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

Commented by chess1 last updated on 05/Dec/19

answer: A

$$\mathrm{answer}:\:\mathrm{A} \\ $$

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

(d^(2019) /dx^(2019) )(X^3 cos(x^2 ))∣_(x=0) = Re{Σ_(k=0) ^(2019) C_(2019) ^k (X^3 )^((k)) .(e^(ix^2 ) )^((2019−k)) } =C_(2019) ^3 .(6).(e^(ix^2 ) )^((2016)) ∣_(x=0) =((2019!)/((2019−3)!)).(e^(ix^2 ) )^((2016)) x=0 e^(ix^2 ) =Σ_(j=0) ^(+∞) (((ix^2 )^j )/(j!)) (d/dx^k )e^(ix^2 ) =Σ_(j=0) ^(+∞) (d/dx^k )(((i)^j x^(2j) )/(j!)) =Σ_(j=E((k/2))) ^(+∞) (2j)(2j−1)...(2j−k+1)(x^(2j−k) /(j!)) (d/dx^(2016) )e^(ix^2 ) =Σ_(j=1008) ^(+∞) (2j)(2j−1).....(2j−2016+1)i^j x^(2j−2018) .(1/(j!)).∣_(x=0) =((2016.(2015).........(1)i^(1008) )/(1008!)) =((2016!)/(1008!)) We get so Re{((2019!)/(2016!)).((2016!)/(1008!))}=((2019!)/(1008!))

$$\frac{\mathrm{d}^{\mathrm{2019}} }{\mathrm{dx}^{\mathrm{2019}} }\left(\mathrm{X}^{\mathrm{3}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\right)\mid_{\mathrm{x}=\mathrm{0}} = \\ $$$$\mathrm{Re}\left\{\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2019}} {\sum}}\mathrm{C}_{\mathrm{2019}} ^{\mathrm{k}} \left(\mathrm{X}^{\mathrm{3}} \right)^{\left(\mathrm{k}\right)} .\left(\mathrm{e}^{\mathrm{ix}^{\mathrm{2}} } \right)^{\left(\mathrm{2019}−\mathrm{k}\right)} \right\} \\ $$$$=\mathrm{C}_{\mathrm{2019}} ^{\mathrm{3}} .\left(\mathrm{6}\right).\left(\mathrm{e}^{\mathrm{ix}^{\mathrm{2}} } \right)^{\left(\mathrm{2016}\right)} \mid_{\mathrm{x}=\mathrm{0}} \\ $$$$=\frac{\mathrm{2019}!}{\left(\mathrm{2019}−\mathrm{3}\right)!}.\left(\mathrm{e}^{\mathrm{ix}^{\mathrm{2}} } \right)^{\left(\mathrm{2016}\right)} \mathrm{x}=\mathrm{0} \\ $$$$\mathrm{e}^{\mathrm{ix}^{\mathrm{2}} } =\underset{\mathrm{j}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(\mathrm{ix}^{\mathrm{2}} \right)^{\mathrm{j}} }{\mathrm{j}!}\: \\ $$$$\frac{\mathrm{d}}{\mathrm{dx}^{\mathrm{k}} }\mathrm{e}^{\mathrm{ix}^{\mathrm{2}} } =\underset{\mathrm{j}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{d}}{\mathrm{dx}^{\mathrm{k}} }\frac{\left(\mathrm{i}\right)^{\mathrm{j}} \mathrm{x}^{\mathrm{2j}} }{\mathrm{j}!} \\ $$$$=\underset{\mathrm{j}=\mathrm{E}\left(\frac{\mathrm{k}}{\mathrm{2}}\right)} {\overset{+\infty} {\sum}}\:\:\:\:\left(\mathrm{2j}\right)\left(\mathrm{2j}−\mathrm{1}\right)...\left(\mathrm{2j}−\mathrm{k}+\mathrm{1}\right)\frac{\mathrm{x}^{\mathrm{2j}−\mathrm{k}} }{\mathrm{j}!} \\ $$$$\frac{\mathrm{d}}{\mathrm{dx}^{\mathrm{2016}} }\mathrm{e}^{\mathrm{ix}^{\mathrm{2}} } =\underset{\mathrm{j}=\mathrm{1008}} {\overset{+\infty} {\sum}}\left(\mathrm{2j}\right)\left(\mathrm{2j}−\mathrm{1}\right).....\left(\mathrm{2j}−\mathrm{2016}+\mathrm{1}\right)\mathrm{i}^{\mathrm{j}} \mathrm{x}^{\mathrm{2j}−\mathrm{2018}} .\frac{\mathrm{1}}{\mathrm{j}!}.\underset{\mathrm{x}=\mathrm{0}} {\mid} \\ $$$$=\frac{\mathrm{2016}.\left(\mathrm{2015}\right).........\left(\mathrm{1}\right)\mathrm{i}^{\mathrm{1008}} }{\mathrm{1008}!} \\ $$$$=\frac{\mathrm{2016}!}{\mathrm{1008}!} \\ $$$$\mathrm{We}\:\mathrm{get}\:\mathrm{so} \\ $$$$\mathrm{Re}\left\{\frac{\mathrm{2019}!}{\mathrm{2016}!}.\frac{\mathrm{2016}!}{\mathrm{1008}!}\right\}=\frac{\mathrm{2019}!}{\mathrm{1008}!} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by vishalbhardwaj last updated on 06/Dec/19

please explain the steps of answer and their notation of symbols used here with depth

$$\mathrm{please}\:\mathrm{explain}\:\mathrm{the}\:\mathrm{steps}\:\mathrm{of}\:\mathrm{answer}\:\mathrm{and}\:\mathrm{their} \\ $$$$\mathrm{notation}\:\mathrm{of}\:\mathrm{symbols}\:\mathrm{used}\:\mathrm{here}\:\mathrm{with}\:\mathrm{depth} \\ $$

Commented by chess1 last updated on 06/Dec/19

thanks

$$\mathrm{thanks} \\ $$

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

ok sir i Will post complet solution later

$$\mathrm{ok}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{Will}\:\mathrm{post}\:\mathrm{complet}\:\mathrm{solution}\:\mathrm{later}\: \\ $$

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