Solved wrong in 𝗖𝗼𝗮𝗰𝗵𝗶𝗻𝗴 𝗧𝗲𝘀𝘁 ❌
Vložit
- čas přidán 6. 09. 2024
- 𝗗𝗼𝘄𝗻𝗹𝗼𝗮𝗱 Our App From Google Play Store :
play.google.co...
For App Related Queries :-
Email Address : rankbuddy.official@gmail.com
Whatsapp Number : 8800825254
----
Join us on Telegram - t.me/bhannatma...
#maths #bhannatmaths
log(a^2)=2log(a) ❌
log(a^2)=2log(|a|)✅
Domain of f(x)=loga(x) ,a>0,a≠1 is (0,♾)
Sir you are "Great" ....Because of you, I developed interest in mathematics...The way you approach the problems is really amazing...Keep teaching in the same way for years and years.........❤❤❤❤
Sir i have paused the video at 10:00 and want to tell the mistake
The mistake is that if power of y comes in front as the coefficient of logarithm then we need to put a modulus sign on y because as per question y can take negative values as well and argument of logarithm can't be negative. This is the problem in the solution.
Ye sirf even powers mai hoga na odd power mai to mod nhi aega ? Confirm kardo
@@ballPool-tj7nq nahi, odd powers me mod nahi ayega.
ln (x^(2n)) = 2n.ln (|x|)
ln (x^(2n+1)) = (2n+1).ln x
The thumbnail though 💀
The Property You Applied, in which you put the power in front, i only valid when we can make sure that after shifting the power the log is still defined, but in this case when you put '2' in front at that time log y is not defined in [-10, 0). So we can't put the power in front, we have proceed it using mod or as it is
Yesss same doubt.
@@uguess3609 That's Not Doubt Bro, Sir asked where did i do wrong, i am telling him where he did wrong.
@@uguess3609 abey pagal h kya.
Kash inka.... PYQ's course thoda sasta me mil jata
VOTE FOR LECTURES ON NUMBER THEORY FOR ISI/CMI 👇👇👇
I also need the same bro as I am an ISI/CMI aspirant.
Yes I also need
Ye hai kya
@@Benzene42 Bhai Kabhi bhi Indian Statistical Institute ka naam suna hai aur Chennai Mathematical Institute ka naam? Jaan lo a6 se taab pata chalega kya cheez hai I can guarantee these 2 institutes>>>>>>>>>>>>>>>>>IIT Bombay.
@@AmlanSarkar-wr2prmat batao bhai in chomuo ko, nhi toh befaltu me competition badh jayega
This concept is used used in one of the problems of black book... But i was not able to identify the mistake
Y ki power ko directly aage nai la sakte,kyuki y to negative bhi hai,given interval mein... Sir aap ne jab vo pehili baar step likha tha na,tab hi maine uss mistake ko identify kar liya tha... You are great sir ❤❤
Someone solve it- (i) log of (x+y) to the base 3 plus log of (x-y) to the base 3 equal to 1.
(ii)2^(y/x+x/y) equal to 32
@@KushalKalita-yf3ydAre you mental? This is not a board exam in which if we solve any problem in easier way then we will lose marks. It is MCQ exam, it is numerical value exam. We can use any method, and graph method is the easiest. Why you think so much? Remember one thing, you will remember it well. Remember many, you will be confused.
I apologise you for my mistake ,you solve it which you think better
10:00
2 hatake mod y likhna prega, 8 values positive vali as well as negative vali side hogi and total solutions 16 honge...jus like x^2>p jab hota hai tab ya to a^2-b^2 se wavy curve lagado, ya direct mod x>sqrt(p) se evaluate krdo...
I am sure NTA answer sheet will have the solution as 8.....
Than even that's wrong
Then the Indian teachers would have to rebel for this
Real bro.. and even if we challenge they will pretend they are correct. This year I lost 15 marks without making a mistake in one exam.
NTA would argue that since the coaching book answer is 8 the answer is 8..... End of conversation.....
The inequality will never be satisfied by exactly 1 value of x,if it is satisfied there will be infinitely many values of x satisfying that inequality....
Question ko teda karke pucha hai
“Atleast”
Sir is solution me jb aapne y² me se 2 ko aage nikal liya tha y ki jo kuch values negative ho skti thi uski possibility khatam ho gyi since tb agr y me negative values nhi dal skte since vo out of domain ho jayega...
Mujhe to yahi galti dikhi sir...ab mai aage ka video dekhne ja rha hu...
Great Sir 🤩🤩 Your Concepts are deeply strong
Sir i did it by perfect square
And then got the ans
Solve this pls in next video:
1)Prove that:
(sinA.sin2A+sin3A.sin6A)/(sinA.cos2A+sin3A.cos6A)=(3tanA+2tan^2A+2tan^3A-6tan^4A-tan^5A)/(1-2tan^2A-6tan^3A-3tan^4A+6tan^5A)
Sl loney 😅
On first screen. When we apply the most common log identity. It should be mod of y instead of just y. So at last step there will be 16 integers
10:03 sir mujhe lg rha h ki question mein y^2 hai to y ki jo values positive mein hai vahi negative mein bhi valid hongi
Sir ive paused the video at 9:51
The main reason why its wrong because whenever we have even powers of a variable in logarithm, while using the property we will have to apply modulus to satisfy the domain of the logarithm😊
Sir iota = iota power 4/4
Iota =1power 1/4
Iota = 1
How is it possible sir i just thought of this when I was doing revision of complex no
Where is the catch??????????????
Mistake is that 1^1/4 is not 1 here
@@Root1729 why??
