JEE Maths Zone - 2018-06-18 (page 1 of 2)

Abcd

2:55 AM

@IceInkberry why dont you ask a follow up question on main

Abcd

3:52 AM

@LeakyNun Are you there?

Leaky Nun

a bit busy

Abcd

fine.

Abcd

4:20 AM

@Ishan @samjoe @Sid Help needed.

samjoe

4:55 AM

@Abcd dude you just post the question

Abcd

2

Q: How to find the finite limit of this function?

Let $f(x) = \dfrac{1-\cos \{x\}}{(x^4 + ax^3 +bx^2 +cx)^2}$. If $l= \lim_{x\to 1^+}f(x), m = \lim_{x\to 2^+}f(x) $ and $n= \lim_{x\to 3^+}f(x),$ where $l,m$ and $n$ non-zero finite then: $a+b+c=? $ $l+m+n=?$ $\lim_{x\to 0^+}f(x)=? $ where {} denotes the fractional part functi...

samjoe

@Abcd for a start write $\{x\}$ in different intervals, like in $(1,2)$ its $x-1$

Anonymous

@Abcd Did. Yesterday.

Abcd

@IceInkberry I know. +1ed

Anonymous

1

Q: Flaw in evaluating limit $\lim_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x$

There was this question asked yesterday here: Link: Evaluating limit $\lim_{x\to \infty}\left(\dfrac{P(x)}{5(x-1)}\right)^x$ Consider $P(x)= ax^2+bx+c$ where $a,b,c \in \mathbb R$ and $P(2)=9$. Let $\alpha$ and $\beta$ be the roots of the equation $P(x)=0$. If $\alpha \to \infty$ an...

calculus algebra-precalculus limits functions limits-without-lhopital

Anonymous

5:06 AM

Okay

samjoe

yesterday, by samjoe

@Ice Hmm, now I think This can never happen with quadratic. And also in general with any polynomial so that question has strange phrasing when it says $\alpha \to \infty$

I agree with gimusi

Anonymous

Strange

samjoe

@Abcd for non zero finite condition on limits you need $x=1$ to be double root of the quartic on denominator

Abcd

@samjoe I dont get you.

samjoe

Hmm write $\cos\{x\} $ as $\cos(x-1)$ for $x\in(1,2)$ for the limit $x\to {1^+}$..

Anonymous

5:16 AM

And then you can use cos expansion

samjoe

or $1-\cos(x) = 2\sin^2(x/2)$

Abcd

@IceInkberry din help

@samjoe @Ice I hope you both agree that $\lim_{x\to 0} \sqrt x$ doesn't exist.

Do you?

Please reply a little fast.

samjoe

No

Abcd

@samjoe What????? Function isn't even defined for 0-

thats why limit doesnt exist.

samjoe

one sided limit has to be considered, as domain is $[0,\infty)$

Anonymous

5:19 AM

Agree with samjoe

Abcd

@samjoe Well, where have you read that rule?

Anonymous

We consider limits in their domains

Abcd

Our sir (FIITJEE Maths HOD) said it doesn't exist.

27 secs ago, by Abcd

@samjoe Well, where have you read that rule?

@Ice^

Anonymous

I mean, we can state that: The limit exists in it's domain

Abcd

@Ice Please cite your reference.

Reference = some standard book.

Anonymous

5:21 AM

6

Q: Limit of $\sqrt x$ as $x$ approaches $0$

What is the limit of $\sqrt x$ as $x$ approaches $0$? I asked a few people and they all gave me two different answers. Some said that the limit is $0$, and other say that the limit is not defined because the right-hand limit is $0$ while the left-hand limit is undefined. I'm confused. please help!!

calculus algebra-precalculus limits radicals

samjoe

@Abcd will ncert do xD

Anonymous

I am searching it in my book as well. But I think SE reference will do.

