In first question: for negative value f(x) is discontinuous as it gives imaginary number .. Right explanation: f(x) is uniformly continuous on interval (a,b) or [a,b] where 0
F(x) =x where f:R........R then if use the concept of lipschitz function f(x) ka deriative 1which is bounded then uniformly continius But in this same question put - infinity and + infinity then fuction does not exist then f not u.c .... aise kaiseho skta h sir
Manish ji if function does not exist -inf. and inf. Then we can not say function is uniform continuous or not but if function is exist both -inf. and inf. then function is uniform continuous.
@@manishkumarverma3486 logic to aap sahi lga rahe hain but jo LF hota hai wo subset of real no. me define hota hai, ya kisi finite interval me define hota hai but aap jis example ki baat kar rahe Hain wo R me define hai
Sir ek cheez batana Jo aapne first ques karaya , us ques mein derivative wala test aapne use kiya but derivative wale mein end points pe limit kaha hi nikalte hain
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In first question: for negative value f(x) is discontinuous as it gives imaginary number ..
Right explanation: f(x) is uniformly continuous on interval (a,b) or [a,b] where 0
[a,b] pe kyun nahi hoga a toh > 0 hai??? Plz explain
Right because option b has perfect interval than to option d
Product of two u.c is not u.c for example take f(x)=g(x)=x in R,here f and g both are u.c but f.g equal to x^2 which is not u.c in R
@@harshparmar1942 use the definition of uniform continuty
@@faruksk399 Sorry ... My reply was sent unknowingly.
@@harshparmar1942 ok
X is not continous on real numbers, it is only continous when it is bounded between two given values
36:46 here in question they have given x>0 ,
Good explanation every line.
Very helpful video thank u sooooo.... Much sir
Sir dill se izaat karta hn ap ki or hamesha krunga jitna aap se seekha he shayad kisi se nh❤️
F(x) =x where f:R........R then if use the concept of lipschitz function f(x) ka deriative 1which is bounded then uniformly continius
But in this same question put - infinity and + infinity then fuction does not exist then f not u.c .... aise kaiseho skta h sir
Wahi to mai bhi keh rha hoon ji
Any 1 condition should be satisfied
Product of Two UC need not UC, example f(x)=x on R
Thanks sir ji 🙏
Sir super explaination excellent performance
Unacadmy pr class li thi same question tha but option D was wrong
Sir question number 6 me problem hai thodi,kyoki f(x)=x, agr ye function le to ye UC on R but infinity pr limit nhi exists krti h
Manish ji if function does not exist -inf. and inf. Then we can not say function is uniform continuous or not but if function is exist both -inf. and inf. then function is uniform continuous.
@@hariomgupta9116 but jara LF ki condition lga k dekho, function LF hai mtlb Lipschitz function hai ye aur LF impliesUC
@@manishkumarverma3486 logic to aap sahi lga rahe hain but jo LF hota hai wo subset of real no. me define hota hai, ya kisi finite interval me define hota hai but aap jis example ki baat kar rahe Hain wo R me define hai
You are right 👍
@@happinessforever4187 thanks 👍
Sir ek cheez batana Jo aapne first ques karaya , us ques mein derivative wala test aapne use kiya but derivative wale mein end points pe limit kaha hi nikalte hain
Thank you sir 🙏🙏🙏🙏🙏🙏
Thank you sir😊
Thnkuuuuuuuuu guru ji🎉🎉🎉
What about f(x) = sinx on [0,infinity)
Thanku so much sir
Sir prod of uc is not uc. Product is uc only if it is bdd , know clarify my doubt sir
Maja aa gaya sir
Thanku sir ❤
sir q2 ky solution ki pdf share kr dy
B option
So plese sir help me to solve this question
To I think option B is true
🙏🌺🌺🌺🙏