help-4

[ivory valley]

Now you check when x+4>0 and x-4>0

That means x>-4 and x>4

[orchid thistle]

hold on [\f(x+4}{x-4} ≥ 0 ]

ow it dosnt work

[orchid thistle]

yez

[orchid thistle]

does ge mean greater equal to?

[ivory valley]

It's my custom preamble so it won't work on you

[orchid thistle]

ow 😢

[ivory valley]

You have good examples how to draw x>-4 and x>4

[orchid thistle]

ts the thing i dont

[ivory valley]

You first draw a line

Take the less value, here -4 and start to draw the line with integers starting from -4

I will do the example with integers first

[orchid thistle]

ty

[ivory valley]

x>-4 would have {-3,-2,-1, ... , 4, 5, 6 ...}

x>4 would start from 4 but excluding it so {5, 6, ...}

Now you can see there are numbers that one list doesnt contain

When we look for the intersection we look for all numbers that both lists contain

The list x>-4 contains also the numbers in x>4, does that make sense?

[orchid thistle]

yes somewhat

x>-4 is like -4→+inf?

[ivory valley]

Yes but without -4 because there is no equal sign

This is where math ppl write (-4,+oo) instead of [-4,+oo)

[orchid thistle]

oo

[ivory valley]

infinity

[north walrus]

∞

[orchid thistle]

i meant to mean say smth like "ooh" xd

[orchid thistle]

So we put a close asymptote at -4? with like an arrow reaching to inf?

[ivory valley]

Drawing is much better because you would only have to figure where both "arrows" overlap

[orchid thistle]

ic

[ivory valley]

The overlap happens when x>4

[orchid thistle]

So thats the answer i needed or is there something that i need as well?

[ivory valley]

Numbers in green are also im blue contained

[ivory valley]

Well there are two more cases

The second is when the expression is 0

[orchid thistle]

very well let us proceed

[ivory valley]

So you check the numerator when that is 0

The third case is when both the numerator and denominator are negative

Because dividing a negative number by a negative number returns a positive number

2. x+4=0
3. x+4<0 and x-4<0

Try the 3. case by drawing

[orchid thistle]

ill bookmark ts i have class ty for help

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[solemn vortex]

Guys I can’t understand the integrers

[sharp whale]

bro is saying it with the hard r

[solemn vortex]

lol yes

[wraith heart]

Do you have homework problems involving integers

[solemn vortex]

I don’t

My bad I waste tour time

[sharp whale]

.close ok

[vale dock]

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[river shale]

wait

.reopen

[vale dock]

✅

[river shale]

What do you mean by this?

[solemn vortex]

Like what is it used for

[stiff flower]

counting usually

[sharp whale]

the integers are 0, 1, -1, 2, -2, 3, -3, and so on

theres a lot of uses for positive and negative whole numbers

[solemn vortex]

Ok

So it’s to count in specific order

[river shale]

Does it solve your doubts?

[sharp whale]

it doesnt need to be in a specific order, it counts

[solemn vortex]

Ok from 0,1,-1…

Ah now I understand

[lavish mortar]

think it also refers to a whole number

[sharp whale]

thats just a list of things, it doesnt mean a specific order

I couldve listed them as ... -3, -2, -1, 0, 1, 2, 3, ...

[solemn vortex]

Ok it will dépend

Yea thanks

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[orchid thistle]

Hello, Rational Inequalities.
(2x/(x-4))≥1

[orchid thistle]

further discussion from my previous one

[magic lynx]

two options to do this

• simplify the right hand side as a mixed fraction
• multiply both sides by x-4. this requires casework on whether x-4 is positive or negative

[orchid thistle]

Case 3 should be
{-4,-oo} and {4, -oo}?

[orchid thistle]

oh im currently in the graphing part but i did get
(x+4)/(x-4)≥0

[magic lynx]

ok so assuming this is correct, this inequality holds when x+4 and x-4 are both positive or negative (or x+4=0)

[orchid thistle]

what does casework men

[stiff lily]

considering what happens under certain conditions/assumptions

[stark wedge]

option 2 is not recommended the instant your denominator becomes something remotely complicated.

and if we're being honest, also not recommended even here.

[orchid thistle]

oh

[stiff lily]

if you want the multiplication route, multiply by (x-4)^2,
and instead of annoying algebraic casework, you can use wavy method

[magic lynx]

btw i meant simplify left hand side, not right hand side

[orchid thistle]

i dont understand what is the wavy method

[magic lynx]

uh you need to deal with possible multiplication by 0 then

oh wait x-4 cannot be 0 or this is undefined

[stiff lily]

wavy method is identifying roots of your polynomial (in this case will be quadratic) and doing a rough sketch which pretty tells you when it's positive/negative

[orchid thistle]

oh thats what it meant

but how do I apply it here

[stiff lily]

multiply both sides by (x-4)^2
rearrange until you get 0 on one side

[toxic ridge]

number line

not graph

[toxic ridge]

what do you not understand in that

[orchid thistle]

like how to use it

but im curr trying what he said tho

[toxic ridge]

if you've understood what it is, its fairly simple to apply it here

[toxic ridge]

hes also saying the same thing

[orchid thistle]

I got x-4=0 and x+4=0 but x-4≠0 bcs it would turn into undefined

oh wait i didnt account for the ²

mb

Btw im doing this right?
2x²-8x-x²+8x-16≥0
=x²-16≥0

[stiff lily]

1sec

sign for the 16 is wrong

[orchid thistle]

oh

[stiff lily]

ok, now factorise x^2 - 16
to identify the roots

[orchid thistle]

(x-4)(x+4)

[stiff lily]

where are the roots of that

[orchid thistle]

x-4=0
x+4=0

[stiff lily]

and x = what?
at those locations

[orchid thistle]

x=4 and x=-4

[stiff lily]

then identify the sign of the leading coefficient to see whether you have an upwards or downwards facing parabola

i.e whether you'll have a
U or
n
shape

and draw one passing through those x intercepts

(those are the only pieces of info that matter for this)

[orchid thistle]

sory 😭 but what is the leading coefficient for this?

