Đồ thị hàm số y = x^4 -2mx^2 +m có 2 điểm cực trị cùng điểm D (0;-4) tạo thành 1 hình thoi . Tìm m 27/09/2021 Bởi Adeline Đồ thị hàm số y = x^4 -2mx^2 +m có 2 điểm cực trị cùng điểm D (0;-4) tạo thành 1 hình thoi . Tìm m
\[\begin{array}{l} y = {x^4} – 2m{x^2} + m\,\,\,\left( C \right)\\ \Rightarrow y’ = 4{x^3} – 4mx\\ \Rightarrow y’ = 0\\ \Leftrightarrow 4{x^3} – 4mx = 0\,\,\,\left( * \right) \Leftrightarrow 4x\left( {{x^2} – m} \right) = 0\\ = \Leftrightarrow \left[ \begin{array}{l} x = 0\\ {x^2} = m \end{array} \right.\\ \Rightarrow hs\,\,\,co\,\,3\,\,\,diem\,\,\,cuc\,\,tri \Leftrightarrow \left( * \right)\,\,\,co\,\,3\,\,nghiem\,\,pb \Leftrightarrow m > 0.\\ \left( * \right) \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \sqrt m \\ x = – \sqrt m \end{array} \right. \Rightarrow \left\{ \begin{array}{l} A\left( {0;\,\,m} \right)\\ B\left( {\sqrt m ;\,\,m – {m^2}} \right)\\ C\left( { – \sqrt m ;\,\,\,m – {m^2}} \right) \end{array} \right..\\ \left( C \right)\,\,\,co\,\,\,3\,\,diem\,\,\,cuc\,\,tri\,\,\,A,\,\,B,\,\,C\,\,\,\,cung\,\,\,D\left( {0; – 4} \right)\,\,tao\,\,thanh\,\,1\,\,hinh\,\,thoi\\ Ta\,\,thay\,\,A,\,\,D \in Oy \Rightarrow ABDC\,\,\,la\,\,\,hinh\,\,thoi \Leftrightarrow AB = CD\\ Ta\,\,co:\,\,\,\overrightarrow {AB} = \left( {\sqrt m ;\,\, – {m^2}} \right);\,\,\,\overrightarrow {CD} = \left( {\sqrt m ;\,\,{m^2} – m – 4} \right)\\ \Rightarrow AB = CD \Leftrightarrow m + {m^4} = m + {\left( {{m^2} – m – 4} \right)^2}\\ \Leftrightarrow {m^4} = {\left( {{m^2} – m – 4} \right)^2}\\ \Leftrightarrow \left[ \begin{array}{l} {m^2} = {m^2} – m – 4\\ – {m^2} = {m^2} – m – 4 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = – 4\,\,\,\left( {ktm} \right)\\ 2{m^2} – m – 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = \frac{{1 + \sqrt {33} }}{4}\,\,\,\left( {tm} \right)\\ m = \frac{{1 + \sqrt {33} }}{4}\,\,\,\left( {ktm} \right) \end{array} \right. \Leftrightarrow m = \frac{{1 + \sqrt {33} }}{4}.\, \end{array}\] Bình luận
\[\begin{array}{l}
y = {x^4} – 2m{x^2} + m\,\,\,\left( C \right)\\
\Rightarrow y’ = 4{x^3} – 4mx\\
\Rightarrow y’ = 0\\
\Leftrightarrow 4{x^3} – 4mx = 0\,\,\,\left( * \right) \Leftrightarrow 4x\left( {{x^2} – m} \right) = 0\\
= \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = m
\end{array} \right.\\
\Rightarrow hs\,\,\,co\,\,3\,\,\,diem\,\,\,cuc\,\,tri \Leftrightarrow \left( * \right)\,\,\,co\,\,3\,\,nghiem\,\,pb \Leftrightarrow m > 0.\\
\left( * \right) \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt m \\
x = – \sqrt m
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
A\left( {0;\,\,m} \right)\\
B\left( {\sqrt m ;\,\,m – {m^2}} \right)\\
C\left( { – \sqrt m ;\,\,\,m – {m^2}} \right)
\end{array} \right..\\
\left( C \right)\,\,\,co\,\,\,3\,\,diem\,\,\,cuc\,\,tri\,\,\,A,\,\,B,\,\,C\,\,\,\,cung\,\,\,D\left( {0; – 4} \right)\,\,tao\,\,thanh\,\,1\,\,hinh\,\,thoi\\
Ta\,\,thay\,\,A,\,\,D \in Oy \Rightarrow ABDC\,\,\,la\,\,\,hinh\,\,thoi \Leftrightarrow AB = CD\\
Ta\,\,co:\,\,\,\overrightarrow {AB} = \left( {\sqrt m ;\,\, – {m^2}} \right);\,\,\,\overrightarrow {CD} = \left( {\sqrt m ;\,\,{m^2} – m – 4} \right)\\
\Rightarrow AB = CD \Leftrightarrow m + {m^4} = m + {\left( {{m^2} – m – 4} \right)^2}\\
\Leftrightarrow {m^4} = {\left( {{m^2} – m – 4} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
{m^2} = {m^2} – m – 4\\
– {m^2} = {m^2} – m – 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
m = – 4\,\,\,\left( {ktm} \right)\\
2{m^2} – m – 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
m = \frac{{1 + \sqrt {33} }}{4}\,\,\,\left( {tm} \right)\\
m = \frac{{1 + \sqrt {33} }}{4}\,\,\,\left( {ktm} \right)
\end{array} \right. \Leftrightarrow m = \frac{{1 + \sqrt {33} }}{4}.\,
\end{array}\]