Question Number 113854 by ayenisamuel last updated on 15/Sep/20 | ||
$${if}\:{f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{10}.\: \\ $$$$\left({i}\right)\:{sketch}\:{the}\:{graph}\:{of}\:{y}=\mid{f}\left({x}\right)\mid\:{for} \\ $$$$−\mathrm{1}\leqslant{x}\leqslant\mathrm{7}. \\ $$$$\left({ii}\right)\:{find}\:{the}\:{set}\:{of}\:{values}\:{of}\:{k}\:{for} \\ $$$${which}\:{the}\:{equation}\:\mid{f}\left({x}\right)\mid={k}\:{has}\:\mathrm{4} \\ $$$${distinct}\:{roots}. \\ $$ | ||
Answered by MJS_new last updated on 15/Sep/20 | ||
$$\mid\mathrm{2}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{10}\mid={k}\:\Rightarrow\:{k}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{10}\right)^{\mathrm{2}} ={k}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{12}{x}^{\mathrm{3}} +\mathrm{46}{x}^{\mathrm{2}} −\mathrm{60}{x}−\frac{{k}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{25}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}={t}+\mathrm{3} \\ $$$${t}^{\mathrm{4}} −\mathrm{8}{t}^{\mathrm{2}} −\frac{{k}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} =\frac{\mathrm{8}\pm{k}}{\mathrm{2}}\:\Rightarrow\:{t}=\pm\sqrt{\frac{\mathrm{8}\pm{k}}{\mathrm{2}}}\:\:\:\left[\mathrm{4}\:\mathrm{solutions}\right] \\ $$$$\Rightarrow\:{x}_{\mathrm{1},\:\mathrm{2}} =\mathrm{3}\pm\sqrt{\frac{\mathrm{8}−{k}}{\mathrm{2}}};\:{x}_{\mathrm{3},\:\mathrm{4}} =\mathrm{3}\pm\sqrt{\frac{\mathrm{8}+{k}}{\mathrm{2}}} \\ $$$$\Rightarrow\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{only}\:\mathrm{if}\:\mathrm{8}−{k}>\mathrm{0}\wedge\mathrm{8}+{k}>\mathrm{0} \\ $$$$\Rightarrow\:−\mathrm{8}<{k}<\mathrm{8}\:\mathrm{but}\:{k}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{0}\leqslant{k}<\mathrm{8} \\ $$ | ||