Abcd

@Ice @sam Our sir even said that he had a debate over this with the person interviewing him for fiitjee teaching

Anonymous

Our teacher had told us that

Abcd

Now what do I do :O

Anonymous

5:23 AM

This is funny :P

Abcd

And the two people had a debate like "you don't know maths"

Anonymous

Trash~~ Not uploading

Abcd

@IceInkberry I know and my sir was so confident :O

samjoe

Generally we only check one sided limits at end points.. but you dont have to debate over this with the teacher

It doesnt make sense (to me) to find limits at points which are not even in domain

Sid

@Abcd Yeah?

Anonymous

5:27 AM

Reference of our little books isn't required, in the SE answer some have mentioned their references

Abcd

32 mins ago, by Abcd

2

Q: How to find the finite limit of this function?

Let $f(x) = \dfrac{1-\cos \{x\}}{(x^4 + ax^3 +bx^2 +cx)^2}$. If $l= \lim_{x\to 1^+}f(x), m = \lim_{x\to 2^+}f(x) $ and $n= \lim_{x\to 3^+}f(x),$ where $l,m$ and $n$ non-zero finite then: $a+b+c=? $ $l+m+n=?$ $\lim_{x\to 0^+}f(x)=? $ where {} denotes the fractional part functi...

calculus limits

Sid

(Sorry, I was asleep. Couldn't see the message earlier)

Abcd

@Sid ^

I dont even have a friend in my batch with whom I can share this link of $\lim_{x\to 0} \sqrt x$:/

So will sir keep teaching wrong thing every year @samjoe

samjoe

I don't know, wish you luck

Abcd

Okay, I know the thing at least....that's what matters at the end.

Anonymous

5:29 AM

You can just look up for controversial stuff, when you feel.

Anonymous

@Abcd Is your batch that big? :0

Anonymous

Or too small?

Abcd

@IceInkberry No, 25 students

Anonymous

@Abcd Same. 24.

Abcd

@IceInkberry are you in top batch of allen

samjoe

5:31 AM

@Abcd Is there a flaw in question? just asking, because I think we will need denominator to have $x=1,2,3$ as roots with 2 multiplicities each, not possible for quartic.. let me check again

Anonymous

@Abcd Nah :P Not eligible to be :P

Abcd

@IceInkberry They dont conduct some "move to top batch" kinda tests?

Anonymous

@Abcd Allen has Kohinoor(top most; almost full scoring students) and average batches. There is Special rankers batch(I am in it), but we don't have other lectures. We sit with our normal classes. The only thing is, we get extra tests and extra things to do and teachers give us more things/teach us some more, keep our tracks, etc.

samjoe

Oops sorry didnt see the square in denominator :\

all cool so denominator has roots $1,2,3$ :)

Abcd

@IceInkberry wow Special rankers ;)

Anonymous

5:37 AM

@Abcd But that doesn't really matter.

Anonymous

We get deadlines like: Assignment given at 8 pm, submit by 9 am tomorrow. When I have planned to so something, I get something else.

Anonymous

I am wondering, what about FIITJEE? Top batch?

Sid

@Abcd don't think I can help you, sorry. Since, the only ideas that come to my mind are Taylor's expansion and sin 1/2 rule.The former isn't working, you said and the latter has already been suggested by Ice and Sam

Anonymous

@Sid Now that I think of the question, I wonder why the limit isn't zero.

Abcd

@IceInkberry Yes but Of classroom program ....but the top rankers are mostly from Pinnacle (=Integrated school program batch) ...so....

samjoe

5:41 AM

arre the root of quartic are $1,2,3$ are you even listening @abcd

Abcd

@samjoe I am not able to follow you. Could you please explain what you are tryna do

Sid

@IceInkberry limit when x is tending to what?

Anonymous

@Abcd There is this guy. Our class 12 school topper this year. Medical student, PCMB in school. Non-integrated, just like you. He got AIR 237 in NEET and 97.8 in boards.