[stiff lily]

look at the term with the highest power

in x^2 - 16

[orchid thistle]

x²?

[stiff lily]

and identify the coefficient

[orchid thistle]

oh is x² itself alone a coefficient?

[stiff lily]

no

the coefficient is
what * x^2 = x^2

[orchid thistle]

oh

then 1?

[stiff lily]

yes

[orchid thistle]

positive?

[stiff lily]

yes

[orchid thistle]

U shape?

[stiff lily]

yes

[orchid thistle]

ic

What next

[stiff lily]

draw an upwards parabola passing through (-4,0) and (4,0)
it can be quite ugly and still serve its purpose

you don't need to worry about numerical values of stuff like y intercepts, vertex or any other coordinates

[orchid thistle]

im struggling 😭

[orchid thistle]

oh

ight

[stiff lily]

with this info, it should take less than 10 seconds

[stiff lily]

yeh

[orchid thistle]

I was worried about the vertex

[stiff lily]

drawing by hand, you don't need any numbers on the y-axis

Pretty much draw a horizontal line
a vertical line for the y-axis (but you can actually ditch it)
dot your roots
and draw your curvy shape

and with this curve, you should be able to see when your function is positive/negative

[orchid thistle]

its negative from -4 to 4

[stiff lily]

yes, where is it positive

[orchid thistle]

-4 to -oo and 4 to oo

[stiff lily]

use inf for infinity

[orchid thistle]

okey, -4 to -inf and 4 to inf

[stiff lily]

now you need to be a little more specific

for the final solution

[orchid thistle]

It took me 4 hours finally its near 😭

[stiff lily]

as it's unclear whether that includes the -4 and 4

note that you've also you've also identified earlier the restriction that x can't be 4

can you try stating the final solution to the inequality now

[orchid thistle]

How do i write a final solution for inequalities

[stiff lily]

interval notion

or inequality notation

[orchid thistle]

The [-4,4)?

[stiff lily]

that's interval notation, but that interval isn't what you want here

[orchid thistle]

[-4,-inf) and (4, +inf)

[stiff lily]

careful with the first interval

lower value comes first

[orchid thistle]

oh

(-inf,-4] and (4, +inf)

[stiff lily]

with interval notation, you should write

$x \in (-\infty, -4] \cup (4,\infty)$

[rocky lotus]

ℝαμOmeganato5

[orchid thistle]

after this its complete 😭?

[stiff lily]

for inequality notation, you'd write
$$x\leq -4 \vee x> 4$$

[rocky lotus]

ℝαμOmeganato5

[stiff lily]

yes

[orchid thistle]

tyy 😭

does' U' mean and?

it is, wo ty

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[orchid thistle]

How to get that; x ≤ -4?

[orchid thistle]

i mean i know why it should be; x ≤ -4

but our teacher based his on an equation like
x-4≥0
x≥4

though cuz of that he also got x≥-4

i wanna know if there are any other ways to get x≤-4 by means of equation

[stiff lily]

annoying casework

[stiff lily]

U means union

[orchid thistle]

yez

[orchid thistle]

that is unfortunate

[stiff lily]

though cuz of that he also got x≥-4
how did he get that?

(because that'd be wrong here)

[orchid thistle]

ts what im confused too

[stiff lily]

do you have a pic of their work

[orchid thistle]

lemme send what we copied on the board

yes i do

this honestly just confuses me cuz it contradicts😭

[stiff lily]

yeh, that middle part with 2x >= x-4 is dodgy,
ignore/erase it

intentions of the bottom right chunk is a little unclear too

[orchid thistle]

im taking notes on my personal notes so its fine

In case our teacher insisted that he is correct then at least when he checked the notebooks I'll still be fine w the grades

[stiff lily]

intentions of the bottom right chunk is a little unclear too
ok, that identifies when x-4 and x+4 are positive (for the sign table)
the sign table is missing
x+4
x-4
on the left column

[orchid thistle]

oh wait lemme draw the graph that our teacher drew as well

i forgot to copy it bcs it did contradict with our graph

Which kind of implies that
[-4,inf) U (4,inf)

hence the x≥-4

[orchid thistle]

oh so thats what it was

[orchid thistle]

Are there any methods?

[stiff lily]

not really,

casework, algebra, graphical
or a mix

[orchid thistle]

this is unfortunate

so we just leave it to logic?

[stiff lily]

wdym

[orchid thistle]

like we would use reasoning to identify the interval notation, like theres no specific methods by equation like ex.
"If x+4≥0 then
x≥-4, thus the lowest interval is greater than -4"

[stiff lily]

yes

[orchid thistle]

ight thx 😭

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[lament flicker]

Can anyone help number 7

[lament flicker]

2^x -3^y = 5

Solution I got (3,1) (5,3) for pair of (x,y)

I couldn't prove that there will be no solution other than that

[vale dock]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vale dock]

Show your work, and if possible, explain where you are stuck.