@@Jee2025-oz1zs there will be 4 values and only iota satisfy , learn complex numbers bro then ask doubts
@@Jee2025-oz1zs bcoz. In x^4=1. Has four roots 1,-1,I,-i
x^(1/4)=1 has 4 roots 1,-1,i,-i
We can't change the domain of function by applying modification to original situation
Sir very good explaination🎉🎉🎉
first inequality mai function is negative should'nt the D be negative then because the fucntion is never zero and always negative? but maybe im wrong kyuki it is not mentioned for all x
edit : i just realised that vice versa of " if discriminant is negative then function is above or below the axis" is not true
Sir log x power even number comes out to be log of mod x and because in this case x can be positive or negative
Sir please start any series for Olympiads ..
Sir jab 2 ko aage lekar jayenga tab mode y aayega this is the mistake in 1st solution
Bhaiya class 12th ka AOD CHAPTER KA. 6 unbelievable hidden lecture kaha par milega. Very need.
Sir aapse questions kaise puch sakte h
Yes sir
ISI number theory classes TOMATO book solutions please🙏
sir i have a doubt please solve it . Sir log is only defined for positive numbers to why will we put mod
Sir y2 me y ko positive hona zaruri nahi tha par jab sirf y hai to y ka positive hona zaruri hai ye condition Q me nahi thi hamne dali
Log(a^n), n is even, then you have to put mod.
Thank you sir. Aaj ek nayi cheez seekhi
sir log mein -ve define hi nahi hai to mod lagane knha jarurat hai
Unacademy ki beizzati
But it is allen test question, the student who asked has commented 🎉🎉yet
Essa too kabhi kisi ne padaya hee nahi sir
Mene solve kiya pause karke too 8 hee aa raha tha
But aage se galti nahi hogi
Thanks a lot of sir ❤❤
sir jab log ko t put kiye the usme log decreasing function h
y ke square ko remove karke hamne use positive assume kar liye aur negative values ko not aaceptable assume kar liya
good solution....
ON TAKING POWER OUT, TAKEN CARE IS Y TO BE POSITIVE
Y can be negative too
Amazing !!
THANKYOU SIR FOR SOLVING THE QUESTION ASKED BY ME , BTW IM IN ALLEN KOTA , ITS THE QUESTION OF ALLEN KOTA LOWER BATCH MINOR TEST
Edit : YES , THE TEACHERS OF ALLEN KOTA PROVIDED US WRONG SOLUTION 😂🎉🎉🎉
Before watching this lecture, my answer came 16.
Hurray muje samajh me aa gaya tha❤
Sir 1st sol mai aapne jab 2 ko bahar kra tab aapne anjane mai y ko positive assume kr liye
Yea true, y must be positive
Sir apne y² ka domain toh liya hie nhi
Sir this is Sajid Ali from Pakistan Sindh province
M.phil scholar in mathematics
But I got so many concepts from you
And with the help of your concepts
Finally I passed my commission exam
Recently working as subject specialist mathematics bps-17 in Pakistan 🙏🙏🙏
Sir pls upload solution of question: sinπ/7. sin2π/7. sin3π/7.
Already uploaded
√7/8 is the answer right??
@@uttu2008 (√7)/8
@@shashwatorsamriddhithakur1774you are wrong.
Put π/7=0, so our answer comes as sin(0)sin(0)sin(0) = 0
@@FlakernNBhai π/7 constant na ki variable uski value fixed hai.
Thumara statement hi galat hai
π/7=0=> π=0
Which is incorrect
10:01 sir aapne square bahar lia so andar y pe mod lagana tha
exactlyyy
@@tiwary7320 are bhai because maan lo koi log 4 ko likh de log (-2)^2 to aise to log undefined ho jaega power directly bahar nikaal doge to... islie mod lgate hai taki always defined rhe log
@@tiwary7320 ehhh bro, log is not defined like this.... you can't have a non positive number inside log.... i suggest you first study the definition of log.... for short- log y base x means the power to which x should be raised to get y, so thus, and we always take x as positive .... (check the definition online)... thus, x to the power anthing can't come negative hence y cant be negative
@@tiwary7320 welcome
Sir maine bhi same glti kri😂😭
Sir log |y| (MOD aaega) aaega not log(y)
ln (x^(2n)) = 2n.ln (|x|)
ln (x^(2n+1)) = (2n+1).ln x
mistake identified and i solved this question correctly by myself that expression should be -2log(mod(y))
Sir due to you my maths is strong sir thanks sir
Mode
sir y ke square ko aise hata nahi skate kyunki kya pata y negative ho, jo ki log mein possible nahi hai
😂
That is why we have to use the mod sign
@@kdhd100 are i know i just commented kyunki sir ne kaha tha
Sir I solved this question in the first time
Mod of y
Sir ye common mistake bhi ho sakta hai
Mai Manta hu aisi common nahi honi chahiye
But kabhi kabhi ho jati hai
So please don't create unnecessary problem
Very good you are trying to make students correct
But your thumbnail was very immature
Vha par y ki jagah mody aayega
i got 10 lmao
Sir I think this solution wrong bcz when log(m)^n applied property then the correct version is nlog(|m|).
Argument positive nahi kiya
I cant assume 2logy = t..I think that's the mistake
😂😂 bhai galti unhone bata di
No you can
Sur apne 4 ko kat diya hai
Sir you are wrong
No my brother sir is right
Agar koi mistake ki hai to aap batao
Mai apko suggest karta hu aap google par 'desmos' graph par iska graph bhi dekh sakte hai...
Solution bilkul correct hai
Ans:-16🙂🙂
r u from Unacademy?
@@AmbrishTiwari777 no brother
I do not prefer any online platforms