Sid

@Aditya

i messaged

check whatsapp

screw it.

@Aditya

got it

delete it

@Aditya i dont get notifications in this chat thats why im asking

i'll tell u when i recieve it.. u delete it

nope

@Aditya

theres a whatsapp group

they can tell u what to do... a roadmap

and i am one myself

i know a community of iitians and nitians

phone number

send ur #.

ok i thinik we can chat privately here

are u in 12th? my friend has this group to help aspirants... the group has iitians and nitians

ok dude i want to ask ur # but is this a secure place to do so? :P

but nice qstn dude... where is this from? jee mains?

Uhm finding those limits was my doubt. I already checked Wolfram etc and also graphed the given equation. I asked it here thinking I will get a solution (without online calculator).

Anyways, thanks for the help, I'll wait for someone to solve the qstn

Ok but how do u get those limits without a calculator. That's the qstn in hand here

U are not allowed to use Wolfram alpha or any calculator for that matter! Look, if you know the answer post it, if u don't let it be, but please stop spamming, it's annoying.

And it can't be done mentally. Look at that graph above. It increases, decreases and increases. In the region f(3/4)< f(x) < max(f(x)) it attains the same value 4 times in the interval (0.25,1.25). It's not obvious as just putting values of x and claiming those are the only points it's equal to 2√2

No in ur original answer u said some granting. And put x=45 u get 2√2... I agree u do, but where's the proof those are the only solutions. U need to prove that.

this has nothing to do with what im asking. my question boils down to: can $(sinx-cosx)(2+sinxcosx) \leq 2$ be solved in a more intuitive way that what the have provided. Thats all. no calculator, no wolfram, nothing. Please help in this regard if possible.

the result is 3:1... the guy above also confirms it...@Raffaele

yep and how are u sure these are the only 2 value ranges

and no the value ranges are $x \in [0.25, 0.5] \cup [1,1.25]$

see but, just because the function = $2\sqrt2$ when x = 1/4, 1 (by trying out values) doesn't mean those are the only values between [0,2]... we need to prove that those two are in fact the only two values... because for example, the value $f(3\pi/4)$ is attained thrice in the range... look at the graphical solution the other person provided... (which is also not allowed cuz u would need a calculator.), so its not obvious that $f(x) = 2\sqrt2$ has only two solution in the range $x \in [0,2]$

I told in the problem description, no calculator allowed... It's a test qstn... Sorry for the inconvenience.

I can't just put random values. I need a systematic method. And How can u gaurentee f(x)>2√2 in [0.5,1]?

Ya that was my doubt... Does the divisor have to divide all the possible values of the sum??

@mathlove is there still something wrong with the wording of the question? You yourself said if the or is switched to with then the qstn makes sense... who is this person saying like this??

Uhm the question doesn't look like anything how I'd written it... Someone has already eddited it. Check the history

And dude. I asked @mathlove wouldn't the values of x in the AP affect the set S, he said no my mathematical reasoning was wrong. Just take some time to read it and you'll get the answer. If the question is written like this in the test what give me the right to change it???

But ur wrong abt S... Read the comment just above your allegation...

@mathworker21... Read the comments completely... I had the same doubt as you about S... But look what mathlove wrote just about your comment.

@mathlove Consider this simpler statement: x = 1 with S = {x:x<2}, will S = (-$\infty$,2) or will S be solely 1? Because x is already locked to be = 1. Am I wrong?

@mathlove... If or is replaced with 'with' then S won't be those 5 terms because x is restricted to be only 5/3 and 2, so shouldn't S={2}?? If I'm wrong, correct me

@fleablood even if this word can be means that we can take t3/t1 be those functions of x, isn't the solution still wrong? because to find the 3|S| terms of the AP, arn't we restricting the values of x to be only those values satisfying the AP and in that case S ={2}, not those other 4 integers as those values of x dont satisfy the AP? Again if the word can be is being misused, please correct me...

@fleablood Ok now I understand... How can we substitute those values for t1 and t3 if the first condition(AP) need not be true... But one last thing... sir said: "read the last two words of the question, it's written can be, so he said there is a possibility that it is true, and we can substitute... But that leads to another confusion. If we do substitute, arn't we restricting x to be only those 2 integer values of x(namely: 2,5/3)... Please correct if the word can be is being misused...

@fleablood: he said you are right it need not be true: but he said the words can be are key to the qstn... :/, Please clear this up. Thanks

I argued that: so your question is saying either that AP is true or S is true but sir said no: or need not mean you have to choose only one. You can chose both also. He said a+b= true means either a is true or b is true or both are true... and he said indepently find the values of x that satisfy the AP and then find the set S... Im confused

Yes yes... I'm confused. Please explain clearly why it's wrong so I can argue

@fleablood why is S=Z if t3/t1<=2 ... Only 5 integers satisfy this inequality (as provided in their solution)

Wait I'm confused... How is S=Z if t3/t1<=2? We must satisfy t3/t1<=2 and extract the integer solutions... That's what I understood from S...

jee advaced anyone?