Problem Solving Strategies

Yashas

04:27

Yogi govt. abolishes quota for SC/ST, OBC students for post-graduation courses in private medical colleges

:O

@blue

Madhuchhanda Mandal

06:11

Anyone there?

Madhuchhanda Mandal

06:36

I am having a very basic doubt !!! I never thought at it

If f'(x) =g(f(x)) and f'(x0)=0. Will the graph of f(x) be constant from x=x0 to x=infinity??

Madhuchhanda Mandal

09:41

?

Koolman

12:18

@Yashas I wish it could be for every field

Yashas

You are from UP?

Koolman

ah na

jaipur

modi should do that

Yashas

It is very difficult to do that.

because people who currently take advantage of it will protest

Koolman

yeah

But I think Modi can only take such bold decision

Are everyone here from general ?

Yashas

I think so

Minimum possible atomic number of the atom which can have a 4d orbital is ______

0o

@blue

Koolman

12:23

39

Yashas

@MadhuchhandaMandal That doesn't look like a basic doubt lol

@Koolman why 39?

The question did not mention that the atom is in ground state.

Koolman

is it correct

Yashas

even a hydrogen atom has 4d orbitals

empty 4d orbitals

if you give enough energy you can excite the 1s electron to 4d

Anonymous

@Yashas You mean filled with atleast 1 electron, right? (In ground state)

Yashas

I have no idea

I have stated the question as it was given

Anonymous

12:27

From where did you get it?

Anonymous

JEE?

Koolman

@Yashas oh , thats interesting

Anonymous

I think they mean that only

Yashas

Resonance mock test

Anonymous

Ah, they mean it has atleast 1 electron

Anonymous

12:28

In ground state

Yashas

I don't know the answer

I haven't hit the FINISH button yet

Anonymous

Don't chat during mock tests :-P

Yashas

chemistry

Koolman

man you are cheating

Yashas

I complete it in 30 mins

and spend another 1.5 hours chatting

@Koolman I did not understand the question

I marked one and I wouldn't change the answer

Anonymous

12:29

They will give solutions after the test

Yashas

becaz I don't want to get a fake mock test rank

Anonymous

Don't worry

Yashas

yea but I am cross checking all my answers

I have 20 mins more

Anonymous

Then what is the use of giving a mock test if you check the answers from other sources ? :-P

Anonymous

Click on the finish button

Anonymous

12:30

That is a bad habit!!

Yashas

I am not checking answers from other sources lol

I start cross checking my answers

and write down the list of topics I am not comfortable with

and do stuff like that

overanalyze weakness

Anonymous

Okay =P

Yashas

The shape of the orbital is given by magnetic quantum number.

I don't know if I should mark it as incorrect or not

value of m decides the shape of the orbitals

but not all by itself

it depends on l too

Anonymous

It decides orientation...

Anonymous

That's debatable

Anonymous

12:37

For jee mark it as incorrect

Yashas

that is not good

you shud mark the most accurate answer according to ur knowledge

Anonymous

@Yashas The jee guys follow ncert

Anonymous

In ncert it is written that m decides orientation

Anonymous

l decides shape

Anonymous

n decides size

Yashas

12:38

in case, JEE expected you to know that m decides the shape, then you are in trouble

you are going to lose marks

if they say m does not decide the shape then you can argue

you cannot argue in the other case

Anonymous

@Yashas They won't listen

Anonymous

lol

Yashas

You can move the court if they don't listen and screw everything up :D

Anonymous

ok, whatever :P

Anonymous

it is a stupid question

Yashas

12:40

btw JEE Advanced does not follow NCERT imo

NCERT isn't even close to JEE Advanced

JEE Advanced syllabus is way ahead of NCERT

Do you have Crystal field theory in NCERT?

Anonymous

I can solve 60 percent of the paper just using NCERT. Checked the 2016 paper

Yashas

or Ligand Field Theory?

@blue you would get a three digit rank (less than 600) if you score 60%

Anonymous

@Yashas Well, I am not studying only NCERT :P

Yashas

in The h Bar, 8 mins ago, by rob

@Yashas Bad question. "Have an orbital" is different from "have that orbital filled when the atom is in the ground state". The second version requires you to know something interesting about how the periodic table is shaped. The first version is an annoying trick question, which you've answered correctly.

Yashas

12:57

How many electrons in a chromium atom at ground state have $n+l+m = 4$ and clockwise spin?

wth

another ambiguous question

the answer can be 2 or 3 or 4

so many stupid questions

@blue

Anonymous

13:10

@Yashas I think they want max no.

