Problem Solving Strategies

Koolman

02:47

How they have written the first equation .

Yashas

05:08

Guess the function

Anonymous

05:40

@Yashas $cx|\sec(kx)|$

Anonymous

$c$ and $k$ are constants.

Yashas

not bad

Anonymous

:D

Yashas

Your's is a bit steeper

Anonymous

Adjust $c$

Yashas

05:43

Anonymous

I was close :P

Yashas

ye

Anonymous

Your function was not guessable

Yashas

I did not want you to guess the exact function

If you notice the graph carefully, you will see that the line joining the maximum and minimum is x = y

slope = 1

first approximation = x

as the curve goes to infinity periodically, you need a term which can do that

it can be tan or sec or cot or cosec or .... or you need a zero in the denominator

Anonymous

yeah :)

Yashas

05:54

The first infinity happens near 3pi/4

~4.7

It is hard lol :P

JEE APEX BOARD SECRETARIAT
JOINT ENTRANCE EXAMINATION (MAIN) - 2017 RESULT

Enter your Roll Number
Enter Date of Birth
( Type DOB in dd/mm/yyyy format or Click icon )

Note: Other than Indian National/OCI/PIO are not allowed to fill choices online for counseling in NITs/IIITs/GFTIs because there is a separate DASA channel for Foreign National candidates.

Anonymous

Where???????

Yashas

I was joking :D

Anonymous

-_-

Anonymous

I'm very annoyed already

Yashas

Sorry :P

Solve this integral

$$\int_0^{\frac{\pi}{3}} \frac{x}{1 + \sec x}$$

jeeadv.ac.in

lol

Anonymous

06:05

1/18(-2sqrt(3)pi + pi^2) + log(4/3) ?

Anonymous

@Yashas ??

Yashas

yes

what did you do to get the answer in such a bad form lol?

Anonymous

yes :D

Anonymous

@Yashas Indefinite...

Yashas

did u use sec x as 1/cos and then use tan x/2 substitution?

or did u do parts?

I used parts

Anonymous

06:10

@Yashas Nope. Just (1+cos(x)-1)/(1+cos(x))

Yashas

integrated 1/(1 + secx)

Anonymous

And then by parts

Anonymous

yes

Yashas

I think tan x/2 substitution won't work

sec ^2 comes in the denom

Anonymous

It works..it works..

Anonymous

06:11

Check

Anonymous

It always works in this form

Yashas

mby I shud do it on paper :P

Anonymous

Give me a tougher one :P

Yashas

I wanted to know if there was a shorter way

I did using indefinite too

Anonymous

No idea....

Anonymous

06:19

It's pi/3

Yashas

yea

pi/2 is nice :P\

Anonymous

For such cases even King's rule doesn't work

Yashas

*pi

what's King's rule? o0

Anonymous

f(x)=f(a+b-x)

Anonymous

that one

Yashas

06:19

.

why are obvious stuff given names -_-\

Anonymous

Queen is f(2a-x)=f(x)

Yashas

...

Anonymous

And the periodicity rules are Jack rules

Yashas

.......

Is there a pesant's rule?

minister's rule?

Anonymous

Joker's rule is indefinite integration :P

Anonymous

06:21

@Yashas Let's say by parts

Yashas

.................

Anonymous

@Yashas Substitution is peasant's rule :P

Anonymous

(I'm making them up)

Yashas

$\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x + \cos x}$

King's rule

but I am getting the wrong answer

Do you get? $$2I = \frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x}$$

ATHARVA

@Yashas @blue which books are you using (have used) for jee advanced for maths. Is rd sharma (objective mathematics , 2 volumes) up to that level?

Yashas

06:28

R.D. Sharma JEE book is good for complex numbers

Anonymous

@ATHARVA I didn't use any book. Coaching notes or Etoos notes suffice along with study material

Yashas

For trig I use S L Loney

For coordinate geo, I use S L Loney

Anonymous

R.D. Sharma is far easier...

Yashas

calculus Edwards

algebra? uhm... do you really need?

I used Higher Algebra

but that is minimal

Anonymous

@Yashas seems fine

Yashas

06:30

$$2I = -2\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{dt}{t^2 -2t - 1}$$

$t = \tan \frac{x}{2}$

Anonymous

wtf...

Anonymous

change the limits

Yashas

ignore the limits

I later substitute t back and use limits for x

the rest is correct, right?

Anonymous

i think so

Anonymous

i'm lazy to check on paper

Yashas

06:33

$$2I = -2\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{dt}{(t - 1)^2 - 2}$$

Anonymous

it should give correct answer

Yashas

ignore limits again

Anonymous

then?

Yashas

$$2I = -2\frac{\pi}{2} \left[\frac{1}{2\sqrt{2}} \ln\left(\frac{t - 1 + \sqrt{2}}{\sqrt{2} - t + 1}\right)\right]$$

4 mins ago, by Yashas

$t = \tan \frac{x}{2}$

at x = pi/2, t = 1

at x = 0, t = 0

Anonymous

That thingy inside ln might be changing sign in 0 to pi/2

Anonymous

06:35

check once

Yashas

Hmm, intersting

I ignored the | | for ln

Anonymous

Does it?

Yashas

that cud be causing rpoblems

let me see

Anonymous

No it doesn't

Anonymous

It's positive throughout

ATHARVA

06:38

@Yashas @blue thanks! Is ML Khanna good? And please can you tell which books you have used for chemistry?