Abcd

@IceInkberry Wow, #Inspiring.

samjoe

For $l$, Break $\{x\}$ in intervals like $(1,2)$ etc and find $\lim_{x\to1^+}$. Then we need 0/0 form for nonzero finite limit, so $x=1$ is a root of denominator quartic.. similarly for $m,n$

@IceInkberry non-integrated means a separate school?

Anonymous

5:45 AM

@samjoe Yes, our school(we have both integrated and non integrated) and Aakash, we had tie up with Rao IIT/Allen.

Sid

@Abcd Pro Tip: whenever someone seems very confident on something, 7/10 times, they are wrong

Avnish Kabaj

@Sid I disagree

Making fake facts up :P

samjoe

I want to know what has come of the quora neet girl who solved this years Mains and got 210+ without attempting maths..

Avnish Kabaj

Technically according to your confident statement

Your statement becomes a loop 7/10

Becomes

49/100

49/100*49/100

It will tend to 0

So your statement is a paradox

samjoe

stoppp

Sid

5:48 AM

@AvnishKabaj Hehe. Okay, that number was made up. But, the idea is that whenever someone appears very confident, they tend to be wrong.(at least in my experience)

samjoe

@Sid even $9.99/10$ will tend 0 if you apply cabbage logic

Avnish Kabaj

samjoe

@AvnishKabaj Please explain your philosophy

@AvnishKabaj sounds so familiar

Sid

@AvnishKabaj Okay. How did you find that faster than me? :P

Anonymous

Maths room. Maths room.

Anonymous

5:52 AM

Philosophy room. Philosophy room.

Avnish Kabaj

@Sid i saw it in a meme

And i know all the memes

Anonymous

@AvnishKabaj It's real.

Abcd

@samjoe Please show for x-> 1+ , i will get it better with symbols.

Avnish Kabaj

@IceInkberry who said memes aren't real

O.o

Abcd

@samjoe @Ice @sid what's the formal proof that $f(x+y) = f(x)f(y)$ is satisfied by $f(x)= a^x$.

Anonymous

5:53 AM

I was seeing a TED video in that they explained how less experienced are most confident :P and the problem with experts is that they think everyone is expert.

Avnish Kabaj

@Abcd it's in arihant

Anonymous

@AvnishKabaj I didn't

Anonymous

@Abcd I will search in my notebook, I think we had done that..

Anonymous

Or was it obvious? hmm.. searching

Anonymous

Got ittt

samjoe

5:56 AM

@Abcd use difference quotient form of derivative.. ah first assume its differentiable

Anonymous

I have it for f(xy) = f(x) + f(y)

Abcd

@IceInkberry that must be ln x

Anonymous

Page 1

Abcd

@sam do you know

Anonymous

5:58 AM

Anonymous

Page 2

Abcd

@IceInkberry thanks

Anonymous

Welcome and good bye! (When do you people have lunch? I have it now. Yummm)

samjoe

Hmm $f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = f(x)f'(0)$ where I assumed $f(0) = 1$

integrate now ..

Sid

@IceInkberry 2/3PM

samjoe

6:01 AM

Same as sid

@IceInkberry lunch before noon? :)

Sid

@samjoe I wonder why just equating LHS and RHS not work?

We know, $f(x)=a^x$

Anonymous

......

samjoe

No I think he was saying to show it

Sid

Oh, we have to find out the function for which that is true?

Anonymous

Yes, lunch before noon. Hungryyy. For me 2/3 is too late. That's because of my breakfast timings as well.

Anonymous

6:06 AM

Typing with one hand is '_'

Anonymous

@Sid Yes

Sid

Yeah, do what is comfortable for you. I mean, I sleep at 4/5 and don't take breakfast. I wouldn't advise anyone to follow my schedule.

Anonymous

(See my proof, for example)

Anonymous

@Sid 4/5 haha. I sleep at 10. I need lot of sleep.

Anonymous

No, 10 at night.

samjoe

6:08 AM

@IceInkberry im on pc, your answer gave me [neck exercise]

Abcd

@samjoe f(xy)= f(x)f(y), f(x)= ?

