Toughest Question of JEE Advanced | Quadratic Equations | Arvind Kalia Sir
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- Опубликовано: 9 июл 2023
- Toughest question of JEE Advanced from the topic Quadratic Equations in Last 10 Years.
Best Question on "Relation between Roots & Coefficients in JEE Advanced"
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The JEE Advanced is a highly competitive engineering entrance exam in India, taken by students aspiring to pursue undergraduate programs in prestigious Indian Institutes of Technology (IITs). The exam is known for its challenging nature and tests students on various subjects, including mathematics.
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this question was quite a beauty.....learned a lot ..thank you arvind sir
Sir is too much smart calm and intelligent 😊😊
I'm already in IIT kgp but still I love watching these problems, I found this problem relatively easier than the problems which other people call normal or average
Who asked 🙄
Me@@user-jaintarun
Kisne pucha?
SIR REALLY WANT MORE VIDEOS LIKE THIS👌👌😊😊😎😎
love the way u teach 😢
agar humare time mai aap hotey tohum b iit ki preparation kr rhe hotey
abhi job mai agya to ye dekh kr afsos hota ki kash JEE ka prep kr lia hota to life thori behtr ho skti thi 😅
Man these questions are really helpful for the preparation
WHICH CLASS
8:03, could be directly done by separating den (a-b) to each term of num, and then multiplying a/a, den becomes a² -ab =a²+1 [ a and b here are denoting alpha beta respectively]......for the c part , we have by newton sum an-2 = an - an-1 puting n= 3 to n+2 will telescope the sum
Mein bhi wahi bola raha tha
This Can be done very easily by substituting n=1 or 2 in all options and satisfying LHS and RHS....
sir for option no b so much manipulation wasnt required,we could open and take lcm and then put value of a and b and get same
this was really tough and the option which gets in the way is B
I solved it at once
No it's actually easy. Instead of starting with LHS to get RHS to tick option , make LHS and RHS equal as the option says and then simplify them. If the equality boils down to a true equality , just tick the option. The difference between the two method is that first method requires thinking and creativity because you need to 'think' which relations you should use to reach LHS while in second , you can substitute the known relations wherever you see them in the process of simplification.
To clarify , consider option B .
Let LHS = RHS
Then , we have , ( assume that I mean 'p' as alpha and 'q' as beta)
[ p^(n-1) -q^(n-1) + p^(n+1)+q^(n+1) ]/[ p-q ] = p^(n) + q^(n)
Now , take the (p-q) in division to RHS and use the relation that pq =-1 ( because product of root is -1) to immediately see that equality indeed holds. Hence , B is correct.
Utna tough nhi tha yaar b , just put n=2 and solve
You'll get alpha^n - beta^n = 3
@@ArthBachhuka you are STUPID. is sir stupid hes finding all values himself. Clearly they have to satisfy for all n>=1. Then also you just put 1 value and try to solve? You never know what will happen if other values do satisfy or not, bruh
You did not have to go for the roots of the equation. Instead of replacing (1+α^2) by (α+2), you could replace (1+α^2) by (-αβ+α^2) and (1+β^2) by (-αβ+β^2) since αβ=-1 and similarly in (C). And (A) can be easily calculated using the already calculated value in (D) and the given recurrence relation. Lastly, this problem reminds me of the properties of the golden ratio, the Fibonacci sequence, and their relationship. a_n is the Fibonacci sequence, and the given expression is Binet's formula, a closed-form expression for the Fibonacci numbers.
Yes you are definitely right
I did it with this concept . Fibonacci Series made it quiet easier than Arbind sir's solution . ( This is my opinion . those who are unfamiliar to the golden ratio , can solve in other ways )
Love you sir ji
You are the best teacher
Here we observe that a1,a2,a3,a4……are the terms of Fibonacci Sequence 1,1,2,3,5,8,13,21,34….So all the options can be solved easily .
Can you elaborate further??
