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May 22, 2019, 10:39:07 pm

Author Topic: VCE Methods Question Thread!  (Read 2282763 times)  Share 

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peachxmh

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Re: VCE Methods Question Thread!
« Reply #17925 on: May 10, 2019, 09:35:29 pm »
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I need help with the following question:

Let f : R->R  where f(x) = ex + k, where k is a real constant. If f and f -1 have two points of intersection then:
A. k < -1
B. k < 0
C. k > 1
D. k ≤ 0
E. k ≤ 1

The answer is A. I'm thinking you need to equate f and f -1 and then use the discriminant (set it as > 0) to find k, however I'm having trouble rearranging the resulting equation into a quadratic form. I've tried equating f and f -1 to x respectively (since they should share a point of intersection on the line y=x), as well as the change of base rules for logarithms and exponential with no result.

Would greatly appreciate if you could explain in detail how to get the answer! Thank youuuu :)
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DBA-144

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Re: VCE Methods Question Thread!
« Reply #17926 on: May 10, 2019, 10:42:42 pm »
+1
I need help with the following question:

Let f : R->R  where f(x) = ex + k, where k is a real constant. If f and f -1 have two points of intersection then:
A. k < -1
B. k < 0
C. k > 1
D. k ≤ 0
E. k ≤ 1

The answer is A. I'm thinking you need to equate f and f -1 and then use the discriminant (set it as > 0) to find k, however I'm having trouble rearranging the resulting equation into a quadratic form. I've tried equating f and f -1 to x respectively (since they should share a point of intersection on the line y=x), as well as the change of base rules for logarithms and exponential with no result.

Would greatly appreciate if you could explain in detail how to get the answer! Thank youuuu :)

This is not worth the trouble of finding a mathematical solution. Sketch it on your cas for the k values given and you should be able to get the answer. Intuitively, it cannot include 0, as e^x and in(x) do not have 2 solutions, and the same thing applies for any value of k that is greater than 0- there are not 2 solutions. If you sketch this, you can discern this. Hence, we have either B or A. Now, sketching again we get that there are 2 sols when k<-1. Hence A.
Like I said, there is almost certainly a mathematical solution, but this is not recommended given how long this would take. Another way would just be to sketch y=x and y=e^x + k. From here, it is really easy to see that if I move the exponential up, it won't intersect with the linear graph, but moving it down will cause it to intersect. Then just eliminate B like I did above.
Quadratic unlikely to work as you can't really convert between e^x and x and get a nice quadratic/other expression.

I hope this helps. I know that I did not provide a full detailed solution, but I hope this helps nonetheless.
Edit: y intercept of e^x is at 1,  of y=x is 0 hence must move down by one unit for one sol and by more than one for 2. hence k<-1.
« Last Edit: May 11, 2019, 12:02:06 pm by DBA-144 »

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17927 on: May 11, 2019, 03:52:18 pm »
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I'm thinking you need to equate f and f -1 and then use the discriminant (set it as > 0) to find k, however I'm having trouble rearranging the resulting equation into a quadratic form. I've tried equating f and f -1 to x respectively (since they should share a point of intersection on the line y=x), as well as the change of base rules for logarithms and exponential with no result.

You actually can't write a "quadratic equation" resulting from any of those equations since it's not possible to isolate \(x\).

This is not worth the trouble of finding a mathematical solution.

I actually believe the following solution is quicker and more economical given that it can be applied to most increasing functions.

The equation  \(f(x)=f^{-1}(x)\), where  \(f(x)=e^x+k\),  will have one solution for \(x\) if the graphs of \(f\) and \(f^{-1}\) are tangential to each other, and so we have \[\begin{cases} f(x)=x\\ f'(x)=1\end{cases} \implies k=-1\ \text{ and }\ x=0.\] Since the equation  \(f(x)=f^{-1}(x)\)  has no solution for  \(k=0\),  we have two solutions only for  \(k<-1\).

Edit: y intercept of e^x is at 1,  of y=x is 0 hence must move down by one unit for one sol and by more than one for 2. hence k<-1.

On its own, this actually isn't a sufficient reason for the answer and requires more discussion.

For example, if we instead had  \(f(x)=e^{2x}+k\),  then the correct answer would be  \(k<\dfrac{-2}{e}\)  even though the \(y\)-axis intercept of  \(y=e^{2x}\) is at  \((0,\,1)\).