Anonymous

@samjoe ¯\_(ツ)_/¯

samjoe

$0,1,x$?

Anonymous

@samjoe Just $x$ would do, 0 and 1 are included in that.

Abcd

@Ice^

samjoe

6:14 AM

@IceInkberry nope

Anonymous

@samjoe Why

Anonymous

Ohhh got it

Anonymous

No, why?

samjoe

You are saying $f(x)=x$ and $g(x) = 0$ are same

Avnish Kabaj

@IceInkberry I thought that I was the only one who had lunch @11:30

We went on a school trip to an ashram

They used to serve lunch @ 12

Somehow my breakfast time became my lunchtime

Anonymous

6:19 AM

@samjoe Oh yeah, got it.

Anonymous

@AvnishKabaj Are you the only one in your family then?

Anonymous

(I have dinner around 7:30-8, you?)

samjoe

you dont eat for approx 8-9 hours o_o

Anonymous

Uploading vertical picture of proof to avoid neck exercises:

samjoe

i mean just snacking

Anonymous

6:22 AM

@samjoe Yes, that's why when I get up I am toooo hungry.

samjoe

@IceInkberry I mean after lunch. Nobody eats while sleeping :P

unless you sleep after having lunch for 8-9 hrs

Anonymous

@samjoe I thought you meant about dinner, my friends ask me if I get hungry because I do my dinner early.

Anonymous

I have food after almost 11 hours after my dinner

Anonymous

samjoe

6:25 AM

btw neck exercises are good.. for health

Anonymous

And for afternoon, I do eat snacks at 3 and have my milk at 5. I am not hungry then.

Anonymous

@samjoe I am imagining you doing your neck exercise on PC (you don't have face btw; in imagination lol).

Abcd

@samjoe @Sid @Ice f(xy) = f(x)f(y). f(x)=? ,given f(1)=1. f(x) is not a constant function

samjoe

@Abcd then $f(x) = 1,x$ doesnt this satisfy

Abcd

@samjoe Please tell the method of finding it

@sam please reply

samjoe

6:35 AM

cant find a general method, there can be many more like $x^2$ etc.

Anonymous

@Abcd $$f(1)=1$$ $$Differentiate$$ $$xf'(xy)=f(x)f'(y)$$ $$y=1$$ $$xf'(x)=f(x)f'(1)$$ $$Integrate$$ $$ f(x) = klnx + c$$ $$[k=f'(1)]$$$$f(1)=1$$ Putting it in above equation, and solving, $$c=1$$ $$f(x)= klnx +1$$

Anonymous

Is this correct?

Anonymous

Ah mistake. Correcting.

samjoe

No you have $\ln(f(x))$

so we already get $x^k$ as general solution, easily guessed :P

Avnish Kabaj

@IceInkberry yea

Dinner is at normal timings tho

8-9

samjoe

6:43 AM

My dinner at 9:30 generally

Anonymous

$$f(1)=1$$ $$Differentiate$$ $$xf'(xy)=f(x)f'(y)$$ $$y=1$$ $$xf'(x)=f(x)f'(1)$$ $$Integrate$$ $$lnf(x) = klnx + c$$ $$[k=f'(1)]$$ $$f(1)=1$$ Putting it in above equation, and solving, $$lnf(1) = kln1 + c$$ $$c=0$$ $$lnf(x)= klnx $$ $$ f(x) = x^k$$

Abcd

I got same thing Ice Thanks.

