Google
Câu hỏi: Cho các hàm số ${{f}_{0}}\left( x \right),{{f}_{1}}\left( x \right),{{f}_{2}}\left( x \right),...$ thỏa mãn:
A. 6063.
B. 6059.
C. 6057.
D. 6058.
Xét hàm số ${{f}_{0}}\left( x \right)=\ln x+\left| \ln x-2019 \right|-\left| \ln x+2019 \right|$
Đặt $t=\ln x,$ ta có ${{f}_{0}}\left( x \right)=g\left( t \right)=t+\left| t-2019 \right|-\left| t+2019 \right|$
$g\left( t \right)=\left\{ \begin{aligned}
& t+4038khit\le -2019 \\
& -tkhi-2019<t<2019 \\
& t-4038khit\ge 2019 \\
\end{aligned} \right.$
${{f}_{n+1}}\left( x \right)=\left| {{f}_{n}}\left( x \right) \right|-1\ge -1,\forall x>0;\forall n\in \mathbb{N}$
Ta có 3 trường hợp
+ Nếu ${{f}_{n+1}}\left( x \right)=-1\Leftrightarrow \left| {{f}_{n}}\left( x \right) \right|-1=-1\Leftrightarrow {{f}_{n}}\left( x \right)=0.$
+ Nếu ${{f}_{n+1}}\left( x \right)=0\Leftrightarrow \left| {{f}_{n}}\left( x \right) \right|-1=0\Leftrightarrow {{f}_{n}}\left( x \right)=1\Leftrightarrow \left[ \begin{aligned}
& {{f}_{n}}\left( x \right)=-1 \\
& {{f}_{n}}\left( x \right)=1 \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& {{f}_{n-1}}\left( x \right)=0 \\
& {{f}_{n-1}}\left( x \right)=2 \\
\end{aligned} \right.$
+ Nếu ${{f}_{n+1}}\left( x \right)=k>0\Leftrightarrow {{f}_{n}}\left( x \right)=k+1$
Khi đó, ${{f}_{2020}}\left( x \right)=0\Rightarrow {{f}_{2018}}\left( x \right)\in \left\{ 0;2 \right\}\Rightarrow {{f}_{2016}}\left( x \right)\in \left\{ 0;2;4 \right\}\Rightarrow {{f}_{2014}}\left( x \right)\in \left\{ 0;2;4;6 \right\}$
...
$\begin{aligned}
& \Rightarrow {{f}_{2}}\left( x \right)\in \left\{ 0;2;4;...;2018 \right\}\Rightarrow {{f}_{1}}\left( x \right)\in \left\{ -1;1;3;5;...;2019 \right\} \\
& \Rightarrow \left| {{f}_{0}}\left( x \right) \right|\in \left\{ 0;2;4;...;2020 \right\}\Rightarrow {{f}_{0}}\left( x \right)\in \left\{ 0;\pm 2;\pm 4;...;\pm 2020 \right\} \\
\end{aligned}$
Từ đồ thị suy ra:
Mỗi phương trình ${{f}_{0}}\left( x \right)=k;k\in \left\{ 0;\pm 2;\pm 4;...;\pm 2018 \right\}$ có 3 nghiệm.
Mỗi phương trình ${{f}_{0}}\left( x \right)=-2020;{{f}_{0}}\left( x \right)=2020$ có 1 nghiệm.
Vậy số nghiệm phương trình ${{f}_{2020}}\left( x \right)=0$ là $2019.3+2=6059$
${{f}_{0}}\left( x \right)=\ln x+\left| \ln x-2019 \right|-\left| \ln x+2019 \right|,{{f}_{n+1}}\left( x \right)=\left| {{f}_{n}}\left( x \right) \right|-1,\forall n\in N.$
Số nghiệm của phương trình ${{f}_{2020}}\left( x \right)=0$ làA. 6063.
B. 6059.
C. 6057.
D. 6058.
Xét hàm số ${{f}_{0}}\left( x \right)=\ln x+\left| \ln x-2019 \right|-\left| \ln x+2019 \right|$
Đặt $t=\ln x,$ ta có ${{f}_{0}}\left( x \right)=g\left( t \right)=t+\left| t-2019 \right|-\left| t+2019 \right|$
$g\left( t \right)=\left\{ \begin{aligned}
& t+4038khit\le -2019 \\
& -tkhi-2019<t<2019 \\
& t-4038khit\ge 2019 \\
\end{aligned} \right.$
${{f}_{n+1}}\left( x \right)=\left| {{f}_{n}}\left( x \right) \right|-1\ge -1,\forall x>0;\forall n\in \mathbb{N}$
Ta có 3 trường hợp
+ Nếu ${{f}_{n+1}}\left( x \right)=-1\Leftrightarrow \left| {{f}_{n}}\left( x \right) \right|-1=-1\Leftrightarrow {{f}_{n}}\left( x \right)=0.$
+ Nếu ${{f}_{n+1}}\left( x \right)=0\Leftrightarrow \left| {{f}_{n}}\left( x \right) \right|-1=0\Leftrightarrow {{f}_{n}}\left( x \right)=1\Leftrightarrow \left[ \begin{aligned}
& {{f}_{n}}\left( x \right)=-1 \\
& {{f}_{n}}\left( x \right)=1 \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& {{f}_{n-1}}\left( x \right)=0 \\
& {{f}_{n-1}}\left( x \right)=2 \\
\end{aligned} \right.$
+ Nếu ${{f}_{n+1}}\left( x \right)=k>0\Leftrightarrow {{f}_{n}}\left( x \right)=k+1$
Khi đó, ${{f}_{2020}}\left( x \right)=0\Rightarrow {{f}_{2018}}\left( x \right)\in \left\{ 0;2 \right\}\Rightarrow {{f}_{2016}}\left( x \right)\in \left\{ 0;2;4 \right\}\Rightarrow {{f}_{2014}}\left( x \right)\in \left\{ 0;2;4;6 \right\}$
...
$\begin{aligned}
& \Rightarrow {{f}_{2}}\left( x \right)\in \left\{ 0;2;4;...;2018 \right\}\Rightarrow {{f}_{1}}\left( x \right)\in \left\{ -1;1;3;5;...;2019 \right\} \\
& \Rightarrow \left| {{f}_{0}}\left( x \right) \right|\in \left\{ 0;2;4;...;2020 \right\}\Rightarrow {{f}_{0}}\left( x \right)\in \left\{ 0;\pm 2;\pm 4;...;\pm 2020 \right\} \\
\end{aligned}$
Từ đồ thị suy ra:
Mỗi phương trình ${{f}_{0}}\left( x \right)=k;k\in \left\{ 0;\pm 2;\pm 4;...;\pm 2018 \right\}$ có 3 nghiệm.
Mỗi phương trình ${{f}_{0}}\left( x \right)=-2020;{{f}_{0}}\left( x \right)=2020$ có 1 nghiệm.
Vậy số nghiệm phương trình ${{f}_{2020}}\left( x \right)=0$ là $2019.3+2=6059$
Đáp án B.