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Let a vector r=a+bi z=c+di Given, |r|=2 |z|=2×2^(1÷2) Find range of r.z=? r.z=2×2×2^(1÷2) cos (angle bwn r and z)=4÷2^(1÷2) cos(the angle) cos(angle bw r and z)=[-1,1] r.z=[-4×2^(1÷2),4×2^(1÷2)]
@malaygupta216 Absolutely correct But question mein ek value nhi Range of all values manga hain. Toh solution ki lower range ki value kaha hain??? Hence method not great
Did it by 7 methods 1.Coordinate geometry 2.Trigonometry 3.Vectors 4. Algebra 5. cauchy shwards inequality 6. Maxima minima using differentiation 7. Pure geometry Last vale ka 18 hai uttar You can find the max value of ac by using vectors or trigonometry and then you have to make a quadratic in xy to find the minimum value either use -D/4a or use derivative or just put -b/2a to the eqn formed. Right now I am trying to find the minimum value of x²+y² using the distance formula ❤
in the given question since there is no condition given on a b c d , we can assume any values for abcd which satisfy the given equation we can even assume a = b , c = d if needed , since there is no condition on a,b,c,d let us assume a = b = root 2 , and c = d = 2 , hence ac + bd = 4root2
Guys,I think we can do it more easily,I am Bipc student so just tell me if we can do it like this- a^2+b^2=4 It is like Pythagoras eq for right triangle. Here we don't know a or b, but we know hypotenuse is "2" from root of 4. So, right angle triangle of sides a, b, 2. Similarly, c^2+d^2=8 can also be considered a right triangle of sides-c, d, 2√2(from root of 8). Then we can expand the first triangle to get a like triangle. Multiply all sides with √2.The new triangle we get with sides- √2a , √2b , 2√2(hypotenuse). This new triangle is similar to 2nd triangle. We can join both the hypotenuse to form a rectangle. This rectangle has sides c, d, √2a and √2b. Taking √2a=c and √2b=d.(cause opp. Sides of triangle are equal). ac+bd= √2a*a+√2b*b =√2a^2+√2b^2 =√2(a^2+b^2) ;[taking √2 common] We know a^2+b^2=4. Then ac+bd=√2(a^2+b^2)=√2(4) =4√2.
This question was not that much hard but I thought we have to find the value of ac+bd... I had already thought for A1, A2, A3 & A4... I didn't know we have to find the range otherwise it was fairly a question for a student preparing for Olympiads in class 7th or 8th... BTW I am also in class 8th.... Also there was nothing that much typical to see in A3, I made that conclusion just by looking at the equation an in fact it was my first thought... But according to me this question made a fun of me as I thought we can find the actual value of ac+bd, I was repeatedly looking the question if a, b, c & d were Integers or we had to find all possible values of ac+bd...then this question was very easy in fact
Whenever there is a question asking the value of an expression, it's wrong to assume it will have just one single value, bcuz finding one of all possible values is quite simple. But to analyse in all direction is where true beauty lies
a²+b²= 9 9(cos²p + sin²p) = 9 .·. 9cos²p = a² and 9sin²p = b² .·. a = 3cosp and b = 3sinp Similarly, c = 2cosq and d = 2sinq ad - bc = 6cospsinq - 6cosqsinp ad-bc = 6 (·.· given) cospsinq -sinpcosq =1 sin(p-q) = 1 .·. p-q=90 p=90+q ..(1) Replace (1) in value of a and b, we get a = -3sinq and b = -3cosq Now, ac = -6sinqcosq = -6((sin2q)/2) = -3sin(2q) Max value of sin2q = 1 Max value of ac = -3(1) = -3 .·. K = -3 Bhaiya iske aage kya karu??? 25:27
@@kaustavbhowal2109 bro but sin function ka max value 1 hota hai and sin2q = 2sinqcosq, isse max value of sin2q =1 aayega. Please explain me what I am doing wrong over here.
i reached until 6:04 , this student doesn’t understand the core meaning of the question and based upon any fact that they think is wright they are doing , i didn’t use any formula basic understanding the opposite of square is root ok ((a*a)*(c*c))+(b*b)*(d*d))= 12 opposite is Root so 12 root is 3.46410161… may be i am wrong , but i would like to understand why i am wrong?
