Question Number 1147 by prakash jain last updated on 04/Jul/15 $${f}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=? \\ $$ Commented by prakash jain last updated on 04/Jul/15 $${f}\left({f}\left(\mathrm{0}\right)\right)=\mathrm{1}…
Question Number 132198 by benjo_mathlover last updated on 12/Feb/21 $$\mathrm{If}\:\begin{cases}{\mathrm{16}^{{a}+{b}} \:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{16}^{{b}+{c}} \:=\:\mathrm{2}}\\{\mathrm{16}^{{a}+{c}} \:=\:\mathrm{2}\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\:\mathrm{then}\:\mathrm{c}\:=\:\_\_\: \\ $$ Answered by EDWIN88 last updated on 12/Feb/21 $$\mathrm{eq}\left(\mathrm{1}\right)\:\mathrm{16}^{\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}…
Question Number 132180 by bounhome last updated on 12/Feb/21 $${solve}\:: \\ $$$$\mathrm{2}{sec}^{\mathrm{2}} {x}+\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right){secx}−\mathrm{3}\sqrt{\mathrm{2}}=\mathrm{0}\:;\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}} \\ $$ Answered by benjo_mathlover last updated on 12/Feb/21 $$\:\mathrm{3}\sqrt{\mathrm{2}}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right)\mathrm{cos}\:\mathrm{x}−\mathrm{2}=\mathrm{0} \\…
Question Number 66619 by mr W last updated on 17/Aug/19 $${solve}\:{for}\:{x},{y}\in{R} \\ $$$$\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}=\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{\mathrm{ln}\:\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$ Answered by Smail last updated on…
Question Number 1041 by tera last updated on 22/May/15 $${if}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:{ax}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{2}=\mathrm{0}\:{is}\:\mathrm{2}\:{and}\:{b}. \\ $$$$ \\ $$$${then}\:\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{ab}+{b}^{\mathrm{2}} =… \\ $$ Answered by prakash jain last updated…
Question Number 66546 by Sayantan chakraborty last updated on 17/Aug/19 Commented by Sayantan chakraborty last updated on 17/Aug/19 $$\mathrm{urgent}\:\mathrm{need} \\ $$ Commented by ~ À…
Question Number 1007 by tera last updated on 13/May/15 $$\left[{x}−\mathrm{2}\right]>−\mathrm{2}\:.\:\:{x}=….. \\ $$ Answered by prakash jain last updated on 13/May/15 $$\left({x}−\mathrm{2}\right)>−\mathrm{2}\Rightarrow{x}−\mathrm{2}+\mathrm{2}>−\mathrm{2}+\mathrm{2}\Rightarrow{x}>\mathrm{0} \\ $$ Terms of…
Question Number 989 by Madan pd gupta last updated on 13/May/15 $$\mathrm{3}+\mathrm{4} \\ $$ Answered by rpatle69@gmail.com last updated on 13/May/15 $$\mathrm{7} \\ $$ Terms…
Question Number 984 by 112358 last updated on 13/May/15 $${Show}\:{that}\:\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{3}{x}−\mathrm{1}}>\mathrm{1}\:{given}\:{that}\:{x}>\frac{\mathrm{1}}{\mathrm{3}}\:. \\ $$ Commented by prakash jain last updated on 13/May/15 $$\mathrm{For}\:{x}=\mathrm{1}\:\mathrm{LHS}=\frac{\mathrm{1}×\mathrm{2}}{\mathrm{2}}=\mathrm{1}\ngtr\mathrm{1} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{inequality}\:\mathrm{should}\:\mathrm{be} \\ $$$$\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{3}{x}−\mathrm{1}}\geqslant\mathrm{1}…
Question Number 66508 by Masumsiddiqui399@gmail.com last updated on 16/Aug/19 Commented by Prithwish sen last updated on 16/Aug/19 $$\left(\sqrt{\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}}\right)^{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}} =\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{x}+^{\mathrm{3}} \sqrt{\mathrm{2x}+\mathrm{4}}} \\ $$$$\because\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\:=\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\therefore\:\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}=\mathrm{x}+^{\mathrm{3}} \sqrt{\mathrm{2x}+\mathrm{1}}\:\:\Rightarrow\mathrm{x}=\mathrm{2}…