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March 28, 2024, 11:53:18 pm

Author Topic: 3U Maths Question Thread  (Read 1230239 times)  Share 

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spnmox

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Re: 3U Maths Question Thread
« Reply #4005 on: March 26, 2019, 09:49:58 am »
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Find the area enclosed by x=0 and x=3 between the x-axis and the curve y=3x(x^2 -1)^5.

I got 65 535.75, answers say 65 536.25. Answers wrong or I'm wrong?

fun_jirachi

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Re: 3U Maths Question Thread
« Reply #4006 on: March 26, 2019, 07:35:23 pm »
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Nope, you're right.

If there's a discrepancy, probably check an integral calculator/Desmos/GeoGebra.

Not sure why they added on the 1/4 though ???
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spnmox

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Re: 3U Maths Question Thread
« Reply #4007 on: March 26, 2019, 09:10:55 pm »
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Thanks for verifying that!

not a mystery mark

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Re: 3U Maths Question Thread
« Reply #4008 on: March 28, 2019, 12:16:35 pm »
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Heyy, how would you approach this question?

By restricting the domain of each of the following functions to a monotonic increasing curve, find the inverse function, and state the domain and range of the inverse function.
b) 1/(x^2 -1)
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georgebanis

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Re: 3U Maths Question Thread
« Reply #4009 on: March 28, 2019, 01:06:54 pm »
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Hey guys,

I have attached two induction divisibility questions. I have two answers for them but are not sure if they are right. Could you please show me how to do them.

Thanks

fun_jirachi

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Re: 3U Maths Question Thread
« Reply #4010 on: March 28, 2019, 08:30:29 pm »
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Heyy, how would you approach this question?

By restricting the domain of each of the following functions to a monotonic increasing curve, find the inverse function, and state the domain and range of the inverse function.
b) 1/(x^2 -1)


Here note that the original function is undefined at x=1 or -1, but on the new function it's totally okay (endpoint at -1, 0). Also notice that if you sub in a negative number for x between -1 and 0, the thing inside the square root becomes negative, so we have a discontinuity between -1 and 0. For all purposes though, you can just reflect the graph in y=x and infer the domain, since the range is given and you've found the function.

Hey guys,

I have attached two induction divisibility questions. I have two answers for them but are not sure if they are right. Could you please show me how to do them.

Thanks

Doing the full induction proof is pretty long to type (coming up to half yearlies as well! so i don't have that much time for now), and it doesn't really help your personal understanding of the topic. Do you have any working to show us? If so, it would make my job a lot easier :)

EDIT: If you're stuck, it should involve something about using k's property of being even for both proofs. In the inductive step, the first step should be expanding, then subbing in the assumption, then subbing in k = 2c for some integer c since k is even. It should come out reasonably simply after that. :)
« Last Edit: March 28, 2019, 08:44:59 pm by fun_jirachi »
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georgebanis

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Re: 3U Maths Question Thread
« Reply #4011 on: April 02, 2019, 08:21:51 pm »
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Hey guys, just need a hand with part B of both of these applications of calculus questions.

Thanks

RuiAce

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Re: 3U Maths Question Thread
« Reply #4012 on: April 02, 2019, 08:38:34 pm »
+3
Hey guys, just need a hand with part B of both of these applications of calculus questions.

Thanks
\[ \text{Both follow the usual process so for now I'll just do Q7.}\\ \text{If you have trouble with Q5, please post relevant working.} \]
Note that Q7 has an extra conversion at the end.
\[ \text{Proven in part a) is that }\frac{dx}{d\theta} -\frac{3}{2\sin^2\theta}\\ \textbf{Given in the question}\text{ is that }\frac{dx}{dt} = -650. \]
This is just the velocity of the plane.
\[ \text{Hence following the standard procedure, by the chain rule we have}\\ \begin{align*} \frac{dx}{dt} &= \frac{dx}{d\theta} \, \frac{d\theta}{dt}\\ 650 &= -\frac{3}{2\sin^2\theta} \frac{d\theta}{dt} \end{align*}\]
\[ \text{When }\theta = \frac\pi3\text{, we then have}\\ \begin{align*}-650 &=-2\frac{d\theta}{dt}\\ \frac{d\theta}{dt} &= 325 \end{align*}\\ \text{measured in radians per hour.}\]
\(325\) rad/hr means \(\frac{325}{3600} \) rad/s. Then multiply by \( \frac{180^\circ}{\pi}\) to get something with deg/s.
« Last Edit: April 02, 2019, 08:58:28 pm by RuiAce »

terassy

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Re: 3U Maths Question Thread
« Reply #4013 on: April 03, 2019, 07:05:04 pm »
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I've been stuck of part (iii) for a long time and I'm getting nowhere



For part (i) I was able to show it was equal to 1 - (1-a)n and I was also able to show the second part, but how do I do the third part?

Thank you

RuiAce

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Re: 3U Maths Question Thread
« Reply #4014 on: April 03, 2019, 09:05:21 pm »
+2
I've been stuck of part (iii) for a long time and I'm getting nowhere

(Image removed from quote.)

