good day sir . Sir i have been watching you from your video of ramanujan paradox . sir i am in 11 th class preparing for jee . sir can you please upload an worksheet of quality questions of each chapter . sir i will be very grateful to you
i solved this question in 5 lines, RTF, ab + bc +ac = ((a+b+c)^2 - (a^2+b^2-c^2))/2 = (-a^2-c^2 -8b + 16)/2 = (4a-a^2)/2 + (4c-c^2)/2 [substituted 2b] = 2 + 2[Used differentiation/Quadratic eq concept to find maxima and put in reqd equation] = 4
ab+bc+ca=b(a+c)+ac Now Put the value of b from given expression (4-a-c)/2 ×(a+c) + ac Solve it. We get (4a-a^2+4c-c^2)/2 =4- (a-2)^2 -(c-2)^2 Which is max when a=2, c=2
I didn't understand the solution. Anyway we can do like this. (a+b)+(b+c)=4 As we know Am is greater or equal to GM so (a+b)+(b+c)/2>=(a+b)(b+c)^1/2 4/2>=((ab+bc+ca)+b2)^1/2 Squaring we get 4>=(ab+bc+ca)+b2 4-b2>=ab+bc+ca As b is real so max value of ab+bc+ca is 4.
@@UbiytsaEditz At first , for using am and gm inequality, it's not like a,b,c>0 , here it's a+b,b+c>0 and here it is , so it is true , also can be done by cauchy schwartz or vector method
I kinda took a different approach. Here's how I solved it : a+2b+c=4 a= 4-2b-c x = ab+bc+ca = ab + (4-2b-c)(a+b) = -a^2 -2b^2 -2ab +4a +4b = -(a+b)^2 -b^2 +4(a+b) Let a+b=d x =4d -d^2 -b^2 For x to be maximum, 4d-d^2 should be maximum and b^2 should be minimum. b^2 is minimum at b = 0 Let f(d) = 4d-d^2 f'(d) = 4-2d f''(d) = -2 f'(d) =4-2d=0 d=2 As f''(d)
I did it as follows: a+2b+c=4 ---(1) a+2b+c = (a+b) + (b+c). Squaring results, (x+y)² = 16. Maximum value will be when x=y, where x=a+b & y=b+c, i.e., when a+b=b+c implies a=c. Putting this in (1), gives us a+b=2=b+c. Now, (a+b)(b+c)= 4 implies, ab+bc+ca=4-b². So the value of ab+bc+ca is maximum when b=0 and max(ab+bc+ca)=4. -Nihar Tripathy
(x+y)² = 16 is just an equation representing a relation between x and y which forms a pair of straight lines. x² + 2xy + y² -16 =0 and factorising it: (x+y-4)(x+y+4) = 0 is combined equation of two straight lines and how tf can you take maxima as x=y, totally a wrong solution, but somehow got you to the right answer lol
After getting the quadratic equation in variable c, we can go in another way also by making this equation in two parts. One with equal to zero and the other one
There's a alternate solution also,if rewrite a+2b+c as (a+b)+(b+c) and consider a+b=x and b+c =y. Then x+y=4 .Now applying AM>=GM we can write x+y/2 >=√(xy). Now put x+y=4 we get 2>=√xy Now since both lhs and rhs are positive we can square them and we get 4>=xy. Now put x=a+b and y=b+c we get xy=ab+bc+ac+b^2. Therefore 4>=ab+bc+ac+b^2 and therefore 4-b^2>=ab+bc+ac Now the max value of lhs=4 therefore we get ab+bc+ac=4
I also solved by this method. But this method is applicable for positive values of a+b and b+c. In the given question, the condition is that a, b and c belong to set of all real numbers. Thanks
Sir my approach was quite different. Let a+b = p and b+c = q then ab+bc+ca = pq-[b]sq for max value of ab+bc+ca. [b]sq must be least and pq must be max therefore I concluded that p,q are positive and b=0. then the question simply became a,c are positive with sum 4 and find their max product which was 2*2=4. :]
Is this solution right= a+c=4-2b ,a+c/2=2-b 2-b>=√ac_ b^2-4b+4>=ac for ac max it will be equal ac+bc+ac= b(a+c)+ac. B(4-2b)+ac =4b-2b^2+b^2-4b+4= -b^2+4 so max=4
Easiest approach: ab + bc + ca = b(a+c) + ca = b(4-2b) + ca {By AM GM inequality ca (max) = ((c+a)/2)^2 = (2-b)^2} --> 4b - 2b^2 + 4 + b^2 - 4b = 4 - b^2 So max value is 4
[ Use Partial Derivatives (I am not sure if that is part of the 11 / 12 Syllabus though) ] Using partial derivatives and the concept of Hessian matrix to check for the maximum, such questions can be straightforward !!
