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Câu hỏi: Cho hàm số $y=f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e \left( a\ne 0 \right)$ có đồ thị $\left( C \right)$. Biết rằng $\left( C \right)$ cắt trục hoành tại bốn điểm phân biệt là $A\left( {{x}_{1}};0 \right)$, $B\left( {{x}_{2}};0 \right)$, $C\left( {{x}_{3}};0 \right)$, $D\left( {{x}_{4}};0 \right)$ ; với ${{x}_{1}}, {{x}_{2}},{{x}_{3}},{{x}_{4}}$ theo thứ tự lập thành cấp số cộng và hai tiếp tuyến của $\left( C \right)$ tại A, B vuông góc với nhau. Khi đó, giá trị của biểu thức $P={{\left[ {f}'\left( {{x}_{3}} \right)+{f}'\left( {{x}_{4}} \right) \right]}^{2022}}$ bằng
A. ${{\left( \dfrac{4}{3} \right)}^{1011}}.$
B. ${{\left( \dfrac{4}{3} \right)}^{2022}}.$
C. ${{\left( \dfrac{4a}{3} \right)}^{1011}}.$
D. ${{\left( \dfrac{4a}{3} \right)}^{2022}}.$
A. ${{\left( \dfrac{4}{3} \right)}^{1011}}.$
B. ${{\left( \dfrac{4}{3} \right)}^{2022}}.$
C. ${{\left( \dfrac{4a}{3} \right)}^{1011}}.$
D. ${{\left( \dfrac{4a}{3} \right)}^{2022}}.$
Gọi $g$ là công sai của cấp số cộng, khi đó:
$f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)$
$\Rightarrow \left\{ \begin{matrix}
{f}'\left( x \right)=a\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)+\left( x-{{x}_{1}} \right){{\left[ a\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right) \right]}^{\prime }}\Rightarrow {f}'\left( {{x}_{1}} \right)=-6a{{g}^{3}} \\
{f}'\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)+\left( x-{{x}_{2}} \right){{\left[ a\left( x-{{x}_{1}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right) \right]}^{\prime }}\Rightarrow {f}'\left( {{x}_{2}} \right)=2a{{g}^{3}} \\
{f}'\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{4}} \right)+\left( x-{{x}_{3}} \right){{\left[ a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{4}} \right) \right]}^{\prime }}\Rightarrow {f}'\left( {{x}_{3}} \right)=-2a{{g}^{3}} \\
{f}'\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)+\left( x-{{x}_{4}} \right){{\left[ a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right) \right]}^{\prime }}\Rightarrow {f}'\left( {{x}_{4}} \right)=6a{{g}^{3}} \\
\end{matrix} \right.$
Do tiếp tuyến tại $A\left( {{x}_{1}};0 \right)$, $B\left( {{x}_{2}};0 \right)$ vuông góc nhau nên ${f}'\left( {{x}_{1}} \right){f}'\left( {{x}_{2}} \right)=-1\Leftrightarrow {{a}^{2}}{{g}^{6}}=\dfrac{1}{12}$
Ta có $P={{\left[ {f}'\left( {{x}_{3}} \right)+{f}'\left( {{x}_{4}} \right) \right]}^{2022}}={{\left( 4a{{g}^{3}} \right)}^{2022}}={{\left( 16{{a}^{2}}{{g}^{6}} \right)}^{1011}}={{\left( \dfrac{4}{3} \right)}^{1011}}$.
$f\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)$
$\Rightarrow \left\{ \begin{matrix}
{f}'\left( x \right)=a\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)+\left( x-{{x}_{1}} \right){{\left[ a\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right) \right]}^{\prime }}\Rightarrow {f}'\left( {{x}_{1}} \right)=-6a{{g}^{3}} \\
{f}'\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right)+\left( x-{{x}_{2}} \right){{\left[ a\left( x-{{x}_{1}} \right)\left( x-{{x}_{3}} \right)\left( x-{{x}_{4}} \right) \right]}^{\prime }}\Rightarrow {f}'\left( {{x}_{2}} \right)=2a{{g}^{3}} \\
{f}'\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{4}} \right)+\left( x-{{x}_{3}} \right){{\left[ a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{4}} \right) \right]}^{\prime }}\Rightarrow {f}'\left( {{x}_{3}} \right)=-2a{{g}^{3}} \\
{f}'\left( x \right)=a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right)+\left( x-{{x}_{4}} \right){{\left[ a\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)\left( x-{{x}_{3}} \right) \right]}^{\prime }}\Rightarrow {f}'\left( {{x}_{4}} \right)=6a{{g}^{3}} \\
\end{matrix} \right.$
Do tiếp tuyến tại $A\left( {{x}_{1}};0 \right)$, $B\left( {{x}_{2}};0 \right)$ vuông góc nhau nên ${f}'\left( {{x}_{1}} \right){f}'\left( {{x}_{2}} \right)=-1\Leftrightarrow {{a}^{2}}{{g}^{6}}=\dfrac{1}{12}$
Ta có $P={{\left[ {f}'\left( {{x}_{3}} \right)+{f}'\left( {{x}_{4}} \right) \right]}^{2022}}={{\left( 4a{{g}^{3}} \right)}^{2022}}={{\left( 16{{a}^{2}}{{g}^{6}} \right)}^{1011}}={{\left( \dfrac{4}{3} \right)}^{1011}}$.
Đáp án A.