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which q

Last one x = y = z = 4

any question that u like...these r the question which i wsn't able to solve

@ParthKohli how did u get x=y=z=4

It's a symmetric situation. Apply all inequalities you please, but you'll finally get to this.

first q doesnt look easy

Good questions by the way!

thanks @Empty
@ParthKohli i can't get to the result x=y=z.
can u tell how to start ..

wait m doing the 1st one

yes so how do we know that in which case x=y gives the max

i feel #1 is still a stupid q, you're not gona learn anything frm it

wait sry yes we can put a^5...

yes it is @ganeshie8 :D

Isn't that the standard way to do it though? How did you originally do it?

I factored it

the 1st one is solved thanks guys :)

To be fair, that solution was most likely obtained by working backwards. :P

that is okay for a proof

Yes of course, but you can't really think of that on your own.

that all goes as "scratch work"
:P

I'm ashamed that I have no other ideas other than telescoping for #3.

It wants to telescope, why force it to be anything it's probably not? :P

When you separate it out, the coefficients are both 1, it's truly a beautiful result

I'm either dumb or too much drunk to notice it but why do you think telescoping is a bad idea ?

Yeah you're not drunk enough ganeshie, cause this baby wants to see the moon

my mind is jst nt wrkin normal today :(

OK, it's actually telescoping really nicely.

no i mean hw did u split that

No, I _created_ this

There's a typo.

Supposed to be\[= \frac{3^{k+1}}{2^{k+1} - 3^{k+1}} - \frac{3^k}{2^k - 3^k}\]

OK, so we're sorta done with all of the questions.

whats the ansur for q2

Are you asking me or asking imqwerty?

m calculating the answer...

@Empty Did you check out the solution to the problem I gave you yesterday?

Oh no I didn't, also I ended up getting 2 to that, I'll go check real fast

2 is correct.

#2 im getting 5/144

my answer the #2 is 121/144

I am familiar with that problem
\[\sum_{n=1}^{\infty}F_nx^n= \frac{x}{1-x-x^2}\]
plugin \(x=1/2\)

they just want the exams to be tough, they dont bother to look at the syllabus of 12th grader

at my home m nicknamed - 'silly mistake'

trust me, you and PK are way better than me when i was in my 12th

It's like teachers want people to believe math is useless AND ugly.

humblebrag

lol duuuude gotta go for that \(2 \pi\) score man!

the paper wasn't enough easy to get 360/360. the guys are too crazy...they are like 24 x 7 studying.

Their parents must be awful people

Ok ok maybe I'm being pessimistic here haha

his parents don't interfere his personal life

Woah he must be a genius :O

this guy got 1st rank in 2/4 test held till date

Kaunsa institute hai ye?

allen

Kota wala?

han

Tumhe internet mil jaata hai?

mai to kota ka hi hu

Oh, badiya yaar.

Guest kyun aa jaate hain?

pata nai... btw mere relatives paas me hi rahte hai to wo dadi dada se milne aa jate hai.

btw @ganeshie8 did u got the answer to the question -
dIzI/dz z=complex number

common sense :
common lannguage guys, when others are there in the thread

ok :) sry

thats okay, just saying..

あなたは何について話していますか？理解できません。

dIzI/dz
derivative doesnt exist

What kind of derivative is this?

z is a complex number
f(z) = |z| is a real valued function that takes complex number as input

Exactly, the "interval on real number line" becomes a "ball in argand plane"

@imqwerty knows all this?! o.O

not too much..

Well, there goes my self-esteem.

if you're in 12th grade and if you say you know all this, then you're just lying

He's in 11th grade!

So am I.

Me too...lol

i asked my teacher to explain this but he refused :( he said - its of no use

That's stupid. "It's not there in the JEE examination" should not be restated as "it's of no use".

^true that

I think complex derivatives are a hell lot more useful that the four questions you gave us.

yes!! the teaching methodology nd the mindset of ppl must change.

Wow, I just searched him up and he looks pretty young, like a passout.

I think it's better to not ask for it...

If you get a satisficatory explanation, please do share wid us

haahaha once i got kicked out (frnkly) fr asking such doubts..
nd yes i'll definately share it.

pk have u joined any institute?

Yeah, Vidyamandir.

oh nice :)

Eh, this problem isn't that hard.

It may be calculative, but it's not hard.

looks that fraction evaluates to \(\left(\dfrac{n+1}{n}\right)^2\)

Yes it has lil bit calculation prt

You can do the gauss fliperoo method there I think.

oops, the second should be the sum up to \(n-1\), actually

ok i am getting 151 as the answer

wait no m getting 150 as the answer

anyways, \(|z|=z\bar z\) is not differentiable on \(\mathbb{C}\) because it depends on \(\bar z\)