Question Number 1952 by prakash jain last updated on 25/Oct/15 $$\mathrm{Inequality}\:\mathrm{relation}\:\mathrm{starting}\:\mathrm{a}\:\mathrm{new}\:\mathrm{thread} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{p}}\geqslant\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{q}} \\ $$$${p}=\mathrm{2},\:{q}=\mathrm{1},\:{x}=\mathrm{1} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\frac{{x}^{{p}}…
Question Number 1937 by Rasheed Soomro last updated on 25/Oct/15 $$\bullet{Is}\:\:\:'\Leftrightarrow'\:\:{necessary}\:{and}\:{suficient}\:{for}\:{two} \\ $$$${inequalities}\:{to}\:{be}\:{equivalent}? \\ $$$$\bullet{If}\:\:\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}\:: \\ $$$${Are}\:\:\boldsymbol{\mathrm{A}}>\boldsymbol{\mathrm{B}}\:{and}\:\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{a}}\:>\:\boldsymbol{\mathrm{B}}+\boldsymbol{\mathrm{b}}\:{equivalent}? \\ $$ Answered by 123456 last updated on…
Question Number 133009 by shaker last updated on 18/Feb/21 Commented by liberty last updated on 18/Feb/21 $$?\:=\:\mathrm{1} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 1930 by Rasheed Soomro last updated on 24/Oct/15 $${f}\:'\left({x}\right)−{g}\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)+{g}'\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$$${g}\left({x}\right)=? \\ $$ Answered by prakash jain last updated…
Question Number 1902 by Rasheed Soomro last updated on 23/Oct/15 $${f}^{\:\mathrm{2}} \left({x}\right)−{f}\left({x}^{\mathrm{2}} \right)=\mathrm{2}\:,\:{f}^{\:\mathrm{2}} \left({x}\right)\:{stands}\:{for}\:\left[{f}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${f}\left({x}\right)=? \\ $$$$\left({If}\:{possible}\:{solve}\:{stepwise}\right) \\ $$ Commented by 123456 last updated…
Question Number 1899 by Yozzy last updated on 22/Oct/15 $${Consider}\:{the}\:{system}\:{of}\:{equations} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{yz}+{zx}−\mathrm{5}{xy}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{yz}−{zx}+\mathrm{2}{xy}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{yz}−\mathrm{2}{zx}+\mathrm{6}{xy}=\mathrm{3}. \\ $$$${Show}\:{that}\:{xyz}=\pm\mathrm{6}\: \\ $$$${and}\:{find}\:{the}\:{possible}\:{values} \\ $$$${of}\:{x},{y}\:{and}\:{z}. \\ $$ Commented…
Question Number 132961 by danielasebhofoh last updated on 17/Feb/21 Answered by Ar Brandon last updated on 17/Feb/21 $$\mathrm{2n}−\mathrm{3}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{3}} {\sum}}\mathrm{k}=\mathrm{2n}−\mathrm{3}+\frac{\left(\mathrm{n}−\mathrm{3}\right)\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\frac{\mathrm{4n}−\mathrm{6}+\mathrm{n}^{\mathrm{2}} −\mathrm{5n}+\mathrm{6}}{\mathrm{2}}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\overset{\mathrm{n}}…
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Question Number 1839 by RasheedAhmad last updated on 11/Oct/15 $${Sum}\:{to}\:{n}\:{terms}\:{the}\:{series}: \\ $$$$\mathrm{0}\centerdot\mathrm{3}+\mathrm{0}\centerdot\mathrm{33}+\mathrm{0}\centerdot\mathrm{333}+… \\ $$$$ \\ $$ Answered by 112358 last updated on 12/Oct/15 $${We}\:{may}\:{begin}\:{by}\:{searching}\:{for} \\…
Question Number 1838 by 112358 last updated on 11/Oct/15 $${Show}\:{that} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{1}}{{sinz}}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$$\left(\mathrm{2}\right)\:{cotz}=\frac{\mathrm{1}}{{z}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{z}+{n}\pi}+\frac{\mathrm{1}}{{z}−{n}\pi}\right) \\ $$$${where}\:{z}\neq{m}\pi,\:{m}\in\mathbb{Z}\:,\:{given}\:{the}\:{fact}\:{that} \\ $$$${cosax}=\frac{\mathrm{2}{sina}\pi}{\pi}\left[\frac{\mathrm{1}}{\mathrm{2}{a}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}}…