Menu Close

Category: Others

Ques-5-Prove-that-if-a-b-are-any-elements-of-a-group-G-then-the-equation-y-a-b-has-a-unique-solution-in-G-Ques-6-a-Show-that-the-set-G-of-all-non-zero-complex-numbers-i

Question Number 193759 by Mastermind last updated on 19/Jun/23 $$\mathrm{Ques}.\:\mathrm{5}\: \\ $$$$\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{a},\mathrm{b}\:\mathrm{are}\:\mathrm{any}\:\mathrm{elements}\:\mathrm{of}\:\:\mathrm{a}\: \\ $$$$\mathrm{group}\:\left(\mathrm{G},\:\ast\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{y}\ast\mathrm{a}=\mathrm{b} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{in}\:\left(\mathrm{G},\:\ast\right). \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{6}\: \\ $$$$\left.\:\:\:\:\:\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{set}\:\mathrm{G}\:\mathrm{of}\:\mathrm{all}\:\mathrm{non}-\mathrm{zero}\: \\ $$$$\mathrm{complex}\:\mathrm{numbers},\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{under} \\…

Ques-3-Show-that-Z-4-is-a-group-Hence-find-the-order-of-the-group-and-of-the-element-2-Z-4-if-it-exists-Ques-4-Prove-the-if-a-b-are-any-elements-of-a-group-G-then-the-e

Question Number 193712 by Mastermind last updated on 18/Jun/23 $$\mathrm{Ques}.\:\mathrm{3}\: \\ $$$$\:\:\:\:\:\mathrm{Show}\:\mathrm{that}\:\left(\mathbb{Z}_{\mathrm{4}} ,\:+\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}.\:\mathrm{Hence} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{the}\:\mathrm{group}\:\mathrm{and}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{element}\:\mathrm{2}\in\mathbb{Z}_{\mathrm{4}} ,\:\mathrm{if}\:\mathrm{it}\:\mathrm{exists}. \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{Prove}\:\mathrm{the}\:\mathrm{if}\:\mathrm{a},\mathrm{b}\:\mathrm{are}\:\mathrm{any}\:\mathrm{elements}\:\mathrm{of}\:\mathrm{a}\: \\…

Ques-1-Let-G-be-a-group-and-b-a-fixed-element-of-G-Prove-that-the-map-G-into-G-given-by-x-bx-is-bijective-Ques-2-Let-G-be-a-group-and-g-be-an-element-of-G-Prove-that-a-g-1-1

Question Number 193711 by Mastermind last updated on 18/Jun/23 $$\mathrm{Ques}.\:\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{Let}\:\mathrm{G}\:\mathrm{be}\:\mathrm{a}\:\mathrm{group}\:\mathrm{and}\:\mathrm{b}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{element} \\ $$$$\mathrm{of}\:\mathrm{G}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{map}\:\mathrm{G}\:\mathrm{into}\:\mathrm{G}\:\mathrm{given} \\ $$$$\mathrm{by}\:\mathrm{x}\rightarrow\mathrm{bx}\:\mathrm{is}\:\mathrm{bijective} \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{2}\: \\ $$$$\:\:\:\:\:\mathrm{Let}\:\mathrm{G}\:\mathrm{be}\:\mathrm{a}\:\mathrm{group}\:\mathrm{and}\:\mathrm{g}\:\mathrm{be}\:\mathrm{an}\:\mathrm{element}\: \\ $$$$\mathrm{of}\:\mathrm{G}.\:\mathrm{Prove}\:\mathrm{that}\: \\…

Question-193612

Question Number 193612 by Noorzai last updated on 17/Jun/23 Answered by aba last updated on 17/Jun/23 $$\mathrm{x}^{\mathrm{x}^{\mathrm{2}} } =\mathrm{3}\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} \mathrm{lnx}=\mathrm{ln3} \\ $$$$\Rightarrow\mathrm{2lnx}.\mathrm{e}^{\mathrm{2lnx}} =\mathrm{2ln3}\:\Rightarrow\:\mathrm{2lnx}=\mathrm{W}\left(\mathrm{2ln3}\right) \\ $$$$\Rightarrow\:\mathrm{x}=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{W}\left(\mathrm{2ln3}\right)}…

