Author Topic: Mathematics Question Thread (Read 1303894 times) Tweet Share

RuiAce · « **Reply #330 on:** August 20, 2016, 08:49:49 am »

Quote from: anotherworld2b on August 20, 2016, 01:48:05 am

Thank you for your help i was able to get the answer for the second one by following your example.

I also wanted to ask for a question like this how should you approach it? Im not sure where i should and how to begin

$\text{The key characteristic of your unaltered exponential is that }\lim_{x\to -\infty}a^x=0\\ \text{This is all we need to find the vertical translation.}$
$\text{So since the first one features }\lim_{x\to -\infty}a^x=2\text{ we have }y=a^{x-b}+2$
$\text{Then all we do is use the grid (which is quite rare for an HSC maths question)}\\ \text{The graph passes through }(0,5)\text{ and }(-1,3)\\ \text{So we sub these points in}\\ \begin{align*}3&=a^{-1+b}+2\\ 5&=a^b+2\end{align*}\\ \text{Solving this gives }a=3, b=1\text{ which you can sub back in}$

anotherworld2b · « **Reply #331 on:** August 20, 2016, 10:15:46 am »

How do you know that lim a^x = 2?
I have never used or hear about lim before

Quote from: RuiAce on August 20, 2016, 08:49:49 am

$\text{The key characteristic of your unaltered exponential is that }\lim_{x\to -\infty}a^x=0\\ \text{This is all we need to find the vertical translation.}$
$\text{So since the first one features }\lim_{x\to -\infty}a^x=2\text{ we have }y=a^{x-b}+2$
$\text{Then all we do is use the grid (which is quite rare for an HSC maths question)}\\ \text{The graph passes through }(0,5)\text{ and }(-1,3)\\ \text{So we sub these points in}\\ \begin{align*}3&=a^{-1+b}+2\\ 5&=a^b+2\end{align*}\\ \text{Solving this gives }a=3, b=1\text{ which you can sub back in}$

RuiAce · « **Reply #332 on:** August 20, 2016, 11:10:05 am »

Quote from: anotherworld2b on August 20, 2016, 10:15:46 am

How do you know that lim a^x = 2?
I have never used or hear about lim before

This does get forgotten by heaps of 2U students to be fair but it's absolutely crucial.

The exponential a^x (doesn't have to specifically be e^x) has a horizontal asymptote at y=0, for very large NEGATIVE x. We say that the limit as x goes to negate infinity is therefore 0.

Because fact is if you raise a number to a really negative number it just goes down to 0

However, if you've never heard of what a limit is, then there's something that you weren't taught that belongs in 2U. I do suspect you're a VCE student but in here I use anything that belongs to the 2U course.

jamonwindeyer · « **Reply #333 on:** August 20, 2016, 12:37:01 pm »

Quote from: RuiAce on August 20, 2016, 11:10:05 am

This does get forgotten by heaps of 2U students to be fair but it's absolutely crucial.

The exponential a^x (doesn't have to specifically be e^x) has a horizontal asymptote at y=0, for very large NEGATIVE x. We say that the limit as x goes to negate infinity is therefore 0.

Because fact is if you raise a number to a really negative number it just goes down to 0

However, if you've never heard of what a limit is, then there's something that you weren't taught that belongs in 2U. I do suspect you're a VCE student but in here I use anything that belongs to the 2U course.

Another world is a WACE student

you can think about limits intuitively without the notation! Like, for the limit of something as x approaches infinity, you mean, What happens as you pop in MASSIVE values of x. It's just a formal notation for saying, when x is HUGE, or x is HUGELY NEGATIVE (-infinity), or (less important for you), when x is REALLY CLOSE to some other value

RuiAce · « **Reply #334 on:** August 20, 2016, 12:49:38 pm »

Quote from: jamonwindeyer on August 20, 2016, 12:37:01 pm

Another world is a WACE student you can think about limits intuitively without the notation! Like, for the limit of something as x approaches infinity, you mean, What happens as you pop in MASSIVE values of x. It's just a formal notation for saying, when x is HUGE, or x is HUGELY NEGATIVE (-infinity), or (less important for you), when x is REALLY CLOSE to some other value

Hmm. Yeah I thought that VCE also do have limits.

But I had a look in the WACE courses. There's an explicit reference to limits in relation to the definition of the derivative in the methods course.

jamonwindeyer · « **Reply #335 on:** August 20, 2016, 03:02:46 pm »

Quote from: RuiAce on August 20, 2016, 12:49:38 pm

Hmm. Yeah I thought that VCE also do have limits.

But I had a look in the WACE courses. There's an explicit reference to limits in relation to the definition of the derivative in the methods course.

Eh, maybe not covered yet, s'all gee in the bee

olivercutbill · « **Reply #336 on:** August 24, 2016, 05:09:24 pm »

Got a foolish question.

Indefinite integral of (3)^(2x-1)

Thanks guys

RuiAce · « **Reply #337 on:** August 24, 2016, 05:26:41 pm »

Quote from: olivercutbill on August 24, 2016, 05:09:24 pm

Got a foolish question.

