Question Number 75456 by peter frank last updated on 11/Dec/19
$${Find}\:{the}\:{domain}\:{and}\: \\ $$$${range}\:{of}\:{relation} \\ $$$$\left({a}\right){R}=\left\{\left({x},{y}\right):{y}=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}}\:\right\} \\ $$$$\left({b}\right){R}=\left\{\left({x},{y}\right):{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{1}\:\right\} \\ $$
Answered by Kunal12588 last updated on 11/Dec/19
![(a) x∈(−∞,−(√6)]∪[(√6),∞) y∈[0,∞) or R_+ (b) x∈(−∞,−1]∪[1,∞) y∈(−∞,∞) or R](Q75460.png)
$$\left({a}\right)\:{x}\in\left(−\infty,−\sqrt{\mathrm{6}}\right]\cup\left[\sqrt{\mathrm{6}},\infty\right) \\ $$$${y}\in\left[\mathrm{0},\infty\right)\:{or}\:\mathbb{R}_{+} \\ $$$$\left({b}\right)\:{x}\in\left(−\infty,−\mathrm{1}\right]\cup\left[\mathrm{1},\infty\right) \\ $$$${y}\in\left(−\infty,\infty\right)\:{or}\:\mathbb{R} \\ $$
Answered by vishalbhardwaj last updated on 11/Dec/19
$$\left(\mathrm{a}\right)\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{6} \\ $$$$\Rightarrow\:\mathrm{x}\:=\:\pm\:\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{6}} \\ $$$$\Rightarrow\:\mathrm{x}\:\geqslant\:\sqrt{\mathrm{6}},\:\forall\:\mathrm{y}\in\:\mathrm{R} \\ $$$$\Rightarrow\:\mathrm{x}\in\:\left[\sqrt{\mathrm{6},}\:\infty\right) \\ $$$$\mathrm{and}\:\mathrm{y}\in\:\left[\mathrm{0},\:\infty\right) \\ $$