Question Number 196183 by cortano12 last updated on 19/Aug/23
Answered by dimentri last updated on 19/Aug/23
$$\:\:\underbrace{\Subset} \\ $$
Answered by mr W last updated on 19/Aug/23
$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}−{k} \\ $$$${f}'\left({x}\right)=\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1},\mathrm{2}} =−\mathrm{2},\mathrm{1} \\ $$$${f}\left({x}_{\mathrm{1}} \right)=\mathrm{2}\left(−\mathrm{2}\right)^{\mathrm{3}} +\mathrm{3}\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{12}\left(−\mathrm{2}\right)−{k}=\mathrm{20}−{k} \\ $$$${f}\left({x}_{\mathrm{2}} \right)=\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{12}\left(\mathrm{1}\right)−{k}=−\mathrm{7}−{k} \\ $$$${such}\:{that}\:{f}\left({x}\right)=\mathrm{0}\:{has}\:\mathrm{3}\:{distinct}\:{real}\: \\ $$$${roots},\:{f}\left({x}_{\mathrm{1}} \right){f}\left({x}_{\mathrm{2}} \right)<\mathrm{0} \\ $$$$\left(\mathrm{20}−{k}\right)\left(−\mathrm{7}−{k}\right)<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{7}<{k}<\mathrm{20} \\ $$$$\left.\Rightarrow{answer}\:{A}\right) \\ $$
Commented by mr W last updated on 19/Aug/23