@@SoumalyaPatra07. a1,a2,a3,a4…….are1,1,2,3,5,8,13,…..Here a1+a2=a4-1. a1+a2+a3=a5-1 and so on .So option C is correct . b1,b2,b3,b4 ……are 1,3,4,7,11,18…..So option B is also correct .For option D ,let S =1/10+1/100+2/1000+3/10000+……..Now find 10S .Subtract to get 9S .Multiply by 10,get 90S .Now subtract S .So S=10/89. As all an>bn ,so option A is incorrect .May God bless you dear Patra Ji .May success attend you in all spheres of works ❤❤❤
@@satyapalsingh4429nice bro
Is fibonacci sequence taught in jee preparation. I reckoned ,i was not taught
@@satyapalsingh4429 Are are thanks bhai.❣️ kya aap bhi jee ki tayyari kar rhe he?
thanks for your efforts
Legends will leave this question and will move on to the next question 😅😊
Our teacher at allen, BS sir has expanded our thinking so much, that these questions, feel like class illustrations, I am so thankful to him, that my maths is improving, although I like arvind sir very much
You study at which Allen, kota?
Physics ke one shots bhi lao sir....
You can see this question in Aakash module
8:35 sir isme maine 1 ko -alpha * beta kar diya tha jisse alpha minus beta common aa kar katt gya aur alpha power -1 and beta power -1 bhi kat gye the jisse calculation aur zyada choti ho gyi
The 2nd expression is looks like L'hopital form a^n-b^n/a-b can we do this by lim
B me A^n+B^n will be a factor of Bsquare
Very much trendy question...this question repeats with its varieties...
Sir maybe you dont reply but please still can you help me with a thing…. i directly put n-1 and n+1 for solving bn and got A^n +B^n but my problem is that for (n-1) is the function defined for n=1 …if yes then answer is 2 which is correct answer………. But i thought that a(n-1) is defined for n>=2 …. So my answer is coming bn=A^n+B^n for n>=2 Consider A as alpha. B as Beta a as A
Sir this question was really easy if you find value of a1,a2 and find relation with b1, b2 so this question can be done in just 3min.
You cannot generalize like that. By taking particular cases, the options which dont match, can be rejected. But options which match in particular case, cannot be accepted....
Thanks a lot sir for correcting me
In exam. We can solve this by putting n = 2 or 3 etc in option 2 and 3
Sir physics kb sai start hoga??
An easy way to solve option C is using Newton's identity -
If you consider a^n - b^n = Pn (let's say)
Then an = Pn/a-b
if you expand the terms a1+a2+a3 ... And take 1/(a-b) as common you'll get (P1+P2+P3...)/(a-b)
On the other side
A(n+2)-1 = P(n+2)/(a-b) -1
Take all the LHS terms on the right and send 1 to the left
Then 1 = P(n+2)/(a-b) - {P1+P2+P3...}/(a-b)
Take 1/(a-b) as common and expand P(n+2) as P(n+1) + Pn (which will get cancelled by the Pn that you brought from left) then expand P(n+1) as Pn+P(n-1) ( which will again get cancelled) and like this every term will get cancelled expect for P2 which will cancel the (a-b) below and leave just (a+b) which is just 1 thus solving the problem
How did u expand P(n+2)=P(n+1)+P(n)??
Because P(n+2) = a^(n+2) - b^(n+2) which is not equal to P(n+1)+P(n)
@@DhruvaS-gx7tqit's the Newton's formula
In the question alpha and beta are the roots of the equation x²-x-1 = 0 according to Newton's formula [coefficient of x² × P(n)] + [coefficient of x × P(n-1)] + [constant × P(n-2)] = 0
So according to x²-x-1 = 0
1×P(n) - 1×P(n-1) - 1×P(n-2) = 0
Therefore P(n) = P(n-1) + P(n-2)
That's what I did
@ayuu4952 satisfy both roots
You'll get 2 eqn in alpha and beta
Multiply both equation by alpha^n+-beta^n
Now add both eqn
Rhs will be 0(as root satisy the eqn)
Lhs will be Newton's identity
@ayuu4952watch Sachin sir's lecture on quadratic equations . He has taught it. Very easy.
