if we assume everything to be greater than 1 and x less than x1 x2 x3....xn then half term will open with positve and half will open with negative in LHS and all will open with positive in RHS (mod) after than just manupilate and we get x1.x2.x3.x4.x5......xn in a quadratic form with constant 0 since we assumed all to be greater than 1 then their multiplication cant be 0 but will tend to 1 or be 1 if we assume them to be >= 1 lekin ek problem hai ki baaki cases ka kya hoga toh my solution is inconsistent
If log(xx1) is negative then log(x/x1) is always positive and if log(x/x1) is positive then log(xx1) is always negative. whatever possibility is applied, both of these functions will break out of the mod in opposite signs unless they are both equal to 0. hence it's positive to say that |log(xx1)| +|log(x/x1)| = |log(xx1) - log(x/x1)| = |2log(x1)|. Once you solve this further, you can see that the only possible number is 1.
bhaiya mene is question ko iss tare ke se kia lag bhag same he par thoda sa long hai Step 1 : separate all the terms |logx+logx1|+|logx+logx2|+.............+|logx+logxn| + |logx-logx1|+|logx-logx2|+.............+|logx-logxn| = |logx1+logx2+....+logxn| Step 2 : Separate the equation in 2 cases and solve them Case1 logx+logx1+logx+logx2+.............+logx+logxn + logx-logx1+logx-logx2+.............+logx-logxn = nlogx+nlogx = logx^2n log(x)^2n=log(x1.x2.x3...xn) (x)^2n=(x1.x2.x3...xn) So this is only possible when all the numbers same equal to 1 Case2 logx+logx1+logx+logx2+.............+logx+logxn + logx1-logx+logx2-logx+.............+logxn-logx= log(x1.x2.x3...xn)+ log(x1.x2.x3...xn) = log(x1.x2.x3...xn)^2 log(x1.x2.x3...xn)^2 = log(x1.x2.x3...xn) (x1.x2.x3...xn)^2 = (x1.x2.x3...xn) And this is only possible when all the numbers same equal to 1 So answer is 1
In case 2 , you removed the modulus without - ve sign. And when (x)^2n=(x1.x2.x3...xn) [ not given that x=x1=x2=... =xn ] there will be infinite solutions. eg:- Let n=2, so there exist x, x1 and x2 only. Now x^4 = (x1.x2) is satisfied by x=2, x1=8 and x2= 2
did it vg sir also covered it in his classes not the exact question but the concept good question i after solving quite some questions on log i can say that this is not the most difficult one but yeah the most conceptual one but anyway it was fun to see you doing it love your channel btw keep it up ....
@@codexyz2887 check his apni kaksha lectures in one of them you will find him explaining these concepts after watching all letures you will understand that each and every concept was already covered in vg sirs class not direectly but yeah its same
Maine logxxi ko a mana Tha aur logx/xi ko b Tho idhar lal + lbl =l (a-b) l de rakha Tha question me, ab dono side square karke solve kare tho humko finally (lal+lbl)²=0 and this is possible only when lal=0 and lbl=0, x/xi=1 and xxi=1, this is only possible when x and xi both are 1 Iss method me property ka use bhi nahi hua Anything wrong in this method?
Mera bhi 1 aa gaya tha , lekin apne us solution ko galat bola tha . Kyunki mod me kaat nhi sakte , vhi solution 10 bacho ne or bheja tha😂 ( allen vale group me ) .
thought of this approach initially solve karne baitha same logic same formula lagaya but further aage jaake confuse ho gaya ... my bad...will try better next time ...but got the solution instantly
bhaiya mene is question ko iss tare ke se kia lag bhag same he par thoda sa long hai Pin my answerrr plz Step 1 : separate all the terms |logx+logx1|+|logx+logx2|+.............+|logx+logxn| + |logx-logx1|+|logx-logx2|+.............+|logx-logxn| = |logx1+logx2+....+logxn| Step 2 : Separate the equation in 2 cases and solve them Case1 logx+logx1+logx+logx2+.............+logx+logxn + logx-logx1+logx-logx2+.............+logx-logxn = nlogx+nlogx = logx^2n log(x)^2n=log(x1.x2.x3...xn) (x)^2n=(x1.x2.x3...xn) So this is only possible when all the numbers same equal to 1 Case2 logx+logx1+logx+logx2+.............+logx+logxn + logx1-logx+logx2-logx+.............+logxn-logx= log(x1.x2.x3...xn)+ log(x1.x2.x3...xn) = log(x1.x2.x3...xn)^2 log(x1.x2.x3...xn)^2 = log(x1.x2.x3...xn) (x1.x2.x3...xn)^2 = (x1.x2.x3...xn) And this is only possible when all the numbers same equal to 1 So answer is 1
Ans: x=1 Approach: Using Logarithmic Properties We can Combine all logs on LHS as log{(x^2n)(x1x2x3....xn)/(x1x2x3...xn)} = log(x^2n) RHS can also be written as log(x1x2x3x4x5...xn) Now if we equate LHS = RHS After removing logs: x^2n=x1x2...xn From here we can infer x = 1 or 0 0 is neglected as it's not in the domain of Log(x) This is my approach.