@@MAYANKGUPTA-t3f Yes , I agree ! We can also use the Cauchy-Schwarz inequality to get : (a^2+b^2)(c^2+d^2)≥(ac+bd)^2 Substituting the values, we have: min{ac + bd} = √(32) = 4√2
I was getting answer 4√2 in positive but not in negative here is my way of tackling it a^2+b^2=4 C^2+D^2=8 Here I find that any number minimum value lies when bothe number are equal so I got value a,b as √2 and c,d as 2 so by normal multiplication I got answer 4√2
I am also at class 10th I done it very differently tho. I used right angle triangle but I didn't use trigonometry. I used pythogoreas theorem. a²+b²= 2² c²+d² = (✓8)² they are following pythogoreas theorem so we can make right angle triangle having hypotenuse 2 and ✓8 let make triangle ABC for a²+b² 2² and PQR for c² + d² = (✓8)². where 2 and ✓8 is hypotenuse of ABC and PQR respectively now draw perpendicular bisector for both traingle and I think you can solve next.
This question cannot jave a definite answer put a=2,b=0, c=0,d=root 8. So ac+bd=0. Esse infinite tarike s ehumm kars hakte hain And a,b,c,d are real so i can take irrational numbers also so infnite possibiliy is there
while the possibilities are infinite, that does not mean "it can take any real value"... there are infinite numbers between 0 and 1, does that mean that every single real number is between 0 and one? no.... So, if you can't determine exact value you need to determine interval...
if the min value of a=-2 and max value=2 then b=0 similarly, if c min value-2*root2 and max value =2*root2 then d=0 ac=put max valur of one and min value of another so ac=-4root2 also, cd=0 if ac=-4root2 therefore ac+cd ki min value -4 root 2 and max= 4root2 im in class 10 and im able to do it under 5 minutes all alone
Let's analyze the equations: *Given Equations* 1. a² + b² = 4 2. c² + d² = 8 3. ac + bd = ? a = 0, b = c = d = 2 *Verification* 1. a² + b² = 0² + 2² = 4 (valid) 2. c² + d² = 2² + 2² = 8 (valid) 3. ac + bd = 0_2 + 2_2 = 4 (valid) Why does it have to be 4√2 when it can be 4
Bhaiya abcd real hai toh a²+b²=4 hai toh isse just a possibility max value ke liye (✓2)²+(✓2)²likh do n similarly sabko c² ko (✓4)² compare kardo sabko correct me if i am wrong and minimum ke liye -ve input krdo root ke bhar
Mujhe a²+b²=4 se standard circle ka eqn chamak rha thha xd and mujhe ek aur approach dikh gya aur geometry se bhi ban gya ye que..😅 aur maine is que ko AM GM se try kia thha but mera answer kuchh aur aa rha thha
Two equations for finding value of 'x' and 'y' 1) ax + by = C 2) px + qy =C (If "C" is the Real Number as 0,1,2,3......). Can we write or say. "ax+by = px+qy = C" ("ax+by = px+qy") ( Because both are equal to "C"). Is it true or not. please help with example 🙏
Why can't we go a²+b²-4 = 0 and then (a-b)² = 0 => a=b=√2? Then same way we can get c=d=±2 So ac+bd = ±4√2 Am I doing something inherently wrong, or oversimplifying?
Simplest question I've ever seen. The number of equations= the number of values of variables that are fixed. So we can choose any two variables whose values we can suppose as fixed. And rest can take any value. So i take b=d =0. This gives 4√2🤷🏻♂️
@MathsUnpluggedIndia if there are 4 variables and 2 equations, then the values of only 2 variables can be fixed. So we can randomly put any real value in the rest of two variables. And that's what I did. Even if i had taken different real values for b and d we would still have gotten the same answer.
@@MathsUnpluggedIndia I didn't see the whole video. I just saw the thumbnail and answered it. I didn't we had to find a range of values. Ig option elimination may help after finding multiple values.
