Author Topic: Mathematics Question Thread (Read 1626449 times) Tweet Share

itssona · « **Reply #1695 on:** April 16, 2017, 06:22:24 pm »

Quote from: RuiAce on April 16, 2017, 05:51:48 pm

$\text{One possible way of solving }|x-2| < x\text{ algebraically is by splitting the cases}\\ \text{Note that}\\ |x-2|=\begin{cases}x-2 & x\ge 2\\ -(x-2) & x< 2\end{cases}$
$\text{For }x\ge 2\text{ we have}\\ \begin{align*}x-2 &< x\\ -2 &< 0\end{align*}\\ \text{Trivially true for all }x\ge 2\text{, so }x\ge 2\text{ belongs to the solution}$
$\text{For }x< 2\text{ we have}\\ \begin{align*}2-x &< x\\ x&> 1\end{align*}\\ \text{Hence }1 < x < 2\text{ belongs to the solution}$
$\text{Combining the cases, the solution is } x > 1$

thank you so much!!!!

I love this algebraic way you solved it :O I never knew this, thank you!

JuliaPascale123 · « **Reply #1696 on:** April 17, 2017, 07:45:01 pm »

Whoa!!

RuiAce · « **Reply #1697 on:** April 17, 2017, 07:50:53 pm »

Quote from: JuliaPascale123 on April 17, 2017, 07:45:01 pm

Whoa!!

(Image removed from quote.)

This problem is not suitable for 2U purposes. It requires the use of a 4U technique known as implicit differentiation
$\text{Remark: If the result of part a) can be assumed then part b) is straightforward}\\ \text{When }(x,y)=(-1,1), \, \frac{dy}{dx}=\frac{-2-1}{-1-2}=1$
$\text{This is just a simple substitution into the point-gradient form }y-y_1=m(x-x_1)$

itssona · « **Reply #1698 on:** April 18, 2017, 10:51:07 am »

hey would you be able to see if im doing this problem right- im not sure entirely
5log(a) - log_a(a)^4

so I would say 5loga-4log_a(a)

but now im not sure,

like do i do 5/4 log_aa

thank you

RuiAce · « **Reply #1699 on:** April 18, 2017, 10:57:37 am »

Quote from: sssona09 on April 18, 2017, 10:51:07 am

hey would you be able to see if im doing this problem right- im not sure entirely
5log(a) - log_a(a)^4

so I would say 5loga-4log_a(a)

but now im not sure,

like do i do 5/4 log_aa

thank you

$\text{Hint: }\log_a a = 1\text{ so that second term is just equal to 4.}$

itssona · « **Reply #1700 on:** April 18, 2017, 11:09:41 am »

Quote from: RuiAce on April 18, 2017, 10:57:37 am

$\text{Hint: }\log_a a = 1\text{ so that second term is just equal to 4.}$

oh!! that passed me! so the answer is simply 5loga-4?

cant simplify it further yeah?

itssona · « **Reply #1701 on:** April 18, 2017, 11:11:48 am »

Quote from: RuiAce on April 18, 2017, 10:57:37 am

$\text{Hint: }\log_a a = 1\text{ so that second term is just equal to 4.}$

oh wait, so 5-4

because that means both terms are 1
so the answer is 1?

RuiAce · « **Reply #1702 on:** April 18, 2017, 11:38:37 am »

Quote from: sssona09 on April 18, 2017, 11:11:48 am

oh wait, so 5-4

because that means both terms are 1
so the answer is 1?

$\text{If your first term was meant to be }5\log_a a\text{ then it's 1}\\ \text{If it's just }5\log a \text{ then that's }5\log_e a\text{ so your first answer is right.}$

itssona · « **Reply #1703 on:** April 18, 2017, 11:50:27 am »

Quote from: RuiAce on April 18, 2017, 11:38:37 am

$\text{If your first term was meant to be }5\log_a a\text{ then it's 1}\\ \text{If it's just }5\log a \text{ then that's }5\log_e a\text{ so your first answer is right.}$

ah that makes sense, thank you a tonne

itssona · « **Reply #1704 on:** April 18, 2017, 04:13:29 pm »

hey how would i find the value of |x-4| when x<4

thanksss!

why am i finding that 3u is easier than 2u for prelim HYs lol

RuiAce · « **Reply #1705 on:** April 18, 2017, 04:32:13 pm »

Quote from: sssona09 on April 18, 2017, 04:13:29 pm

hey how would i find the value of |x-4| when x<4

thanksss!

why am i finding that 3u is easier than 2u for prelim HYs lol

$\text{This will just be }4-x$
$\text{By replacing }x\text{ by }x-4\text{ in the definition:}\\ |x|=\begin{cases}x& x\ge 0\\ -x & x<0\end{cases}$
$|x-4|=\begin{cases}x-4 & x-4\ge 0\\ -(x-4) & x < 0\end{cases}\\ \text{so we are just considering the second case}$

jamonwindeyer · « **Reply #1706 on:** April 18, 2017, 05:03:03 pm »

Quote from: sssona09 on April 18, 2017, 04:13:29 pm

hey how would i find the value of |x-4| when x<4

thanksss!

why am i finding that 3u is easier than 2u for prelim HYs lol

I found the same thing for most of Prelim, it's because 3U doesn't throw that much at you in Year 11 - Doesn't really get gnarly until Year 12 (Extension Trig was the only prelim topic that gave me strife)

Sukakadonkadonk · « **Reply #1707 on:** April 18, 2017, 06:45:11 pm »

Hi, could someone please help me with proving this?

Thanks

itssona · « **Reply #1708 on:** April 18, 2017, 07:09:04 pm »

Quote from: RuiAce on April 18, 2017, 04:32:13 pm

$\text{This will just be }4-x$
$\text{By replacing }x\text{ by }x-4\text{ in the definition:}\\ |x|=\begin{cases}x& x\ge 0\\ -x & x<0\end{cases}$
$|x-4|=\begin{cases}x-4 & x-4\ge 0\\ -(x-4) & x < 0\end{cases}\\ \text{so we are just considering the second case}$

oh! That makes sense, thank you so much

itssona · « **Reply #1709 on:** April 18, 2017, 07:12:59 pm »

Quote from: jamonwindeyer on April 18, 2017, 05:03:03 pm

I found the same thing for most of Prelim, it's because 3U doesn't throw that much at you in Year 11 - Doesn't really get gnarly until Year 12 (Extension Trig was the only prelim topic that gave me strife)

argh I wish they spaced it out better so they teach harder 3U concepts in yr11 too! Well I was looking at HSC trig questions and gave up right away XD

now I wonder- do they test you prelim topics in HSC?? Like in 2U HSC, is there absolute value and that kinda thing?

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