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Author: Tinku Tara

sin-x-y-cos-x-y-tanx-tany-1-x-y-

Question Number 201427 by mathlove last updated on 06/Dec/23 $$\begin{cases}{{sin}\left({x}+{y}\right)={cos}\left({x}−{y}\right)}\\{{tanx}−{tany}=\mathrm{1}}\end{cases} \\ $$$$\left({x},{y}\right)=\left(?,?\right) \\ $$ Answered by Rasheed.Sindhi last updated on 06/Dec/23 $$\begin{cases}{{sin}\left({x}+{y}\right)={cos}\left({x}−{y}\right)…\left({i}\right)}\\{{tanx}−{tany}=\mathrm{1}………\left({ii}\right)}\end{cases} \\ $$$$\left({x},{y}\right)=\left(?,?\right) \\…

Question-201452

Question Number 201452 by tri26112004 last updated on 06/Dec/23 Answered by Calculusboy last updated on 06/Dec/23 $$\int\boldsymbol{{x}}^{−\mathrm{2}} \boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}} \\ $$$$\boldsymbol{{Solution}}:\:\:\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{IBP}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \:\:\:\boldsymbol{{du}}=−\mathrm{4}\boldsymbol{{e}}^{−\mathrm{4}\boldsymbol{{x}}} \boldsymbol{{dx}}\:\:\boldsymbol{{dv}}=\boldsymbol{{x}}^{−\mathrm{2}}…

2025-2025-x-mod-17-

Question Number 201418 by cortano12 last updated on 06/Dec/23 $$\:\:\:\:\:\:\mathrm{2025}^{\mathrm{2025}} \:=\:\mathrm{x}\:\left(\mathrm{mod}\:\mathrm{17}\:\right) \\ $$ Answered by mr W last updated on 06/Dec/23 $$\mathrm{2025}^{\mathrm{2025}} \:\left({mod}\:\mathrm{17}\right) \\ $$$$=\left(\mathrm{119}×\mathrm{17}+\mathrm{2}\right)^{\mathrm{2025}}…

0-1-Li-3-x-2-1-x-dx-

Question Number 201473 by MrGHK last updated on 06/Dec/23 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{Li}}_{\mathrm{3}} \left(−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Let-f-x-and-g-x-be-given-by-f-x-1-x-1-x-2-1-x-4-1-x-2018-and-g-x-1-x-1-1-x-3-1-x-5-1-x-2017-Prove-that-f-x-g-x-gt-2-for-any-non-in

Question Number 201441 by dimentri last updated on 06/Dec/23 $${Let}\:{f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{be}\:{given}\:{by}\: \\ $$$$\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}−\mathrm{2}}\:+\frac{\mathrm{1}}{{x}−\mathrm{4}}\:+\:…\:+\frac{\mathrm{1}}{{x}−\mathrm{2018}} \\ $$$$\:{and}\: \\ $$$$\:\:{g}\left({x}\right)=\frac{\mathrm{1}}{{x}−\mathrm{1}}\:+\frac{\mathrm{1}}{{x}−\mathrm{3}}\:+\frac{\mathrm{1}}{{x}−\mathrm{5}}\:+…+\:\frac{\mathrm{1}}{{x}−\mathrm{2017}}. \\ $$$$\:\:{Prove}\:{that}\:\:\mid\:{f}\left({x}\right)−{g}\left({x}\right)\mid\:>\mathrm{2} \\ $$$$\:\:{for}\:{any}\:{non}−{integer}\:{real}\:{number} \\ $$$$\:\:{x}\:{satisfying}\:\mathrm{0}<{x}<\mathrm{2018}.\: \\ $$ Answered…