Anonymous

@Abcd

Anonymous

You should have told earlier that you had got :P

Anonymous

@AvnishKabaj Oh that's normal.. people here have it at 11 or so on.. Everyone's normal is different :P

samjoe

6:47 AM

No having dinner early is good, around 7-8 is generally recommended

Abcd

@IceInkberry you try the actual question:

Anonymous

4 hours before you sleep is the recommended dinner time

samjoe

@IceInkberry :|

Anonymous

@Abcd Throw it :P

Anonymous

@samjoe Recommended dinner

Anonymous

6:48 AM

xD

Abcd

Catch it @sam @avatarshiny @ice

Sid

@IceInkberry I have dinner at 10:30-11 if I am at home. And at 9-9:30 when I am away from home

Anonymous

@Abcd 1st is definitely true

Anonymous

And 2nd is false

Abcd

@IceInkberry Y

Anonymous

6:53 AM

For $x^k$ max value in $[0,1]$ is $1$

Anonymous

And $1/x^k$ is bounded in (0,1]

samjoe

what if $k<0$

Abcd

^

samjoe

and they didnt specify interval for (b)

Anonymous

Oh yes, no then

Anonymous

6:54 AM

B is correct.

samjoe

Well wait, we need this condition $(1+x)^k = 1+x(1+g(x))$ to find a value of $k$ right

Anonymous

Yes, but that doesn't put a restriction on 1 and 2 options

Anonymous

It does, on 3rd one.

samjoe

no if we get $k>0$ from there then?

Third is obviously wrong, imo, it cant be $e/2$

Avnish Kabaj

@Abcd man I don't get pings for avatarshiny

Anonymous

6:57 AM

@samjoe Okay, doesn't put restriction on 1st option :P

Anonymous

@samjoe Yeah, it would be e to the power something

samjoe

and fourth option $ f(x)/x$ is never continuous at $x=0$

Anonymous

Option 4 also depends on value of $k$ if it is positive, it is true, otherwise not.

samjoe

@IceInkberry Yeah it does, if $k>0$ then (a) is true

Anonymous

@samjoe Why? If $k$ is positive, it become $x^{k-1}$ which exists at $x=0$

samjoe

6:59 AM

@IceInkberry No in calculus, at least here in india, we treat $x^2/x$ as $x^2/x$ and not $x$

Anonymous

Like.. oh yeah

samjoe

I mean $x^2/x = x$ only if $x\neq 0$

Anonymous

I had learnt it in class 11 basics. Correct.

samjoe

@avnish eww that thing sends chills down the spine

Anonymous

@AvnishKabaj Do you see insects eating insects? Gross.

Avnish Kabaj

7:02 AM

@IceInkberry that's the good shit

@samjoe mine too

Mine too Buddy

Anonymous

Good shit sending chills down the spine lol

samjoe

just imagine that thing showing up at your home

Avnish Kabaj

Just imagine you're walking in a park

Anonymous

Yuck

Avnish Kabaj

And then

Aaaghhhhh

Anonymous

7:04 AM

It's not so horrific in imagination though

Anonymous

By the way, how are we supposed to find $k$

Sid

my friend eats insects. It's not so horrific as you all think

Anonymous

@Sid Insects are nutritious, full of proteins.

Anonymous

In some parts of the world, they dry the insects, crush them and use the powder in cooking.

Avnish Kabaj

A bit of Kalu Sarai lore

samjoe

7:07 AM

@Sid no thank you

Anonymous

Ancient people used to eat insects, but since insects started destroying our crops, we disregarded them and eating them became gross over time.

Avnish Kabaj

Told to every new kid eating momos

Is that

They knead the dough with their feet

Anonymous

Heard that for bread :P

Avnish Kabaj

But they actually do that

samjoe

@IceInkberry ancient people were eating insects then who were they growing crops for? xD

Avnish Kabaj

7:08 AM

I've legit seen

samjoe

ancient people must be crazie

Sid

@samjoe Variety is the spice of life

Jasmine

These messeges might be flagged by some vegiterian :o

Although I am not

Anonymous

@samjoe Themselves :P You can't eat like a bucket of insects, right? xD

Anonymous

I am a vegetarian

Avnish Kabaj

7:09 AM

@IceInkberry yuk

Anonymous

@AvnishKabaj I hope with some feet glove thing on

Anonymous

@AvnishKabaj -_-

Sid

@Jasmine Don't worry, I am >10k user. I can decline flags here.

