Room for Yuvraj and robjohn

robjohn

05:23

@Yuvraj: yes?

Yuvraj

05:58

@robjohn question no 25 same link sir

robjohn

looking

$a^2-8a+67=(a-8)^2+3=b^2$

so $(b-(a-8))(b+(a-8))=3$

So... $b-(a-8)=\pm1$ and $b+(a-8)=\pm3$ or $b-(a-8)=\pm3$ and $b+(a-8)=\pm1$

4 things to check

comes down to two: $a-8=\pm1$

both work, so the sum is $16$

Yuvraj

sir why it is equal to +1-1

@robjohn\

robjohn

Think of the 4 factorizations of $3$

Jack Rod

ok got it sir

can you confirm some of the answers because there no proper answer key for this

@robjohn starting from 5 th

option is C

i use the concept of common root

robjohn

06:17

Explain more.

Jack Rod

there is one local maxima

between two roots

robjohn

can you plot a fourth degree polynomial with two roots and two local extrema?

Jack Rod

and 3 local maxima between the four roots rigth?

@robjohn i just mean out of four root we have only 3 unique roots

robjohn

If there are 4 roots, there will be 3 local extrema.

Jack Rod

yes

robjohn

06:27

Try plotting $(x+1)^2x(x-1)$

Jack Rod

it ihas three root 1 -1 and 0

robjohn

how many roots and how many local extrema?

Jack Rod

wolframalpha.com/input/?i=%28x%2B1%29%5E2x%28x-1%29

@robjohn

robjohn

?

Jack Rod

i mean only one local maximum

3 root

robjohn

06:32

you just mentioned 3 roots

If you have 3 roots, what is the minimum number of local extrema?

you need a local extremum between two adjacent roots

Jack Rod

yes

16 mins ago, by Jack Rod

there is one local maxima

16 mins ago, by Jack Rod

between two roots

robjohn

AT LEAST one local extremum

Jack Rod

yes

robjohn

a multiple root of even degree is also a local extemum

Jack Rod

sorry at least one

robjohn

06:40

so count again for $(x+1)^2x(x-1)$

Jack Rod

so 2

@robjohn

robjohn

count again

Jack Rod

4 roots can have maximum three local extremum

and minimum can be 1

@robjohn

robjohn

count for $(x+1)^2x(x-1)$

Jack Rod

it has 3 roots so 1 local extremum

robjohn

06:52

I see three of each

Jack Rod

between 0 and -1

then local minimum

robjohn

between $0$ and $1$, between $-1$ and $0$, and at $-1$

three local extrema

Jack Rod

but between 0 and 1 function value approaches to minimum

robjohn

yes, that is a local extremum

Jack Rod

oh

robjohn

06:56

Maxima and minima

In mathematical analysis, the maxima and minima (the respective plurals of maximum and minimum) of a function, known collectively as extrema (the plural of extremum), are the largest and smallest value of the function, either within a given range (the local or relative extrema), or on the entire domain (the global or absolute extrema). Pierre de Fermat was one of the first mathematicians to propose a general technique, adequality, for finding the maxima and minima of functions. As defined in set theory, the maximum and minimum of a set are the greatest and least elements in the set, respectively...

Jack Rod

i am stupid i thought that question is asking about local maxima

so i had that i mage in my mind

@robjohn

robjohn

So, (B)

Jack Rod

@robjohn yes

sir one request if you allow me to make it?

@robjohn last q 30

robjohn

what is your answer?

Jack Rod

07:12

i use the volume concept

n volumes of cubes = volume of rectangle

robjohn

if there are 30 cubes in a rectangular solid block, what are possible arrangements?

Jack Rod

possible arrangements mean how we can arrange them right..?like linearly or any other shape?

robjohn

$1\times1\times30$, $1\times2\times15$, ...

what is the smallest number of cubes so that $a\times b\times c$ cubes are hidden?

Jack Rod

(l-2)(b-2)(c-2)=30

@robjohn considering them as heigth length breadth

where 30 can split into 2.3.5

robjohn

that is not the only way

Jack Rod

07:25

okay..?

robjohn

$1\times3\times10$

Jack Rod

@robjohn sorry a request I find it very hard to concentrate on the screen I lose concentration and answer arbitrary

if i seat on the screen for long time '

@robjohn that is also one of the case

robjohn

so you need to find the minimum $(a+2)(b+2)(c+2)$ where $abc=30$ and $a,b,c\in\mathbb{Z}$

Jack Rod

using AM gm

@robjohn right..?

robjohn

I don't see how to use that here, but perhaps.

Jack Rod

07:30

@robjohn give me a seocnd i prepared the list

i mean if l b h is 2,3,5 we have

4,5,7 as the new

which gives minmum 140

@robjohn

rest have volume above 140

robjohn

did you prove that or are you assuming that is the minimum because they are closest to a square?

$1\times1\times30$ gives $3\cdot3\cdot32=288$

Jack Rod

just took all the possible values and find corresponding to that

robjohn

$1\times5\times6$ gives $3\cdot7\cdot8=168$

Jack Rod

1.6.5 or

robjohn

Ah, so brute force? looked at all possibilities

Jack Rod

07:35

yes

robjohn

that will work.

