^{-mx} = -3$

$e^{-mx} = \frac{3}{m}$

Taking logarithm of each side:

$\log_e(e^{-mx}) = \log_e\left(\frac{3}{m}\right)$

$-mx = \log_e\left(\frac{3}{m}\right)$

$x = -\frac{1}{m}\log_e\left(\frac{3}{m}\right)$

$x = \frac{1}{m}\log_e\left(\frac{3}{m}\right)^{-1}$

$x = \frac{1}{m}\log_e\left(\frac{m}{3}\right)$ (1)

ii. We want to find the values of $m$ for which the value of $x$ for which $f'(x) = 0$ is greather than zero i.e. $\frac{1}{m}\log_e\left(\frac{m}{3}\right)>0$

We have $\frac{1}{m}>0 \ \text{for all} \ m>0$

And $\log_e\left(\frac{m}{3}\right) \ \text{for all} \ m>3$

Therefore $m>3$ (1)

b. The interpretation of this question is that the line of the tangent to the graph $f(x)$ at the point $x = -6$ will go through the point $(0,0)$

Firstly find the gradient of the tangent:

$f'(-6) = -me^{-m(-6)} + 3 = -me^{6m}+3$

One point we know the tangent goes through is $(0,0)$ , another can be found by substituting $x = -6$ into the original equation $f(x)$

$f(-6) = e^{-m(-6)} + 3(-6) = e^{6m} - 18$ (1)

We can now use the equation:

$y-y_1 = m(x-x_1)$ (where m is the gradient)

$y - (e^{6m} - 18) = (-me^{6m}+3)(x-(-6))$

We want it to go through the point (x,y) = (0,0) so

$0 - (e^{6m} - 18) = (-me^{6m}+3)(0+6)$ (1)

$-e^{6m} + 18 = -6me^{6m}+18$

$(6m-1)e^{6m} = 0$

$e^{6m} =0$ has no solutions, so solve for $6m-1=0$

$6m-1 = 0$

$6m = 1$

$m=\frac{1}{6}$ (1)
Title: Re: Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: D.H on November 07, 2012, 04:57:20 pm: For question 5a) would I lose marks if I didn't include the co-ordinates of the cusp?
Since technically it is not an axis intercept nor an end point..
Title: Re: Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: himesh95 on November 07, 2012, 05:05:42 pm: i think you're working out the wrong antiderivative for question 2, the function is 1/(2x-1)^3 and also question 3 should have a cube root in it. other than that looks fine except Q4c which i still dont know who to believe so ill just wait until december 17th for that
Title: Re: Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: HERculina on November 07, 2012, 05:11:12 pm: Nice! :D
You know for q. 9b) it was worth three marks, where what would the third mark be awarded for?
Title: Re: Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: Phy124 on November 07, 2012, 05:18:37 pm: Quote from: himesh95 on November 07, 2012, 05:05:42 pm
i think you're working out the wrong antiderivative for question 2, the function is 1/(2x-1)^3 and also question 3 should have a cube root in it. other than that looks fine except Q4c which i still dont know who to believe so ill just wait until december 17th for that
LOL, I went to my old notes to get the formula to show and accidentally took the question from below the formula :P - Thanks

Quote from: HERculina on November 07, 2012, 05:11:12 pm
Nice! :D
You know for q. 9b) it was worth three marks, where what would the third mark be awarded for?
Thank you, the search function said I had 39 (1)'s and I couldn't work out where the other mark was supposed to go :P

I edited it. However, it's pretty hard to guess where they will and won't award marks. I'm sure if a few of the other past students had input the scheme could be a lot more accurate.
Title: Re: Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: HERculina on November 07, 2012, 05:26:01 pm: Thanks for fixing it! Ok I may have just lost one mark for that q. Cause I think i got step 2 right, but when I went to step 3, i misred the first pi as an x :o I wrote (1/2)x -pi/12 -sqrt(3)/2 and thought the thing they were looking for was somewthing with an x plus a constant which was my -pi/12 -sqrt(3)/2 ... :'(
Title: Re: 2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: Nick Nack 94 on November 11, 2012, 01:46:38 pm: Quote from: rangaaaaaa on November 07, 2012, 04:47:22 pm

Taking logarithm of each side:

$\log_e(e^{6m}) = \log_e(0)$

$6m = 1$

$m=\frac{1}{6}$ (1)

log_e(0) is undefined, not 1 haha. log_e(1) = 0 :p

The problem is:

-e^6m + 18 = -6me^6m + 18
5e^6m = 0.

If you do this though it works:

-e^6m + 18 = -6me^6m + 18
(-e^6m)(1) = (-e^6m)(6m)
1 = 6m
m = 1/6

Might want to just give your answers that little tweak! The rest is great though, thanks :)
Title: Re: 2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: Phy124 on November 11, 2012, 05:49:30 pm: Quote from: Nick Nack 94 on November 11, 2012, 01:46:38 pm
Quote from: rangaaaaaa on November 07, 2012, 04:47:22 pm
$\log_e(e^{6m}) = \log_e(0)$

$6m = 1$

Might want to just give your answers that little tweak! The rest is great though, thanks :)
LOL oh wow... I worked out what the answer had to be on inspection and must've just forced myself to it :P I assume I read $log_e(e^{6m}) = e^0$

It should have read $(6m-1)e^{6m} = 0$ rather than $5me^{6m}$ (or whatever I had) anyway.
Title: Re: 2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: BubbleWrapMan on November 11, 2012, 05:54:25 pm: I wonder if anyone did that on the exam
Title: Re: 2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: Phy124 on November 11, 2012, 06:24:57 pm: It wouldn't be all that surprising under exam conditions, but let's hope they didn't after doing all that previous working :P
Title: Re: 2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: Jordzs on November 24, 2012, 08:02:40 am: Shouldn't question 1a be to the power of 3 instead of 2?
Title: Re: 2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: BubbleWrapMan on November 24, 2012, 12:13:24 pm: Yeah
Title: Re: 2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions
Post by: Phy124 on November 24, 2012, 04:24:30 pm: Quote from: Jordzs on November 24, 2012, 08:02:40 am
Shouldn't question 1a be to the power of 3 instead of 2?
Yep, typo, thx.