Avnish Kabaj

"paneer samajh ke kha ja"

samjoe

@IceInkberry best. theory. ever.

Sid

7:10 AM

As long as there's no swearing and stuff

Jasmine

@Sid oh :D

Anonymous

@AvnishKabaj I have no problem in eating non veg, it's just that my mom is vegetarian so we kids became vegetarian as well.

Anonymous

I have tasted some non veg stuff at times

Anonymous

@samjoe Why the best?

Avnish Kabaj

@IceInkberry man it's from one of AIB's videos

Jasmine

7:12 AM

@IceInkberry then that makes you nonvegiterian

Avnish Kabaj

I thought everyone would get the reference

@Jasmine non vegetarian

Anonymous

@AvnishKabaj I didn't because I don't see AIB

Avnish Kabaj

@IceInkberry wut wut

Y y

Anonymous

I used to earlier, I just lost interest.

Avnish Kabaj

It's the only good indian utoob channel

Anonymous

7:13 AM

¯\_(ツ)_/¯

Jasmine

It's vegetarian right my phone is bad

I have no opltion to turn auto wrong off

Anonymous

@Jasmine Haha, it corrected it again xD

samjoe

@AvnishKabaj nice joke

Avnish Kabaj

@Jasmine waaa

@samjoe wait what

Was the joke

Anonymous

Tooblight

Anonymous

7:15 AM

Jk haha

samjoe

Uthoob has much better content

to each his own

@IceInkberry maybe we can have $g(x)=0$ and $k=1$?

Anonymous

Isn't Saying Uthoob carry minati thing?

Anonymous

@samjoe I thought about that.. but how can we be sure about g(x)

samjoe

idk how carryminati and pewdiepie get so many views, most of their vids are not worthy of so many views, I mean I have been watching pewds from class 8-9 and his content back then was awesome .. now its only reaction to posts/vids

Anonymous

Wait, there is this much discussion in this room for the first time.

samjoe

7:19 AM

Everyone came here instead of physics room

Anonymous

I don't understand what's so nice in their videos. Not worthy of time.

samjoe

what is that one word which everyone is using over the internet and nobody knows its exact meaning?

yes I am referring to "cringe"

Avnish Kabaj

Carry minati/bhuvan bam

Like very bad jokes

Sid

@samjoe Felix has earned his lot though. It's only natural that the content isn't as good as it once was because he's been there for so long.

samjoe

@Sid Yeah but how is he showing trash to people and people are watching it, even "upvoting" it!

I really liked his old gameplay type vids

Hmm I guess now thats what people want him to do

Anonymous

7:24 AM

Most of the time my videos revolve around song artists' videos. Hehehe.

Anonymous

@samjoe Are you Quora person?

Jasmine

$|x|+|x-1|+|x-3|+|x-6|+.....+|x-(1+2+3+...101)|$

Sid

@samjoe Cognitive biases. The votes usually revolve around,"He's popular and was good. Surely, this one is good too"

Anonymous

They use the word 'upvote'

Sid

@IceInkberry Upvote is a normal word. They use it in SE as well

Anonymous

7:26 AM

Oh yeah, here too

Anonymous

No, people usually use those words which they use the most. Upvotes is used a lot on Quora as well

Jasmine

In the question I am supposed to find number of integral points for which f(x) is minimum

Avnish Kabaj

‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍ ‍

Jasmine

$|x|+|x-1|+|x-3|+|x-6|+.....+|x-(1+2+3+...101)|$

Anonymous

@AvnishKabaj Message disappeared

Anonymous

7:27 AM

O_O

Jasmine

I would like to know a shorter method for it as it's a test question where answers are tricky

Avnish Kabaj

O.o o.O

Anonymous

What is the longer method?