Indefinite integral of (3)^(2x-1)

Thanks guys

$\text{We use the fact that }e^{\ln x}=x\text{ for all }x>0\\ \begin{align*}\int 3^{2x-1}dx&=\int e^{\ln \left(3^{2x-1}\right)}dx\\ &= \int e^{(2x-1)\ln 3}dx\\ &= \frac{1}{2\ln 3}e^{(2x-1)\ln 3}+C\\ &= \frac{1}{2\ln 3}3^{2x-1}+C\end{align*}$
$\text{Observe that }\ln 3\text{ is constant and our integral was of the form }\int e^{ax+b}dx = \frac{1}{a}e^{ax+b}+C\\ \text{and that we reverse engineer steps at the end.}\\ \text{Recall that }\log a^m = m \log a$

olivercutbill · « **Reply #338 on:** August 24, 2016, 07:32:14 pm »

Quote from: RuiAce on August 24, 2016, 05:26:41 pm

$\text{We use the fact that }e^{\ln x}=x\text{ for all }x>0\\ \begin{align*}\int 3^{2x-1}dx&=\int e^{\ln \left(3^{2x-1}\right)}dx\\ &= \int e^{(2x-1)\ln 3}dx\\ &= \frac{1}{2\ln 3}e^{(2x-1)\ln 3}+C\\ &= \frac{1}{2\ln 3}3^{2x-1}+C\end{align*}$
$\text{Observe that }\ln 3\text{ is constant and our integral was of the form }\int e^{ax+b}dx = \frac{1}{a}e^{ax+b}+C\\ \text{and that we reverse engineer steps at the end.}\\ \text{Recall that }\log a^m = m \log a$

ah - more complicated than I thought but I got it. Awesome, you're a legend Rui!

jamonwindeyer · « **Reply #339 on:** August 24, 2016, 11:23:22 pm »

Quote from: olivercutbill on August 24, 2016, 07:32:14 pm

ah - more complicated than I thought but I got it. Awesome, you're a legend Rui!

Yeah that was definitely not a foolish question (not that there are any foolish questions, but that one is actually pretty tough to do unless you know the trick)

olivercutbill · « **Reply #340 on:** August 27, 2016, 12:00:15 pm »

Quote from: jamonwindeyer on August 24, 2016, 11:23:22 pm

Yeah that was definitely not a foolish question (not that there are any foolish questions, but that one is actually pretty tough to do unless you know the trick)

Yeah 100% - that is one of those things to look out for when you are stuck I guess. Goes into the list of:
-equations reducible to quadratics
-factoring
-simultaneous equations
-the limiting sum
-graphing the problem
etc etc

That being said, I have another question for probability:

a) Given a standard deck of cards, a player is dealt five cards. What is the probability that 4 aces are drawn?
b) Dealt five cards, what is the probability of all cards being of the same suit?

I've been trying a four of a kind +1 kind of approach but it hasn't worked.

Cheers guys

RuiAce · « **Reply #341 on:** August 27, 2016, 12:36:17 pm »

Quote from: olivercutbill on August 27, 2016, 12:00:15 pm

Yeah 100% - that is one of those things to look out for when you are stuck I guess. Goes into the list of:
-equations reducible to quadratics
-factoring
-simultaneous equations
-the limiting sum
-graphing the problem
etc etc

That being said, I have another question for probability:

a) Given a standard deck of cards, a player is dealt five cards. What is the probability that 4 aces are drawn?
b) Dealt five cards, what is the probability of all cards being of the same suit?

I've been trying a four of a kind +1 kind of approach but it hasn't worked.

Cheers guys

$\text{Assuming no replacement.}$
$\text{If all four aces must be drawn, then in a deck of 52 we have }\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}\times\frac{1}{49}\\ \text{And then the last card can be any of the remaining. }\left(\frac{48}{48}\right)$
$\text{Within one suit, if all five cards are of the same suit then we have }\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{49}\times\frac{9}{48}\\ \text{But we have to multiply this by 4 because there are four suits in a deck of cards.}$

olivercutbill · « **Reply #342 on:** August 27, 2016, 02:48:53 pm »

for a) the answer the textbook gives is 1/54145. I had the same logic as you and I'm not sure how to do it. On my calculator, it gives a much smaller answer.

Weird

RuiAce · « **Reply #343 on:** August 27, 2016, 02:54:11 pm »

$\text{The card that isn't an ace can be drawn before, after, or in between }\\ \text{successive draws of the aces. There are thus 5}\\ \text{positions to draw the aces.}$
$\text{So the correct answer multiplies my incorrect expression by }5.$
The point: When ordering isn't specified, the number of favourable outcomes is changed. (I had a feeling my answer was wrong but I couldn't see where so blindly.)

olivercutbill · « **Reply #344 on:** August 27, 2016, 05:19:26 pm »

Quote from: RuiAce on August 27, 2016, 02:54:11 pm

$\text{The card that isn't an ace can be drawn before, after, or in between }\\ \text{successive draws of the aces. There are thus 5}\\ \text{positions to draw the aces.}$
$\text{So the correct answer multiplies my incorrect expression by }5.$
The point: When ordering isn't specified, the number of favourable outcomes is changed. (I had a feeling my answer was wrong but I couldn't see where so blindly.)

Ok! I thought so but didn't full grasp the logic.

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