Well one thing tha could be done is square both the sides and arrange eqn as:(a+c)²+4(b²+ab+bc)=16,noww,put a+c as 4-2b here, u clearly see that on simplyfing b=0 emerges as like its trying to yell is some thing and that is -"I am the soln" rest needa be thinked by u being an ambitious maths student,So on putting b=0 boom---the ans has arrived,,yay....Done!!----So HW find mistakes(iff any...)
sir can it also be done by vectors considering the equation for the maximum as 2 separate vectors as ; (ai+bj+ck).(bi+cj+ak) taking their dot product and writing them in terms of cos and squaring both sides and putting the whole equation as less than or equal to 1 (because of cos^2max) ... similar to Cauchy-Schwarz inequality ?
Now I am studying in 10th standard I have made small research in mathematics I just want to confirm that it is useful In mathematics field a²=[(a-1)×2+1]+(a-1)²
Bro your ff pic is fire op🔥 Bhai it's ok ,everyone can become great take it as a joke . Frankly this is one of the infinite combination you are showing It's like saying I discovered water in my bottle Good luck ,I love your enthusiasm 💪
@@vipul_IITian bro it is said maximum, so first of all , we can't take negative number as the value of a,b or c because it might not give us the maximum value
Sir plzz solve my confusion... Sir I tried this same question by using am gm inequality.. and sir i got the answer but my answer was not 4.. it was in fraction... Plzz reply sir jii
Sir isme maine a + b + b + c = 4 kar diya fir (a+b) aur (c+b) ko group kar diya fir am greater than gm laga diya to fir ab+bc+ ca = 4 - b² aaya..ab uski value max tab hi hogi tab b² min i.e 0 sor 4...please coreect me if my method is wrong
Ap log sirf 12th tak ya college ke 2nd year tak aise problems solve krte rahoge. Uske baad agr ap engineering ya medical loge tab kya apse aise algebra ke questions puchenge kya? Aisa math solving practice ka future kya he?
I solved it like: The below expression can be written as abc(1/a+1/b+1/c)= a+b+c Now from above expression, a+b+c=4-b, so max value is 4 (I.e. b will be zero in case of max value)
Sir I have a question, shouldn't the quantity of X be just less that 4, instead of X less that equal to 4? Because if x is equal to four then it will subtract from four and cancel out just leaving ( c-2 )² less than equal to zero , where for example C is equal to 1 ,then (1-2)² = (-1)² = 1 which is not lesser than or equal to zero , right???
Sir directly aa rha 2b hai toh b ko kam se kam rakhna hai Toh b=0 Since, multiple of 2,2 is more than 1,3 , we take a = c = 2 Therefore , we get the maximum value as 4
@@diffrant6209 see put value of a from 1st eqn in 2nd and then partial differentiate it 1 wrt b and 2nd wrt C then solve these 2 eqn and then put it back in the question asked you will get 4 as answer
SIR I DID IT WITH ARITHMETIC MEAND AND GEOMETRIC MEAN CONCEPT IS IT CORRECT , MINE WAS , [A+B] +[B+C] / 2 > root[a+b][b+c]-----> ab+bc+ca< 4 - [b]squared
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Sir Ravi substitution padha aur explain kar de gi a ek video Mai pls pls pls pls sir🙏🏻
Sir pls make video on Basic Arithmetic and Algebra and Geometry
Sir goat grazing math problem solve karwa dijiye
Sir 🙏🙏
Sir Class 11th के video बना दीजिए महान कृपा होगी सर जी
Namaste sir aap koi series laaiye n plz basic maths to advance cat ke liye plz
May everyone get their desired ranks in jee❤❤❤❤
You mean you are desiring for everyone to get AIR 1 💀
@@basically-ky2rpeveryone doesn't go for the Air 1
Everyone can't get AIR 1
Bro wants likes!! 💀
kya hua be ajkal indians videshi comment copy kar ke chep rhe hai ye haddii wala emoji bada famous ho raha kya hai ye sab @@ShashankChoudhary-vl8kb
I think you should bring a practice session on all the chapters but it shouldn't be a PYQ's practice session as we r doing it by ourselves
Sir, please explain the solution of this question:-
Prove that sin(2π/7) + sin(4π/7) + sin(8π/7) = √(7)/2.