Avnish Kabaj

@Jasmine I don't see the series

1 3 6

Jasmine

@AvnishKabaj how to bring in line

@AvnishKabaj it's 123456....101

@IceInkberry like opening modulus

|x|+|x-1|+|x-3|+|x-6|+.....+|x-(1+2+3+...101)|

the answer is 8 whenever anyone gets a short solution then share a hint

Anonymous

7:38 AM

There is a pattern

Anonymous

Investigating further.

Jasmine

Yeah they are adding terms

Anonymous

No

Anonymous

Pattern as in the number of integral points for min value

Jasmine

I mean 1,1+2,1+2+3

@IceInkberry oh!

Anonymous

7:39 AM

For |x| then |x| + |x-1| then |x| + |x-1| + |x-3|

samjoe

ARre its simple!

Hmm @jasmine is it from 0 to 101?

Jasmine

@samjoe how?

@samjoe yeah you can add 0 to 1 no harm

@samjoe yes

samjoe

See, all these modulus terms from 0 to 101 are 102 in total

Anonymous

@samjoe Every question for you is simple, I wonder that your rank in JEE shouldn't be around what yours is.. it should be even better.

Jasmine

@samjoe yes

Anonymous

7:42 AM

I am not getting 8 but

Jasmine

@IceInkberry strongly agree

samjoe

If you break open these modulus, you will get coefficient of $x$ as negative for first 51 terms

I mean you will get coefficient of $x = 0$ only for $x\in \sum_{i=0}^{50}i$ to $\sum_{i=0}^{51}i$

frogeyedpeas

what is the question

i'm curious

samjoe

so answer is $f(\sum_{i=0}^{50}i)$ or $f(\sum_{i=0}^{51}i)$

Anonymous

21 mins ago, by Jasmine

$|x|+|x-1|+|x-3|+|x-6|+.....+|x-(1+2+3+...101)|$

Sid

7:48 AM

@frogeyedpeas ^

Anonymous

21 mins ago, by Jasmine

In the question I am supposed to find number of integral points for which f(x) is minimum

Jasmine

@samjoe it talks about minima so yea for that interval

Anonymous

We have to find the number of values of $x$ for which we get that minima

Anonymous

I get the answer as 51, ridiculous :P

samjoe

Oh so you need to only find number of integral points, they are the ones lying between $\sum_{i=0}^{50}i$ to $\sum_{i=0}^{51}i$ that is $51$ :)

@IceInkberry same :)

Jasmine

7:50 AM

@samjoe answer is 8

Anonymous

@samjoe That's not the answer

frogeyedpeas

ah hmm

samjoe

@Jasmine sorry earlier I thought you were asking minimum value

@IceInkberry no its 51 only

Anonymous

@samjoe I mean according to her answer key

Anonymous

My pattern also gives the answer as 51

Anonymous

7:52 AM

The pattern is $1-2-1-3-1-5.......1-51$

samjoe

Oh you found pattern, but its easily shown that it is indeed minimum for these points

Anonymous

As the terms increase

Anonymous

@samjoe Yeah, didn't think about that

frogeyedpeas

Its not even clear to me that $f(x)$ is constant for multiple points, let alone that that there are multiple integral points for which it takes on a global minimum

samjoe

say you have $|x|+|x-1|+|x-(0+1+2)| + |x-(0+1+2+3)|$. Then we have four points where we need to break the modulus $x=0,1,3,6$. Now on the left of $x=0$ all terms have negative coefficient of $x$ so function is decreasing.

frogeyedpeas

7:56 AM

hmm this is going to be a good puzzle

so how many minutes do you guys have to crack this in, when ya'll are taking the test

Anonymous

3-4 minutes

Anonymous

More if you invest less time on other questions.

frogeyedpeas

@samjoe what do you mean by "break up the modulus" do you mean, re-express it without absolute value bars?

samjoe

Just use fact that $|x-a| = -x+a$ for $x<a$. Using this we can say coefficient of $x$ is zero only when $x$ lies betweeen zero points of the middle terms (when no. of terms is even, which is case for 0-101)

@frogeyedpeas yeah

Transcript for

$Mathematics$ JEE Maths Zone