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Author: Tinku Tara

let-f-x-cosx-1-x-1-prove-that-f-x-1-x-2-x-2-8-x-0-2-ptove-that-f-x-2-pi-e-1-x-1-ln-cosx-x-pi-2-

Question Number 66322 by mathmax by abdo last updated on 12/Aug/19 $${let}\:{f}\left({x}\right)=\left({cosx}\right)^{\frac{\mathrm{1}}{{x}}} \:\left(\:\mathrm{1}\right)\:\:{prove}\:{that}\:{f}\left({x}\right)\sim\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{8}}\:\:\left(\:{x}\rightarrow\mathrm{0}\right) \\ $$$$\left(\mathrm{2}\right){ptove}\:{that}\:{f}^{'} \left({x}\right)\sim−\frac{\mathrm{2}}{\pi}\:{e}^{\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right){ln}\left({cosx}\right)} \:\:\left({x}\rightarrow\frac{\pi}{\mathrm{2}}\right) \\ $$ Terms of Service Privacy Policy…

r-1-0-pi-2-5-2-2-x-2r-dx-

Question Number 787 by 123456 last updated on 22/Mar/15 $$\underset{{r}=\mathrm{1}} {\overset{+\infty} {\sum}}\sqrt{\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{5}+\Gamma\left(\frac{\mathrm{2}}{\mathrm{2}+{x}^{\mathrm{2}{r}} }\right){dx}} \\ $$ Commented by 123456 last updated on 13/Mar/15 $$\underset{{r}=\mathrm{1}}…

Prove-that-if-two-numbers-are-chosen-at-random-then-the-probability-that-their-sum-is-divisible-by-n-is-1-n-

Question Number 785 by rishabh last updated on 12/Mar/15 $${Prove}\:{that}\:{if}\:{two}\:{numbers}\:{are}\:{chosen} \\ $$$${at}\:{random}\:{then}\:{the}\:{probability}\:{that} \\ $$$${their}\:{sum}\:{is}\:{divisible}\:{by}\:{n}\:{is}\:\frac{\mathrm{1}}{{n}}. \\ $$ Answered by prakash jain last updated on 12/Mar/15 $$\mathrm{Sum}\:\mathrm{mod}\:{n}={k},\:\mathrm{where}\:\mathrm{0}\leqslant{k}\leqslant{n}−\mathrm{1}.…

analysis-II-evaluate-1-10-x-2-d-x-x-fractional-part-of-x-

Question Number 131852 by mnjuly1970 last updated on 09/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:\:{analysis}\:\left({II}\right)… \\ $$$$\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\varnothing=\int_{\mathrm{1}} ^{\:\mathrm{10}} {x}^{\mathrm{2}} {d}\left(\left\{{x}\right\}\right)=? \\ $$$$\:\:\:\:\:\:\:\left\{{x}\right\}\:::\:{fractional}\:{part}\:{of}\:{x}\:… \\ $$$$ \\ $$ Answered by…