Anonymous

My net is sooooooooooooooo slow today

Anonymous

ugh

Anonymous

50 kbps

Yashas

I have 8Mbps today :d

50kbps or 50kBps?

Anonymous

Kbps

Anonymous

13:20

=/

Anonymous

This jio sim is shit

Anonymous

They give only 1 GB per day

Anonymous

Can't even buy more

Yashas

Are all metamers positional isomers?

Anonymous

@Yashas Don't think so

Anonymous

13:31

In position isomerism the principal carbon chain stays intact

Yashas

Anonymous

@Yashas Ah, in that the principal carbon chain is intact. Isn't it?

Madhuchhanda Mandal

@Yashas The doubt starts from there

Yashas

@blue do you treat the amine as a substituent?

Madhuchhanda Mandal

Should I discuss the doubt?

Anonymous

13:36

@Yashas IIRC NH2 has higher priority than alkyl

Anonymous

I'm checking it

Yashas

yea but it is a substituted amine

I remember writing N, N` - alkyl alkane

I think the amine is still a substituent

Madhuchhanda Mandal

Can I once discuss the doubt?

Yashas

yea

7 hours ago, by Madhuchhanda Mandal

If f'(x) =g(f(x)) and f'(x0)=0. Will the graph of f(x) be constant from x=x0 to x=infinity??

Anonymous

For secondary amines (of the form R-NH-R), the longest carbon chain attached to the nitrogen atom becomes the primary name of the amine; the other chain is prefixed as an alkyl group with location prefix given as an italic N: CH3NHCH2CH3 is N-methylethanamine. Tertiary amines (R-NR-R) are treated similarly: CH3CH2N(CH3)CH2CH2CH3 is N-ethyl-N-methylpropanamine. Again, the substituent groups are ordered alphabetically.

Yashas

13:39

The $g$ appears from nowhere :d

200 Puzzling problems in physics has weird questions.

If you have to balance a cube between two fingers, find the minimum value of coefficient of friction.

Madhuchhanda Mandal

g can be any function.

Yashas

The question does not give you any numbers but you end up with a number for $\mu$ lol

Madhuchhanda Mandal

That must come from properties of cube

Yashas

ye

Madhuchhanda Mandal

I think

Yashas

13:41

but it still gives me goosebumps

Madhuchhanda Mandal

That's why "weird" :P

Ok.. Lets get back to the question.

g can be any function

Yashas

f(x) has a critical point hmm

oh

it is an inflection point

Madhuchhanda Mandal

(Ok the doubt is created by me Lol)

Yashas

f`(x) = g(f(x))

differentiate both sides

Madhuchhanda Mandal

Ok then.

?

Yashas

13:43

f``(x) = f`(x) [g(f(x))]/[df(x)]

for x_0, f`(x) is zero

therefore, f``(x) is also zero

Madhuchhanda Mandal

Then?

Yashas

so x_0 is a point of inflection

it could locally be a straight line

Madhuchhanda Mandal

Why not globally?

I mean from x0 to infinite

Yashas

need to think

ok so

we differentiate f``(x) again

you get

f```(x) = f(x) [dg(f(x))] + g(f(x)) [f(x)]

at x_0, f```(x) is also zero

so it remains as an inflection point for ever

it should be a straight line

Madhuchhanda Mandal

Absolutely

Yashas

13:46

OMG no

wait

Madhuchhanda Mandal

Ok

Yashas

that does not prove it

Madhuchhanda Mandal

Let me say how I am thinking

Yashas

I need to show that ntn derivative is zero for x0

AHHAHA ]

nth derivative of f`(x) at x_0 is zero

so the slope never changes

you can pull the f`(x) outside

then apply the product rule

for the first time f`(x) kills the term (df/dx * (some derivative))

the second term will be f``(x)

Madhuchhanda Mandal

Or simply can't we proceed like this : f(x+dx)=f(x) so f'(x+dx)=f'(x)=0. Hence the samething loops : f(x+2dx)=f(x+dx) and f'(x+2dx)=f'(x+dx)=f'(x)=0...

Yashas

13:52

o0

Madhuchhanda Mandal

So f'(x) remains 0 forever from x0 to infinity

Yashas

f(x + dx) = f(x)?

how?

Madhuchhanda Mandal

f'(x)=0

Yashas

AHAA

Madhuchhanda Mandal

By definition of derivative

Yashas

13:52

I have some weird feeling that it is wrong

Madhuchhanda Mandal

Got it?

Yashas

I understood your logic but I think it is wrong

Madhuchhanda Mandal

Due to 2dx I think?

Yashas

I don't know why but it sounds wrong

the first step

take a parabola

say it has a maxima at x_0

Madhuchhanda Mandal

Then?

