Question-74997

Question Number 74997 by chess1 last updated on 05/Dec/19

Commented by chess1 last updated on 05/Dec/19

answer : A

Answered by mind is power last updated on 05/Dec/19

\frac{d^{2019}}{{dx}^{2019}} (X^{3} \cos (x^{2})) ∣_{x = 0} =

Re {\underset{k = 0}{\sum^{2019}} C_{2019}^{k} {(X^{3})}^{(k)} . {(e^{{ix}^{2}})}^{(2019 - k)}}

= C_{2019}^{3} . (6) . {(e^{{ix}^{2}})}^{(2016)} ∣_{x = 0}

= \frac{2019!}{(2019 - 3)!} . {(e^{{ix}^{2}})}^{(2016)} x = 0

e^{{ix}^{2}} = \underset{j = 0}{\sum^{+ \infty}} \frac{{({ix}^{2})}^{j}}{j!}

\frac{d}{{dx}^{k}} e^{{ix}^{2}} = \underset{j = 0}{\sum^{+ \infty}} \frac{d}{{dx}^{k}} \frac{{(i)}^{j} x^{2 j}}{j!}

= \underset{j = E (\frac{k}{2})}{\sum^{+ \infty}} (2 j) (2 j - 1) \dots (2 j - k + 1) \frac{x^{2 j - k}}{j!}

\frac{d}{{dx}^{2016}} e^{{ix}^{2}} = \underset{j = 1008}{\sum^{+ \infty}} (2 j) (2 j - 1) \dots . . (2 j - 2016 + 1) i^{j} x^{2 j - 2018} . \frac{1}{j!} . \underset{x = 0}{∣}

= \frac{2016 . (2015) \dots \dots \dots (1) i^{1008}}{1008!}

= \frac{2016!}{1008!}

We get so

Re {\frac{2019!}{2016!} . \frac{2016!}{1008!}} = \frac{2019!}{1008!}

Commented by vishalbhardwaj last updated on 06/Dec/19

please explain the steps of answer and their

notation of symbols used here with depth

Commented by chess1 last updated on 06/Dec/19

thanks

Commented by mind is power last updated on 06/Dec/19

ok sir i Will post complet solution later