Yashas

@ATHARVA J.D. Lee for Inorganic Chem, Ranjeet Shai + Morrison Boyd for Organic, Whiley Publications Physical Chem book

I also had Resonance study material

Anonymous

@ATHARVA Etoos study material only

Anonymous

And NCERT

Anonymous

ML Khanna is faltu

Yashas

omg

I took sqrt(2) - 1/sqrt(2) + 1 as 1 -_-

I was getting 0 - 0 all this time

Anonymous

06:40

:P

Yashas

30 mins wasted

.....

Anonymous

What would you do if that ln changed sign?

Yashas

still the wrong answer

@blue that minus is wrong :P

Anonymous

@Yashas Na, I'm just asking

Yashas

you get -t^2 in the denom

Anonymous

06:43

In such cases how would you write it?

Yashas

I wanted to make it positive

so I multiplied and divided by -1

I get 50% of the problems wrong when I try to complete the square with -t^2

:P

Anonymous

Suppose $\ln\frac{x-a}{x+a}$ changed sign at x=b when integrated from x=c to x=d how would you write the final answer?

Yashas

oh you were asking that lol

I thought you were talking about how the minus came out in my integral

Anonymous

nope

Yashas

integrate from c to b

then b to d

the integral of 1/x is ln |x|

Anonymous

06:45

Good :)

Yashas

uh... u were testing me....huh

how do u remember the integral for 1/(x^2 - a^2) and 1/(a^2 - x^2)

I have a weird way to remember

ATHARVA

Thanks a lot ! @Yashas @blue . All the best for you JEE mains result :)

Yashas

arctanh is $\frac{1}{2} \ln \left(|\frac{1 + x}{1 - x}\right|)$

the range of tanh is -1 to 1

take it as 0 to 1

it is less than 1 :p

so it must be 1 - x

Anonymous

@Yashas They are the same... with a - sign :P

Yashas

you can switch easily using the minus sign

but how do you remember at least one of them?

if u know one of them, u get the other one

Anonymous

06:57

I remember just like that :P

Anonymous

It takes 5 seconds to derive it

Anonymous

1/(x-a)(x+a)

Anonymous

Partial fraction

Anonymous

I don't even need to remember

Anonymous

@Yashas

Yashas

06:58

I remember it too but I forget after a few months

until I revise again

Anonymous

@Yashas What is there to remember?

Anonymous

It is just partial fraction

Anonymous

Takes 5 seconds to derive

Yashas

I take 10-20 seconds

Anonymous

I can clearly see 1/(x-a)-1/(x+a) written

Anonymous

07:01

I can mentally derive it

Madhuchhanda Mandal

07:45

@Yashas @blue @Koolman Would you like a mechanics problem as a time challenge?

I will give all a pulley-block system. Let's see who can solve fastest and in what time

Anonymous

I'm very busy now. But go ahead. Can't make any promises.

Madhuchhanda Mandal

Nope.. That won't help. You have to participate. Its a game type. If you are busy then how can you take part.

Anonymous

Sorry then...

Madhuchhanda Mandal

It happens.. No prob.

Yashas

People forcing me to take CET -_-

Anonymous

07:56

f*ck

Anonymous

I hate state exams

Anonymous

I was forced to give WBJEE :/

Yashas

2 days are going to be wasted

Anonymous

08:06

What is equivalent weight of CO2 ?

Anonymous

@Yashas

Yashas

depends on the reaction?

Anonymous

Metal Carbonate to CO2

Yashas

Ranks are out

cbseresults.nic.in

JEE Main page trolled me

nope

Anonymous

Where??/

Anonymous

08:08

Can't see any rank

Yashas

jeemain.nic.in/webinfo/Public/Home.aspx

Anonymous

-_- It redirects to the cbse page

Anonymous

What nonsense

Yashas

lol

and the CBSE page does not work

Anonymous

Wht abt the CO2 ?

Yashas

08:09

It was CO3 2-

3 equivalent resonance structures

Anonymous

CO3(2-) to CO2

Yashas

the oxidation state of C did not change

Anonymous

What is the equivalent wt of CO2 ?

Yashas

O's state did not change either

Anonymous

Yeah, that's why I'm asking

Yashas

08:10

you can't use equivalent weights here

Anonymous

Oh

Anonymous

I see

Yashas

there must be a compound whose oxidation state changed

Anonymous

I need to revise this chapter

Yashas

*atom

You can take it as -2 though

Anonymous

08:11

Gotcha

Yashas

I do it

Anonymous

@Yashas What? How?

Yashas

nah u can't

LOL

cbseresults.nic.in/jee_main/Jee_cbse_2017.htm

from last year's link

I guessed the link

Brought to you by National Informatics Centre

JEE APEX BOARD SECRETARIAT
JOINT ENTRANCE EXAMINATION (MAIN) - 2017 RESULT
Access denied, pl. click here to visit the site http://results.gov.in

Anonymous

Does it work?

Yashas

nope

Access Denied

Yashas

08:47

they updated @blue

:(

my rank is worse than last year -__________________________-

Kunal Pawar

Is it working for sure lol?

Yashas

yes

Kunal Pawar

Pakka you aren't trolling me? I'm sick of refreshing the page -_-

Yashas

I am not trolling

cbseresults.nic.in/jee_main_zxc/Jee_cbse_2017.htm

Kunal Pawar

Okay thanks

Yashas

08:54

my rank doubled wth

4 marks bonus?

Kunal Pawar

I'm in no place to say this but lol

What's the cutoff?

Yashas

81

Kunal Pawar

Isn't that too low

Yashas

ye

Kunal Pawar

Like too too low:|

Everyone was saying it's gonna be high this time

Transcript for