Question Number 118275 by bramlexs22 last updated on 16/Oct/20
Commented by bramlexs22 last updated on 16/Oct/20
$${f}\left({x}\right)={x}^{\mathrm{2}} \:,\:{when}\: \\ $$$$\:\left(\frac{\mathrm{3}}{{f}\left(\mathrm{1}\right){f}\left(\mathrm{2}\right)}+\frac{\mathrm{5}}{{f}\left(\mathrm{2}\right){f}\left(\mathrm{3}\right)}+\frac{\mathrm{7}}{{f}\left(\mathrm{3}\right){f}\left(\mathrm{4}\right)}+\frac{\mathrm{9}}{{f}\left(\mathrm{4}\right){f}\left(\mathrm{5}\right)}+\frac{\mathrm{11}}{{f}\left(\mathrm{5}\right){f}\left(\mathrm{6}\right)}\right).{x}\:<\:\mathrm{70}\: \\ $$$${Find}\:{the}\:{largest}\:{complete} \\ $$$${solution}\:{of}\:{the}\:{inequality}\:. \\ $$$$\left({a}\right)\mathrm{64}\:\:\:\:\:\left({b}\right)\mathrm{55}\:\:\:\:\:\:\:\left({c}\right)\mathrm{71} \\ $$$$\left({d}\right)\:\mathrm{69}\:\:\:\:\left({e}\right)\:\mathrm{74} \\ $$
Answered by 1549442205PVT last updated on 16/Oct/20
$$\left(\frac{\mathrm{3}}{\mathrm{f}\left(\mathrm{1}\right)\mathrm{f}\left(\mathrm{2}\right)}+\frac{\mathrm{5}}{\mathrm{f}\left(\mathrm{2}\right)\left(\mathrm{f3}\right)}+\frac{\mathrm{7}}{\mathrm{f}\left(\mathrm{3}\right)\mathrm{f}\left(\mathrm{4}\right)}+\frac{\mathrm{9}}{\mathrm{f}\left(\mathrm{4}\right)\mathrm{f}\left(\mathrm{5}\right)}+\frac{\mathrm{11}}{\mathrm{f}\left(\mathrm{5}\right)\mathrm{f}\left(\mathrm{6}\right)}\right)\mathrm{x}<\mathrm{70} \\ $$$$\Leftrightarrow\left(\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{36}}+\frac{\mathrm{7}}{\mathrm{144}}+\frac{\mathrm{9}}{\mathrm{400}}+\frac{\mathrm{11}}{\mathrm{900}}\right)\mathrm{x}<\mathrm{70} \\ $$$$\Leftrightarrow\frac{\mathrm{35}}{\mathrm{36}}\mathrm{x}<\mathrm{70}\Leftrightarrow\mathrm{x}<\frac{\mathrm{70}.\mathrm{36}}{\mathrm{35}}=\mathrm{72} \\ $$$$\left.\Rightarrow\mathrm{x}_{\mathrm{max}} =\mathrm{71}.\mathrm{Hence}\:,\mathrm{choose}\:\mathrm{c}\right) \\ $$