@@tusharmotwani987watch from 40:10 on this video by Sachin sir quadratic equations ruclips.net/user/liveiOwi85y9oSE?si=VSTdBV9cCQ1ku7vS
1st ques me D and C toh hogya tha…but B me wo root ki value put krne wala nhi soch paya
\alpha > \beta gives you a hint that either values might be needed or the sign of the difference of the roots !! It's always best to write all information implied in the question first !! For quadratic equations, you have Equation with roots, sum of roots, product of roots, and additionally roots (if they are straightforward to find, and if something in the question seems unnecessary without this information) !!
c option because it required the manipulation sir told while B option could be done via taking the nth power
I would like to see the IIT professor comment f)who created this problem) on this video when we started by brute force method in this video.
May be many made mistake in option b
B option is really tough because patience and thinking is required.
Sabse asaan wahi tha😂
@@AvikGupta-ch7oh ha 🤣 jidha Rahe the !!
Sir, yeh data kahan se lia? Ki kitne bacho ne correct kiya, kitno ne galat?
In a long run we should solve it not try values of alpha and beta
Sir indefinite integration ka lec pls
Sahi m bohot khatarnak ques h , mujhe bohot time laga phr bhi smjh nhi aaya
Sir can u pls explain me how {-h}=1-h pls sir
Sir alpha > 1 then how did u used the formula of a(1-r^n)/(1-r),it would be a(r^n -1)/(r-1)
c is correct, but since alpha is (1+ \sqrt{5})/2, it's greater than 1, so the formula for the series should be \frac{a(a^n-1)}{a-1}. Either wayy, when i calculated it, i got the same as the statement, can somenone explain why? it's quite shocking...
Sir itf ka one shot pliz
Bas itna kahunga maza aagaya
Sir can't we just directly use the info of n>_1 for option b and sub 1 in place of n.
No it can't be used as it may be a coincidence.
Solved it in a single go
I dont know if i was lucky enough to make the required changes or it was my subconscious observation
option B ne difficulty di hogi sir max bachon ko kyuki exam pressure me though hota hai manipulation ke bare me sochna, to max bache use manipulate hi nhi kr paye honge
Sir ki smile at 17:22 was very funny 🤣 I can't stop laughing after seeing that
Bhai chote se galti😢 pura question kharab kar sakte the 😂😂
sir PnC one shot please !!!
How beauty looks in maths
Option B indeed was the most critical option
This question is closely related to fibonacci sequence.
Why can’t we use newton’s formula for quadratics ?
IT'S only for sum related terms when provided the the base actually depends on n, but in this question if used Newton's formula it would give cubic root of inequalities
Charan esparsh ✍️🙌🙌
Who will teach physics in this channel
Bhai Kuchh Bhi Ho Ye Toughest Questions Ki Series Rukni Nahi Chahiye! 🔥
The easiest and fastest way is to put n=2(or any other) and check the options.
Sir maf Karo hamse na ho paye ga
Class Notes Kidhar hai
Pls someone Help
Nhi Mil rha hai
Unke site pe jate hi unacademy download Karo Aisa option aa rha hai
Sir b and c to hogya d nehi aa rha
I am in class 10th. I used a different method to find option B. Sir I solved the entire question correctly but took some where close to 30-35 min.😅😅
1q = 20 minutes is all I could calculate
i was so wrong after thinking newtons formula will be used here
From where you get the data that how many attempt question, how many are wrong , how many are correct, if anybody knows than please answer
jee advanced website, go to archives section then to jee reports
Ye sab karne ka confidence tab ata hai jab pata ho ki ye answer correct Hain...
binomial theorem sir
This question also appeared in csir net
Small idea: an forms fibonacci series. an=an-1 + an-2. a0 =0, a1=1
In 9:10 we can also put 1 = -alpha beta
Bhai konse iit me ho
@@UmarSayyed-fx2pd 12 yh
@@murgeshnyamagoud244 bhai tumhara syllabus kab Tak khatam hogaya tha
I did it jn first attempt 🎈🎈
Sir, Physics Tharun Bhaiya 😊😊padhangae kya?
