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Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{x-2023}{\left( 2022-x \right)\left( 2021-x \right)\left( 2020-x \right)...\left( 2-x \right)\left( 1-x \right)}$ có đồ thị $\left( C \right)$. Gọi đường thẳng $d$ là tiếp tuyến của $\left( C \right)$ tại điểm có hoành độ bằng $2023$. Diện tích hình phảng giới hạn bởi $d$, trục hoành và hai đường thẳng $x=2022$, $x=2024$ bằng
A. $\dfrac{1}{2022!}$.
B. $\dfrac{1}{2024!}$.
C. $\dfrac{1}{2023!}$.
D. $2022!$.
A. $\dfrac{1}{2022!}$.
B. $\dfrac{1}{2024!}$.
C. $\dfrac{1}{2023!}$.
D. $2022!$.
Ta có ${f}'\left( 2023 \right)=\underset{x\to 2023}{\mathop{\lim }} \dfrac{\dfrac{x-2023}{\left( 2022-x \right)\left( 2021-x \right)\left( 2020-x \right)...\left( 2-x \right)\left( 1-x \right)}-f\left( 2023 \right)}{x-2023}$
$=\underset{x\to 2023}{\mathop{\lim }} \dfrac{\dfrac{x-2023}{\left( 2022-x \right)\left( 2021-x \right)\left( 2020-x \right)...\left( 2-x \right)\left( 1-x \right)}}{x-2023}$
$=\underset{x\to 2023}{\mathop{\lim }} \dfrac{1}{\left( 2022-x \right)\left( 2021-x \right)\left( 2020-x \right)...\left( 2-x \right)\left( 1-x \right)}$
$=\underset{x\to 2023}{\mathop{\lim }} \dfrac{1}{\left( 2022-2023 \right)\left( 2021-2023 \right)\left( 2020-2023 \right)...\left( 2-2023 \right)\left( 1-2023 \right)}=\dfrac{1}{2022!}$.
Khi đó $d:y={f}'\left( 2023 \right)\left( x-2023 \right)+f\left( 2023 \right)=\dfrac{1}{2022!}\left( x-2023 \right)$.
$\Rightarrow S=\int\limits_{2022}^{2024}{\left| \dfrac{1}{2022!}\left( x-2023 \right) \right|\text{d}x}=\dfrac{1}{2022!}\int\limits_{-1}^{1}{\left| t \right|\text{d}t}=\dfrac{2}{2022!}\int\limits_{-1}^{1}{t\text{d}t}=\dfrac{1}{2022!}$.
$=\underset{x\to 2023}{\mathop{\lim }} \dfrac{\dfrac{x-2023}{\left( 2022-x \right)\left( 2021-x \right)\left( 2020-x \right)...\left( 2-x \right)\left( 1-x \right)}}{x-2023}$
$=\underset{x\to 2023}{\mathop{\lim }} \dfrac{1}{\left( 2022-x \right)\left( 2021-x \right)\left( 2020-x \right)...\left( 2-x \right)\left( 1-x \right)}$
$=\underset{x\to 2023}{\mathop{\lim }} \dfrac{1}{\left( 2022-2023 \right)\left( 2021-2023 \right)\left( 2020-2023 \right)...\left( 2-2023 \right)\left( 1-2023 \right)}=\dfrac{1}{2022!}$.
Khi đó $d:y={f}'\left( 2023 \right)\left( x-2023 \right)+f\left( 2023 \right)=\dfrac{1}{2022!}\left( x-2023 \right)$.
$\Rightarrow S=\int\limits_{2022}^{2024}{\left| \dfrac{1}{2022!}\left( x-2023 \right) \right|\text{d}x}=\dfrac{1}{2022!}\int\limits_{-1}^{1}{\left| t \right|\text{d}t}=\dfrac{2}{2022!}\int\limits_{-1}^{1}{t\text{d}t}=\dfrac{1}{2022!}$.
Đáp án A.