Evaluate-I-s-x-3-dydz-x-2-ydzdx-where-S-is-the-closed-surface-consis-ting-of-the-cylinder-x-2-y-2-a-2-0-z-b-and-the-cylinder-disks-z-0-and-z-b-x-2-y-2-b-x-2-y-2-a-Help-

Question Number 193513 by Mastermind last updated on 15/Jun/23 $$\mathrm{Evaluate}\:\mathrm{I}=\underset{\mathrm{s}} {\int}\int\mathrm{x}^{\mathrm{3}} \mathrm{dydz}\:+\:\mathrm{x}^{\mathrm{2}} \mathrm{ydzdx}. \\ $$$$\mathrm{where}\:\mathrm{S}\:\mathrm{is}\:\mathrm{the}\:\mathrm{closed}\:\mathrm{surface}\:\mathrm{consis}− \\ $$$$\mathrm{ting}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cylinder}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} ,\:\mathrm{0}\leqslant\mathrm{z}\leqslant\mathrm{b} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{cylinder}\:\:\mathrm{disks}\:\mathrm{z}=\mathrm{0}\:\mathrm{and}\:\mathrm{z}=\mathrm{b}, \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…

Question-193323

Question Number 193323 by fit last updated on 10/Jun/23 Answered by MM42 last updated on 10/Jun/23 $$\left.\mathrm{9}\right)\:{n}\left({s}\right)=\mathrm{80}\:\&\:{n}\left({m}\right)=\mathrm{40}\:\&\:{n}\left({e}\right)=\mathrm{65} \\ $$$${n}\left({s}\right)={n}\left({m}\right)+{n}\left({e}\right)−{n}\left({m}\cap{e}\right) \\ $$$$\mathrm{80}=\mathrm{40}+\mathrm{65}−{n}\left({m}\cap{e}\right)\Rightarrow{n}\left({m}\cap{e}\right)=\mathrm{15} \\ $$$$\left.\mathrm{10}\right)\frac{\mathrm{15}!}{\mathrm{3}!\mathrm{2}!\mathrm{2}!\mathrm{2}!\mathrm{2}!\mathrm{2}!} \\ $$$$\left.\mathrm{11}\right)\mathrm{10}×\mathrm{9}×\mathrm{8}×\mathrm{7}…

1-Prove-that-a-b-c-a-b-c-2-Find-all-x-R-that-satify-the-follow-ing-inequalities-i-x-2-4-lt-5-ii-x-x-2-lt-5-Help-

Question Number 193137 by Mastermind last updated on 04/Jun/23 $$\left.\mathrm{1}\right)\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{a}+\mathrm{b}+\mathrm{c}\mid\geqslant\mid\mathrm{a}\mid−\mid\mathrm{b}\mid−\mid\mathrm{c}\mid \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Find}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}\:\mathrm{that}\:\mathrm{satify}\:\mathrm{the}\:\mathrm{follow}− \\ $$$$\mathrm{ing}\:\mathrm{inequalities}\: \\ $$$$\left.\mathrm{i}\right)\:\mid\mathrm{x}^{\mathrm{2}} −\mathrm{4}\mid<\mathrm{5} \\ $$$$\left.\mathrm{ii}\right)\:\mid\mathrm{x}\mid+\mid\mathrm{x}+\mathrm{2}\mid<\mathrm{5} \\ $$$$…

Show-that-for-all-a-b-R-i-ab-1-2-a-2-b-2-ii-a-b-2-2-a-2-b-2-iii-ab-1-2-a-b-for-a-b-0-such-that-a-and-b-have-square-roots-Help-

Question Number 193138 by Mastermind last updated on 04/Jun/23 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b}\in\mathbb{R} \\ $$$$\left.\mathrm{i}\right)\:\mathrm{ab}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{ii}\right)\:\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{iii}\right)\:\sqrt{\mathrm{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}+\mathrm{b}\right),\:\mathrm{for}\:\mathrm{a},\mathrm{b}\geqslant\mathrm{0}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{have}\:\mathrm{square}\:\mathrm{roots}. \\ $$$$…