填充第 11 題:
令 \(f(x)=x^3+ax^2+x+1\),則 \(f\,'(x)=3x^2+2ax+1\)
設通過原點的切線與 \(y=f(x)\) 切於 \((x_0,y_0)\)
則 \(\displaystyle f\,'(x_0)=\frac{y_0-0}{x_0-0}\) 且 \(y_0=f(x_0)\)
\(\displaystyle\Rightarrow 3x_0^2+2ax_0+1=\frac{y_0}{x_0}\) 且 \(y_0=x_0^3+ax_0^2+x_0+1\)
\(\Rightarrow x_0\left(3x_0^2+2ax_0+1\right)=x_0^3+ax_0^2+x_0+1\)
\(\Rightarrow x_0\left(3x_0^2+2ax_0+1\right)=x_0^3+ax_0^2+x_0+1\)
\(\Rightarrow 2x_0^3+ax_0^2-1=0\)
令 \(g(x)=2x^3+ax^2-1\),
\(g\,'(x)=6x^2+2ax=0\Rightarrow x=0\) 或 \(\displaystyle x=\frac{-a}{3}\)
依題意,可知 \(g(x)=0\) 有三相異實根,
所以,\(\displaystyle g(0)\cdot g(\frac{-a}{3})<0\)
\(\Rightarrow a>3.\)
相似類題:請參見 99台中二中填充第 5 題
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