Yashas

13:53

f`(x_0) = 0

f(x_0 + dx) = f(x)

Madhuchhanda Mandal

Absolutely

Yashas

same logic

It proves that a parabola is a straight line

Madhuchhanda Mandal

Nope

You can't write the second step

f'(x+dx) not = f'(x)

Because f'(x) for a parabola is not g(f(x))

Its g(x)

I mean g(x,y)

Yashas

I am confused

brb in 15 mins

Madhuchhanda Mandal

So for a parabola although in vicinity of x0 , y is constant but x changes. So g(x,y) changes.. Hence & etc.

Ok. No problem

Anonymous

13:59

@Yashas Those two will be only position isomers and not metamers.

Anonymous

I checked my notes

Anonymous

14:29

@MadhuchhandaMandal Your claim doesn't seem correct. Check with $f(x)=\cos(x)$ and define $g(x)$ such that it is $\sqrt{1-x^2}$ when $f'(x)$ is positive and $-\sqrt{1-x^2}$ when $f'(x)$ is negative. $f'(0)=0$, but f won't be a constant function.

Madhuchhanda Mandal

(Please can you send the non- formatted (I mean non Latexed or so)) version?

I don't understand it

Anonymous

@MadhuchhandaMandal Your claim doesn't seem correct. Check with f(x)=cos(x) and define g(x) such that it is sqrt{1-x^2} when f'(x) is positive and -sqrt{1-x^2} when f'(x) is negative. f'(0)=0, but f won't be a constant function.

Madhuchhanda Mandal

Hmm. I see

Yashas

@MadhuchhandaMandal meta.stackexchange.com/a/220976

@blue It is assumed that the function is continuous and differentiable in the whole range. Otherwise, he wouldn't have taken the derivate for a general number.

Madhuchhanda Mandal

Cos x is continuous although

Anonymous

14:40

@Yashas cos(x) is continuous and differentiable in the whole range.

Yashas

g(x) isn't

Madhuchhanda Mandal

g(x) is continuous

I mean he is making so by defining two branches of g(x)

Yashas

it is not differentiable

Anonymous

Why does g need to be differentiable?

Anonymous

He took the derivative of f and not g

Yashas

14:43

my proof requires that it is differentiable

1 hour ago, by Yashas

f(x) has a critical point hmm

read through those msgs

Madhuchhanda Mandal

Wait... Yes I understand. Absolutely. You cannot define two branches

It needs to be differentiable

Anonymous

Can you explain why?

Anonymous

The transcript is too long to read

Madhuchhanda Mandal

Assume that g(f(x)) is a non-branched function

I mean a single function

Anonymous

g(f(x)) is a non-branched function in the way I defined

Anonymous

14:49

g is branched function

Madhuchhanda Mandal

Yes. Assume g(f(x)) to be non-branched

Anonymous

Ok. So say f(x)=cos(x) and g be sqrt(1-x^2) or -sqrt(1-x^2) so that g(f(x))=-sin(x)

Anonymous

-sin(x) is non-branched

Madhuchhanda Mandal

"or"

Anonymous

g is branched

Madhuchhanda Mandal

14:51

Ok. Let me see

Anonymous

oh wait

Anonymous

ok, go on

Madhuchhanda Mandal

So can we conclude?

Anonymous

@MadhuchhandaMandal Conclude what?

Madhuchhanda Mandal

Constancy

Anonymous

14:53

I still don't see why g has to be differentiable

Yashas

if g is diff-able then it is constant

Madhuchhanda Mandal

The fundamental condition is that : g is a non-branched function. Nothing more. Can we conclude then?

Yashas

nope

Madhuchhanda Mandal

Ok.. Why?

Yashas

blue gave an example of such a case

Madhuchhanda Mandal

14:56

That was branched

Anonymous

@Yashas Can you show your proof?

Yashas

ok

8 hours ago, by Madhuchhanda Mandal

If f'(x) =g(f(x)) and f'(x0)=0. Will the graph of f(x) be constant from x=x0 to x=infinity??

Observations:

1. f(x) has a critical point at 'a'

Anonymous

what is a?

Yashas

some constant

Anonymous

ok

Yashas

14:57

where f`(x) goes to zero

Let f, g and g(f(x)) be differentiable

f`(x) = g(f(x))

differentiate both sides

f``(x) = f`(x) * [dg(f(x))/d(f(x))]

Madhuchhanda Mandal

Why my argument don't hold?

Yashas

As f`(x) is zero, f``(x) is zero

let us differentiate f``(x)

Transcript for