good day sir . Sir i have been watching you from your video of ramanujan paradox . sir i am in 11 th class preparing for jee . sir can you please upload an worksheet of quality questions of each chapter . sir i will be very grateful to you
Go for cengage
Go to Coaching material
If you want Best option : "Competishun" is best 😁
Sb chupate hai but me share kr rha hu 😆😊
@@GarvitGupta-xx5hyonly dpp kar cengage ka our integer type wala excercise
Kuch nahi sab achha hoga
Jee ADV k pyq kar le beta
i solved this question in 5 lines,
RTF,
ab + bc +ac
= ((a+b+c)^2 - (a^2+b^2-c^2))/2
= (-a^2-c^2 -8b + 16)/2
= (4a-a^2)/2 + (4c-c^2)/2 [substituted 2b]
= 2 + 2[Used differentiation/Quadratic eq concept to find maxima and put in reqd equation]
= 4
Great
ab+bc+ca=b(a+c)+ac
Now Put the value of b from given expression
(4-a-c)/2 ×(a+c) + ac
Solve it. We get
(4a-a^2+4c-c^2)/2
=4- (a-2)^2 -(c-2)^2
Which is max when a=2, c=2
😮
Sir इस question को calculus वाले method से भी solve कर दीजिए
I didn't understand the solution. Anyway we can do like this.
(a+b)+(b+c)=4
As we know Am is greater or equal to GM so (a+b)+(b+c)/2>=(a+b)(b+c)^1/2
4/2>=((ab+bc+ca)+b2)^1/2
Squaring we get 4>=(ab+bc+ca)+b2
4-b2>=ab+bc+ca
As b is real so max value of ab+bc+ca is 4.
No u cant....this is only valid when a,b,c>0 is given in question
even if b is real, it can have a negative value
@@UbiytsaEditz At first , for using am and gm inequality, it's not like a,b,c>0 , here it's a+b,b+c>0 and here it is , so it is true , also can be done by cauchy schwartz or vector method
@@DayerKing oh yea....thanks bud
Cool... Looked like a simpler and more efficient approach. Kudos to rivun, dayerking and ubistaeditz
I kinda took a different approach. Here's how I solved it :
a+2b+c=4
a= 4-2b-c
x = ab+bc+ca
= ab + (4-2b-c)(a+b)
= -a^2 -2b^2 -2ab +4a +4b
= -(a+b)^2 -b^2 +4(a+b)
Let a+b=d
x =4d -d^2 -b^2
For x to be maximum, 4d-d^2 should be maximum and b^2 should be minimum.
b^2 is minimum at b = 0
Let f(d) = 4d-d^2
f'(d) = 4-2d
f''(d) = -2
f'(d) =4-2d=0
d=2
As f''(d)
Bruh... the first step is wrong... u took c as common and substituted the value of a in c
@@balasudalaimuthu5151yes you're right
@@mandar_desai34 Yess bruh...😉
Sir cengage ke aache Qs lao sir plz. Maza aagya ❤
Sir plzz saare ch ki ques solving ki series baniye naa
I am very happy😊😊 because i solve this in different way by only 30second😊😊🎉🎉
Please tell how
I did it as follows: a+2b+c=4 ---(1)
a+2b+c = (a+b) + (b+c). Squaring results, (x+y)² = 16. Maximum value will be when x=y, where x=a+b & y=b+c, i.e., when a+b=b+c implies a=c. Putting this in (1), gives us a+b=2=b+c. Now, (a+b)(b+c)= 4 implies, ab+bc+ca=4-b². So the value of ab+bc+ca is maximum when b=0 and max(ab+bc+ca)=4.
-Nihar Tripathy
Bhai socha kese
how did you conclude that the point of maxima is x=y for the equation (x+y)^2=16?
@@a_minorNo this solution isn't entirely correct, that is not the point of maxima.
(x+y)² = 16 is just an equation representing a relation between x and y which forms a pair of straight lines. x² + 2xy + y² -16 =0 and factorising it: (x+y-4)(x+y+4) = 0 is combined equation of two straight lines and how tf can you take maxima as x=y, totally a wrong solution, but somehow got you to the right answer lol
Very simple ques.
Okay very good
Sir you're great. Thank you so much sir. Maths seems like a game solving these questions w you. You really make things interesting. 🙏
Bro inka koi paid course hai kya?
