so here if we take as a general complex number that is know, let a = x+iy, and assume that we know x and y, now take z = w+iy, which is unknown,, now for set A when we convert it into cartesian form we get, W/(x+y) - V/(x+y) < 0 and similarly for Set B we get W/(x+y) - V/(x+y) > 0, Now draw the regions occupied by A and B and draw it for (x+y)>0 and (x+y)
Very easy with idea of locus, two area separated by line A represent y x + (a+b) Slope is 1 and y intercept is a+b By graph it can be easily seen that neither A nor B can occupy the complete x axis for any value of a or b
Bhai tumhara ye solution mujhe bilkul samajh nhi aaya. Abhi 40 din baad mains dena h. Ye topic agar easy h and paper m high weightage h to please topic ka name btado
@khuranapranshu Complex - II se coaching mein padhate hai kisi bhi book ke complex part ke last part mein rehta hai. Locus of complex number topic ka naam....
Answer: (b)_Both are false All I did was skip the above info Just read the two statements, Its obvious that if a Complex number (say A) has a constraint on only its Img,Re part, that wont effect its nature of being purely real or complex Given that here they are always attaining values either greater or less that zero Hence both the statements are wrong I mean, I hope its a correct explanation..
I had this doubt So in short what does the statement A contains all the real numbers mean, is it that the real part can be any real number or that the set of all the imaginary and real part will have every single real number or that the set of magnitudes of A will have every single real number Secondly If it is already given that if imaginary part of A and real part of A are > 0 so isnt it already enough to say that the -ve real numbers are missing and likewise in set B positive real numbers are missing?
SO60 is live >> courses.jeesimplified.com/courses
so here if we take as a general complex number that is know, let a = x+iy, and assume that we know x and y, now take z = w+iy, which is unknown,, now for set A when we convert it into cartesian form we get, W/(x+y) - V/(x+y) < 0 and similarly for Set B we get W/(x+y) - V/(x+y) > 0, Now draw the regions occupied by A and B and draw it for (x+y)>0 and (x+y)
If im not wrong the actual Q in S1 and S2 is Re(a) and Im(a) has an inequality
Very easy with idea of locus, two area separated by line A represent y x + (a+b)
Slope is 1 and y intercept is a+b
By graph it can be easily seen that neither A nor B can occupy the complete x axis for any value of a or b
Nice observation!
I did the same
Bhai tumhara ye solution mujhe bilkul samajh nhi aaya. Abhi 40 din baad mains dena h. Ye topic agar easy h and paper m high weightage h to please topic ka name btado
@khuranapranshu Complex - II se coaching mein padhate hai kisi bhi book ke complex part ke last part mein rehta hai. Locus of complex number topic ka naam....
@asmitsarkar526 easy topic h to btado padh lunga fir
Answer: (b)_Both are false
All I did was skip the above info
Just read the two statements, Its obvious that if a Complex number (say A) has a constraint on only its Img,Re part, that wont effect its nature of being purely real or complex
Given that here they are always attaining values either greater or less that zero
Hence both the statements are wrong
I mean, I hope its a correct explanation..
exactly its simple af takes 1 min both are false
Its pretty logical to say option c both are false
Arey bhai sahab 🤯🤯 mtlb bilkul chakma dediya mains ne
I had this doubt
So in short what does the statement A contains all the real numbers mean, is it that the real part can be any real number or that the set of all the imaginary and real part will have every single real number or that the set of magnitudes of A will have every single real number
Secondly
If it is already given that if imaginary part of A and real part of A are > 0 so isnt it already enough to say that the -ve real numbers are missing and likewise in set B positive real numbers are missing?
Bhaiyya please make videos for 11th class also to guide us......🙏🏻🙏🏻🙏🏻🙏🏻
3:16 mein mujhe samaj nahi aaya
Piece of chit brother ..
❤
What if z=6+0i is s solution for this set A for some a other than a=5-8i?
Bro just one question, if imaginary part is not equal to zero then how can it contain any real number
Easy to visualise this using graphs.
Bhaiya ye question geometry se bhi ho Shakti hai…
Aagayi nazare bhaiii
Bhaiyaa please make video on last month revision for jee mains mathsss
Bhaiya what does Re(A) mean? how can you define real part of a set?
Re means real part of number A
Bhai complex no. Me padhoge
Bhai complex number ke basics hai , pehle complex number padho jaake
I think its a typo, Re(a) kehena chah rhe the.. A is a set you don't have a real and imaginary part of it
@@Titan_x_god_orion_001 how do you define real part of a set?
did it in 2 to 3 min its been pretty easy
6th and 7th wasted 😂
😂
Life wasted bhai 😂😂😂😂
Khatam, nothing can happen
Kal hi same ques kiya hai
Wo backlog tha 😂😂
Got the answer B
Have Done this question before in a mock test similar logic
Sir meri 6th waste ho gayi hai kya abhi bhi jee mains me air 1 aur iit bombay me common sense branch mil sakti hai kya?
Yes bro 😊 keep working hard, by the time you come in 11th from 6th, you will get 2 Cr+ package, and after that, IIT Bombay also (CS/CSE)🎉
ayien baigan
Lmao same reaction
Answer b?
mujhe bhi whi lg raha hai
I'm getting B
B
Farji sawal tha ye to
First
B