That's not how real maths work. If I ask x^2=4 , x=? , now will you say oh x=2 is the answer? No, that's not complete, the answer will be +-2 , so ,even if you just read the thumbnail that also means all values of ac+bd. But good to see you r into maths ritik😅
i used method that a^2 >= 0 therefore max of a = root 4 , min = - root 4 max of c = root 8 , min = - root 8 so multiplying range is root 32 to - root 32 but very fascinating method, never thought of this approach
Tum sb bot ho A²+B²=4 ..hai C²+D²=8 Ac+bd= ..find krna hai So agar ham root 2²+ root 2 ² krne to 4 hoga Aur dusre me 2² rakhdo do 8 hoga To a or b ki value hame root 2 mil gyi or c or d ki 2 To a ×c =root 2 ×2 =2root2 Same with bd 2root 2 than we add it 2root2+2root2 =4root2 .. I do it in my mind 4
bruh yeh jee wala sawal tha? Mera dost mujhe yeh question ka min and max value find out karne ka challenge diya tha, and I actually solved just by inserting random values of a,b,c,d
@@Doraemon-jm5oo 🤦♂️ that's not how you solve problem. Just getting the answer and feeling oh I solved it is worst that can happen. Develop problem solving and conceptual clarity as to why what you did was wrong or right
⚡BIG BANG to JEE Maths!! A session you just can't miss! : ua-cam.com/video/WUmWcMDWYzE/v-deo.htmlfeature=shared
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Let a vector r=a+bi
z=c+di
Given, |r|=2
|z|=2×2^(1÷2)
Find range of r.z=?
r.z=2×2×2^(1÷2) cos (angle bwn r and z)=4÷2^(1÷2) cos(the angle)
cos(angle bw r and z)=[-1,1]
r.z=[-4×2^(1÷2),4×2^(1÷2)]
Nice man!
which is the equivalent of the trigonometric solution he gave.
Maximum value dekh kar he pata chal gayi,
To maximize x+y=c
X=y
Wese he (√2)²+(√2)²=4
2²+2²=8
So Max ad +bc =2√2+2√2
Guess nhi karna hai correct hona hai
That's correct answer bro, he is not guessing. Its a cyclic equation so you would get the maximum at equal value of variables
@malaygupta216
Absolutely correct
But question mein ek value nhi
Range of all values manga hain.
Toh solution ki lower range ki value kaha hain???
Hence method not great
Do correct me if I'm wrong
Did it by 7 methods
1.Coordinate geometry
2.Trigonometry
3.Vectors
4. Algebra
5. cauchy shwards inequality
6. Maxima minima using differentiation
7. Pure geometry
Last vale ka 18 hai uttar
You can find the max value of ac by using vectors or trigonometry and then you have to make a quadratic in xy to find the minimum value either use -D/4a or use derivative or just put -b/2a to the eqn formed.
Right now I am trying to find the minimum value of x²+y² using the distance formula ❤
Bhai, solution to likh deta ese kar diya use kardiya, banda sikha raha ha , flex kar lo bas
Idhar likhna mushkil hai varna likh toh dunga mai@@maxnoob7612
Bhai phys chem ke saath itna kese manage kar rhe ho
Phekne ka tareeka thora kezual h🤣🤣🤣
@@maxnoob7612 L , mathematics dont need solutions, he is right so he can flex
in the given question since there is no condition given on a b c d , we can assume any values for abcd which satisfy the given equation we can even assume a = b , c = d if needed , since there is no condition on a,b,c,d let us assume a = b = root 2 , and c = d = 2 , hence ac + bd = 4root2
That is why they have not asked a particular value
They have asked a range in which all values of ac + bd will lie definitely.
Which is 4√2,-4√2.
Like who nailed using trigo😅
Vector se kiya
👍
2a²=c²
2b²=d²
√2a=c
√2b=d
a×√2a+b×√2b=?
√2(a²+b²)=?
√2(4)=4√2
Same answer right????
Bro please explain me how you solved using trigo
@@ArijeetBanerjee-md1qw Bhai video dekh lo last me wahi bataye hain
1. given a,b,c,d are reals
2. we have to find range
take two sets {a,b},{c,d}
now apply CAUCHY
(a^2 +b^2)(c^2+d^2)>= (ac+bd)^2 ==> (ac+bd)^2 -4root2
bro literally i got gosebumps😯 after that 5th approach 🔥🔥🔥
I was in ioqm foundation class once and i have seen rhis exact quesiton once before
We solved it for an entire class, thinking and discussing
@@atomicbaby-w4e that's an actual class!