Yes!
Option b
B
Personally I did not find b tricky I am not getting c it is going very lengthy in c
yeah boii
meanwhile I'm applying induction
Values daal ke dekh lo easily ho jayega if nothing clicks
Want more questions ⁉️
B is difficult to find
4:54
• how 10(alpha+beta) became 1?
•how alpha×beta became -1?
Someone help
sum of roots products of roots. alpha + beta= -b/a = 1, alpha x beta = c/a = -1
Bro just do n=1 in option b 😂😂
Pehla wala to easy tha
Newton formula left chat
Still easier than geometry
THE QUESTION WAS REALLY TOUGH ... HATS OFF TO THE TEACHER
BUT A TWIST IS THAT , I PERSONALLY DON'T LIKE THE OPTIONS
this question can simply be done by putting N=2 for satisfying the B option and subsituting value of "alpha and beta" from the given equation .
I think question could be improved by not giving the conditions on "N" and write it is as any natural number or N>100 or any bigger number . So that intriguity of question can me mentioned. :))
ALWAYS NOT NEEDED TO BE PERFECT, IF VALUE TENDS ACCURATE FOR NEGATIVE INTEGERS THEN THE EQUATION WOULD TEND TO -INFINITY WHICH IS NOT POSSIBLE
If you put n=2 and the answer matches with the option, you can not say that it will be true for all n , still you would need to use induction.
Bhai paper me itna nahi dimag chalta
Bruh ek question nahi aata maths me 18 aate hai 18
@@sayantanpaul2982exactly
jo log sahi kiye hai vo tukka maare hai 🗿
Bhai itni देर मे तो jee wale 10 question kr देते hain
The best thing about Kalia sir is that the way he explains a very difficult problem. Solving a tough problem is ok but explaining to average students it is very hard.
BC ye kya hai 😭...
Is any 11th's student getting this ?
Abe yeh vedantu chhor diya kya
Respect karo bhai
Sir EK hi Question mein 20 minutes
Physics kon padhayega
Tharun
Time wasting question.
Sir actually i solved option b without the manipulation by just simplifing in just few steps
😊
Still i think the manipulation thing was great
In future definitely gonna use it in suitable cases
Edit-(alpha-a beta-b an-An bn-Bn)
I did -
( took lcm ab )
(1/a-b).
(a^n.b + a^a.ab - b^n.a - b^n.b.ab) / ab
ab=-1 from eq .
That gives
(1/a-b)(a^n.b - a^n.a - b^n .a +b^n.b) /-1
a^n(b-a)+ b^n(b-a).1/b-a
Bn= a^n+b^n .
Did you telescope the equation or did you substitute an=a^n-b^n/a-b properly?
@@TheHellBoy05 i just simplified it i.e. expanded the terms and used ab=-1 within 1-2 max
@@roshaninayak257 yeah I was attempting this problem and I saw for myself what you meant. On multiplying a and b in the numerator and denominator, you get a²+1 and b²+1 both places when you take a^n and b^n common
Who will teach Physics????
Tharun
Well, this aₙ is actually Fibonacci Sequence
Cuz aₙ⁺₁ = aₙ + aₙ⁻₁
And
a₁ = 1
a₂ = 1
a₃ = 2
a₄ = 3
It's fibonacci Sequence, and I think its beautiful
WHEN YOU MANIPULATE WITH THE OPTIONS, YOU WILL REMAIN with Beta(n) which is not possible
Mere hisaab se C option tha
We can put values of n to calculate few options😂😂