Star Problem # 3, 4, 5 from Functions & Logarithm | Problem Series for JEE Aspirants

𝗧𝗲𝗹𝗲𝗴𝗿𝗮𝗺 𝗖𝗵𝗮𝗻𝗻𝗲𝗹 𝗟𝗶𝗻𝗸 : t.me/rankbuddy
𝗗𝗼𝘄𝗻𝗹𝗼𝗮𝗱 𝗼𝘂𝗿 𝗔𝗽𝗽 : play.google.com/store/apps/de...
𝗗𝗼𝘄𝗻𝗹𝗼𝗮𝗱 𝗪𝗶𝗻𝗱𝗼𝘄𝘀 𝗔𝗽𝗽𝗹𝗶𝗰𝗮𝘁𝗶𝗼𝗻 : drive.google.com/file/d/1fYt3...
𝗛𝗲𝗹𝗽𝗹𝗶𝗻𝗲 𝗡𝘂𝗺𝗯𝗲𝗿 : 8800825254
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Free Maths Content For JEE Aspirants
#bhannatmaths #jee2025 #jee #functions #logarithm

[@amanbhaskar9960]

Sir you are great 😢
For providing quality education with no cost 🙏🙏🙏🙏

[@SarveshSingh-kw8xb]

❤️

[@arjunphaneesh6051]

Thank you sir i loved the second problem of functions

[@Dheeraj.dheera]

Thanks so much sir for giving the solution of problem 3

[@amanbhaskar9960]

Thanks sir 🙏🙏

[@neerajgupta2386]

I solved the #4 problem using a different approach.
So, what I did was that I substituted x=sin^2(t) and x belongs to [0,1] therefore t belongs to [0, pi/2], (so that the substitution is bijective). Substituting x=sin^2(t) in f(x)=4x(1-x) we get f(x) =sin^2(2t), f(f(x)) =sin^2(4t), f(f(f(x))) =sin^2(8t). Now we are required to solve f(f(f(x))) =x/3. It therefore becomes equivalent to solving sin^2(8t) =(sin^2(t))/3. And lastly draw both the graph for t belongs to [0, pi/2] they will cut at 8 points, we get 8 values of t➡we get 8 values of x (This is true because for a particular value of t there is only one value of x as I mentioned above)

[@gametimewitharyan6665]

I haven't learnt trigonometry till now but the rigorousness of your solution is very beautiful to me
Good work 🎉
By any chance are you are Maths stackexchange user or Olympiad participant?

[@dexroner]

Loved the visualization of taking sin²t.. Nice approach

[@jagithgunti]

Learning a lot from questions sir thank u ,increase the level sir for learning more from individual question

[@SarveshSingh-kw8xb]

Aur sir math me feel bhi do please

[@user-qp9gj2hf2d]

Same question tarun sir ne brain teaser me Diya tha

[@UDAYPRATAPSINGH-hg7ts]

real legend

[@saipratiksamal]

Sir please give the solution of this question
In the Fibonacci series prove that
[a(n+1)]²-a(n)×a(n+1)= (-1)^n
Question from the book challenge and thrills of pre college mathematics, chapter 2, arithmetic of integers, topic principle of mathematical induction

[@mathmaticianboyjain]

Put n something as a convergence value

[@saipratiksamal]

@@mathmaticianboyjain can u please elaborate.

[@MantuKumar-hd8ze]

Sir how to draw graphs of multiple composite functions...... Ex:-sin(Q(x)).

[@LGFZ136]

New video kab aayega?

[@santoshanand8661]

Sir naya question kab dolege qsn 8 ke baad apne qsns nhi bheje and qsn 5 ke baaad apne soln nhi bheja

[@ayushdevdhar2178]

Problems are (Nice)²⁰⁰⁰

[@SarveshSingh-kw8xb]

Aap easy sawal la rahe ho

[@ssgamer5693]

Q6 ans 30?

[@SarveshSingh-kw8xb]

Sir aur khatin lao sawal

[@vivekkushwaha1457]

Sir ? We know that a^x = a a^y = a then powers x=y. If we suppose powers 1^1 =1 ; 1^0= 1. Then powers 1=0 not possible .
It means either 1^1=1 or 1^0=1
is wrong. Please explain me.
1^infinity = 1 I am 100% sure but other says that it is not possible why?

[@neerajgupta2386]

1)U cannot apply a^x=a^y➡ x=y for a equal to 1. For the condition to hold true, a belongs to (0, infinity) - {1}. 1^1=1 and 1^0=1 are 100% correct.
2) 1 raised to any power of a real number equals 1, take a note that here the number 1 is exactly 1, there are no limits involved. SO, u can say that limit of 1 to the power infinity is 1. But if something tending to 1 is raised to a very large power then the answer is a limit which can be different in different situations. So to sum it up as long as you mean exactly 1 by the number 1 you are right.

[@toofaaniHINDU]

I couldn't solve #4 star problem 😢

[@op-zy2if]

Sir pls increase level of question these question can be easily found in standard books such as arihant or cengage

[@AJ-io5xd]

Yes

[@AnshPathak2005]

​@@jiveshdonode1271kuch galat nhi kaha . Tera level khrab h to kya kr sakte h . Star problem ka level high hona chahiye

[@Atharva-jf7cq]

Sir problem 6 ka answer hain 30

[@SarveshSingh-kw8xb]

Bhai kaise solution bhejo

[@SarveshSingh-kw8xb]

Please

[@MATHSBABA-tw7mw]

I do first question just by observation without using pen and paper while travelling in abus😊

[@NewMonaJewellers]

Mujhe to solve karte hi nhi aata

[@skkirar7699]

Modiii

[@gametimewitharyan6665]

Kiee guranteeeee

[@neerajgupta2386]

I solved the #4 problem using a different approach.
So, what I did was that I substituted x=sin^2(t) and x belongs to [0,1] therefore t belongs to [0, pi/2], (so that the substitution is bijective). Substituting x=sin^2(t) in f(x)=4x(1-x) we get f(x) =sin^2(2t), f(f(x)) =sin^2(4t), f(f(f(x))) =sin^2(8t). Now we are required to solve f(f(f(x))) =x/3. It therefore becomes equivalent to solving sin^2(8t) =(sin^2(t))/3. And lastly draw both the graph for t belongs to [0, pi/2] they will cut at 8 points. (This is true because for a particular value of t there is only one value of x as I mentioned above)

[@photongaming6325]

Excellent 👏👏

[@taggy9117]

wow nice solution 👍