Mod of Dot product of
ai + bj and ci+ dj ≤ √(a2+b2)(c2+d2)
|ac + bd|≤√(32)
Yup aise kiya maine
👍🏽
Guys,I think we can do it more easily,I am Bipc student so just tell me if we can do it like this-
a^2+b^2=4
It is like Pythagoras eq for right triangle.
Here we don't know a or b, but we know hypotenuse is "2" from root of 4.
So, right angle triangle of sides a, b, 2.
Similarly, c^2+d^2=8 can also be considered a right triangle of sides-c, d, 2√2(from root of 8).
Then we can expand the first triangle to get a like triangle. Multiply all sides with √2.The new triangle we get with sides- √2a , √2b , 2√2(hypotenuse).
This new triangle is similar to 2nd triangle. We can join both the hypotenuse to form a rectangle. This rectangle has sides c, d, √2a and √2b. Taking √2a=c and √2b=d.(cause opp. Sides of triangle are equal).
ac+bd= √2a*a+√2b*b
=√2a^2+√2b^2
=√2(a^2+b^2) ;[taking √2 common]
We know a^2+b^2=4.
Then ac+bd=√2(a^2+b^2)=√2(4)
=4√2.
What bout min value
you cannot say that these 2 triangles are similar or whether they could form a triangle at all far all the values of a,b,c,d
The triangles are not similar just because they have the same hypotenuse
Even without log. We could clearly see the value of A and B to under root 2 . And c and D to be under root 4. On first glance
The comment section is giving me heart attack
😂
Same bro yeh log jyada hosiyar hein yaa hum jyada gadhe doubt ho rha hein😂
@@BirikhSaikia bhai ham hi gadhe h
This question was not that much hard but I thought we have to find the value of ac+bd... I had already thought for A1, A2, A3 & A4... I didn't know we have to find the range otherwise it was fairly a question for a student preparing for Olympiads in class 7th or 8th... BTW I am also in class 8th.... Also there was nothing that much typical to see in A3, I made that conclusion just by looking at the equation an in fact it was my first thought... But according to me this question made a fun of me as I thought we can find the actual value of ac+bd, I was repeatedly looking the question if a, b, c & d were Integers or we had to find all possible values of ac+bd...then this question was very easy in fact
Whenever there is a question asking the value of an expression, it's wrong to assume it will have just one single value, bcuz finding one of all possible values is quite simple. But to analyse in all direction is where true beauty lies
Bhaiya please upload the full unbeatable crash course on UA-cam. Really need it 😭
Sol a=2/1/2,b=a,d=2,c=d
Bhaiya apni jee mains pyqs series band mat kariye please continue kariye . yahi. Apse mera nivedan hai 🙏
You won't believe I found approach 5 by myself (somewhat)
Fr love you bhai ❤❤ after I saw u on unacademy first time I came to your channel literally bhai kya kaint chij batadi op 🔥🔥🔥
Thanks and welcome😎
Polar coordinates 🗣️💃, dekhte hi approach hit krgyi yayyy
a²+b²= 9
9(cos²p + sin²p) = 9
.·. 9cos²p = a² and 9sin²p = b²
.·. a = 3cosp and b = 3sinp
Similarly, c = 2cosq and d = 2sinq
ad - bc = 6cospsinq - 6cosqsinp
ad-bc = 6 (·.· given)
cospsinq -sinpcosq =1
sin(p-q) = 1
.·. p-q=90
p=90+q ..(1)
Replace (1) in value of a and b, we get
a = -3sinq and b = -3cosq
Now, ac = -6sinqcosq = -6((sin2q)/2) = -3sin(2q)
Max value of sin2q = 1
Max value of ac = -3(1) = -3
.·. K = -3
Bhaiya iske aage kya karu??? 25:27
sin(2q) = -1 k liye max of ac = +3 hoga
@@kaustavbhowal2109 bro but sin function ka max value 1 hota hai and sin2q = 2sinqcosq, isse max value of sin2q =1 aayega. Please explain me what I am doing wrong over here.
