Question

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Answer 1

\(y'=3x^2-6mx=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2m\end{matrix}\right.\)

Hàm có 2 cực trị \(\Rightarrow m\ne0\)

\(\Rightarrow A\left(0;3m^3\right)\) ; \(B\left(2m;-m^3\right)\)

Do A thuộc trục tung \(\Rightarrow OA=\left|y_A\right|=\left|3m^3\right|\)

\(d\left(B;OA\right)=\left|x_B\right|=\left|2m\right|\)

\(S_{OAB}=\dfrac{1}{2}OA.d\left(B;OA\right)\Leftrightarrow48=\dfrac{1}{2}.\left|3m^3.2m\right|\)

\(\Leftrightarrow m^4=16\Rightarrow m=\pm2\)