Love you Aman Sir ji ♥️🙏🏻
After getting the quadratic equation in variable c, we can go in another way also by making this equation in two parts. One with equal to zero and the other one
sir please take class on Olympiad level of diophontine equation
Video quality is great and your teaching methods
((a+b)+(b+c))/2 >= ((a+b)(b+c))^0.5....am>= gm
b² goes to lhs ...for max value b becomes 0...therefore max values is 4
am gm only for positive values, they gave abc belong to real numbers
Sir please tell solution of that 35 years old question from subjective paper of iit
I am very curious about that
Please support of you also want
Which question can you tell plz
What's that 'out of the box' method
Given expression b(a+c)+ac
Now put b= (4-a-c)/2 from given expression and solve, ee get
4-(a-2)^2 -(c-2)^2 which is max when a=2, c=2
Sir please make a video on other methods also
There's a alternate solution also,if rewrite a+2b+c as (a+b)+(b+c) and consider a+b=x and b+c =y. Then x+y=4 .Now applying AM>=GM we can write x+y/2 >=√(xy). Now put x+y=4 we get 2>=√xy Now since both lhs and rhs are positive we can square them and we get 4>=xy.
Now put x=a+b and y=b+c we get xy=ab+bc+ac+b^2. Therefore 4>=ab+bc+ac+b^2 and therefore 4-b^2>=ab+bc+ac Now the max value of lhs=4 therefore we get ab+bc+ac=4
This will not be true for negative values . As we are dealing in real number values of a b and c
we can also do this " [(a+b) + (b+c)]/2 >= [(a+b)(b+c)]^1/2, a.m >= g.m formula
by simplifying this [(a+b)(b+c)]^1/2
That is only valid for positive values of a,b &c.
This method is kind of contradictory
@@_lost._.in._.space_ yeah true, unfortunately works here, but not in the all the cases, thanks for correcting me
If you want to apply am GM
Then square the numbers and then add
I did it in this way and got ab+bc+ac≤4-b^2
I also solved by this method. But this method is applicable for positive values of a+b and b+c. In the given question, the condition is that a, b and c belong to set of all real numbers. Thanks
Sir my approach was quite different.
Let a+b = p and b+c = q
then ab+bc+ca = pq-[b]sq
for max value of ab+bc+ca. [b]sq must be least and pq must be max
therefore I concluded that p,q are positive and b=0.
then the question simply became a,c are positive with sum 4 and find their max product which was 2*2=4.
:]
Really great approach
@@AmanGupta-fi4wb thanks
Great approach I loved it bro !
a^b+b^a=32 and a+b=6 discuss in detail explanation
a+2b+c=4
Square both side
a^2+4b^2+c^2+4ab+4bc+2ac=16
Rewrite
4(ab+bc+ca)=16- (a-c)^2-4b^2
Max when a=c and b=0
4 answer kaise aayega fir ??
Aayega, check kar
Is this solution right= a+c=4-2b ,a+c/2=2-b 2-b>=√ac_ b^2-4b+4>=ac for ac max it will be equal ac+bc+ac= b(a+c)+ac. B(4-2b)+ac =4b-2b^2+b^2-4b+4= -b^2+4 so max=4
👋👋👋👋👋👋👋👋👋Sir here one equation I have to solve for x
How do we solve for x
a = x(1 - b/(sqrt(x ^ 2 + b ^ 2)))
sir i solved it by breaking it into (a+b)+(b+c)=4 and then applying AM>=GM so 4>=ab+bc+ca+b^2 and so ab+bc+ca is max at 4
Sir, please bring more problems from Cengage 🙏🙏🙏🙏
If I break (a+b)+(b+c) then we apply AM>GM it will be very easy
Your method is also good 🙏
but you'll be left with a b² term, what about that?
2≥√ab+ac+bc+b²
oh wait I got it, ab+ac+bc≤4-b² and b= 0 for max value
nice method 👍🏻
@@AlmostMathb2 zero kaise kiya 🗿
@@skies_06 b² is always greater or equal 0 hoga na
Toh whole expression less or equal 4 hoga
A two liner question using vector dot product
Kaise bhai??
😅
i can guess what youre trying to say, (a i cap + b j cap).(b i cap + c j cap) will indeed give us ab + bc + ca but how do you proceed further?
Sir iota wala question ka Sol samja dijia
Sir hm isme a,b,c ko let bhi kr skte hai (a=2, b=0, c=2)
Now,
X=ab+bc+ca
X= 2×0+0×2+2×2
X= 4
Maximum value of (ab+bc +ca )= 4
Answer dekhne ke baad toh koi bhi let kar lega
Great.a very good and elegant solution. Superb explanation 🙏🙏🙏🙏
राम राम ❤❤❤गुरू देव ❤❤😊
AM>= GM
((a+b)+(b+c))÷2 >= √((a+b)(b+c))
so
2>=✓(ab+bc+ac+b²)
so
ab+bc+ac
Ye hua n jee students waala method....
