Author Topic: Mathematics Question Thread (Read 1625958 times) Tweet Share

amelia20181 · « **Reply #3540 on:** July 21, 2018, 11:16:20 pm »

how do you do this

amelia20181 · « **Reply #3541 on:** July 21, 2018, 11:22:03 pm »

also how do you differentiate this

envisagator · « **Reply #3542 on:** July 21, 2018, 11:48:55 pm »

Quote from: amelia20181 on July 21, 2018, 11:16:20 pm

how do you do this

$y=\frac { lnx }{ x } \\ f^{ ' }\left( x \right) =\quad \frac { 1-lnx }{ { x }^{ 2 } } \quad (using\quad quotient\quad rule)\\ solve\quad f^{ ' }\left( x \right) =0\quad to\quad determine\quad stationary\quad points\\ i.e.\quad 1-lnx=0\rightarrow lnx=1\\ using\quad log\quad laws\quad x=e\\ To\quad determine\quad the\quad type\quad of\quad stationary\quad pt.\quad sub\quad values\quad greater\quad and\quad smaller\quad than\quad e\quad into\quad the\quad first\quad derivative\quad (note\quad that\quad e=2.71828.....),\\ subbing\quad in\quad smaller\quad values\quad gives\quad a\quad positive\quad gradient,\quad while\quad greater\quad values\quad gives\quad a\quad negative\quad gradient,\quad this\quad implies\quad at\quad x=e\quad there\quad is\quad a\quad maximum\quad turning\quad point.\\ sub\quad x=e\quad into\quad the\quad original\quad function\quad to\quad find\quad the\quad y-coordinate\quad of\quad the\quad maximum\quad turning\quad point.\quad $

skisso · « **Reply #3543 on:** July 22, 2018, 04:38:48 pm »

Quote from: jamonwindeyer on July 21, 2018, 04:45:06 pm

Hey hey! So they conclude that $q$ must be negative, because the point Q is to the left of the origin. So, it's $x$ coordinate must be negative.

So:

$2q<0\\q<0$

Since Q is left of the origin, then $q$ must be a negative number

Ohh i see, thank you

btw your lecture was superrr helpful so thank you!

jamonwindeyer · « **Reply #3544 on:** July 22, 2018, 04:42:49 pm »

Quote from: amelia20181 on July 21, 2018, 11:22:03 pm

also how do you differentiate this

Hi! Chain rule says that you differentiate the inside of the function:

$\frac{d}{dx}\left(e^x+x\right)=e^x+1$

Then this goes out the front of the sine function, which you then turn into a cosine function because of differentiation:

$\frac{d}{dx}\sin{\left(e^x+x\right)}=\left(e^x+1\right)\cos{\left(e^x+x\right)}$

I hope this helps!! Also, try not to post a heap of questions as separate posts - Put them all in one post! You can actually go back and add to your last post by clicking 'Modify' just above the post you made

jamonwindeyer · « **Reply #3545 on:** July 22, 2018, 04:44:49 pm »

Quote from: skisso on July 22, 2018, 04:38:48 pm

Ohh i see, thank you
btw your lecture was superrr helpful so thank you!

You are so welcome!! Really glad it helped

infectmarshroom · « **Reply #3546 on:** July 22, 2018, 10:26:34 pm »

Hey,
Came across this exciting (not!

) log and exp question and was stumped for part 3 and hence was not able to complete part 4. I know it probably has something to do with the features of y=x like gradients etc but can't get any conclusive answer with my maths of a coordinate purely in terms of 'a' without x's. Help is greatly appreciated!

Q16. part b
For some a>0 two curves f(x)=a^x and g(x)=loga(x) are drawn on the same axes so that they touch on y=x.
i) Write down expressions for f'(x) and g'(x) (2mks)
ii) Write down an equation involving natural logarithms whose solution is the x value at their point of contact. (1mk)
iii) Find the coordinate of the point of contact (2mks)
iv) What is the value of a? (1mk)

Btw Jamon your 2u and 3U lectures were amazing, love what you guys do and what you stand for.