You were lucky that it so happens to be that  \(e^0=\left.\dfrac{d}{dx}\big[e^x\big]\right|_{x=0}=1\).
« Last Edit: May 11, 2019, 10:08:55 pm by AlphaZero »
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I\'m Not A Robot

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Re: VCE Methods Question Thread!
« Reply #17928 on: May 12, 2019, 12:55:41 pm »
0
Does anyone know how to answer this Q worth 9MKS:
A friend bets you $100 on a game involving two six-sided dice, one red and one green.
You choose the number of times the pair of dice will be rolled. You win if the number of
times a red 6 is rolled is at most 2 and the number of times a green 6 is rolled is at least 2.
a) How many times should the dice be rolled to maximise your chance of winning?
b) With that number of rolls, what are your expected winnings?

Thanks in advance

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17929 on: May 12, 2019, 05:21:57 pm »
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Does anyone know how to answer this Q worth 9MKS:
A friend bets you $100 on a game involving two six-sided dice, one red and one green.
You choose the number of times the pair of dice will be rolled. You win if the number of
times a red 6 is rolled is at most 2 and the number of times a green 6 is rolled is at least 2.
a) How many times should the dice be rolled to maximise your chance of winning?
b) With that number of rolls, what are your expected winnings?

Thanks in advance

This is quite an interesting question.

Part a
First, it's important to realise that the results of each dice are independent of each other, and that the probability of rolling a 6 on any given roll on either dice is constant.

Let \(n\) be the number of times the pair of dice are rolled, where \(n\geq 2\).
Let \(R\) be the number of observed \(6\)'s on the red dice from \(n\) rolls.
Let \(G\) be the number of observed \(6\)'s on the green dice from \(n\) rolls.

Note: \(R\) and \(G\) have the same distribution, so you could just define a single variable \(X\), but I chose to separate them for clarity.

That is, \[R\sim \text{Bi}(n,\ 1/6)\quad\text{and}\quad G\sim\text{Bi}(n,\ 1/6).\] We are told that you win the game so long as \(R\leq 2\) and \(G\geq 2\).  Hence, \[\text{Pr}(\text{win})=\text{Pr}(R\leq 2)\times \text{Pr}(G\geq 2)\] From here, you could using your CAS define a function say \[b(n)=\texttt{binomCdf}(n,\,1/6,\,0,\,2)\cdot \texttt{binomCdf}(n,\,1/6,\,2,\,n)\] and then use trial and error to find the value of \(n\) that gives you the highest probability of winning. Or, you could continue as follows: \begin{align*}\text{Pr}(\text{win})&=\text{Pr}(R\leq 2)\times\Big[1-\text{Pr}(G\leq 1)\Big]\\
&=\left[\sum_{k=0}^2\binom{n}{k}\left(\frac16\right)^k\left(\frac56\right)^{n-k}\right]\left[1-\sum_{k=0}^1\binom{n}{k}\left(\frac16\right)^k\left(\frac56\right)^{n-k}\right]\\
&\qquad \vdots\\
&=\frac{1}{250}\left(\frac{5}{36}\right)^n(n^2+9n+50)\big[5\!\times\!6^n-5^n(n+5)\big] \end{align*}
Using a graph of \(\text{Pr}(\text{win})\) against \(n\), it is quite easy to see that  \(\boxed{n=12\,}\)  rolls will maximise the chance of winning.


Part b
When  \(n=12\), \[\text{Pr}(\text{win})\approx 0.419101\quad\text{(6DP)}\] and so your expected earnings from the game is \[E=\$100 \times 0.419101=\$41.91\quad\text{(nearest cent)}.\] Assuming you also bet \(\$100\) to play the game, you expected winnings is \[W=\$41.91-\$100=\boxed{-\$58.09\,}\quad \text{(nearest cent)}\]
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne


-_-zzz

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Re: VCE Methods Question Thread!
« Reply #17930 on: May 17, 2019, 11:44:24 pm »
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Hey guys,

Just wondering what your opinions are on ExamPro for methods, I'm finding their questions very difficult and wanted to know if they're comparable to VCAA questions in any way in terms of difficulty.

Cheers :)
« Last Edit: May 17, 2019, 11:46:00 pm by -_-zzz »
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Remy33

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Re: VCE Methods Question Thread!
« Reply #17931 on: May 19, 2019, 08:48:00 pm »
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When finding inverse functions, do you always reject the negative? Or are there cases where you use both or reject the positive?
Cheers.
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17932 on: May 19, 2019, 08:57:31 pm »
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When finding inverse functions, do you always reject the negative? Or are there cases where you use both or reject the positive?
Cheers.

If we find we have to take a square root, we do not always take the positive one.

We also cannot take both roots as this wouldn't give us a function.

You need to take the sign that is consistent with its range (or the domain of the original function).

Here's an easy example question where we would take the negative root. Try sketching both roots and see which one produces a function that is symmetrical to the original about the line \(y=x\).

Question:    Let  \(f:(-\infty,\,0]\to\mathbb{R},\ f(x)=x^2\).  Find the rule, domain and range of \(f^{-1}\), the inverse function of \(f\).
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