No brother this is a standard gp and can take many values , 1 is just one value ... so u cannot directly infer ...... instead you can assume at first that x>x1,x>x2... and so on and then solving the equation will give you the obtained gp where u would see only those values which takes x>x1 and so on which is not possible except for 1 ...and hence x=x1=x2=...xn=1
I like your videos a lot but in this particular video, there are a lot of false claims. You proved that sum of ln(xi) is 0 does not imply that all xi have to be 1.
So what should others do praise you on how great of a human you are or what.... Ok you solved it, so what its not some ground breaking achievement As the question was not that hard....
@@harshitkunjalwar i mean im sharing my opinion that its not tough . plus im preparing for smth Which is tougher than jee advance in grade 9(not flexing but want to say this since you commented Like this )
![ILLSLICK - WINTER IS COMING [Official Video]](http://i.ytimg.com/vi/dwTMeM1zIhQ/mqdefault.jpg)
Here the link to {SETOF60} = courses.jeesimplified.com
.
Send us the hardest question you have ever tried
forms.gle/vedh4VUh6fFn7eUeA
if we assume everything to be greater than 1 and x less than x1 x2 x3....xn
then half term will open with positve and half will open with negative in LHS and all will open with positive in RHS (mod)
after than just manupilate and we get x1.x2.x3.x4.x5......xn in a quadratic form with constant 0 since we assumed all to be greater than 1 then their multiplication cant be 0 but will tend to 1 or be 1 if we assume them to be >= 1
lekin ek problem hai ki baaki cases ka kya hoga
toh my solution is inconsistent
If log(xx1) is negative then log(x/x1) is always positive and if log(x/x1) is positive then log(xx1) is always negative. whatever possibility is applied, both of these functions will break out of the mod in opposite signs unless they are both equal to 0. hence it's positive to say that |log(xx1)| +|log(x/x1)| = |log(xx1) - log(x/x1)| = |2log(x1)|. Once you solve this further, you can see that the only possible number is 1.
Ab toh yeh bahana bhi nahi maar skta ki mai toh 11th mei hoon 💀
Me 11th me hu to me maar sakta hu 😅
@@Anant_soni512 Maro Maro
Main to 9th mein hoon ☠
Par main bahana nahi marunga
@@Amol-Bhāratvāsi bhai tu jake pogo dekh
@@EmpireX-099 Minecraft khel raha hoon 😁
bhaiya mene is question ko iss tare ke se kia lag bhag same he par thoda sa long hai
Step 1 : separate all the terms
|logx+logx1|+|logx+logx2|+.............+|logx+logxn| + |logx-logx1|+|logx-logx2|+.............+|logx-logxn| = |logx1+logx2+....+logxn|
Step 2 : Separate the equation in 2 cases and solve them
Case1
logx+logx1+logx+logx2+.............+logx+logxn + logx-logx1+logx-logx2+.............+logx-logxn = nlogx+nlogx
= logx^2n
log(x)^2n=log(x1.x2.x3...xn)
(x)^2n=(x1.x2.x3...xn)
So this is only possible when all the numbers same equal to 1
Case2
logx+logx1+logx+logx2+.............+logx+logxn + logx1-logx+logx2-logx+.............+logxn-logx= log(x1.x2.x3...xn)+ log(x1.x2.x3...xn)
= log(x1.x2.x3...xn)^2
log(x1.x2.x3...xn)^2 = log(x1.x2.x3...xn)
(x1.x2.x3...xn)^2 = (x1.x2.x3...xn)
And this is only possible when all the numbers same equal to 1
So answer is 1
In case 2 , you removed the modulus without - ve sign.