[stark wedge]

i was thinking that you could maybe consider both sides of the equation modulo some specific small power of 2 or 3

but i haven't been able to find one that works

[quick rose]

mod 6 gives x must be odd

maybe that is helpful

LTE might also work its often used in questions like these ones

[lament flicker]

X and y must be odd

[lament flicker]

I mean x and y are different

[quick rose]

why should y be odd

[quick rose]

do uk what is LTE btw?

[lament flicker]

Lifting the exponent right

A^n+b^n

[quick rose]

yeah

[lament flicker]

Like that ?

[quick rose]

oh right wont work

i switched a and b and n

[lament flicker]

Since x is odd

Substituting x = 2k+1

Then mod 4

[quick rose]

ah okay thx

[ebon glade]

modulo a small number cant work because there are some solutions and modulo cant tell the difference between those and possibly bigger numbers

[lament flicker]

ah okay

then what should i try here

[ebon glade]

dunno

[quick rose]

okay so i found a solution via approach0

[lament flicker]

okay

[quick rose]

should i hint u or should i send the solution

[lament flicker]

what is approach0

[stark wedge]

LTE?

[quick rose]

its a platform to search for solutions to math problems

ur problem has been posted in an AOPS thread

and i found a solution in that thread

[quick rose]

lifting the exponent lemma

[lament flicker]

yeah

[quick rose]

yeah that thread contains a solution

approach0 helped me find that thread

[lament flicker]

alrr

nice info

[quick rose]

anyhow, back to this

[lament flicker]

Give me the general idea / hint so I can understand it by my self

[quick rose]

well assume that x>5, y>3

then write 5 as 32-27, factorize

then use the fact that x and y are odd to note that x-5 and y-3 are even

then use LTE, then FLT and some computations

and then u arrive at a contradiction for the value of v_3((x-5)/2)

[lament flicker]

i get 32(2^(x-5)-1) = 27(3^(y-3)-1)

the what should i do next

[quick rose]

*y-3

[lament flicker]

mb

[quick rose]

from here, rewrite your expression as 32(4^a-1)=27(9^b-1)

now try to use LTE to find v_3(a)

[lament flicker]

alr

so just v3(3)+v3(a)+v3(32) = v3(27)+v3(8)+v3(b)

[quick rose]

did you apply it directly

[lament flicker]

yeah ...

[quick rose]

theres a much better way to simplify the rhs

LHS is correct

[lament flicker]

how

[quick rose]

and also

[quick rose]

3 must divide x-y for this to work

but in our case x-y=8

[lament flicker]

yep

[quick rose]

so u cant do what u did on the RHS

[lament flicker]

wait \

im dumb ?

should it just be 3

[quick rose]

yes lol

[lament flicker]

😭

forgor

[quick rose]

its okay bro

[lament flicker]

so 1+v3(a)+v3(32) = 3

sorry

so v3(a)+v3(32) = 2

then v3(a) = 2

[quick rose]

good

what does that tell you?

[lament flicker]

that a= 9b

wait

[quick rose]

mhm now go back to my list of hints

[lament flicker]

i want to say that b is not a multiply of 3

[quick rose]

b wont matter trust me

[lament flicker]

alr

[quick rose]

now what can you exactly apply fermats little theorem on

[lament flicker]

i suppose that we plu a = 9b here

plug *

[quick rose]

nope thats wasting time 😐

[quick rose]

u saw that 9|a and u need to apply FLT on something

which term do you think it should be

[lament flicker]

couldnt see it

[quick rose]

well u need to do the following
4^a = 2^2a = (2^18)^(a/9) = 1 mod 19 by FLT

reason?

1. you just derived that 9|a, so clearly you should do something with the term containing 4^a
2. since 9|a, a/9 is an integer which is the motivation for the 2^2a = (2^18)^(a/9) manipulation (remembering u need a power to apply FLT)

[quick rose]

from here, 19 | 4^a - 1 which means 19|9^b-1

now its a bunch of computations

[lament flicker]

yeah

okay what is the next goal then

[quick rose]

do whatever the fuck this is

i have no clue how you're supposed to think of this step

[lament flicker]

this obtained through a computation my ahh

[quick rose]

fr

[lament flicker]

i mean

the fuck 😭

[quick rose]

fr 😭

let me see if theres a better soln

theres this, if u can understand this

[lament flicker]

i mean this proof is somewhat edible

i mean is this problem imo level or just hard national level

[quick rose]

theres also something called hensels lemma which maybe usable?

[lament flicker]

my original intuition is exactly like this

[quick rose]

no clue

[lament flicker]

but

[quick rose]

i see

[lament flicker]

i am stumped in the case that (3^(c-3)-1)=(2^(a-5)-1)

but the more i think

what i was thinking back then bruhh

[quick rose]

isnt that similar to what we derived in the computation soln

[quick rose]

also formally proving this intuition is not easy

[lament flicker]

the only problem that i found is that (2^(a-5)-1) = 27b and (3^(c-3)-1)=32b

but

they couldnt be posibly have the same factor right

or their gcd is 1

[quick rose]

prove it

[lament flicker]

one side is odd

one side is even

or so i thought

[quick rose]

err b is odd ?