@@Veer-q6b is channel ka next video dekhlo jisme iska soln hai, samjh jaoge Kahan galti ki hai
😮😮😮😮
@@Veer-q6b the problem was that you took cospsinq-cosqsinp=1 as sin(p-q)
It is actually sin(q-p)
So it will q=90+p
i reached until 6:04 , this student doesn’t understand the core meaning of the question and based upon any fact that they think is wright they are doing , i didn’t use any formula basic understanding the opposite of square is root ok ((a*a)*(c*c))+(b*b)*(d*d))= 12 opposite is Root so 12 root is 3.46410161… may be i am wrong , but i would like to understand why i am wrong?
Brother your explanation skill is amazing ❤😮
You can also use cauchy Schwartz inequality.
In which chapter is that? Is that not in our syllabus?
@@PranitSumanvector
@@MAYANKGUPTA-t3f Yes , I agree ! We can also use the Cauchy-Schwarz inequality to get :
(a^2+b^2)(c^2+d^2)≥(ac+bd)^2
Substituting the values, we have:
min{ac + bd} = √(32) = 4√2
Cauchy Schwartz inequality ka special case hai toh directly ho Gaya 10 sec me
Same here 😊
Last one was really killer step
Broooooooooooo. Loved iiiitttttt❤❤❤
alternatively u can solve using vectors, complex nos, matrices and determinats,and trigonometry formulas 😊
I was getting answer 4√2 in positive but not in negative here is my way of tackling it
a^2+b^2=4
C^2+D^2=8
Here I find that any number minimum value lies when bothe number are equal so I got value a,b as √2 and c,d as 2 so by normal multiplication I got answer 4√2
I am also at class 10th
I done it very differently tho. I used right angle triangle but I didn't use trigonometry.
I used pythogoreas theorem.
a²+b²= 2²
c²+d² = (✓8)²
they are following pythogoreas theorem so we can make right angle triangle having hypotenuse 2 and ✓8
let make triangle ABC for a²+b² 2² and PQR for c² + d² = (✓8)². where 2 and ✓8 is hypotenuse of ABC and PQR respectively
now draw perpendicular bisector for both traingle and I think you can solve next.
This doesn't make any sense at all
Perpendicular bisect karke kya phir?
This question cannot jave a definite answer put a=2,b=0, c=0,d=root 8. So ac+bd=0. Esse infinite tarike s ehumm kars hakte hain
And a,b,c,d are real so i can take irrational numbers also so infnite possibiliy is there
while the possibilities are infinite, that does not mean "it can take any real value"... there are infinite numbers between 0 and 1, does that mean that every single real number is between 0 and one? no.... So, if you can't determine exact value you need to determine interval...
if the min value of a=-2 and max value=2 then b=0 similarly, if c min value-2*root2 and max value =2*root2 then d=0
ac=put max valur of one and min value of another so ac=-4root2
also, cd=0 if ac=-4root2
therefore ac+cd ki min value -4 root 2 and max= 4root2
im in class 10 and im able to do it under 5 minutes all alone
Bhaiya pyq jee mains ka karadoo vector and comics ka ple😢
🚬 Charas method
Let a=b= real
So, 2a²=4
a=√2=b
Let c=d=real
So, 2c²=8
c=2=d
ac+bd=2√2+2√2=4√2
Those who studied chauchy schwarz inequality: laughing in the corner 😈😈
Lol i also immediately recognised it after seeing the problem.
@@arsh_arora10 advanced aspirant umm hmm..
Tu ha kya air 1 bro 🗣️
@@maxnoob7612 shh 🤫 aise sabke samne mat bolo
Not preparing for jee but I recently prepared for IOQM in class 10 only. But will prepare from 11th for jee. @@18AyushKumar-gy4ml
Let's analyze the equations:
*Given Equations*
1. a² + b² = 4
2. c² + d² = 8
3. ac + bd = ?
a = 0, b = c = d = 2
*Verification*
1. a² + b² = 0² + 2² = 4 (valid)
2. c² + d² = 2² + 2² = 8 (valid)
3. ac + bd = 0_2 + 2_2 = 4 (valid)
Why does it have to be 4√2 when it can be 4
Bro maximum value is asked!
Bhaiya abcd real hai toh a²+b²=4 hai toh isse just a possibility max value ke liye (✓2)²+(✓2)²likh do n similarly sabko c² ko (✓4)² compare kardo sabko correct me if i am wrong and minimum ke liye -ve input krdo root ke bhar
i did it the same way as approach 3 and solved it and got the answer
Mera bhi aise hi aya tha...
