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for-x-R-satisfy-the-equation-f-x-3x-f-1-x-2-x-1-find-f-2019-




Question Number 83956 by jagoll last updated on 08/Mar/20
for x ∈ R satisfy the equation   f(x)+3x f((1/x)) = 2(x+1)  find f(2019) .
$$\mathrm{for}\:\mathrm{x}\:\in\:\mathbb{R}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{3x}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:=\:\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right) \\ $$$$\mathrm{find}\:\mathrm{f}\left(\mathrm{2019}\right)\:.\: \\ $$
Commented by mr W last updated on 08/Mar/20
f(x)+3x f((1/x)) = 2(x+1)   ...(i)  f((1/x))+(3/x) f(x) = 2(((1+x)/x))   ...(ii)  (i)−3x(ii):  ⇒f(x)=((1+x)/2)  ⇒f(2019)=((1+2019)/2)=1010
$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{3x}\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:=\:\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)\:\:\:…\left({i}\right) \\ $$$$\mathrm{f}\left(\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{3}}{{x}}\:\mathrm{f}\left({x}\right)\:=\:\mathrm{2}\left(\frac{\mathrm{1}+{x}}{{x}}\right)\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\mathrm{3}{x}\left({ii}\right): \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}+{x}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left(\mathrm{2019}\right)=\frac{\mathrm{1}+\mathrm{2019}}{\mathrm{2}}=\mathrm{1010} \\ $$
Answered by john santu last updated on 08/Mar/20
replace x by 2019 in eq (1)  f(2019)+3(2019) f((1/(2019))) = 2(2020) (eq 2)  replace (1/x) by 2019 in eq (1)  f((1/(2019))) +(3/(2019)) f(2019) = 2(((2020)/(2019)))  now you are will get the value  f(2019)
$$\mathrm{replace}\:\mathrm{x}\:\mathrm{by}\:\mathrm{2019}\:\mathrm{in}\:\mathrm{eq}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{f}\left(\mathrm{2019}\right)+\mathrm{3}\left(\mathrm{2019}\right)\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2019}}\right)\:=\:\mathrm{2}\left(\mathrm{2020}\right)\:\left(\mathrm{eq}\:\mathrm{2}\right) \\ $$$$\mathrm{replace}\:\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{by}\:\mathrm{2019}\:\mathrm{in}\:\mathrm{eq}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{2019}}\right)\:+\frac{\mathrm{3}}{\mathrm{2019}}\:\mathrm{f}\left(\mathrm{2019}\right)\:=\:\mathrm{2}\left(\frac{\mathrm{2020}}{\mathrm{2019}}\right) \\ $$$$\mathrm{now}\:\mathrm{you}\:\mathrm{are}\:\mathrm{will}\:\mathrm{get}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{f}\left(\mathrm{2019}\right)\: \\ $$$$ \\ $$

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