Question Number 206571 by mr W last updated on 19/Apr/24
Answered by cortano21 last updated on 19/Apr/24
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Commented by mr W last updated on 19/Apr/24
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Answered by MM42 last updated on 19/Apr/24
$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} \:\:\:\:;{A}\left({a},\mathrm{2}{a}^{\mathrm{2}} \right)\:,{f}'\left({a}\right)=\mathrm{4}{a} \\ $$$${g}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}\:\:\:;\:{B}\left({b},\mathrm{2}{b}^{\mathrm{2}} −\mathrm{8}{b}+\mathrm{16}\right)\:\:,{g}'\left({b}\right)=\mathrm{4}{b}−\mathrm{8} \\ $$$${f}'\left({a}\right)={g}'\left({b}\right)\Rightarrow{a}={b}−\mathrm{2}\:\:\:\left({i}\right) \\ $$$${m}_{{AB}} =\frac{\mathrm{2}{b}^{\mathrm{2}} −\mathrm{8}{b}+\mathrm{16}−\mathrm{2}{a}^{\mathrm{2}} }{{b}−{a}}=\mathrm{4}{a}\:\left({ii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\Rightarrow{a}=\mathrm{1}\:\&\:{b}=\mathrm{3} \\ $$$$\Rightarrow{m}_{{AB}} =\mathrm{4}\Rightarrow{L}:\:{y}=\mathrm{4}{x}−\mathrm{2} \\ $$$${f}={g}\Rightarrow{x}=\mathrm{2} \\ $$$${s}_{\mathrm{1}} =\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\right){dx}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${s}_{\mathrm{2}} =\int_{\mathrm{2}} ^{\mathrm{3}} \left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{16}−\mathrm{4}{x}+\mathrm{2}\right){dx}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{s}=\frac{\mathrm{4}}{\mathrm{3}}\:\checkmark \\ $$$$ \\ $$
Commented by mr W last updated on 19/Apr/24
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Answered by MathematicalUser2357 last updated on 28/Apr/24