[lament flicker]

b is odd right right

[quick rose]

m not telling m asking

[lament flicker]

yeah

[quick rose]

why r u agreeing 💀

[lament flicker]

cuz it right ?

nah

im dumbing again

[quick rose]

i am very clueless about what you're doing rn

[lament flicker]

im also clue less

[lament flicker]

haow can the author decide to only work on 3^(c-3) -1 = 32

like

[quick rose]

why would i know 😭

i dont understand that solution xd

i posted it to see if u could make sense of it

[lament flicker]

the only part of it where i couldnt understand

[quick rose]

https://math.stackexchange.com/questions/1781722/prove-that-2x-3-cdot-9m5-has-no-positive-integer-solutions-for-m-geq-2?
heres some more mad math solutions to your question btw

[lament flicker]

hell no

they all use like mod 41 or shi

[quick rose]

https://math.stackexchange.com/questions/3256556/solve-for-integer-m-n-2m-3n-5
this ones actually good

u need to do some order computation when it comes to exp diophantine

but this is much better imo

[lament flicker]

MY GOD

i found it

yeah as i assume

they both are coprime

(2^(a-5) -1), (3^(c-3)-1)

[quick rose]

👏

but what exactly are a and c again..?

[lament flicker]

i mean diffrent proof diffrent a and c

[quick rose]

write your entire proof in 1 message

so that i can understand it

(please )

[lament flicker]

ill try to prove that 2^a... part is co prime with 3^c... part

since you can digest all else before it

[quick rose]

oh

[lament flicker]

2^a ... part = 27b

3^c... part = 32b

[quick rose]

(why?)

[lament flicker]

since 32(2^a..) = 27(3^c....)

hence 2^a...... = 27b

3^c.... = 32b

i aint writing all at

then we know that 2^a..... is odd

then b must be odd

[quick rose]

???

[lament flicker]

you know what

this prove is flawed

proof*

even i cant write proeprly 😭

cooked

[quick rose]

r u self prepping?

[lament flicker]

yeah

[quick rose]

thats tough

https://math.stackexchange.com/questions/3256556/solve-for-integer-m-n-2m-3n-5

[lament flicker]

i mean

[quick rose]

check this.

[lament flicker]

alr

hell no

[quick rose]

whats bad about the first proof 😭

[lament flicker]

the second proof is hell no

[quick rose]

whos asking u to read the second proof

it does explain the 757 btw

[lament flicker]

well this proof is good

not the 757

the first proof

[quick rose]

yeah

but it seems the 2nd proof generalizes better

but ofc u cant use it in an olympiad

[lament flicker]

then what it is use for

[quick rose]

solving for solvings sake, i guess

[lament flicker]

i mean

this question shoulndt be in any olympiad right ??

[quick rose]

i think they've been asked before

[lament flicker]

man

[quick rose]

i could be wrong though u'll have to do some hunting yourself

not every one is as tough or involved as this one

some are easy enough that u can argue by simply properties of primitive roots and stuff

https://math.stackexchange.com/questions/1941354/elementary-solution-of-exponential-diophantine-equation-2x-3y-7
for example this

[lament flicker]

yeah

[quick rose]

what i meant to say is dont be sad, a question like this in an oly is unlikely to be as involved

because olympiads usually give problems that require u to think more than crunch numbers

[lament flicker]

yeah i understand

[quick rose]

if ur done, coose it!

[lament flicker]

.close

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[small crystal]

f(x)=(x+a)(x+b), a and b are positive constants, f(-21)>0 and f(-15)>0 and f(-18)<0, what could be a+b

[vale dock]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[woeful trench]

(Is this a multiple choice question?)

[small crystal]

i get that x=-21, -15 are outside the roots and -18 are inside the interval of roots so i got -21<-a<-18<-b<-15 or -21<-b<-18<-a<-15 but im kinda stuck from here

free response

so like i can enter any value

i solved it

.close

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[midnight pier]

@stark wedge sorry i just noticed that you have crossed my interpretation of this expression. can you please tell me what the correct conclusion would be?

Referance (this was my statement you crossed) : "so basically for even degree polynomials, its a_(n-k)/a_n (without the negative)"

[stark wedge]

(-1)^k will still be there

regardless if n is even or not

the product of all roots [which means k=n] is (-1)^n a_0/a_n, and in that case for n even the (-1)^n will become 1 and disappear

[midnight pier]

but since this was a quintic we were talking about, it would be odd, so the (-1) remains right?

[stark wedge]

yes (-1)^5 = -1

[midnight pier]

okay now i understand that term in your answer

[stark wedge]

(-1)(-2)(-3)(-4)(-5) + 6 = -120+6=-114 and then you multiply that by -1 for an answer of +114

[midnight pier]

can you give me a couple scenarios to make that expression in? so i know i have it down correctly this time

[stark wedge]

hmm sure ig

Let f and g be defined as follows:

• f(x) = x^7 - 4x^6 - 2x^5 + 60x^4 - 9x^3 + x^2 + 28x - 40.
• g(x) = 5x^3 + 8x^2 + 50x - 210.

Find the sums of the following:
a) roots of f taken 3 at a time
b) roots of g taken 2 at a time
c) roots of fg taken 1 at a time (ie just the sum of the roots)

[midnight pier]

a) -60
b) 10
c) I dont understand how this works

you mean the terms -40 and -210?

[stark wedge]

i want the sum of the roots of the equation f(x)*g(x)=0.

@midnight pier

[midnight pier]

4

[vale dock]

@midnight pier Has your question been resolved?

[stark wedge]

incorrect! but show your work.

