Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 624 by 123456 last updated on 15/Feb/15

  f(x,y)=(√(2(x+(√(x^2 −y)))))  what is the domain of f(x,y)

$$ \\ $$$${f}\left({x},{y}\right)=\sqrt{\mathrm{2}\left({x}+\sqrt{{x}^{\mathrm{2}} −{y}}\right)} \\ $$$${what}\:{is}\:{the}\:{domain}\:{of}\:{f}\left({x},{y}\right) \\ $$

Commented by prakash jain last updated on 12/Feb/15

y<x^2                      ...(i)  x+(√(x^2 −y)) ≥0     (√(x^2 −y)) ≥ −x     ...(ii)  From (ii)      x<0 ,y≤0  From  (i)      x≥0, y<x^2

$${y}<{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{i}\right) \\ $$$${x}+\sqrt{{x}^{\mathrm{2}} −{y}}\:\geqslant\mathrm{0}\:\:\: \\ $$$$\sqrt{{x}^{\mathrm{2}} −{y}}\:\geqslant\:−{x}\:\:\:\:\:...\left(\mathrm{ii}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{ii}\right)\:\:\:\:\:\:{x}<\mathrm{0}\:,{y}\leqslant\mathrm{0} \\ $$$$\mathrm{From}\:\:\left(\mathrm{i}\right)\:\:\:\:\:\:{x}\geqslant\mathrm{0},\:{y}<{x}^{\mathrm{2}} \\ $$$$ \\ $$

Answered by prakash jain last updated on 15/Feb/15

{(x,y)∈R^2 :x^2 ≥y and x+(√(x^2 −y )) ≥0}

$$\left\{\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} :{x}^{\mathrm{2}} \geqslant{y}\:\mathrm{and}\:{x}+\sqrt{{x}^{\mathrm{2}} −{y}\:}\:\geqslant\mathrm{0}\right\} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com