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Question Number 87784 by john santu last updated on 06/Apr/20
$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{y}\:\mathrm{f}\left(\mathrm{y},\mathrm{x}\right)\:+\mathrm{x} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{1},\mathrm{2}\right)\:. \\ $$
Commented by john santu last updated on 06/Apr/20
![(i) f(1,2) = 2f(2,1)+1 (ii) f(2,1)=1.f(1,2)+2 ⇒f(1,2) = 2[f(1,2)+2]+1 f(1,2) = 2f(1,2) + 5 f(1,2) = −5](Q87787.png)
$$\left(\mathrm{i}\right)\:\mathrm{f}\left(\mathrm{1},\mathrm{2}\right)\:=\:\mathrm{2f}\left(\mathrm{2},\mathrm{1}\right)+\mathrm{1} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{f}\left(\mathrm{2},\mathrm{1}\right)=\mathrm{1}.\mathrm{f}\left(\mathrm{1},\mathrm{2}\right)+\mathrm{2} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{1},\mathrm{2}\right)\:=\:\mathrm{2}\left[\mathrm{f}\left(\mathrm{1},\mathrm{2}\right)+\mathrm{2}\right]+\mathrm{1} \\ $$$$\mathrm{f}\left(\mathrm{1},\mathrm{2}\right)\:=\:\mathrm{2f}\left(\mathrm{1},\mathrm{2}\right)\:+\:\mathrm{5} \\ $$$$\mathrm{f}\left(\mathrm{1},\mathrm{2}\right)\:=\:−\mathrm{5}\: \\ $$
Answered by jagoll last updated on 06/Apr/20
Commented by john santu last updated on 06/Apr/20
$$\mathrm{typo}\:\mathrm{in}\:\mathrm{line}\:\mathrm{2} \\ $$
Commented by jagoll last updated on 06/Apr/20
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$