Question Number 132358 by Engr_Jidda last updated on 13/Feb/21 $${find}\:{the}\:{equation}\:{whose}\:{roots}\:{are} \\ $$$$\alpha\:{and}\:\beta.\:{find}\:\alpha\:{and}\:\beta\:{if}\:\:\alpha−\beta=\mathrm{2}\:{and}\:\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} =\mathrm{3} \\ $$ Commented by Dwaipayan Shikari last updated on 13/Feb/21 $${x}^{\mathrm{2}}…
Question Number 1280 by Rasheed Soomro last updated on 19/Jul/15 $$\:{f}\:\left(\:\frac{\mathrm{1}}{{f}\left({x}\right)}\right)={f}\left({x}\right)\: \\ $$$${f}\left({x}\right)=? \\ $$ Commented by 123456 last updated on 19/Jul/15 $${f}\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$…
Question Number 1278 by Rasheed Soomro last updated on 19/Jul/15 $${Determine}\:{f}\left({x}\right)\:{when}\:{f}\left(\:{f}\left({x}\right)\right)={x}\:\:. \\ $$ Commented by prakash jain last updated on 19/Jul/15 $${f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}} \\ $$$${f}\left({f}\left({x}\right)\right)=\frac{{a}\frac{{ax}+{b}}{{cx}+{d}}+{b}}{{c}\frac{{ax}+{b}}{{cx}+{d}}+{d}}={x} \\…
Question Number 1260 by Rasheed Ahmad last updated on 18/Jul/15 $$\mathrm{What}\:\mathrm{is}\:\mathrm{degree}\:\mathrm{of}\:\frac{\mathrm{x}}{\mathrm{y}}\:?\:\left(\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are}\:\mathrm{both}\:\mathrm{variables}\right) \\ $$$$\mathrm{Is}\:\frac{\mathrm{x}}{\mathrm{y}}\:\:\mathrm{and}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{of}\:\mathrm{same}\:\mathrm{degree}? \\ $$ Commented by prakash jain last updated on 18/Jul/15 $$\mathrm{Degree}\:\mathrm{of}\:\mathrm{rational}\:\mathrm{expression}\:\mathrm{is}\:\mathrm{the}\:\mathrm{degree} \\…
Question Number 1239 by Rasheed Ahmad last updated on 17/Jul/15 $${Rasheed}\:{Ahmad}\:\left({Rasheed}\:{Soomro}\right) \\ $$$$\bullet\mathrm{For}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{where}\:\mathrm{x}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{both}\:{are} \\ $$$$\mathrm{real}\:\mathrm{the}\:\left(\mathrm{x},{f}\left({x}\right)\right)\:{can}\:{be}\:{plotted}\:{as} \\ $$$${a}\:{point}\:{easily}.\:\bullet{Now}\:{consider}\:{F}\left({X}\right) \\ $$$${where}\:{X}\:{and}\:{F}\left({X}\right)\:{are}\:{complex}\: \\ $$$${numbers}.\:{How}\:{can}\:\left({X},{F}\left({X}\right)\right)\:{be} \\ $$$${plotted}?\:{For}\:{a}\:{particular}\:{example}:\:\left(\mathrm{3}+\mathrm{2}{i},\mathrm{4}−\mathrm{5}{i}\right) \\ $$$${how}\:{can}\:{be}\:{plotted}?…
Question Number 66769 by John Kaloki Musau last updated on 19/Aug/19 $${simplify} \\ $$$$\frac{{x}+\mathrm{4}}{{x}−\mathrm{4}}−\frac{\mathrm{5}{x}+\mathrm{20}}{{x}^{\mathrm{2}} −\mathrm{16}} \\ $$ Commented by Prithwish sen last updated on 19/Aug/19…
Question Number 1231 by Rasheed Soomro last updated on 16/Jul/15 $${i}^{\mathrm{2}} ={i}.{i}=\sqrt{−\mathrm{1}}.\sqrt{−\mathrm{1}}=\sqrt{−\mathrm{1}×−\mathrm{1}}=\sqrt{\mathrm{1}}=\mathrm{1}??? \\ $$$${i}^{\mathrm{2}} =\mathrm{1}\Rightarrow−\mathrm{1}=\mathrm{1}???\: \\ $$$${Resolve}\:{the}\:{contradiction}. \\ $$ Commented by prakash jain last updated…
Question Number 1229 by Rasheed Soomro last updated on 16/Jul/15 $${f}\left(\:{f}\left({x}\right)\:\right)={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$${f}\left({x}\right)=? \\ $$$$\left({Modification}\:{of}\:{Q}\:\mathrm{1147}\right) \\ $$ Commented by 123456 last updated on 17/Jul/15…
Question Number 66760 by John Kaloki Musau last updated on 19/Aug/19 $${By}\:{writing}\:{your}\:{answer}\:{in}\:{the} \\ $$$${form}\:{a}^{{y}} \:{simplify} \\ $$$$\left(\mathrm{3}^{\mathrm{5}{x}} ×\mathrm{5}^{\mathrm{2}{x}} ×\mathrm{3}^{−{x}} \boldsymbol{\div}\mathrm{5}^{−\mathrm{2}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$ Commented by…
Question Number 132295 by shaker last updated on 13/Feb/21 Commented by mathmax by abdo last updated on 13/Feb/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{logx}−\sqrt{\mathrm{5x}−\mathrm{5}}\:\:\:\mathrm{f}\:\mathrm{defined}\:\mathrm{on}\:\left[\mathrm{1},+\infty\left[\right.\right. \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5x}−\mathrm{5}}}\:=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}}\:=\frac{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}−\sqrt{\mathrm{5}}\mathrm{x}}{\mathrm{2x}\sqrt{\mathrm{x}−\mathrm{1}}} \\ $$$$=\frac{\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}−\sqrt{\mathrm{5}}\mathrm{x}\right)\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{5}}\mathrm{x}\right)}{\mathrm{2x}\sqrt{\mathrm{x}−\mathrm{1}}\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{5}}\mathrm{x}\right)}\:=\frac{\mathrm{4}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{5x}^{\mathrm{2}} }{\left(…\right)}=\frac{−\mathrm{5x}^{\mathrm{2}}…