Mujhe a²+b²=4 se standard circle ka eqn chamak rha thha xd and mujhe ek aur approach dikh gya aur geometry se bhi ban gya ye que..😅 aur maine is que ko AM GM se try kia thha but mera answer kuchh aur aa rha thha
AM GM sirf +ve real no.s par work karta ha
@@VARUN......z ha islie shayad answer nhi aaya usse
@@VARUN......z😮😮😮😮
Two equations for finding value of 'x' and 'y'
1) ax + by = C
2) px + qy =C
(If "C" is the Real Number as 0,1,2,3......).
Can we write or say.
"ax+by = px+qy = C"
("ax+by = px+qy") ( Because both are equal to "C").
Is it true or not. please help with example 🙏
Why can't we go a²+b²-4 = 0
and then (a-b)² = 0 => a=b=√2?
Then same way we can get c=d=±2
So ac+bd = ±4√2
Am I doing something inherently wrong, or oversimplifying?
Is this channel useful for class 10th students
Simplest question I've ever seen. The number of equations= the number of values of variables that are fixed. So we can choose any two variables whose values we can suppose as fixed. And rest can take any value. So i take b=d =0. This gives 4√2🤷🏻♂️
Elaborate
@MathsUnpluggedIndia if there are 4 variables and 2 equations, then the values of only 2 variables can be fixed. So we can randomly put any real value in the rest of two variables. And that's what I did. Even if i had taken different real values for b and d we would still have gotten the same answer.
This will give you just one of the many possible values, how do you get the exact range of values of ac+bd?
@@MathsUnpluggedIndia I didn't see the whole video. I just saw the thumbnail and answered it. I didn't we had to find a range of values. Ig option elimination may help after finding multiple values.
That's not how real maths work. If I ask x^2=4 , x=? , now will you say oh x=2 is the answer? No, that's not complete, the answer will be +-2 , so ,even if you just read the thumbnail that also means all values of ac+bd. But good to see you r into maths ritik😅
i used method that a^2 >= 0 therefore max of a = root 4 , min = - root 4
max of c = root 8 , min = - root 8
so multiplying range is root 32 to - root 32
but very fascinating method, never thought of this approach
Cauchy schwarz inequality and game over❤
Comments m sb JEE k topper h😊
Sir your video is most useful
BHAIYA STRAIGHT LINE KE BAD VECTOR 3D START KAR DENA AUR USKE BAD STATISTICS PLEASE....CIRCLE BAD MEIN KARA DENA
14:14 Your ideology perfectly matches with AG Sir , you should definitely check him out
1 liner ans using cuachy schwarz inequalities
We can also use the Cauchy-Schwarz inequality to get :
(a^2+b^2)(c^2+d^2)≥(ac+bd)^2
Substituting the values, we have:
min{ac + bd} = √(32) = 4√2
😮😮😮
it is most easily done by vectors
I found the answer in 10 seconds only by trial and error
Solved it using Coordinate geometry(2nd approach) and Functions(D and R) only after taking some hints from comments 😭😭😭
Am I cooked?
Use Cauchy Schwarz instead,uska proof bhi quadratic equation assume krke aata hai
Ye konsa software hein jisme ap solve kar rahe ho
Just use cauchy Schwartz and case work
It was a pretty simple problem I am in class 10th and with help of mata rani I was able to do it at my first try 😅
using which approach?
Sir please continue the pyq series
M ssc aspirant hu fir bhi dekh ra hu....mne 10th class wali approach lagai or fir dimag hang ho gya😂😂😂
God level heat and trail ...
My pg sir teaches same way like you.. I'm in kota he's one of best teacher 🎉
Prashant Jain sir?
@@shyamparikumsri8891 nah..
Maine to compare karke aur identity use karke 2√2+2√2
It can be solved as vector product only...
cauchy scharz theorem 2step qsn
@@darkking9381 Did you mean :
(a^2+b^2)(c^2+d^2)≥(ac+bd)^2
Substituting the values, we have:
min{ac + bd} = √(32) = 4√2
u are in which class???
