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Question Number 196183 by cortano12 last updated on 19/Aug/23

Answered by dimentri last updated on 19/Aug/23

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Answered by mr W last updated on 19/Aug/23

f(x)=2x^3 +3x^2 −12x−k  f′(x)=6x^2 +6x−12=0  ⇒x^2 +x−2=0  ⇒x_(1,2) =−2,1  f(x_1 )=2(−2)^3 +3(−2)^2 −12(−2)−k=20−k  f(x_2 )=2(1)^3 +3(1)^2 −12(1)−k=−7−k  such that f(x)=0 has 3 distinct real   roots, f(x_1 )f(x_2 )<0  (20−k)(−7−k)<0  ⇒−7<k<20  ⇒answer A)

$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}−{k} \\ $$$${f}'\left({x}\right)=\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow{x}_{\mathrm{1},\mathrm{2}} =−\mathrm{2},\mathrm{1} \\ $$$${f}\left({x}_{\mathrm{1}} \right)=\mathrm{2}\left(−\mathrm{2}\right)^{\mathrm{3}} +\mathrm{3}\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{12}\left(−\mathrm{2}\right)−{k}=\mathrm{20}−{k} \\ $$$${f}\left({x}_{\mathrm{2}} \right)=\mathrm{2}\left(\mathrm{1}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{12}\left(\mathrm{1}\right)−{k}=−\mathrm{7}−{k} \\ $$$${such}\:{that}\:{f}\left({x}\right)=\mathrm{0}\:{has}\:\mathrm{3}\:{distinct}\:{real}\: \\ $$$${roots},\:{f}\left({x}_{\mathrm{1}} \right){f}\left({x}_{\mathrm{2}} \right)<\mathrm{0} \\ $$$$\left(\mathrm{20}−{k}\right)\left(−\mathrm{7}−{k}\right)<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{7}<{k}<\mathrm{20} \\ $$$$\left.\Rightarrow{answer}\:{A}\right) \\ $$

Commented by mr W last updated on 19/Aug/23

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