And when (x)^2n=(x1.x2.x3...xn) [ not given that x=x1=x2=... =xn ] there will be infinite solutions. eg:-
Let n=2, so there exist x, x1 and x2 only. Now x^4 = (x1.x2) is satisfied by
x=2, x1=8 and x2= 2
Watching it 5 minutes after upload
Easy problem... Feeling genius 😊
Which class
@@shubharmy5527 12
did it vg sir also covered it in his classes not the exact question but the concept good question i after solving quite some questions on log i can say that this is not the most difficult one but yeah the most conceptual one but anyway it was fun to see you doing it love your channel btw keep it up ....
U r in 11th or 12th?
I am also enrolled in 11th class course of Vg sir
And which question u r talking about?
@@codexyz2887 hmm
@@codexyz2887 check his apni kaksha lectures in one of them you will find him explaining these concepts after watching all letures you will understand that each and every concept was already covered in vg sirs class not direectly but yeah its same
@@codexyz2887 he did it in basic maths or compound i actually forgot it has been few months
Maine logxxi ko a mana Tha aur logx/xi ko b
Tho idhar lal + lbl =l (a-b) l de rakha Tha question me, ab dono side square karke solve kare tho humko finally (lal+lbl)²=0 and this is possible only when lal=0 and lbl=0, x/xi=1 and xxi=1, this is only possible when x and xi both are 1
Iss method me property ka use bhi nahi hua
Anything wrong in this method?
Bro You can't directly equate any equation to zero
This will definitely help me in neet, thanks youtube 👆😃👍
ek aur approach lag raha hia generating function complex number ka bhi use ho sakta hia ek aisa question olympiad me bhi aaya tha Kuch time pehle
THANKYOU SO MUCH FOR THIS VIDEO
Mera bhi 1 aa gaya tha , lekin apne us solution ko galat bola tha . Kyunki mod me kaat nhi sakte , vhi solution 10 bacho ne or bheja tha😂 ( allen vale group me ) .
allenites
Not so complex . all you need to know is triangle inequality
thought of this approach initially solve karne baitha same logic same formula lagaya but further aage jaake confuse ho gaya ...
my bad...will try better next time ...but got the solution instantly
I solved this question exactly same approach as you take in this question, but after 1 hour of
solving.
😅😅😎😎
3:12 Bhaiya ek doubt tha ki jab apne yeah property lagai to sare terms mod ke ander plus me kese gye mtlb ki | log(x1) | + |logx2| hoga na
Jo aapne likha same vo hi likha hua hai...
Log x1|+. |Log x2.....
Nhi Bhai bhaiya ne mod ke ander hi sabhi ko add Kiya hua hai
Appreciate your work , waited a long time ❤
Free for all,Doubt : integral of root
1+tanx dx
Just Wolfram alpha
Bprp has a video on it
Bhaiya kya yee permutation and combination wale method se bhee hoskata hai solve !!?🫠
The approach was far clearer than what I did and performed
Adore Your work in this field.😁
I've got a doubt, as yk Log value can never be negative as it is not defined so we can just remove the mod and then solve
Bro log value can be negative
Only its base and number can't be negative.
Example : Log(1/2) base 2
Is equal to -1
Hope it helps😊
bhaiya mene is question ko iss tare ke se kia
lag bhag same he par thoda sa long hai
Pin my answerrr plz
Step 1 : separate all the terms
|logx+logx1|+|logx+logx2|+.............+|logx+logxn| + |logx-logx1|+|logx-logx2|+.............+|logx-logxn| = |logx1+logx2+....+logxn|
Step 2 : Separate the equation in 2 cases and solve them
Case1
logx+logx1+logx+logx2+.............+logx+logxn + logx-logx1+logx-logx2+.............+logx-logxn = nlogx+nlogx
= logx^2n
log(x)^2n=log(x1.x2.x3...xn)
(x)^2n=(x1.x2.x3...xn)
So this is only possible when all the numbers same equal to 1
Case2
logx+logx1+logx+logx2+.............+logx+logxn + logx1-logx+logx2-logx+.............+logxn-logx= log(x1.x2.x3...xn)+ log(x1.x2.x3...xn)
= log(x1.x2.x3...xn)^2
log(x1.x2.x3...xn)^2 = log(x1.x2.x3...xn)
(x1.x2.x3...xn)^2 = (x1.x2.x3...xn)
And this is only possible when all the numbers same equal to 1
So answer is 1
Jee ka bohot questions karliya aapne yaar abb thoda olympiad ke questions bhi start karo maths or physics ke?
Summation 1/n^2 (n is a natural number) ise solve krna
= z(2) = π²/6 . Where, z : zeta function.
Le bhai, kar diya solve.