Assume a=0, b=c=d=2
0*0+2*2= 4
2^2 + 2^2 = 8.
0*...+ 2^2 = 4.
Answer is 4.😂😂
He is so handsome 😮❤
😂
Girl?
just 3 months remaining
@@drkhush lesbian
@@Medha-h6j💀
Baaki sab badhiya hai bhai bas thoda dheere bola karo
Last year ka paid Batch ? Mil skta h abhi please ? Btao maths me Especially For Geometry
I mean Brahmagupta's identity ka direct use hai.
Cauchy is the most apparent approach.
Dude I have a beautiful method, you can assume these points to be on a circle and solve it further
Uh guys, i just did it by algebra.
Im now confused if i did it correctly or i just got it by some calc. mistake.
Cauchy Schwarz To the rescue two line me answer, there is a reason top ranker's are usually olympiad aspirants
Merko toh trigo wale part maza aagya
This question is from adv test in allen (different data though).
Can be solved using circles
Bhai koi please ye explain krdo
I literally randomly took
a = 1, b = √3, c = √2, d = √6,
you get ac + bd = 4√2
Why and how???
Maine to value assume ki 4root2 tk pahoch gya tha 😂
good afternoon sir
day 1 of begging you to make pyq video on APPLICATION OF DERIVATIVES
Last question ka solution kaha milega???
Bhaiya please substitution pe ek specific video banao jo aapne Approach 5 mei use kiya. Please 🥺🙏
Tum sb bot ho
A²+B²=4 ..hai
C²+D²=8
Ac+bd= ..find krna hai
So agar ham root 2²+ root 2 ² krne to 4 hoga
Aur dusre me 2² rakhdo do 8 hoga
To a or b ki value hame root 2 mil gyi or c or d ki 2
To a ×c =root 2 ×2 =2root2
Same with bd 2root 2 than we add it
2root2+2root2 =4root2 ..
I do it in my mind
4
I solved it by trignometry
bruh yeh jee wala sawal tha? Mera dost mujhe yeh question ka min and max value find out karne ka challenge diya tha, and I actually solved just by inserting random values of a,b,c,d
@@Doraemon-jm5oo 🤦♂️ that's not how you solve problem. Just getting the answer and feeling oh I solved it is worst that can happen. Develop problem solving and conceptual clarity as to why what you did was wrong or right
Isliye to tu Doraemon hi reh gaya, kabhi asli zindagi jee kar dekh chotte
Last ka answer 3√2
In black book trigonometry chp multiple qs are there on assuming a=4 cos thetha valw
Typewise pyq when??
Bhai I m a commerce man. Could solve it mentally in 3 min...
Becoz clearly a and b was under root 2
And B and d were under root 4
it wasnt about getting answer
Mai Jo Ise 10 Second mai solve kr diya answer bhi correct aagya😂
Achha..kya guarantee h ki aur solution possible nhi h😂
Proper solution krna hota h naki tukka baji
Cauchy Schwartz died
How do u use it.
It's derived from the dot product of any two vectors
|a.b|
Bro i solved it in 10 seconds ,
±√2 = a =b
±√4 =c=d
So , min = (-√2 ×√4)+(-√2×√4)
=-4√2
Max = (√2×√4)+(√2×√4)
= 4√2
Easy
are how kindly explain
Did it by trigo woh substitution instantly click kr gya tha
Maximum value toh cauchy Schwartz inequality se bhi nikal sakte hein usse ek line mein ajata hein
Not getting the point of this series. We are learning problem solving skills , new ideas
Hahahahah solved in 10 minutes, I am non math student in 11th but a 10th topper of math
9:56 sir yaha directly (ad-bc)² ban raha hai
bhaiya ye question vos ke channel pe rmo wali series me hai sir ne geometry se solve kiya tha
IAM A 10TH CLASS STUDENT 16:04 WALA SAMAJ NI AAY PLEASE EK BAAR SAMJHA DO SHORTS MAI HI SAMJAHO DO PLEASE!
Cauchy Schwarz Inequality 🗿🗿
a²-b²=4
a²+b²=16
😅😊
Me after getting 4 . root 2
10th class concept, 26 minutes? Maybe if you had used 12th class concept, 2 minutes?
bro missed the point of the video and this whole series