RuiAce · « **Reply #3547 on:** July 23, 2018, 08:28:31 am »

Quote from: infectmarshroom on July 22, 2018, 10:26:34 pm

Hey,
Came across this exciting (not! ) log and exp question and was stumped for part 3 and hence was not able to complete part 4. I know it probably has something to do with the features of y=x like gradients etc but can't get any conclusive answer with my maths of a coordinate purely in terms of 'a' without x's. Help is greatly appreciated!

Q16. part b
For some a>0 two curves f(x)=a^x and g(x)=loga(x) are drawn on the same axes so that they touch on y=x.
i) Write down expressions for f'(x) and g'(x) (2mks)
ii) Write down an equation involving natural logarithms whose solution is the x value at their point of contact. (1mk)
iii) Find the coordinate of the point of contact (2mks)
iv) What is the value of a? (1mk)

Btw Jamon your 2u and 3U lectures were amazing, love what you guys do and what you stand for.

$\text{This question has }\textbf{lots}\text{ of subtleties.}$
$\text{Firstly, the curves touch along the line }y=x.\\ \text{Hence, at the point of intersection, }f(x)=g(x)=x.\\ \text{i.e. }\boxed{a^x = \frac{\ln x}{\ln a} = x}$
$\text{Then, the curves }\textbf{touch}.\\ \text{The significance of the word 'touch' is that they must be}\\ \textbf{tangential }\text{to one another at the intersection point.}\\ \text{Hence, at this point, }f^\prime(x) = g^\prime(x)\\ \text{so }\boxed{a^x \ln a = \frac{1}{x\ln a}}$
______________________________________________________
$\text{This gives us heaps of simultaneous equations.}\\ \text{Firstly, considering }a^x = \frac{\ln x}{\ln a}\text{ and }a^x \ln a = \frac{1}{x\ln a}\text{ gives}\\ \boxed{\ln x = \frac{1}{x\ln a}}\implies \boxed{\frac{\ln x}{\ln a} = \frac{1}{x(\ln a)^2}}$
$\text{Then combining that with }\frac{\ln x}{\ln a} = x\text{ gives}\\ \boxed{x = \frac{1}{x (\ln a)^2}}\implies \boxed{x = \pm \frac{1}{\ln a}}$
______________________________________________________
$\text{Now, firstly note that if }a > 1\text{, then the curves }a^x\text{ and }\log_a x\text{ do not intersect at all.}\\ \text{Therefore, }0 < a < 1.$
Note that $a\neq 1$ for the sake of logarithms.
$\text{Since }a < 1\text{, we note that }\ln a < 0.\\ \text{However, clearly }x > 0\text{ due to the domain of }g(x)=\log_a x.\\ \text{Thus, the required }x\text{-coordinate is }\boxed{x = -\frac{1}{\ln a}}$
or equivalently, $x = \left(\ln\left( a^{-1}\right)\right)^{-1} $.
$\text{As for the }y\text{-coordinate, a subtle change-of-base law gives}\\ \begin{align*}y&=a^x\\ \therefore y&= a^{-\frac{1}{\ln a}} \\ &= a^{-\frac{\log_a e}{\log_a a}}\\ &=a^{-\log_a e}\\ &= a^{\log_a e^{-1}}\\ &= e^{-1} \end{align*}\\ \text{and thus the point of intersection is}\\ \boxed{\left( -\frac{1}{\ln a}, \frac1e \right)}$
______________________________________________________
$\text{Finally, backtracking to }\boxed{\frac{\ln x}{\ln a} = x}\text{, upon subbing in our }x\text{-coordinate,}\\ \begin{align*} \frac{\ln \left( -\frac1{\ln a}\right)}{\ln a}&= -\frac{1}{\ln a} \\ \ln \left( -\frac1{\ln a} \right) &= -1\\ -\frac1{\ln a} &= \frac1e\\ \ln a &= -e\\ a &= e^{-e}\end{align*}$

jamonwindeyer · « **Reply #3548 on:** July 23, 2018, 10:39:26 am »

Quote from: infectmarshroom on July 22, 2018, 10:26:34 pm

Btw Jamon your 2u and 3U lectures were amazing, love what you guys do and what you stand for.

Hey! Welcome to the forums! Super glad you found the lectures helpful!

