July 22, 2019, 11:10:32 am

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#### peachxmh

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##### Re: VCE Methods Question Thread!
« Reply #17925 on: May 10, 2019, 09:35:29 pm »
0
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 »
0
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 • MOTM: DEC 18 • Forum Obsessive • Posts: 291 • $\int_a^b f(x)\,\text{d}x=F(b)-F(a)$ • Respect: +121 ##### Re: VCE Methods Question Thread! « Reply #17929 on: May 12, 2019, 05:21:57 pm » +2 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?

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]\\
&=\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 $W=\41.91-\100=\boxed{-\58.09\,}\quad \text{(nearest cent)}$
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#### -_-zzz

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##### Re: VCE Methods Question Thread!
« Reply #17930 on: May 17, 2019, 11:44:24 pm »
0
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 »
+2
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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#### f0od

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##### Re: VCE Methods Question Thread!
« Reply #17933 on: May 24, 2019, 10:36:17 pm »
0
For the question (1aiii) (and worked out answer) attached, I'm confused on how they converted the sin(x) to a cos(x)?

Additionally, could someone please explain the identity cos2(x) - sin2(x) = cos(2x)?

Thanks!
class of 2019

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17934 on: May 24, 2019, 11:31:43 pm »
+3
For the question (1aiii) (and worked out answer) attached, I'm confused on how they converted the sin(x) to a cos(x)?

Additionally, could someone please explain the identity cos2(x) - sin2(x) = cos(2x)?

Thanks!

That has got to be one of the stupidest solutions to a problem I've ever seen.

Clearly, the rule of $d$ has the form  $d(t)=a+b\cos(nt)$.  The amplitude gives the magnitude of $b$,  the period gives $n$,  and $a$ is the average of the minimum and maximum distances.

Anyway, we can still learn something from this. They use complementary angle identities. The main complementary angle identities you should know are $\boxed{\sin\left(\frac\pi 2-\theta\right)=\cos(\theta)\\ \cos\left(\frac\pi 2 -\theta\right)=\sin(\theta)}$ Then, you can just apply periodicity.

For a more intuitive outlook, sketch the graph of  $y=\sin(x)$  and translate it $\dfrac{3\pi}{2}$ units to the left. You'll find that it becomes exactly the graph of  $y=\cos(x)$.

The identity  $\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$  is known as the cosine double angle formula, and is not a part of the Methods course.

I won't go into the details of how it can be used nor present a derivation here (since this is a Methods questions thread), but feel free to chuck a question in the Specialist Questions Thread if you're curious
« Last Edit: May 24, 2019, 11:33:27 pm by AlphaZero »
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#### DBA-144

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##### Re: VCE Methods Question Thread!
« Reply #17935 on: May 26, 2019, 09:05:18 am »
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1. How formal do we need to be with our working out for methods? Not just prove and show that questions btw.
2. Do we get penalised 1 mark per paper or one mark per error for simplification?
3. How much working is required in exam 2, for non-show that/prove questions? For a 3 mark question, can't I just write 3 things, and so long as they fulfill the marking guide, I should get full marks right? I have heard that for 3 marks we should be showing plenty of working, but I am not sure if this is required given that essentially, 3 marks is just 3 things that we need to show/do.

Thanks.

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17936 on: May 26, 2019, 02:39:24 pm »
+2
1. How formal do we need to be with our working out for methods? Not just prove and show that questions btw.

You don't need to be able to present solutions is if they were for an academic paper, but you are expected to be able to communicate clearly what you are doing so that an examiner can follow your train of thought.

2. Do we get penalised 1 mark per paper or one mark per error for simplification?

Algebraic simplification and calculations are usually assigned to the final answer mark of a question. If you do not obtain the correct answer in the required form, you cannot be awarded the mark.

Please note, you are never penalised for mistakes. Depending on the error, you will just not obtain the mark that the working/answer is assigned to.

3. How much working is required in exam 2, for non-show that/prove questions? For a 3 mark question, can't I just write 3 things, and so long as they fulfill the marking guide, I should get full marks right? I have heard that for 3 marks we should be showing plenty of working, but I am not sure if this is required given that essentially, 3 marks is just 3 things that we need to show/do.

You need to show enough working to:
2.  show the examiner that you understand what the question is testing.

How much work you need to show depends on the question, but here's a good general rule:  Suppose you are showing your friend, who does not understand how to answer a question, your solution to it. Without you talking them through it, would they be able to understand the method just from what you wrote? If not, you probably haven't shown enough.
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#### Ansaki

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##### Re: VCE Methods Question Thread!
« Reply #17937 on: June 02, 2019, 02:22:56 pm »
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hey guys! i got a question that is in conjunction with related rates of change, i dont quite understand how to tackle the question and also work through it.
Question:
"A spherical balloon is being inflated so that its radius is increasing at a rate of 5 cm/min. Find the rate at which the surface area is increasing when the radius is 10cm given the equation for the surface area is given by SA=4*pi*r^2.

thanks

#### undefined

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##### Re: VCE Methods Question Thread!
« Reply #17938 on: June 02, 2019, 02:37:49 pm »
+1
hey guys! i got a question that is in conjunction with related rates of change, i dont quite understand how to tackle the question and also work through it.
Question:
"A spherical balloon is being inflated so that its radius is increasing at a rate of 5 cm/min. Find the rate at which the surface area is increasing when the radius is 10cm given the equation for the surface area is given by SA=4*pi*r^2.

thanks
This is a related rates question which is not on the study design anymore but

We know that dr/dt = 5 cm/min
and dSA/dr=8 pi r
and we are trying to find dSA/dt
So dSA/dt = dSA/dr * dr/dt (chain rule)
= 8 pi r * 5
so when the radius = 10cm
dSA/dt = 8 pi 10 * 5
= 400pi cm^2 / min
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#### Ansaki

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##### Re: VCE Methods Question Thread!
« Reply #17939 on: June 02, 2019, 03:48:31 pm »
0
This is a related rates question which is not on the study design anymore but

We know that dr/dt = 5 cm/min
and dSA/dr=8 pi r
and we are trying to find dSA/dt
So dSA/dt = dSA/dr * dr/dt (chain rule)
= 8 pi r * 5
so when the radius = 10cm
dSA/dt = 8 pi 10 * 5
= 400pi cm^2 / min

thanks so much! your help is much appreciated!!!