@Ishan which book

@samjoe glad that you're not wasting time like I did a Year ago

@Jasmine That GRB book by Sameer Bansal

@Ishan can you suggest me a good book for calculus for JEE

Our teacher finished with the 9 important gormulae of integration in the 1st class.

Oh god had not i learned little on my own i would have gone blank

He was so fast that i could copy little that he wrote

@Jasmine Depends on what type of book you're asking for. I used to use the Arihant one by Amit Agarwal for Calculus for learning tricks needed for JEE problems. However I read other books too on Calculus which helped in letting me go beyond what was mentioned in the Arihant book. If you're looking for a book with only very hard questions, the Sameer Bansal book is the best. However, it is folly to use the Sameer Bansal book unless you are very, very, very good.

@Ishan i am using calculus by vikas rahi

Ia that ok

For practice, I would first solve the worksheets given, then go to archives, then solved examples after each chapter in the Arihant book (these solved examples were much harder than the illustration problems in the chapter). And finally, I would use the Sameer Bansal book

@Jasmine Sorry, don't know about that book

@Ishan i badly want to reach a good level in integration i dont know what to do should i go and buy sameer bansal and amit agarwal book

@Jasmine Buy the Amit Agarwal book first and see if you can solve all the illustration problems (they are quite good for learning tricks used in the JEE). And then you could follow the order I suggested. I don't really think you'd get much time left after solving all these. If you do get time, then move to Sameer Bansal's book.

@Ishan ok but i also bought calculus by vikas rahi recently but i can see in stores still how is amit agarwal

Having said all this, let me add that I by no means am an expert in all this. The only difference between you all and me and Samjoe is that we've gone through the process, and that you all are going through the process. The best persons to ask are your professors; they would know your levels better, and they have much, much, much more experience than Samjoe and me

@Ishan exactly

My teacher told us go for vikas rahi

He said that if you need a book for tough questions then go for calculus by amarnath anand

@Jasmine Hmmmm... Then stay with it unless you feel that the book is too simple for you, and that it isn't really helping you solve harder JEE level questions.

@Jasmine Again, haven't heard of this one.

I like books which contain tricky question and not lenghthy or out of syllabus stuffs

@Sid just now I was doodling on ms paint :) heres the masterpiece

I will like you all FIITJEE students to go for GMP it is awesome it os not very hard but very idealistic for tricky questions

@samjoe i feel like Mr. sun is trying to ignore newton who is just going to discover gravity with his friend

lol nice interpretation, its just a random sketch :P

@samjoe Wow, it seems @Jasmine is absolutely correct:D

But still its great work @samjoe

@jas thanks! I was getting bored :-)

@samjoe Sun looks like someone about to commit some evil deed

@Jasmine you meant "ideal", I believe.

@Sid oh yes 'ideal' i know these days not reading english

@samjoe But, at least you have started reading calculus. At this time, last year, I was doing crosswords and Sudoku.

You are utilizing your time much better than me

@Sid do not forget that I am one year behind :P

@Ishan You there?

Yeah, for the moment

Got to go soon though

@Abcd

@Ishan Please stay for only 5 minutes

Okay...

@Abcd

@Ishan you can ping in one message itself

@Ishan The question isnt printed properly. Please tell me whats the question's equation

Not sure, I think this is what it means though:

$f(x)=|(x^2+(a-2)\times|x| -2a)|$

I see. Okay.

@Abcd Anything else?

@Ishan then it doesnt fit the question

@Abcd Doesn't fit the question as in?

@Ishan Nevermind

@Abcd Okay...

thanks, that's it. Had a question though but you gotta go so...later

@Abcd first break into two parts, analyse for $x\ge 0$ and $x< 0$. For both quadratics, you need dicriminant must be positive, and roots of same sign

@samjoe Wont read, NO HINTS NEEDED

@Abcd What's the question? I'm going to study Maths itself, so I might have a shot.

@Abcd oops sorry

@Ishan this one

Try breaking the sum into partial sums. Might work

Yeah but first do long division perhaps

@samjoe No need for doing long division first.

$4^4r^4=[(2r+1)+(2r-1)]^4$

@samjoe I dont think the question's equation is correct. (Previous question of quadratic one)

There can be at max 3 points of differentiability in worst case

@Abcd You need hints?

@samjoe No, just tell me is the question's equation correct.

Seems correct to me

its $|(|x|-2)(|x|+a)|$

@Abcd yeah

@samjoe Samjoe please give my graph a look and give nothing more than a Very SMALL hint, please.