Ashish sir ke voice mein
Behen log toh bajut asan hove behan 😂😂
Ans: x=1
Approach: Using Logarithmic Properties
We can Combine all logs on LHS as log{(x^2n)(x1x2x3....xn)/(x1x2x3...xn)} = log(x^2n)
RHS can also be written as log(x1x2x3x4x5...xn)
Now if we equate LHS = RHS
After removing logs:
x^2n=x1x2...xn
From here we can infer x = 1 or 0
0 is neglected as it's not in the domain of Log(x)
This is my approach.
Similar approach but how could you infer that x=0 or 1?
@@NahwShi-g5l By trial and error, 0 and 1 are the only options
Grimace ohio and fanum tax
Damn mate that's a pretty nice approach , Thank you❤😊
No brother this is a standard gp and can take many values , 1 is just one value ... so u cannot directly infer ...... instead you can assume at first that x>x1,x>x2... and so on and then solving the equation will give you the obtained gp where u would see only those values which takes x>x1 and so on which is not possible except for 1 ...and hence x=x1=x2=...xn=1
Saw this Question on your telegram channel 🥲. Couldn't even understand it (in 11th).
Kitna padh liya
@@consuknight1163 uhh maths me sequence and series, physics me nlm, chem me chemical bonding
Easy he bhai , recommend arvind kaliya sir for lectures
@@asheshgupta4845 unka trigno ka lect dekh raha hoo
Approach sahi laga liya tha but last me solution nhi aane par give up kar diya
may be wrong soln as lhs mai mod ki property use krne ke baad individual elements par mod hai par appne pue par mann liya
Yes I have the same doubt
Bhaiya ne galti se summation mod ke andar laga diya, bahar laga do use mod ke. Rest answer is correct.
At 3:50 ig its wrong. How you applied summation in between mod here?
Sahi to hai
Sahi tha!!! 🧸
Hiii😊😊
@@gaurav-no1qhBhai padh le, ladkiyo ke pechhe bhagne se ghar nhi chalta
@@Search957Gyan bhi de rha hai aur follow bhi nhi kr rha 😂
Hypocrisy 🤡
@@Search957😂 boys be strong
Btw hiii beautiful
I like your videos a lot but in this particular video, there are a lot of false claims.
You proved that sum of ln(xi) is 0 does not imply that all xi have to be 1.
bhai to log ke peeche pad gaye hai xd
Finally upload ho gyaa!! Approach thoda alag tha
Ise cases bana kai bhi kar bhi sakte hai
Erm what the sigma
Mere soln. Ke according final answer aaya h :
x1.x2.x3.....xn = x^2n toh iska mtlb kya h?? Please btao. Kya yehi final answer h??
Bhai same process kia tha...bas wait karraha tha approval ke lie
Good question 🗿
What is Eren Doin Here.
Usse ab freedom ki jagah air 1 chahiye
Bhaiya 0 invalid h, solution R+ ME hai
Bhai 0 not fr x, but fr the whole log term.
where entities inside will be 1
@@jeesimplified-subject right bhaiya 😅 ye solution woh trig prob se tracable h jisme hum 2A pe comment krne k liye hum usse A+B and A - B likhte h
Nice 👍
utna bhi hard nhi tha but tricky tha thoda sa I could solve it though......
7 crore...
Sab aise hi bolte hai
Seriously did in 40 sec . no lie , i had a hint it has only 1 sol
So what should others do praise you on how great of a human you are or what....
Ok you solved it, so what its not some ground breaking achievement
As the question was not that hard....
@@harshitkunjalwarwhy so rude?
Kaun se IIT or branch Loge bhaiyya discuss karle ...
@@harshitkunjalwar Bhai have some 🦅🦅
@@harshitkunjalwar i mean im sharing my opinion that its not tough . plus im preparing for smth Which is tougher than jee advance in grade 9(not flexing but want to say this since you commented Like this )
ye title deka dekha lak rha hae
Arey yaar solve karne ke liye time Diya kro
pause krna ata h?
Usse kya hota h?@@sidharthjaiswal221
Solve karne ke liye time milta hai 🙂@@bikki7524
usse kuch nhi hota jitna time bhaiya dete hai usme kro sab usme hi krte hai to increase our speed
@@ashutosh4189 pause karne se data khatam nhi hojayega?
|a|+|b| = |a-2| + |b-2| = |a+1| + |b+1|
Yeh solve Karo ab bas 0:59
Reply toh dedo bhaiya
x1 x2 .... Sabko 1 daldo
Fir x =1 aajayega
Aapki air 1 pakki
Pooor explanation... Made us confused
In the end, Yes a little 🙂↕️