(sorry for the delays on my end -- i am with family rn and might not respond immediately)

[midnight pier]

wow still:((

[midnight pier]

no worries, more than willing to wait to receive a response

[stark wedge]

a and b are correct btw, only your c is wrong

[midnight pier]

I mostly just reasoned that the highest degree term here would be x^10 (theres only one combination of terms whose product results to that term: x^7 and 5x^3, so the a_10 (a_n) would be 5.
and i need product of tersm one at a time, k=1. a_9 (is the coefficient of the term x^9: which also happens to have one way: -4x^6 * 5x^3 : so a_9 is -20.
(-1)^1 a_9/a_10 = -(-20)/5 = 4

ohhh theres two ways

[stark wedge]

x^7 and x^2

[midnight pier]

a_9 = 8 - 20 = -12

[stark wedge]

but also you don't need to work out fg actually

it's just sum of roots of f + sum of roots of g that i want!

4 + (-8/5)

that's all

[midnight pier]

oh wait they'll all still have all those factors. so the multiplied polynomial will still have all those roots

is that a valid way of reasoning?

[stark wedge]

it's not just valid, it's exactly what i wanted from you.

or like

the recommended route

[midnight pier]

this reasoning would also work for the product of all the roots too

but not the other cases, right?

[stark wedge]

well actually would you like a more complicated general problem

[midnight pier]

where i do things like product of roots taken 2 at a time. because it would miss terms

[midnight pier]

yes please

[stark wedge]

right

gimme some time

let $f$ and $g$ be monic polynomials of degree $m$ and $n$ respectively, i.e.

\begin{gather*}
f(x)=x^m + a x^{m-1} + bx^{m-2} + \dots, \ g(x) = x^n + p x^{n-1} + q x^{n-2} + \dots
\end{gather*}

and let $h(x)=f(x)g(x)$.

without expanding $h(x)$, work out the sum of products of its roots taken two at a time, in terms of $a$, $b$, $p$ and $q$.

[rocky lotus]

Ann

[stark wedge]

[you won't need any coefficients aside from the ones named!]

[midnight pier]

wait, i think i could do this!

q + ap + b ?

(btw what is a monic polynomial?)

[light saddle]

Leading coefficient is 1

[stark wedge]

correct

[midnight pier]

thank you soooo much for all your help:)

have a nice day:))

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[hearty belfry]

<@&268886789983436800>, I think this counts as advertising?

[static ivy]

uhh, is this not allowed?

[hearty belfry]

I don't think so, from #rules

[mighty trellis]

I'm afraid we don't allow surveys like this. If you'd like to appeal that you should ask @feral pebble

I'm going to remove your poll now, apologies.

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[static ivy]

.close

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[grave kernel]

@ruby sleet hey im back can you please teach me what you were teaching me yesterday sorry for the ping

[grave kernel]

for the table

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[fiery estuary]

can someone explain to me why shifting the negative exponents from the denominator to the numerator works? I understand dropping the n^-2 to the denominator, because n^-2 is just 1/n^2, but i dont understand how m^-4 and n^-3 shifted to the numerator and became m^4n^3

[hot tide]

$\frac{1}{\frac{1}{x}} = x$

[rocky lotus]

artemetra

[fiery estuary]

so basically this

[hot tide]

also there is a typo here, there's one m missing

[fiery estuary]

o yeah ur right

m at the end

[hot tide]

indeed

[hot tide]

yup

[fiery estuary]

okay thanks

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[hearty belfry]

!status

[vale dock]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pure mirage]

Substituting u = 1 + sin(x) might be a good start

[left fiber]

2

[midnight pier]

Show ur work

What did u try

[left fiber]

from here i could probably try by-parts right?

u=1+sinx

[wraith heart]

or try v = sqrt(u) and then partial fractions

[tribal tiger]

i think its better way

[left fiber]

Yep i've gotten onto the last part now

[tribal tiger]

its classic integal

[left fiber]

okay

thx guys

how do i close this now

.close

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[solar flame]

How many months would this take to finish?

[wraith heart]

Not a math question. Please ask in #study-discussion

[light saddle]

more than 6 weeks lol

[solar flame]

The textbook is a maths textbook. Wouldn’t that qualify as a math textbook?

[wraith heart]

There's no math in your question. Read this

[solar flame]

My bad if it doesn’t

[stark wedge]

help channels are for specific math problems

[solar flame]

Oh alright

[stark wedge]

youre being redirected to a channel where your question is more appropriate

[solar flame]

.close

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[midnight pier]

how to do this integral

[magic lynx]

oh god

doesn't look pretty

[midnight pier]

ye

[magic lynx]

my first step would be to go to integral-calculator.net for advice

uh in seriousness though

let us try u-sub=(e^x+x+1)^2

[midnight pier]

not gonna work

we dont have derivative in numerator

[magic lynx]

we may have something related to the derivative in the numerator? (im not sure yet need to see myself)

uhh du=2(e^x+x+1)(e^x+1)

ohh wait

here

i give you related integral

Calculate $\int\frac{(e^x+x+1)^2-(x^2+x)}{(e^x+x+1)^2}\mathrm{d}x$

[rocky lotus]

Arnavutköy

[magic lynx]

now try u-sub

with u=(e^x+x+1)^2

[somber sequoia]

judging by the structure this was the result of some sort of quotient rule application

[midnight pier]

yes

i tried that kind of reverse engineerign

but that didnt work

[midnight pier]

still

[somber sequoia]

[ \int \frac{x^2}{(e^x + x + 1)^2} \dd{x} + \int \frac{e^x + x + 1}{(e^x + x + 1)^2} \dd{x} - \int \frac{e^x + 1}{(e^x + x + 1)^2} \dd{x} ]

[midnight pier]

doesnt work kinda

[jade ivy]

From this observation, it is just a matter of finding the right ax+b that gives you a useful derivative when you divide by e^x + x + 1 and do the quotient rule.

