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Câu hỏi: Cho hàm số $y=f\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ có đồ thị như bên dưới.
Hỏi đồ thị hàm số $y=\dfrac{\left( {{x}^{2}}-2x \right)\sqrt{2-x}}{\left( x-3 \right)\left[ {{f}^{2}}\left( x \right)-f\left( x \right) \right]}$ có bao nhiêu đường tiệm cận đứng
A. $3$.
B. $5$.
C. $4$.
D. $6$.
A. $3$.
B. $5$.
C. $4$.
D. $6$.
Điều kiện: $\left\{ \begin{aligned}
& x\le 2 \\
& x\ne 3 \\
& {{f}^{2}}\left( x \right)-f\left( x \right)\ne 0 \\
\end{aligned} \right.$
Ta xét: ${{f}^{2}}\left( x \right)-f\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& f\left( x \right)=0\left( 1 \right) \\
& f\left( x \right)=1\left( 2 \right) \\
\end{aligned} \right.$
$\left( 1 \right)\Leftrightarrow \left[ \begin{aligned}
& x={{x}_{1}}<0 \\
& x={{x}_{2}}\in \left( 0;2 \right) \\
& x={{x}_{3}}>2 \\
\end{aligned} \right.\Rightarrow f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)$
$\left( 2 \right)\Leftrightarrow \left[ \begin{aligned}
& x=0\left( k\acute{e}p \right) \\
& x={{x}_{4}}>{{x}_{3}} \\
\end{aligned} \right.\Rightarrow f\left( x \right)-1=a{{x}^{2}}\left( x-{{x}_{4}} \right)$
Khi đó$$ $y=\dfrac{\left( {{x}^{2}}-2x \right)\sqrt{2-x}}{\left( x-3 \right)\left[ {{f}^{2}}\left( x \right)-f\left( x \right) \right]}=\dfrac{\left( {{x}^{2}}-2x \right)\sqrt{2-x}}{\left( x-3 \right)f\left( x \right)\left[ f\left( x \right)-1 \right]}$ $=\dfrac{x\left( x-2 \right)\sqrt{2-x}}{\left( x-3 \right)a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)a{{x}^{2}}\left( x-{{x}_{4}} \right)}=\dfrac{\left( x-2 \right)\sqrt{2-x}}{{{a}^{2}}x\left( x-3 \right)\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)}$
Do $x\le 2$ nên đồ thị hàm số có 3 đường tiệm cận đứng là $x=0;x={{x}_{1}};x={{x}_{2}}$
& x\le 2 \\
& x\ne 3 \\
& {{f}^{2}}\left( x \right)-f\left( x \right)\ne 0 \\
\end{aligned} \right.$
Ta xét: ${{f}^{2}}\left( x \right)-f\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& f\left( x \right)=0\left( 1 \right) \\
& f\left( x \right)=1\left( 2 \right) \\
\end{aligned} \right.$
& x={{x}_{1}}<0 \\
& x={{x}_{2}}\in \left( 0;2 \right) \\
& x={{x}_{3}}>2 \\
\end{aligned} \right.\Rightarrow f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)$
$\left( 2 \right)\Leftrightarrow \left[ \begin{aligned}
& x=0\left( k\acute{e}p \right) \\
& x={{x}_{4}}>{{x}_{3}} \\
\end{aligned} \right.\Rightarrow f\left( x \right)-1=a{{x}^{2}}\left( x-{{x}_{4}} \right)$
Khi đó$$ $y=\dfrac{\left( {{x}^{2}}-2x \right)\sqrt{2-x}}{\left( x-3 \right)\left[ {{f}^{2}}\left( x \right)-f\left( x \right) \right]}=\dfrac{\left( {{x}^{2}}-2x \right)\sqrt{2-x}}{\left( x-3 \right)f\left( x \right)\left[ f\left( x \right)-1 \right]}$ $=\dfrac{x\left( x-2 \right)\sqrt{2-x}}{\left( x-3 \right)a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)a{{x}^{2}}\left( x-{{x}_{4}} \right)}=\dfrac{\left( x-2 \right)\sqrt{2-x}}{{{a}^{2}}x\left( x-3 \right)\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)}$
Do $x\le 2$ nên đồ thị hàm số có 3 đường tiệm cận đứng là $x=0;x={{x}_{1}};x={{x}_{2}}$
Đáp án A.