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Question Number 26681 by Tanveer48 last updated on 28/Dec/17

f:R×R→R such that f(x+iy)=(√(x^2 +y^2 .))  Then f is  a) many−one and into function  b) one−one and onto function  c) many−one and onto function  d) one−one and into function

$${f}:{R}×{R}\rightarrow{R}\:{such}\:{that}\:{f}\left({x}+{iy}\right)=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} .} \\ $$$${Then}\:{f}\:{is} \\ $$$$\left.{a}\right)\:{many}−{one}\:{and}\:{into}\:{function} \\ $$$$\left.{b}\right)\:{one}−{one}\:{and}\:{onto}\:{function} \\ $$$$\left.{c}\right)\:{many}−{one}\:{and}\:{onto}\:{function} \\ $$$$\left.{d}\right)\:{one}−{one}\:{and}\:{into}\:{function} \\ $$

Commented by prakash jain last updated on 28/Dec/17

many−one as (2,3) and (3,2) map       to same value  into as no pair of (x,y) will map      to −ve values.

$${many}−{one}\:{as}\:\left(\mathrm{2},\mathrm{3}\right)\:{and}\:\left(\mathrm{3},\mathrm{2}\right)\:{map} \\ $$$$\:\:\:\:\:{to}\:{same}\:{value} \\ $$$${into}\:{as}\:{no}\:{pair}\:{of}\:\left({x},{y}\right)\:{will}\:{map} \\ $$$$\:\:\:\:{to}\:−{ve}\:{values}. \\ $$

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