Question Number 150231 by Lekhraj last updated on 10/Aug/21
Answered by Ar Brandon last updated on 10/Aug/21
$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{f}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{1}−{x}} +\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}^{{x}} }{\mathrm{3}+\mathrm{3}^{{x}} \sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\centerdot\frac{\mathrm{3}^{{x}} }{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{1}+\frac{\mathrm{3}^{{x}} }{\:\sqrt{\mathrm{3}}}}{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\centerdot\frac{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}}{\mathrm{3}^{{x}} +\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\left[{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\right]=\frac{\mathrm{6}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Ans}:\sqrt{\mathrm{3}}\underset{{x}=\mathrm{1}} {\overset{\mathrm{6}} {\sum}}\left[{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\right]=\mathrm{6} \\ $$
Commented by Lekhraj last updated on 10/Aug/21
$$\mathrm{Very}\:\mathrm{nice}\:.\:\mathrm{Thank}\:\mathrm{you}. \\ $$
Commented by Ar Brandon last updated on 10/Aug/21
$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome}\:\mathrm{Sir} \\ $$