單選第 4 題:
設\(\alpha,\beta,\gamma\)為方程式\(2x^3+x^2-x-7=0\)的三根,將\(\displaystyle \frac{1}{\alpha-1}+\frac{1}{\beta-1}+\frac{1}{\gamma-1}\)之值四捨五入後可以得到下列何數?
(A)0 (B)1 (C)2 (D)3
[解答]
令 \(f(x)=2x^3+x^2-x-7=2\left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)\)
則 \(f\,'(x)=6x^2+2x-1=2\left(x-\alpha\right)\left(x-\beta\right)+2\left(x-\beta\right)\left(x-\gamma\right)+2\left(x-\alpha\right)\left(x-\gamma\right)\)
\(\displaystyle\Rightarrow\frac{1}{x-\alpha}+\frac{1}{x-\beta}+\frac{1}{x-\gamma}=\frac{f\,'(x)}{f(x)}\)
\(\displaystyle\Rightarrow\frac{1}{1-\alpha}+\frac{1}{1-\beta}+\frac{1}{1-\gamma}=\frac{f\,'(1)}{f(1)}=\frac{7}{-5}\)
\(\displaystyle\Rightarrow\frac{1}{\alpha-1}+\frac{1}{\beta-1}+\frac{1}{\gamma-1}=\frac{7}{5}=1.4\)
四捨五入得近似值為 \(1\) 。
註:感謝 thepiano 老師提醒筆誤,已修正。