God observation
Sir yeh question bahat badiya hai....Tarun Khendelwal sir ne class me karwaya tha❤❤❤
Kis batch me bhai
SIR IN THE LAST STEP WHEN WE WRITE X
It is right for c=2
It can't be true because square of a no is always positive nd less than contradict
Square can also be zero for which this inequality holds true
Easiest approach:
ab + bc + ca
= b(a+c) + ca
= b(4-2b) + ca
{By AM GM inequality
ca (max) = ((c+a)/2)^2 = (2-b)^2}
--> 4b - 2b^2 + 4 + b^2 - 4b
= 4 - b^2
So max value is 4
min nhi max value Nikalni hai
@@kyro2349 galti se min likh diya tha edit kr diya
Basic concept is the best method
[ Use Partial Derivatives (I am not sure if that is part of the 11 / 12 Syllabus though) ] Using partial derivatives and the concept of Hessian matrix to check for the maximum, such questions can be straightforward !!
Well one thing tha could be done is square both the sides and arrange eqn as:(a+c)²+4(b²+ab+bc)=16,noww,put a+c as 4-2b here, u clearly see that on simplyfing b=0 emerges as like its trying to yell is some thing and that is -"I am the soln" rest needa be thinked by u being an ambitious maths student,So on putting b=0 boom---the ans has arrived,,yay....Done!!----So HW find mistakes(iff any...)
Statics chapter please
Sir please solve it with *calculus* and *out of the box method* too
Sir aap bhot acha pdate h
Padology is ❤
sir can it also be done by vectors considering the equation for the maximum as 2 separate vectors as ; (ai+bj+ck).(bi+cj+ak) taking their dot product and writing them in terms of cos and squaring both sides and putting the whole equation as less than or equal to 1 (because of cos^2max) ... similar to Cauchy-Schwarz inequality ?
🗿Bhaiya ji calculas se batao n kaise bnega
11 me hu
@@skies_06 maxima , minima and monotonicity ho gaya ?
@@ItsMeHawk Maxima minima ata h curiosity me padha h monotonicity nhi padha h
@@skies_06 calculus se karega toh mar jayega Cauchy-Schwarz padhaya hoga na sequence and series me?
@@ItsMeHawk Ayein ye kya h
How b is the root of that equation
Sir sare solutions bataiye please so that ki thinking evolve ho sake
Please calculus method batao
This question is also in fiitjee module megacosm
Sir ek request hai ap please 🙏🙏 sir video banao mathematical physics par please sir
Now I am studying in 10th standard I have made small research in mathematics
I just want to confirm that it is useful In mathematics field
a²=[(a-1)×2+1]+(a-1)²
I have proof
this is literally the most useless thing
😂😂😂😂😂
Bro your ff pic is fire op🔥
Bhai it's ok ,everyone can become great take it as a joke . Frankly this is one of the infinite combination you are showing
It's like saying I discovered water in my bottle
Good luck ,I love your enthusiasm 💪
From Nepal excellent teacher in the world
Sir circle ke properties bataye
Sir please solve
Area bounded by y²= |x²-x⁴|
Soo usefull concept guruji
Sir please solve the problem
How many times we can scramble a Rubik's cube
one line soln if abc are +-- take a+b=x and b+c=y x+y=4 then xy=ab+bc+ca+b^2 for ab+bc+ca maxm b^2 shall be minm i.e 0 by amgm ans 4
Hi , am gm inequality works for positive real nos not for real nos which include negative nos
@@srividhyamoorthy761i have mentioned in my soln if abc are + ve then
Sir pls tell Out of the box method 😊
3 is the answer, I solved the question before watching this video. I am currently studying in 9th grade
It can be done in 2 seconds . Let a=2, b=0and c=2. Then max value will be 4.