[somber sequoia]

@midnight pier try integrating the first integral by parts

I suspect it will cancel out with the second one

or vice versa

[midnight pier]

oh alr

[somber sequoia]

The third integral can be done with a substitution

[rocky lotus]

maison

[midnight pier]

second and third ar easy

[somber sequoia]

the second one by itself is probably non elementary

[midnight pier]

but what should be hte first function adn the second funtion for by parts

of the first one

[distant pulsar]

you can’t integrate the first one

at all

[midnight pier]

how do i see these kinda things

[somber sequoia]

take the second integral and differentiate 1/(e^x + x + 1) and integrate 1 (IBP)

[midnight pier]

like hwo do i check if i can integrate something or not

[somber sequoia]

it should cancel out with the first hopefully

[magic lynx]

uh i am looking at the solution at integral-calculator.net and i can assure you that I would have never come up with this nor half of the people in the mit integration bee

[midnight pier]

hwo to integrate the first

[somber sequoia]

it's a bit of an advanced technique with IBP

[somber sequoia]

we're not going to integrate it

[midnight pier]

o

no

[magic lynx]

do you want to know what to rewrite it as

[midnight pier]

like what should be the first and second function for by parts

in the first term

[somber sequoia]

leave the first term

work on the second one

integrate 1 and differentiate 1/(e^x + x + 1)

[jade ivy]

Its a rational function of x's and e^x's without any weird nth roots or infinite sums. You are not giving people on an MIT Integration Bee enough credit.

[midnight pier]

ok

[jade ivy]

Even if this turned into a messy integral involving a bunch if dilogarithms it would not be out of reach of a group like that

[midnight pier]

lol

[magic lynx]

(in competition conditions i claim)

with like half an hour yes

[distant pulsar]

half an hour? this is doable in less, which is expected of integral problems

[somber sequoia]

it can be done quick

[distant pulsar]

okay now that i certainly disagree with

the approach you suggested is not even correct

[somber sequoia]

on my JEE shift there was a similar problem except the reverse-IBP twice

[somber sequoia]

it doesn't work?

[midnight pier]

I did it but it's doesn't cancel our

[distant pulsar]

this? no

[magic lynx]

bruh i just asked somebody ranked top 10 in all india in jee and they were stuck

[midnight pier]

This is also a jee question

[distant pulsar]

it just comes down to testing things

[magic lynx]

hopefully not mains

[midnight pier]

not mains

[magic lynx]

wait is the answer multiple choice?

[midnight pier]

no

i dont know the options

but my teacher said taht

[somber sequoia]

wait just combine it now

you can factor the x out

and reduce the power in the denom

[magic lynx]

essentially you only get this through guess and check in jee exam format

[somber sequoia]

also idt that should be a - sign right

you get a - from the IBP and one from the derivative

so it becomes a +

[distant pulsar]

you can notice i.e. $\frac{\mathrm{d}}{\mathrm{d}x} \frac{x}{D} = \frac{(1 - x)e^x + 1}{D^2}$, $\frac{\mathrm{d}}{\mathrm{d}x} \frac{1}{D} = -\frac{e^x + 1}{D^2}$

[rocky lotus]

Mqnic_

[midnight pier]

Idts

[fallen mesa]

,rotate

[distant pulsar]

from here, you find $\frac{x^2 + x}{D^2} = (x + 1)\frac{\mathrm{d}}{\mathrm{d}x}\frac{x}{D} + (1 - x^2)\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{D}$

[rocky lotus]

Mqnic_

[distant pulsar]

and now you integrate by parts

[somber sequoia]

it shouldn't be a - sign

[distant pulsar]

you are done

[midnight pier]

It would be

I mean by parts give minus sign for the second term

[somber sequoia]

and the derivative gives you another one

[midnight pier]

Oh yee

I am sorry

Now how do I integrate this

[frozen ledge]

dude isn’t even reading manics messages

[midnight pier]

i am

[somber sequoia]

probably multiplying with e^-x

[distant pulsar]

one way is x = D - e^x - 1

[somber sequoia]

[distant pulsar]

which gives you the answer essentially immediately

[somber sequoia]

ah yeah that works better

[midnight pier]

Yes

Thanks guys

Ik I am disturbing but can I ask another one

like i was trying asinx+bsinx property but still not getting the correct answer

[somber sequoia]

hmm

I have a bit of a crazy idea

[midnight pier]

i tried reverse engineering also by quotient rule

[magic lynx]

if this is what disturbing is i want to be disturbed

[midnight pier]

thx 😁

[somber sequoia]

[ \int \frac{\tan^2 u}{(\sin u \sin \tan u + \cos u \cos \tan u)^2} \dd{u} ]

[rocky lotus]

maison

[somber sequoia]

now the denominator is cos(u - tan u)

tan^2 u = 1 - sec^2 u

badabing badabong

[midnight pier]

u substituted x as tant

[somber sequoia]

yes

I guess it wasn't a crazy idea after all

I've seen the xsin(x) + cos(x) structure in an integration bee before and I used the same technique

[midnight pier]

should nt there be a cos^2x in the denominator

[somber sequoia]

multiply it inside

[somber sequoia]

that's how you get this

[midnight pier]

Yes I got it

Now what should I do

[somber sequoia]

in the denominator use the cos(A - B) formula

and rewrite the top using the pythagorean identity

[midnight pier]

This ??