@Ishan $\sum_{r=1}^{n} \frac{r^4}{(2r-1)(2r+1)} = \sum_{r=1}^{n} \frac{r^2}{4}+\sum_{r=1}^{n}\frac{1}{16(2r-1)(2r+1)}$ where second one telescopes easily

@samjoe I cant see more than 3 points of non differentiability

And that is because a has to be negative

for te quadratic to have two roots

@Abcd Remember that you have to plot a graphs for $|x|$

@abcd think what happens when $a<0$

@samjoe Agreed. But for me polynomial division is quite tiresome, so I usually use it as the last resort, especially since both our methods will yield the same result.

@samjoe a has to be less than 0 for f(x) two have two roots in this case

@Abcd so mate for $x>0$ your quadratic should have 2 roots for $a<0, a\neq -2$ but you have shown only one

And from there take reflection about y axis to get corresponding graph for $x<0$

@samjoe Silly mistake, i was marking the root to be less than 0 while root is greater than 0 and a is less than 0

@Ishan I saw your expression but how would we proceed from there

@samjoe Why cant -a exceed 2 ?

@Abcd it can but just that $a\neq -2$ because if $a=-2$ then theres only one nondifferentiable point

@samjoe I know but answer given is 2.

I think there are some conditions missing in the question

@Abcd i think i can solve it but my method is very lenghthy.

@Jasmine tell me your answer

Seems like you are discussing some other

@Abcd did not work it to the answer yet

@Abcd idk then dude.. I was thinking same as you got, $a<0, a\neq-2$

@samjoe i went to my sir with that infinte critical points question he gave me the weirdest answer

@samjoe I have one more

@Jasmine what did he say

Why is calculus so magnetic?

Just see the right page

@Abcd will try this one

Oh neck pain?

@samjoe he justified 4 critical points by this

He still told be critical points in function is where the graph changes definition and in aod its wheref'x is not defined or $0$

Do you understand what he said

@Jasmine Hmm its just creating more name conventions.. may lead to confusion, never came across such nomenclature

@samjoe did you solve

Most standard definition is simply $f'(x)=0$ or undefined. thats it

@samjoe exactly

@Abcd havent yet, Ill need to try again

Are all those hard questions from GRB?

@Jasmine hmm

@abcd you there?

By induction its shown that $a_r = 2r^2 +r-3$ for $ r\in \Bbb N$

Limit follows easily as $\frac{\sum_{r=0}^{10}4^r}{\sum_{r=0}^{10}2^r}$ by collecting only the $n^2$ coefficients as $n\to \infty$.. so limit is $683$

@samjoe you there now?

@abcd when do you sleep!! its so late :P

@samjoe $$\text{floor}\left(\dfrac{23- 2^{-(x+1)(x-3)}}{\mu}\right)$$ is discontinuous for at least one real value of x for what $\mu$ s

@samjoe I have tthought about this wholllle day

Not getting.

@samjoe Please dont give anything more than a small hint

No dude its not $2^{(x+1)-(x-3)}$ ! you cant cancel

Oh damn

@Jasmine @Sid You have any idea how to do it?

Yeah but my method too lenghty and not at all good

@Jasmine this new problem: (not that r^4 one)

The question is, that ^?

@Sid yes

I just dont get how it can be discontinuous.

continuous+continuous= continuous and we have continuous functions in numerator

Someone please help me do it

or else I cant study chemistry

@Abcd but how can you say no

@abcd wait! suppose we ensure that argument of floor is always between (0,1) then?

@samjoe then it will give 0 only

so this is the range to be excluded from all R

similarly exclude the range for which argument of floor is in (-1,0)

F(g(x) will be discontinuous at all those points where f(x) is discontinuous and g(x) is discontinuous and it's of the form fogx where f(x) is floor x

@Jasmine but g(x) is continuous

Yes the answer is zero then.

@samjoe No, the answer is 253

@Abcd but f(x) is discontinuous

@Jasmine read what you wrote, you said both g and f should be discontinuous

How are you saying 0!

Hmm. I am running into a rock with that question

It is deceptively hard

Because equal number of integers for positive and negative, so I guess sum may be zero, but now I need to check

@Sid I have given it evrything I could today, I thought about it during physics lecture, on my way back home, and what not

Welll I didn't mean the f(x) or g(x) in original question I mean the one you posted before

@Abcd ask your teacher

@Sid we didnt have maths today

@Abcd here I meant the inner one to be g(x) and outer to be f(x)

sorry didnt read $\mu \in \Bbb N$

thats why I got zero lol

Please I want to do it before sleeping

Its $\sum_{i=1}^{22}i$

@samjoe is it regarding abcd question regarding finding A+B+C+D

@Jasmine no its some other questin of limit only

@samjoe which

@samjoe How in the fresh hell do you get that?

