JEE Advanced Mathematics problems seem to be tricky but are really not

[@ultrasteamcarpetcleaning3207]

Absolutely beautiful Dr. PK!!!

[@Min-cv7nt]

Totally agree with ya

[@pkqualitymath1234]

Thanks a lot my friend for your support haha👍👍👍

[@k.somers459]

You did it again. Great video

[@pkqualitymath1234]

Thanks a lot my friend for your support haha👍👍👍

[@domedebali632]

Another great video professor

[@pkqualitymath1234]

Thanks for your support my friend👍👍👍

[@MrGLA-zs8xt]

Very nice solution professor

[@pkqualitymath1234]

Thanks a lot haha👍👍👍

[@iqtrainer]

Again, so great video🎉

[@pkqualitymath1234]

Thanks a lot my friend for your support👍👍👍

[@Min-cv7nt]

Love JEE advanced maths

[@jagadeesann9527]

Thank you sir

[@pkqualitymath1234]

Thanks a lot my friend for your support👍👍👍

[@ChiragBharadwajYT]

Hi Dr. PK, I think if you set k equal to the entire expression inside the logarithm (in this case 1/3sqrt(2) * sqrt(4 - k)), you end up with the much nicer quadratic of 18k^2 = 4 - k, which simplifies into (2k+1)(9k-4) = 0, leading very quickly to k = 4/9 without as much radical manipulation.

[@MrGLA-zs8xt]

That seems like setting k for two times which eventually the same result, could you show your work more?

[@ChiragBharadwajYT]

@@MrGLA-zs8xt Nah meant more that the choice of k can include the constant term out front which simplifies the expression. But sure, if you let k = 1/3s(2) * s(4 - 1/3s(2) * ...) then you can square both sides to get k^2 = (1/3s(2))^2 * (4 - 1/3s(2) * ...), which can be written as k^2 = (1/3s(2))^2 * (4 - k). The 3s(2) becomes 9 * 2 = 18 when squared, so you're left with k^2 = (1/18) * (4 - k), and rearranging gives us the 18k^2 = 4 - k as expected.

[@pkqualitymath1234]

Nice👍👍👍

[@OmarAlHathaf]

Thank you very much for this nice work.
Do you have a published book?
I want to learn more from you.
Thank you again.

[@pkqualitymath1234]

Wow thanks a lot my friend haha. I am actually writing e-textbooks. I will let you know when they are published👍👍👍