Sets Relations & Functions Class 11 | Lecture 2 | JEE Main | JEE Advanced |Arvind Kalia Sir| Vedantu

In mathematics, a function can be defined as a rule that relates every element in one set. In today's session of Sets, Relations & Functions Lecture 2, Arvind Sir is going to teach you clearly the relations and basics of functions. IIT JEE Maths by your favorite teacher Arvind Kalia Sir (Mera Bhai). VJEE Elite LIVE (11th) series at Vedantu JEE to get your dream scores in JEE 2023 (JEE Main 2023) & (JEE Advanced 2023). We are herewith 👉 Sets relations and functions IIT JEE | Sets relations and functions JEE | Sets relations and functions JEE Mains 2023 | Sets relations and functions JEE Advanced 2023 | Introduction to Functions Grade 11 | Introduction to Domains | Relations IIT JEE Maths | Relations JEE session.

About Teacher 👨‍🏫: Following his passion, Arvind Kalia Sir (Mera Bhai) joined this noble profession after doing M.Sc. in Maths from IIT Delhi.

➡️ Avail all Exciting Vedantu Courses 🔥 by clicking on the links below 🤑

-------------------------------------
#SetsRelationsandFunctionsIITJEE #VJEEEliteLIVE ​#JEEMaths #RelationsIITJEEMaths #IntroductiontoFunctions #IntroductiontoDomains ​​#JEEMain2023 #JEEAdvanced2023 #JEEMain #JEE2023 #JEE #OnlineClasses ​#OnlineJEECoaching #OnlineClassesForJEE #StayHomeWithMe #LearningWontStop #DailyLiveJEEClass #ArvindKaliaSir #VLOP​​ #VLIF​

​
Thanks for watching this session on Sets, Relations and Functions L2 | Relations, Introduction to Functions, Domain | IIT JEE Maths (11th) (VJEE Elite LIVE) by Arvind Kalia Sir. Hope you have enjoyed and learned more about the Sets relations and functions IIT JEE | Sets relations and functions JEE | Sets relations and functions JEE Mains 2023 | Sets relations and functions JEE Advanced 2023 for JEE Mains 2023 & JEE Advanced 2023.

[starboii._]

11:56 Cartesian Product of Two Sets
49:26 Functions

[jeewalepandit]

lecture starts at 8:38 and ends at 02:09:38

[agreng]

Bhaisad kya hi explanation tha. Apne aankh khol diye. Jo beginner h wo dek lo. Eddum basic se sad h kuch milegaa. Concept bhaut aacha bataya sir with very easily. Watch in 2x then also understable

[positronman]

I had many doubts and didn't understand basic concepts of functions, but sir you cleared all my doubts, thank you sir .

[VEDANTA]

I had doubt in finding domain for root fractions, but now I qm able to do i very easily

[kavyajonnala]

Sir your love theory in functions is unforgettable 😂😂😁. I won't get confused from now in function definition 🤣🤣. Thanks a lot sir

[NehaKumari-hibq]

I am leaving a comment here so I can see this after 2 years & feel proud of myself how I was dedicated for my dream.,

[pratikchavan]

Oo bhai ... Functions ka explanation OP 😂😂

[AlishaPatel-hdbo]

Kon kon math ka best teacher search karke arvind kaliya sir se padhne aye hai😊😊

[pwkiarmy]

1:50:45 please explainn me how these factors came

[sankalpwallah]

Sir me up bord se hu fir bhi mere pure dobut bilkul clear ho gaye very very thank you sir ji

[rajeshmehta]

1:03:54 made me laugh that word LUCHA
And sir thx for a beautiful lecture..

[manish.m]

Sir is this content is sufficient for jee advanced

[TheMagicDevil]

1:16:36 I didn't get this part, why the power of 5 on LHS and RHS were made equal ???

[omhindalekar]

Mai speaker pe lecture sun raha tha ab gharwalo ko kaise samjahu ki mai maths hi padhraha tha😂🤦🏻‍♂️😂

[alphaAspirantsJEE]

1:18:08 sir jo aapne bola ki x1^5 =x2^5 solve karne ka ek doosra aur simple method hai. sir x1 ki power 5 ko RHS mein bhej ke 1/5 ya root ban jaayega aur x1=x2^5 *1/5=x2 aa jaayega . sir ye apply kar sakte hain aap.

[akshatpandey]

Sir aj relation ka revision krne aya tha aur sach me ab mujhe apni understanding me farak nazar aa rha hai
#VLOP

[JEEVedantu]

✅ Get Vedantu Pro Subscription 10% OFF(Limited Time Offer):

[seemayadav-cwms]

Thankyou so much sir for this amazing video.
I'm a dropper and do not have clarity that how can I find domain of a function but after this video everything is clear

[alphagamer]

Subset of A×B is relation
*Onto* means range=co-domain.
Domain of f(x)+g(x) is intersection of both functions.