[somber sequoia]

yeah

but write it as sec^2 u - 1

[midnight pier]

oh

[somber sequoia]

you probably see what can be done now

also there should be a ^2 over the cos (t - tan t)

[midnight pier]

Oh yes

Sorry

Sorry but I can't think of what should I do next

Tant as v

?

[rocky lotus]

maison

[somber sequoia]

👀

[midnight pier]

I got this

But what should I do after this

[somber sequoia]

not just tan t

[midnight pier]

Tant-t

Ohhh

[somber sequoia]

yes

[midnight pier]

Oh my god

I am so stupid

[somber sequoia]

because the derivative is on top

[midnight pier]

But like how did u thought so much out of the box

I would have never thought in this manner

[somber sequoia]

you'll get there with practice

integration was my favourite chapter in all of JEE i had a bit of an obsession

[midnight pier]

Same

It's my fav too

I am a jee student

[somber sequoia]

good luck lol

[fallen mesa]

I made all my classmates and my teacher to sign a petition to remove integrals chapter from the syllabus

On our blackboard

[midnight pier]

Lol

I love it

I hate chem tho

Other than physical chem

hey @maison canu recommend me some great integration books

i need some guidance in it

i am currentyl doing modules and cengage sometimes

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[brave urchin]

is there some algorithm to follow so i can remember the 20 something integrals formulas? for derivatives it's kind of easy, even with composed functions, but for integrals it just looks random to me. i'm attepmting to learn them on my own during summer break.

[ebon glade]

wdym 20 something integral formulas

[charred nacelle]

Just practice questions

[ebon glade]

the easy integrals are just reverse of basic derivatives

[charred nacelle]

Also some formulas have same things but just a few changes which can be easily remembered

[steady charm]

unlike for derivatives, there isn't a general algorithm that works for all integrals

[ebon glade]

and then its just u-sub and ibp

[ebon glade]

risch cough cough

[steady charm]

-# technically there sort of is one but it's like 100 pages

[fallen mesa]

Integral of tan is ln |cos| and ln |sec|

[ebon glade]

-ln|cosx|

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[humble flicker]

does someone mind helping me find out why the answer is B here? i guessed it

[humble flicker]

they have this solution/working out

but is R the range?

[warm kayak]

they're asking the y coordinate of B, which is the maximum value of the function

[timid wagon]

have you dy/dx yet?

[humble flicker]

i got 2cosx - 3sinx

but what do i do from there

[timid wagon]

great

if it's a turning point what do we do

[warm kayak]

the maximum of acosx + bsinx is root(a^2 + b^2)

[timid wagon]

yeah that works as well

[timid wagon]

if you don't get this i suggest you to do this first

[humble flicker]

set it equal to 0

then solve

then test if it’s a max or min using second diff or sign table

then u get 2 -3tanx right

if u divide everything by cos

[timid wagon]

yes

then you solve for x

[humble flicker]

but x = 0.58800…

oh do u sub it back into y?

[timid wagon]

yes

that'll be your y value

but make sure you sub back into the original equation

[humble flicker]

now i got 3.6055…

yea

[timid wagon]

great

[humble flicker]

oh it’s the same

[timid wagon]

notice how sqrt13 = 3.6

yep

[humble flicker]

yea

[timid wagon]

there ya go

[humble flicker]

my calculator just didn’t convert it to sqrt13

so i didn’t know beforehand

[timid wagon]

yup

no worries

[river shale]

Never seen this method before

Dawg

[timid wagon]

but use this

[humble flicker]

max point?

[timid wagon]

what you've done is a troubling way

[humble flicker]

whats the proof for this

[timid wagon]

its a formula

you can search it up

[humble flicker]

compound angles?

wait no

[timid wagon]

https://www.bbc.co.uk/bitesize/guides/zsr334j/revision/7

or just type the formula out on google

you should get plenty of explanations

[humble flicker]

oh it’s the R-method

forgot abt that

okay thanks

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[grand nova]

eww jets nose pihh eww

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[whole torrent]

sinx (x in radians) = sinx (x in degrees)

What's the value of x (x isn't equal to 0)

[tidal terrace]

convert the angles to either (rad to deg or the reverse)

[whole torrent]

so sin(xpi/180) = sinx right?

[tidal terrace]

so your right x is in radians ?

[whole torrent]

Radians

[waxen scarab]

Then solve for x in the equation
$$\frac{x\pi}{180}=x+2k\pi$$

[rocky lotus]

Sherif Player

[waxen scarab]

Where k is an integer

[whole torrent]

So x = 2kπ ((π/180) - 1) ?

[waxen scarab]

Actually
$$\pi x = 180x+360k\pi$$
$$\pi x - 180x = 360k\pi$$
$$x(\pi-180)=360k\pi$$
$$x=\frac{360k\pi}{\pi-180}$$

[rocky lotus]

Sherif Player

[whole torrent]

Oh thank you, trigonometry is confusing but i understood it

[waxen scarab]

Do you understand why did we solve for that specific equation?

[whole torrent]

Not actually

[waxen scarab]

For any trig function in the units of radians
f(x) = f(x+2kπ)
like
sin(x) = sin(x+2kπ)
And so on
Because 2π resembles a full revolution

So if we revolved the angle around k full revolutions it would still remain the same

Do you understand now?

There are actually more solutions to the system of sin because sin has another property which is
-sin(x) = sin(-x)
Which would be a bit harder to solve

[whole torrent]

Yeah, i understood it but i have a question. Is it because 2pi is period of sinx?