you cant let the values be anything satisfying the equation...take a negative value and you'll get wrong ans
Bhai yeh toh tumne ans dekhne ke baad guess Kiya hai
Aise sawaal nahi hote hai
@@vipul_IITian bro it is said maximum, so first of all , we can't take negative number as the value of a,b or c because it might not give us the maximum value
@@ShivanshWick me answer dekhne ke pehle kiya
@@suryakantamukherjee4669 I can't understand . B=0 so a=b=2 how??😬🫨
Sir can you make a special video on special numbers such as e and π
Sir plzz solve my confusion... Sir I tried this same question by using am gm inequality.. and sir i got the answer but my answer was not 4.. it was in fraction... Plzz reply sir jii
Sir isme maine a + b + b + c = 4 kar diya fir (a+b) aur (c+b) ko group kar diya fir am greater than gm laga diya to fir ab+bc+ ca = 4 - b² aaya..ab uski value max tab hi hogi tab b² min i.e 0 sor 4...please coreect me if my method is wrong
Your mathod is wrong because it is not given that a,b and c are positive numbers. AM >GM only works for positive numbers only
Sir please homegenous equation par detailed video bana dijiye engineering level ka
Ap log sirf 12th tak ya college ke 2nd year tak aise problems solve krte rahoge. Uske baad agr ap engineering ya medical loge tab kya apse aise algebra ke questions puchenge kya? Aisa math solving practice ka future kya he?
Sir, please make a video on integration of root sinx dx
Sir please continue love with graphs series 🙏🙏🙏if next lec won't come then it will not worth
Using AM≥GM method would be an easier eay to do this question, i think , please correct me if i am wrong.
Sir, jee sutra class 12 ka bhi banao please 🙏
Sir please upload the answer of that lost IIT question
Ye to vectors se solve hoga na sir. Usse bahut aasani se hoga
Fibonacci sequence par video upload karo
When x=4, 2(x-4)=0
Then How can be 4 is the answer?The condition won't be true.I think the maximum value is 3.
If iam wrong,forgive me Plz🙏
Sir we can also do it by a = b = c = 1
isn't that just a pure guess, it won't necessarily give the correct answer in every question
Glat h tarika
@@AlmostMathwo to isme bhi sahi nhi de rha 😂
@@sakshamsingh1778 abc each =1 Karne pe tukke se ans 4 aa rha hai
By dot product of vectors and using its inequality we can get its ans verbally
I solved it like:
The below expression can be written as abc(1/a+1/b+1/c)= a+b+c
Now from above expression, a+b+c=4-b, so max value is 4 (I.e. b will be zero in case of max value)
Sir I have a question, shouldn't the quantity of X be just less that 4, instead of X less that equal to 4? Because if x is equal to four then it will subtract from four and cancel out just leaving ( c-2 )² less than equal to zero , where for example C is equal to 1 ,then (1-2)² = (-1)² = 1 which is not lesser than or equal to zero , right???
We can solve it with Lagrange multiplier
Calculus se hoga ??? Sir, please please please please kar digiye.
At 4:35 can anyone explain please how is b root of equatiion?
Exactly I too have this doubt
What sir meant was, treat it as a quadratic equation in b , 2 is the coefficient of b^2, 2(c-2) is coeff. of b and c^2-4c+x is the constant.
@@siddharthavoleti4758 ok thanks alot
Sir directly aa rha
2b hai toh b ko kam se kam rakhna hai
Toh b=0
Since, multiple of 2,2 is more than 1,3 , we take a = c = 2
Therefore , we get the maximum value as 4
Sir partial differentiation se aap a,b aur c tenno ki value nikal sakte ho
Explain
@@diffrant6209 see put value of a from 1st eqn in 2nd and then partial differentiate it 1 wrt b and 2nd wrt C then solve these 2 eqn and then put it back in the question asked you will get 4 as answer
SIR I DID IT WITH ARITHMETIC MEAND AND GEOMETRIC MEAN CONCEPT IS IT CORRECT , MINE WAS , [A+B] +[B+C] / 2 > root[a+b][b+c]-----> ab+bc+ca< 4 - [b]squared
It is not given that the numbers are positive so u can't apply Am Gm
Rotation in physics and maximum, minimum in maths .... Best to drop both these topics... Whatever you do you will not be able to solve in exam.....
sir why did u delete the other video
Bring content on btech engineering mathematics
Being so bad at math even I could solve this problem in 30 secs. Idk what’s hard in this
Why I am using Lagrange multiplier
Sir I'm from Bangladesh . Regularly i watch your wonderful class and enjoy.
Please try to post more QNA videos. ❤❤
Sir pls how with calculus and the other method you said in the begining
Sir x=4not possible as x-4 becomes zero n left hand is thengreater than zero
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Sir vectors se bhi kr skte hai na?
Sir we can use Cauchy Schwarz inequality in this problem
Sir please start pyq series of jee mains please sir,please like this comment
sir please bring btech maths content