Best explanation sir💯😊 Very helpful... Thank you ❤
@nagasaranyanakka550719 күн бұрын
Nice question Sir
@nagasaranyanakka550719 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-jb7ng23 күн бұрын
Great lectures 🔥 , love the complexity of the questions
@mohittyagi-y5e25 күн бұрын
pooniya agar legendry h toh mai bhi arnold hu
@user-op9zg8dx4g28 күн бұрын
Mere bhi bhi yha hi neet me 2024 me usne admission liya hai
@09debabrata29 күн бұрын
Why so under rated channel 😮😮
@VijendraBijarniya-wl4bwАй бұрын
Sir please 🥺 upload integration chapter
@anuragraj3021Ай бұрын
Can we say net work done by static friction on a system is zero???
@abhisheknagar4098Ай бұрын
❤❤❤
@poorvapokharna22Ай бұрын
24:45
@likeaduck2005Ай бұрын
This channel was a pure gold 🏆
@omniInstintАй бұрын
Thank you very much sir I needed this for a revision
@lakshyasolanki6604Ай бұрын
Ideal Spring ma har jagah force same kyu rhti hai?
@umesh9629Ай бұрын
sir you should have moved aside so that we could take a screenshot. Anyway it was quiet a helpful video 👍
@user-ew2kq9vw4tАй бұрын
Uncle ji maine to 1 attempt mein 2 ¼ minute ke andar 150 Tak ki ginti likh Li ab kya karu?😅
@animevision9764Ай бұрын
Thanks
@ankitrana2752 ай бұрын
Bhut baketi krte h
@Tuhinliza2 ай бұрын
not good
@ckjain_maths85942 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.
@mihirpatel73782 ай бұрын
Me confuse hu isko dekh k
@Darsh-tk5rw2 ай бұрын
Thank u sir 😌
@user-uz4te3si3u2 ай бұрын
❤❤❤❤
@navyaparashar95582 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-xy7cx2 ай бұрын
But then you can't take squares right
@IITianAYUXH3 ай бұрын
this was a much needed concept
@mlmeenameena98803 ай бұрын
211
@Hari-pl2em3 ай бұрын
good for jee adv thanks
@nobitches97393 ай бұрын
bro is not ashneer grover
@mohammaddanishsiddiqui60083 ай бұрын
thanks sir
@parasjain12803 ай бұрын
Case 4 example wrong he sir ji
@parasjain12803 ай бұрын
41.23 f(0 ) =2 a2 -1 aa raha he, u write -2
@parasjain12803 ай бұрын
-1
@betterangle93183 ай бұрын
Best lecture on th topic Luv your energy sir!!
@yashikasaluja5113 ай бұрын
gadha pankaj baluuuuujaaaaa.........
@fortimepass55953 ай бұрын
shouldnt it be 2tcostheta
@user-rf6ly9lk3s4 ай бұрын
28:54 (-pie/2-2 pie/2+2 )
@user-rf6ly9lk3s4 ай бұрын
16:37 ans is C
@ayush_50964 ай бұрын
10:17
@greedskith4 ай бұрын
😀
@Jeeaspirant3004 ай бұрын
Sir aap ashnet grover jaise dikhte ho
@Verma.1634 ай бұрын
Kha ha sir link
@examfun83894 ай бұрын
Backwash coaching he ..don't join
@divyakumar81474 ай бұрын
thanks sir
@divyakumar81474 ай бұрын
thanks sir
@divyakumar81474 ай бұрын
thanks sir
@shauryagupta36444 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
@DevTech034 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