Question Number 749 by 123456 last updated on 06/Mar/15
![x=(√(2(√(2(√(2(√(2(√(2(√(2∙∙∙)))))))))))) y=(√(2+(√(2+(√(2+(√(2+(√(2+(√(2+∙∙∙)))))))))))) wich statment is true? a. xy=4∨xy=0∨x+y=4∨x+y=2 b. x∉Z c.xy∉Z d.x+y∉Z](https://www.tinkutara.com/question/Q749.png)
$${x}=\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\centerdot\centerdot\centerdot}}}}}} \\ $$$${y}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\centerdot\centerdot\centerdot}}}}}} \\ $$$${wich}\:{statment}\:{is}\:{true}? \\ $$$${a}.\:{xy}=\mathrm{4}\vee{xy}=\mathrm{0}\vee{x}+{y}=\mathrm{4}\vee{x}+{y}=\mathrm{2} \\ $$$${b}.\:{x}\notin\mathbb{Z} \\ $$$${c}.{xy}\notin\mathbb{Z} \\ $$$${d}.{x}+{y}\notin\mathbb{Z} \\ $$
Answered by prakash jain last updated on 06/Mar/15
![x=(√(2x))⇒x^2 =2x⇒x=2 ∵x>0 y=(√(2+y))⇒y^2 =2+y⇒(y−2)(y+1)=0 ⇒y=2 ∵y>0 a. true xy=4, xy=4 b. false x∈Z c. false xy∈Z d. false x+y∈Z](https://www.tinkutara.com/question/Q750.png)
$${x}=\sqrt{\mathrm{2}{x}}\Rightarrow{x}^{\mathrm{2}} =\mathrm{2}{x}\Rightarrow{x}=\mathrm{2}\:\because{x}>\mathrm{0} \\ $$$${y}=\sqrt{\mathrm{2}+{y}}\Rightarrow{y}^{\mathrm{2}} =\mathrm{2}+{y}\Rightarrow\left({y}−\mathrm{2}\right)\left({y}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{2}\:\because{y}>\mathrm{0} \\ $$$${a}.\:\boldsymbol{\mathrm{true}}\:{xy}=\mathrm{4},\:{xy}=\mathrm{4} \\ $$$${b}.\:\mathrm{false}\:{x}\in\mathbb{Z} \\ $$$$\mathrm{c}.\:\mathrm{false}\:{xy}\in\mathbb{Z} \\ $$$$\mathrm{d}.\:\mathrm{false}\:{x}+{y}\in\mathbb{Z} \\ $$