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@aiims891 2 күн бұрын
❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
@nidhishah6845 3 күн бұрын
Ekdum op
@Varunarun58 4 күн бұрын
🎉❤
@prateekverma1284 16 күн бұрын
Sir aap banaras se h kyaa?🥲
@rehaAlig 17 күн бұрын
Best explanation sir💯😊 Very helpful... Thank you ❤
@nagasaranyanakka5507 19 күн бұрын
Nice question Sir
@nagasaranyanakka5507 19 күн бұрын
SIR if angle A=angle E then angle B is also equal then the triangle is Equilateral triangle so,angle DAE is equal to 60
@ghost-jb7ng 23 күн бұрын
Great lectures 🔥 , love the complexity of the questions
@mohittyagi-y5e 25 күн бұрын
pooniya agar legendry h toh mai bhi arnold hu
@user-op9zg8dx4g 28 күн бұрын
Mere bhi bhi yha hi neet me 2024 me usne admission liya hai
@09debabrata 29 күн бұрын
Why so under rated channel 😮😮
@VijendraBijarniya-wl4bw Ай бұрын
Sir please 🥺 upload integration chapter
@anuragraj3021 Ай бұрын
Can we say net work done by static friction on a system is zero???
@abhisheknagar4098 Ай бұрын
❤❤❤
@poorvapokharna22 Ай бұрын
24:45
@likeaduck2005 Ай бұрын
This channel was a pure gold 🏆
@omniInstint Ай бұрын
Thank you very much sir I needed this for a revision
@lakshyasolanki6604 Ай бұрын
Ideal Spring ma har jagah force same kyu rhti hai?
@umesh9629 Ай бұрын
sir you should have moved aside so that we could take a screenshot. Anyway it was quiet a helpful video 👍
@user-ew2kq9vw4t Ай бұрын
Uncle ji maine to 1 attempt mein 2 ¼ minute ke andar 150 Tak ki ginti likh Li ab kya karu?😅
@animevision9764 Ай бұрын
Thanks
@ankitrana275 2 ай бұрын
Bhut baketi krte h
@Tuhinliza 2 ай бұрын
not good
@ckjain_maths8594 2 ай бұрын
If 1 maps to 1, 2 has to map with 1. Hence, mapping in pairs shall give the desired composite function. There are six cases. Case 1--- when no element maps to itself. Since number of elements are odd. This case will not give desired composition. Case 2--- when only one element maps to itself. Rest of four element shall make two pairs in which elements of each pair maps each other. Identical mapping element can be chosen in 5 ways. Two pairs out of four elements can be chosen in 4C2 ways i.e. 6 ways. There are 6*5= 30 ways. Case 3 --- when only two elements maps to itself. Emphasis is on " only two". Rest three can not the required composite function. Case 4-- when only three elements maps to itself. 3 elements out of 5 can be chosen in 5C2 = 10 ways. Case 5 --- when only four elements maps to itself. Required composite function is not possible. Case 6--- when all five elements maps to itself. One case is possible. Total number of bijections for desired composite function is 30+10+1= 41.
@mihirpatel7378 2 ай бұрын
Me confuse hu isko dekh k
@Darsh-tk5rw 2 ай бұрын
Thank u sir 😌
@user-uz4te3si3u 2 ай бұрын
❤❤❤❤
@navyaparashar9558 2 ай бұрын
8:43 is it compulsary to take 99/9 or any numbers like 44/4 or 110/10 can also be taken ???
@ParthiSingh-xy7cx 2 ай бұрын
But then you can't take squares right
@IITianAYUXH 3 ай бұрын
this was a much needed concept
@mlmeenameena9880 3 ай бұрын
211
@Hari-pl2em 3 ай бұрын
good for jee adv thanks
@nobitches9739 3 ай бұрын
bro is not ashneer grover
@mohammaddanishsiddiqui6008 3 ай бұрын
thanks sir
@parasjain1280 3 ай бұрын
Case 4 example wrong he sir ji
@parasjain1280 3 ай бұрын
41.23 f(0 ) =2 a2 -1 aa raha he, u write -2
@parasjain1280 3 ай бұрын
-1
@betterangle9318 3 ай бұрын
Best lecture on th topic Luv your energy sir!!
@yashikasaluja511 3 ай бұрын
gadha pankaj baluuuuujaaaaa.........
@fortimepass5595 3 ай бұрын
shouldnt it be 2tcostheta
@user-rf6ly9lk3s 4 ай бұрын
28:54 (-pie/2-2 pie/2+2 )
@user-rf6ly9lk3s 4 ай бұрын
16:37 ans is C
@ayush_5096 4 ай бұрын
10:17
@greedskith 4 ай бұрын
😀
@Jeeaspirant300 4 ай бұрын
Sir aap ashnet grover jaise dikhte ho
@Verma.163 4 ай бұрын
Kha ha sir link
@examfun8389 4 ай бұрын
Backwash coaching he ..don't join
@divyakumar8147 4 ай бұрын
thanks sir
@divyakumar8147 4 ай бұрын
thanks sir
@divyakumar8147 4 ай бұрын
thanks sir
@shauryagupta3644 4 ай бұрын
Sir can we solve this question using similar approach? I tried, but wasn't able to do it, please help: Q. If sin(1 deg) * sin (2 deg) * sin (3 deg) * .... * (sin 90 deg) = k * sqrt(10)/ 2^89, then find k
@DevTech03 4 ай бұрын
sin1 sin2 ..... sin90 = sin1 cos 1 sin2 cos2 ... sin88 cos88 * sin(45) [will be using the sin45 at last] =1/2^44 (sin2 sin4 sin6....sin88) similarly this will be equal to - 1/2^66 (sin4 sin8 sin12 ... sin88) [all multiples of 4] = 1/2^66 (cos2 cos6 cos 10 ... cos86) [just converted sin to cos] now just form pairs of 3 to use the idenitiy cos(A)cos(60+A)cos(60-A) = 1/4 cos3A =1/2^66 *cos30 [ cos2 cos62 cos58 * cos6 cos66 cos54 .....] = 1/2^80*cos30 [cos6 cos18 cos30 cos42 cos54 cos66 cos78] apply the same identity again - 1/2^84 *cos30 *cos30 [cos18 cos54] now cos18 * cos 54 = root80 /16 and (cos30)^2 = 3/4 and sin45 = 1/root 2 plug in all these in the final equation = 3 (root40) /2^90 = 3 (root10)/2^89
@AdvaitAtharva 4 ай бұрын
8:50 answer = 13