@samjoe thats the right answerr you GENIUSSSSSSSSSSSSSSSSSS

@Abcd there's a solution manual for the book

@samjoe telll me how you did ittttttttttt

That would save a lot of time

@sid the function given by abcd is nice one, it has supermum of 23

@AvnishKabaj man if I buy it I would be tempted to see every solution. I dont wanna do that. Right now at least I am able to solv 60-70%

@Abcd ¯\_(ツ)_/¯

i mean the numerator function

and numerator is always positive also

@samjoe What on earth is supermum now?

SO for any number $\mu \ge 23$ the whole function lies in $(0,1)$

@Sid xD I meant supremum :x

@samjoe Can you please explain what you are doing?

Please? i have to sleep and study chem in the morning at any cost

@LeakyNun HI!!

First consider the numerator function and find its maximum value

Oh. Of course!!! I get it.

That's beautiful

@samjoe its 7

?

Youll get 23 but function always lies below it

@LeakyNun Please see this question

@Abcd $x\to \infty$ ?

That's a magnificent question

Yeah if we consider negative values then we get zero

as the sum

@Abcd no I mean if either or both

@samjoe Okay, then?

@LeakyNun Did you see

desmos.com/calculator/2qqcbopkxa

@Abcd If mu is greater than 23, then the whole function is <1. True?

@abcd if you divide a number less than $23$ by a number greater or equal to $23$ you get a number in $(0,1)$, that means these numbers cant be our answer

@Sid yes, then?

I think Sam has pretty much shown the answer.

For any mu greater than 23, the function will always be continuous.

Thus, discontinuity occurs at values of mu less than 23

@Sid why?

The range of $f(x)$ is $[7,23)$

@LeakyNun yes, then?

@Sid I mean like [cos x] is continuous at x=pi

Someone please replyy

so when $\mu \ge 23$, the range is contained inside $[0,1)$, so we can safely exclude every $\mu$ greater than or equal to 23

@LeakyNun yes, I got that part, then?

when $8 \le \mu \le 22$, $\frac7\mu < 1$ and $\frac{23}\mu > 1$, so they are included

dint get

when $1 \le \mu \le 7$, $\frac{23-7}\mu > 2$, so they are included (an interval of length 2 must contain an integer)

where [] denotes floor.

@abcd hes checking whether all naturals till 22 satisfy or not

@Abcd well, $\frac7\mu < 1$, so $g$ is 0 at some point; $\frac{23}\mu > 1$, so $g$ is 1 at some point

@LeakyNun This is the thing that is troubling me man

ok, [cos x] is continuous at x=pi

@LeakyNun exactly thats what is troubling me

floor maps $[-1,0)$ to $-1$, so it is continuous

@abcd we did discuss it that day, cos(x) lies in (0,-1) for both neighbourhood of $x=\pi$

@samjoe so thats what quadratic function will do as welll

sorry $(-1,0)$

@Abcd do we agree that the range of $f(x)$ is $[7,23)$?

@LeakyNun yes yes yes

let's say $\mu = 16$ for an example

then range of $\frac{f(x)}\mu$ is $[0.4375,1.4375)$

so floor maps it to 0 and 1

in particular, 0.4375 goes to 0, and 1.4 goes to 1

so it is discontinuous

@LeakyNun Could you tell me a particular point of discontinuity.

desmos.com/calculator/caysu4nr3m

my problem is that for quadratic we have f(a+)= f(a-) for any a

So how can it just be discontinuous

somewhere around 2.092 and somewhere around -0.092

@LeakyNun okay, please address my problem stated above^

I don't know what you mean by a+ and a-

@LeakyNun limit of f(x) as x to a+ and ...

but floor is not a continuous function

@LeakyNun so?

$\displaystyle \lim_{x \to 3^-} \operatorname{floor}(x) = 2$

$\displaystyle \lim_{x \to 3^+} \operatorname{floor}(x) = 3$

i know

I still cannot identify what you're asking

@LeakyNun okay do one thing put x^2 inside floor

done

oh

so x^2 gives different floor value for a- and a+

thats where I was misinterpreting

got it

ok...

@LeakyNun Thanks for replying. Thanks @samjoe . Bye.