[waxen scarab]

Yes

[whole torrent]

But general solution is 360kpi / (pi-180) right? When k is an integer

[waxen scarab]

Yeah that's a general solution

[whole torrent]

Okay i understood it

Thank you so much

[waxen scarab]

Do you have any other questions?

By the way, I tried it here and it worked

[whole torrent]

What's this app?

[whole torrent]

Not at all, thanks for your help

[waxen scarab]

!done

[vale dock]

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[whole torrent]

Thanks

.close

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[thorn kernel]

how do we draw this again?
for k = 1 we get y = +-3z whic hi get the 2 lines
but the parabolas how do we draw those

anyone know how and why?

[chrome beacon]

examine y = 0 and examine z = 0

[thorn kernel]

yeah i get y = +- 1

z = +-1/3

[chrome beacon]

yep so those are the peaks of those hyperbolas

[thorn kernel]

i did that but what does that gotta do with anything

but how do i know to draw parabolas

at thsoe points

is what im asking

[chrome beacon]

they're not parabolas

they're hyperbolas

[thorn kernel]

in the zy plane yea where x = k

[chrome beacon]

but yeah you know they're hyperbolas because they follow the form $az^2 - by^2 = 1$

[rocky lotus]

not haylee

[thorn kernel]

hm ok so this is just memorization

also what the fuck does the k = 1 thing have to do with the full hyperparabola

[chrome beacon]

kinda yeah. it's a common conic section

[chrome beacon]

hyperbola

it's a degenerate hyperbola

these are conic sections, have you seen this before?

3 is a hyperbola

1 is a parabola and 2 is an ellipse

the crossed lines are what happens when the plane in 3 crosses the centre point

[thorn kernel]

ok but we

use teh 2 y=k and x=k

to draw the hyperparabola

right

[chrome beacon]

do i need to automod the word hyperparabola

hyperbola

[thorn kernel]

same thing

yk what i mean

[chrome beacon]

it's degenerate when it's of the form $az^2 - by^2 = 0$

[rocky lotus]

not haylee

[chrome beacon]

so by setting k = 1 the right side becomes 0 and you get the degenerate case

[thorn kernel]

so is this the math rage bait

[chrome beacon]

?

[thorn kernel]

anyways

so for the y=k i understand this right

we draw circles

but im confused about the x=k

how to incorporate what we drew on that 2d plane here

[chrome beacon]

well the crossed lines is what happens when k = 1

[thorn kernel]

so the top and bottom are lik this i get right

[thorn kernel]

yeah i dont understand where to draw these

or the sides

[chrome beacon]

oh i see what you mean

imagine you have that hyperboloid (boomerang looking thing)

[thorn kernel]

oh wait i think i get

the sides are done

[chrome beacon]

and you look at it from the side so kinda like you've drawn indeed yea

[thorn kernel]

just the 2 crosses left

[chrome beacon]

well the shape you've drawn is for k = 0

it'll look different for k = 1

in particular its waist will be fully snatched and drawn in to a point

[thorn kernel]

i mean this is the answer

[chrome beacon]

yes, most of the time it looks like pictured

[thorn kernel]

where is the crosses incorporated here in the pic

[chrome beacon]

i would recommend using desmos 3d to draw this shape

and add a slider for k

[chrome beacon]

oh hang on

[thorn kernel]

this looks nothing like my graph answer what?? in the zy plane

[chrome beacon]

okay i see, i think the crossed lines happen when the plane youre looking at is tangent to the inner waist of the hyperbola

[thorn kernel]

tbh i dont think it matters

just memorize it

thanks tho

[chrome beacon]

if this is where you take your projection

you can see the crossed lines

(i got this by plotting x = 1 and hitting "extend to 3d")

[thorn kernel]

but now u dont see teh side parabolas

or is that cuz desmos cut it off

[chrome beacon]

https://www.desmos.com/3d/6hu2uiiibn

this has a slider

[chrome beacon]

what parabolas

[thorn kernel]

oh wait nvm

i see it

was talking bout this

[chrome beacon]

after playing with it in the graph view, do you see how the hyperbola gets drawn tighter until it becomes a pair of crossed lines?

[thorn kernel]

yep

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@thorn kernel Has your question been resolved?

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[surreal sinew]

Hey everyone

[glass kelp]

hello

[surreal sinew]

How is this a "legal" move:

[surreal sinew]

Can someone explain please?

[glass kelp]

[ \sqrt{x^2} = |x|]

[rocky lotus]

k

[surreal sinew]

Yes ik

[lyric sundial]

Then what? 😅

[surreal sinew]

But |x|≠-x

[jovial edge]

because when it goes to negative infinity, it's negative

[surreal sinew]

↑

[glass kelp]

huh

what

[lyric sundial]

It's the definition of |x| ...

[glass kelp]

Do u mean:
[ |x| = \begin{cases} x & x \geq 0\
-x & x< 0
\end{cases}]

[surreal sinew]

Yeah I corrected it

[jovial edge]

If $a<0$ then $|a|=-a$ isn't it

[rocky lotus]

Alexis_Fx

[surreal sinew]

Oh wait

[rocky lotus]

k

[lyric sundial]

Yes

[surreal sinew]

That's true

Oh I get it

If x is negative then the absolute value is -x

Makes sense

Ok sorry haha

Thank you guys

[surreal sinew]

🙏👍

.close

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[verbal sentinel]

dont undersand

[stark wedge]

show the full question.

[stark wedge]

ok, so then

[final adder]

did you do part a?

[verbal sentinel]

yes

i need help on part b