Was coming back to sit down and work through this this morning, but now I've seen Rui's working, maybe I'm glad he beat me to it

will just note that this question (while it is a fantastic question) is really beyond what would be asked in 2U Exams. Especially don't like how much emphasis is placed on the word 'touch,' if they were to ask anything like this they would want to be more specific about exactly what information you need to take from that.

Super cool challenge though - Be sure to post any others you need help with

priyanka.sharma13 · « **Reply #3549 on:** July 23, 2018, 12:48:26 pm »

Hi! I need help in this superannuation question:

“Abby has set up her superannuation fund and after 10 years she has accumulated $134,000. However due to an accident she is no longer able to work and make further contributions to the fund. Abby is leaving the money in the superannuation fund to accumulate interest at 8% p.a. compounded annually. However she needs to withdraw $24,000 at the end of each year for normal living expenses. Write an expression for the amount in the fund after 3 years.”

Thanks!

RuiAce · « **Reply #3550 on:** July 23, 2018, 12:58:56 pm »

Quote from: priyanka.sharma13 on July 23, 2018, 12:48:26 pm

Hi! I need help in this superannuation question:

“Abby has set up her superannuation fund and after 10 years she has accumulated $134,000. However due to an accident she is no longer able to work and make further contributions to the fund. Abby is leaving the money in the superannuation fund to accumulate interest at 8% p.a. compounded annually. However she needs to withdraw $24,000 at the end of each year for normal living expenses. Write an expression for the amount in the fund after 3 years.”

Thanks!

$\text{Let }A_n\text{ be the amount in the fund after }n\text{ years}\\ \text{counting from the point of accumulation.}$
We don't care about the other pieces given for the sake of the question.
$\text{Now follow the usual process.}\\ \begin{align*} A_0 &= 134000\\ A_1 &= A_0(1.08)-24000\\ &= 134000(1.08) - 24000\\ A_2 &= A_1(1.08) - 24000\\ &= 134000(1.08)^2 - 24000(1.08) - 24000\\ A_3 &= A_2(1.08) - 24000\\ &= 134000(1.08)^3 - 24000(1.08)^2 - 24000(1.08) - 24000 \end{align*}$

amelia20181 · « **Reply #3551 on:** July 23, 2018, 03:10:25 pm »

how do you do this

13. The number of cats to dogs at a pet hotel is in the ratio of 4 to 7. If 3 pets are chosen at random, find the probability that
(a) they are all dogs (b) just one is a dog
(c) at least one is a cat.

15. At City Heights School it was found that 75% of students in year 12 study 13 units, 21% study 12 units and 4% study 11 units. If 2 students are selected at random from year 12, find the probability that
(a) 1 student will study 12 units (b) at least 1 student will study 13 units.

Mate2425 · « **Reply #3552 on:** July 23, 2018, 03:11:50 pm »

Hi when the questions aks to e.g. use osine rule to find..., what is the directive meaning behind deduce?
Thanks.

RuiAce · « **Reply #3553 on:** July 23, 2018, 03:19:18 pm »

Quote from: Mate2425 on July 23, 2018, 03:11:50 pm

Hi when the questions aks to e.g. use osine rule to find..., what is the directive meaning behind deduce?
Thanks.

Deduce is just another way of saying "hence, prove (with the aid of ...)". The cosine rule has nothing to do with it directly; if the cosine rule has been mentioned, it means you need to use it TO prove something.

(In practice, deduce means to draw a conclusion out of something. In the world of maths, conclusions are of course, drawn from proofs.)

fun_jirachi · « **Reply #3554 on:** July 23, 2018, 04:25:00 pm »

dunno if this is right but here goes
13. The number of cats to dogs at a pet hotel is in the ratio of 4 to 7. If 3 pets are chosen at random, find the probability that
(a) they are all dogs (b) just one is a dog
(c) at least one is a cat.

i think you just draw a tree diagram with cats and dogs, and since it doesnt matter what order you pick them in, its kinda like a three step tree diagram with cats and dogs at every step i think?
so for a), the only option is dog dog dog, so that's (7/11)^3
b) one dog means two cats and there are three with that option so the answer i think is 3*((4/11)^2)*(7/11)
c) at least one is a cat means no all dogs, so that equals 1 minus all dogs = 1 - (7/11)^3
can someone pls confirm

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