Jack Rod

@robjohn i need one advice?

robjohn

yes, $140$ is the minimum

Jack Rod

53 secs ago, by Jack Rod

@robjohn i need one advice?

@robjohn

robjohn

what?

Jack Rod

07:39

actually sir in the 17 question I have approached away but it is lengthy

xy.(x+y+1)=5^2018+1

i wrote (4+1)^2018+1

or 3,2

only possible pairs

now where x and y can be 2,2

@robjohn

robjohn

I am not following what your method is

Jack Rod

ok sir if i take lhs

if x and y both are even

(x+y+1) is odd

odd into even is even so it is always divisible by 4

and rhs is (4+1)^2018+1 is not divisble 4

so lhs cannot be equal to rhs so there is no pair of x snd y satisfy the condition

@robjohn

or even if i take x and y odd lhs will always be even

robjohn

You can look mod 4 or mod 5 or even mod 6

$5^{2018}+1\equiv2\pmod4$

$5^{2018}+1\equiv1\pmod5$

$5^{2018}+1\equiv2\pmod6$

perhaps there is a contradiction and then the number is $0$

Yes... mod $4$ there is no $xy(x+y+1)\equiv2\pmod4$

let me check again

I believe that $xy(x+y+1)\in\{0,3\}\pmod4$

Jack Rod

08:03

@robjohn yes

robjohn

So there are no solutions

Jack Rod

@robjohn yes

robjohn

There are only $7$ cases to check

Jack Rod

actually, i will not lie I got this idea from net to a similar question

robjohn

$x,y\in\{1,2,3\}$ except the two cases where $x+y=3$

and we only need to check the cases where $x\le y$

so it's not even $7$ cases

$4$ cases

$\{(1,1),(3,3),(1,3),(2,3)\}$

Jack Rod

08:09

yes sir

@robjohn

robjohn

don't need to check $(2,2)$ since $xy=4$

Jack Rod

which is divisible

robjohn

as I said, $xy(x+y+1)\in\{0,3\}\pmod4$

So this would work for $xy(x+y+1)=5^{2020}+1$ also

just in case they change the year with the same question

Jack Rod

@robjohn yes

@robjohn it takes time on chat to convey what I want to say, because I have no knowledge about math Jax can i connect with you digitally a half an hour session.

?

robjohn

not sure what you mean

Jack Rod

08:18

what I meant was it take an hour to discuss a single question, because of typing and math Jax writing sir ,can I do a video call or something else which saves my time,exam is just few days away

google.com/…

@robjohn

robjohn

@JackRod I can't do that now. Everyone else is asleep. Can you upload images of a page?

You could get past MathJax that way.

You'd need to do that anyway on a videochat

Jack Rod

@robjohn ok it is around 5 AM in usa?

robjohn

it is 1:22 AM

Jack Rod

i can see you the time you give so at least I can make my schedule?

any time which you give?

robjohn

Do you have a way of sending an image of your math? I could go for a while now.

Jack Rod

08:24

i will upload the list of question which I want to discuss before the time? and a small 30 minutes save my as well as your time!

image like hand writing?

robjohn

I can see the problems in the documents, I was trying to get past your difficulty with mathjax

yes, images of handwriting. For your work, etc.

Jack Rod

@robjohn give me a second

robjohn

Of course, for people asking questions on the site, we recommend learning mathjax

Jack Rod

08:42

i am working on this book

just wrote some hints I do not have a camera right now except the pc camer a

robjohn

that is hard to read, fuzzy

ah

Jack Rod

@robjohn yes it clicked a month ago by my friend

do not worry sir i have clear writing on paper

you will understand what I have written

robjohn

okay

Jack Rod

@robjohn when will you be available online..?

robjohn

I might go to sleep in a bit. shouid be back in 6-7 hours.

Jack Rod

08:49

@robjohn around 7 o clock US time, I will send you my email id on this and then will delete it?

robjohn

Rather than just do it here?

Jack Rod

what?

robjohn

I don't know why you would send and email id

(email address, I assumed)

Jack Rod

yes for video call?

robjohn

I don't think a video call will be any faster. we will need to send the written formulas anyway, I can do mathjax and you can send images.

hang on a few

Jack Rod

08:55

As I said, I do not own a smartphone so it is hard to send pics

robjohn

then I don't see how a videochat will do any better than here. Spoken math can be very imprecise. That is why I prefer chat with mathjax, or at least texted math

Jack Rod

I would try

robjohn

you don't have a tablet on which you can draw the math either?

Jack Rod

no sir

since I do not have a phone who can I have a tablet?

robjohn

you can try using codecogs.com/latex/eqneditor.php

some computers allow you to use a stylus. I have no idea what you have.

Jack Rod

09:05

@robjohn I would try my best, but I know this will consume my all-time and I have nothing to be left .

Good night.......

robjohn

that page might help you enter mathjax. I don't think video will help. I've had it fail badly before.

Jack Rod

way we could work is i have white board and markeer

i will write my solution you can guide me on board where i get wrong

this will save time as well as we do not need copy or pen to show again in camera this could be an option

@robjohn

robjohn

you have a camera on your computer? can you not take a picture of your whiteboard with that?

Jack Rod

yes

robjohn

then take a picture of your whiteboard and I can comment on that

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♦ Room for Yuvraj and robjohn