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Author: Tinku Tara

If-1-a-2-1-b-2-1-c-2-1-ab-1-bc-1-ca-then-prove-that-a-b-c-

Question Number 70312 by Shamim last updated on 03/Oct/19 $$\mathrm{If},\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{ca}}\:\mathrm{then}\:\mathrm{prove}\: \\ $$$$\mathrm{that},\:\mathrm{a}=\mathrm{b}=\mathrm{c}. \\ $$ Commented by MJS last updated on 03/Oct/19 $$\mathrm{this}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{a},\:{b},\:{c}\:\in\mathbb{R}…

Let-be-a-binary-operation-on-Z-defined-by-x-y-1-2-x-y-1-1-2-1-1-x-y-Is-associative-

Question Number 4775 by Yozzii last updated on 08/Mar/16 $${Let}\:\ast\:{be}\:{a}\:{binary}\:{operation}\:{on}\:\mathbb{Z} \\ $$$${defined}\:{by}\: \\ $$$${x}\ast{y}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{x}+{y}} \right)\right). \\ $$$${Is}\:\ast\:{associative}? \\ $$ Commented by prakash jain last updated…

Let-z-Ax-2-Bxy-Cy-2-Find-conditions-on-the-constants-A-B-C-that-ensure-that-the-point-0-0-0-is-a-i-local-minimum-ii-local-maximum-ii-saddle-point-

Question Number 4772 by Yozzii last updated on 07/Mar/16 $${Let}\:{z}={Ax}^{\mathrm{2}} +{Bxy}+{Cy}^{\mathrm{2}} .\:{Find}\:{conditions} \\ $$$${on}\:{the}\:{constants}\:{A},{B},{C}\:{that}\:{ensure} \\ $$$${that}\:{the}\:{point}\:\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:{is}\:{a}\: \\ $$$$\left({i}\right)\:{local}\:{minimum}, \\ $$$$\left({ii}\right)\:{local}\:{maximum}, \\ $$$$\left({ii}\right)\:{saddle}\:{point}. \\ $$$$ \\…

Question-135840

Question Number 135840 by abdurehime last updated on 16/Mar/21 Answered by liberty last updated on 16/Mar/21 $${from}\:{p}\Rightarrow\:\backsim{q}\:{False}\:{we}\:{get}\:\begin{cases}{{p}\::\:{T}}\\{{q}\::\:{T}}\end{cases} \\ $$$${so}\:{the}\:{following}\:{statements}\:{is} \\ $$$${True}\::\:\left({p} \sim{q}\right)\Leftrightarrow\left(\sim{r} {r}\right) \\ $$$${answer}\::\:{C}…

Question-135838

Question Number 135838 by abdurehime last updated on 16/Mar/21 Answered by liberty last updated on 16/Mar/21 $${f}\left({x}\right)=\:\frac{{x}}{{x}^{\mathrm{2}} +{k}}\:\Rightarrow\mathrm{ln}\:{y}\:=\:\mathrm{ln}\:{x}−\mathrm{ln}\:\left({x}^{\mathrm{2}} +{k}\right) \\ $$$$\:\frac{{y}'}{{y}}\:=\:\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +{k}} \\ $$$$\frac{{y}'}{{y}}\:=\:\frac{{k}−{x}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{2}}…