For part (i) I was able to show it was equal to 1 - (1-a)n and I was also able to show the second part, but how do I do the third part?

Thank you
\begin{align*}\sum_{r=1}^n rP_r &= \sum_{r=1}^n r\binom{n}{r} a^r (1-a)^{n-r}\\ &= \sum_{r=1}^n n\binom{n-1}{r-1}a^r (1-a)^r\tag{result from ii}\\ &=n\sum_{r=1}^n a \binom{n-1}{r-1} a^{r-1} (1-a)^{(n-1)-(r-1)}\\ &= na \sum_{r=1}^n \binom{n-1}{r-1} a^{r-1} (1-a)^{(n-1)-(r-1)} \tag{a is a constant}\\ &= na \sum_{k=0}^{n-1} \binom{n-1}{k} a^k (1-a)^{(n-1)-k} \tag{subbing k=r-1}\\ &= na \left( a + (1-a) \right)^{n-1} \tag{binom theorem, x=a and y=1-a}\\ &= na \times 1^{n-1}\\ &= na\end{align*}

terassy

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Re: 3U Maths Question Thread
« Reply #4015 on: April 07, 2019, 05:19:55 pm »
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thank you so much :)

How do you do these questions? I've never seen a similar one in our text books and they are really weird cause there are so many happening things happening at once.


RuiAce

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Re: 3U Maths Question Thread
« Reply #4016 on: April 07, 2019, 05:31:38 pm »
+3
thank you so much :)

How do you do these questions? I've never seen a similar one in our text books and they are really weird cause there are so many happening things happening at once.

(Image removed from quote.)
Have you done binomial probabilities with the formula \( \binom{n}{k}p^k (1-p)^{n-k} \) yet?

(Note: At first glance though, Q1 may require more work.)
« Last Edit: April 07, 2019, 05:34:04 pm by RuiAce »

terassy

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Re: 3U Maths Question Thread
« Reply #4017 on: April 07, 2019, 05:44:10 pm »
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Have you done binomial probabilities with the formula \( \binom{n}{k}p^k (1-p)^{n-k} \) yet?

(Note: At first glance though, Q1 may require more work.)

Yes we have, but the ones we did at school are for example red=0.25 and not red=0.75, if there are 20 cars, what is the probability that 3 are red. But here it's really complex so I'm confused on what to do

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Re: 3U Maths Question Thread
« Reply #4018 on: April 07, 2019, 06:11:36 pm »
+2
Yes we have, but the ones we did at school are for example red=0.25 and not red=0.75, if there are 20 cars, what is the probability that 3 are red. But here it's really complex so I'm confused on what to do
I'll stick a sketch solution to Q1 here for now, because it looks a bit annoying. It somewhat requires conditional probability, but it's possible to get away with just a tree diagram on top of binomial probabilities here.
\[ \text{Seeing as though the lot contains }10\%\text{ defectives,}\\ \text{we anticipate that the probability any randomly selected product is defective}\\ \text{is just }\boxed{p=0.1}. \]
This first bit can be easy to overanalyse, but just break it down logically. 10% of all of them are defective. So why wouldn't the probability that a randomly selected one be defective, be also equal to 10%?
\[ \text{Taking }n=10\text{ as we choose ten at random,}\\ \text{firstly if none are defective, the batch is accepted without question.}\\ \text{This occurs with probability }\binom{10}{0} (0.9)^{10}. \]
\[ \text{Now, suppose 1 is defective. This occurs with probability }\binom{10}{1} (0.1) (0.9)^9.\\ \text{Conditioned on this, another batch of 10 is taken.}\\ \text{The only way we can have at most 3 in total being defective}\\ \text{is if within the batch of 10, }\textbf{at most 2}\text{ are defective.}\\ \text{This occurs with probability }\left[ \binom{10}{0} (0.9)^{10} + \binom{10}{1} (0.1)(0.9)^9 + \binom{10}{2} (0.1)^2 (0.9)^8 \right]. \]
So that case yields \( \binom{10}{1} (0.1)(0.9)^9 \left[ \binom{10}{0} (0.9)^{10} + \binom{10}{1} (0.1)(0.9)^9 + \binom{10}{2} (0.1)^2(0.9)^8 \right] \)
\[ \text{Similarly, if originally 2 are defective}\\ \text{we have }\binom{10}{2} (0.1)^2(0.9)^8 \left[ \binom{10}{0} (0.9)^{10} + \binom{10}{1} (0.1)(0.9)^9 \right] \]
\[\text{Similarly, if originally 3 are defective}\\ \text{we have }\binom{10}{3} (0.1)^3 (0.9)^7 \binom{10}{0} (0.9)^{10} \]
Arguably a branch-and-conquer method. But it ultimately should add up to give the required answer.

terassy

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Re: 3U Maths Question Thread
« Reply #4019 on: April 08, 2019, 01:30:29 am »
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Thank you!!!

For the second question part (a) do you do 16C1(0.8 )1(0.2)15. If so why is that, cause it doesn't make sense.