Phương pháp giải:

Đặt \(f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)\), tính đạo hàm của hàm số \(y=f\left( x \right)\)

Xét hàm số \(h\left( x \right)=\frac{f'\left( x \right)}{f\left( x \right)}\) và chứng minh \(f''\left( x \right).f\left( x \right)-{{\left[ f'\left( x \right) \right]}^{2}}<0\,\,\forall x\notin \left\{ {{x}_{1}};{{x}_{2}};{{x}_{3}};{{x}_{4}} \right\}\) .

Lời giải chi tiết:

Đồ thị hàm số \(y=f\left( x \right)\) cắt trục hoành tại bốn điểm phân biệt nên \(f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)\)

\(\begin{align}  \Rightarrow f'\left( x \right)=a\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)+a\left( x-{{x}_{1}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right) \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{4}} \right)+a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right) \\  f'\left( x \right)=f\left( x \right)\left( \frac{1}{x-{{x}_{1}}}+\frac{1}{x-{{x}_{2}}}+\frac{1}{x-{{x}_{3}}}+\frac{1}{x-{{x}_{4}}} \right)\,\,\,\forall x\notin \left\{ {{x}_{1}};{{x}_{2}};{{x}_{3}};{{x}_{4}} \right\}\Rightarrow f'\left( x \right)\ne 0\,\,\forall x\notin \left\{ {{x}_{1}};{{x}_{2}};{{x}_{3}};{{x}_{4}} \right\} \\ \end{align}\)

Đặt \(h\left( x \right)=\frac{f'\left( x \right)}{f\left( x \right)}=\frac{1}{x-{{x}_{1}}}+\frac{1}{x-{{x}_{2}}}+\frac{1}{x-{{x}_{3}}}+\frac{1}{x-{{x}_{4}}}\,\,\forall x\notin \left\{ {{x}_{1}};{{x}_{2}};{{x}_{3}};{{x}_{4}} \right\}\)

Ta có \(h'\left( x \right)=\frac{f''\left( x \right).f\left( x \right)-{{\left[ f'\left( x \right) \right]}^{2}}}{{{f}^{2}}\left( x \right)}=\frac{-1}{{{\left( x-{{x}_{1}} \right)}^{2}}}+\frac{-1}{{{\left( x-{{x}_{2}} \right)}^{2}}}+\frac{-1}{{{\left( x-{{x}_{3}} \right)}^{2}}}+\frac{-1}{{{\left( x-{{x}_{4}} \right)}^{2}}}<0\,\,\forall x\notin \left\{ {{x}_{1}};{{x}_{2}};{{x}_{3}};{{x}_{4}} \right\}\)

\(\begin{align}  \Rightarrow f''\left( x \right).f\left( x \right)-{{\left[ f'\left( x \right) \right]}^{2}}<0\,\,\forall x\notin \left\{ {{x}_{1}};{{x}_{2}};{{x}_{3}};{{x}_{4}} \right\} \\  \Rightarrow g\left( x \right)={{\left[ f'\left( x \right) \right]}^{2}}-f''\left( x \right).f\left( x \right)>0\,\,\forall x\notin \left\{ {{x}_{1}};{{x}_{2}};{{x}_{3}};{{x}_{4}} \right\} \\ \end{align}\)

 Khi \(f\left( x \right)=0\Rightarrow f'\left( x \right)\ne 0\Rightarrow g\left( x \right)={{\left[ f'\left( x \right) \right]}^{2}}-f''\left( x \right).f\left( x \right)\ne 0\)

Vậy đồ thị hàm số \(y=g\left( x \right)={{\left[ f'\left( x \right) \right]}^{2}}-f\left( x \right).f''\left( x \right)\) không cắt trục Ox.

Chọn A.