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

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