Đáp án đúng: A
Giải chi tiết:\(\begin{align} & \,\,\,\,\,\,\,{{\left( 10x \right)}^{m+\frac{\log x}{10}}}\ge {{10}^{\frac{11}{10}\log x}}\,\,\,\forall x\in \left( 1;100 \right) \\ & \Leftrightarrow \log \left[ {{\left( 10x \right)}^{m+\frac{\log x}{10}}} \right]\ge \log \left( {{10}^{\frac{11}{10}\log x}} \right)\,\,\,\forall x\in \left( 1;100 \right) \\ & \Leftrightarrow \left( m+\frac{\log x}{10} \right)\left( 1+\log x \right)\ge \frac{11}{10}\log x\,\,\,\forall x\in \left( 1;100 \right) \\ & \Leftrightarrow \left( 10m+\log x \right)\left( 1+\log x \right)\ge 11.\log x\,\,\,\forall x\in \left( 1;100 \right) \\ & \Leftrightarrow 10m\left( 1+\log x \right)+\log x+{{\log }^{2}}x\ge 11.\log x\,\,\,\forall x\in \left( 1;100 \right) \\ & \Leftrightarrow {{\log }^{2}}x-10.\log x\ge -10m\left( 1+\log x \right)\,\,\,\forall x\in \left( 1;100 \right) \\\end{align}\)
Ta có \(x\in \left( 1;100 \right)\Rightarrow \log x>\log 1=0\Leftrightarrow 1+\log x>0\)
\(\Rightarrow \frac{{{\log }^{2}}x-10.\log x}{1+\log x}\ge -10m\,\,\,\forall x\in \left( 1;100 \right)\)
Đặt \(t=\log x;\,\,t\in \left( 0;10 \right)\Rightarrow \frac{{{t}^{2}}-10t}{1+t}\ge -10m\,\,\forall t\in \left( 0;10 \right)\)
Xét hàm số \(f\left( t \right)=\frac{{{t}^{2}}-10t}{1+t}\Rightarrow -10m\le \underset{\left[ 0;10 \right]}{\mathop{\min }}\,f\left( t \right)\)
Ta có
\(f'\left( t \right) = \frac{{\left( {2t - 10} \right)\left( {1 + t} \right) - {t^2} + 10t}}{{{{\left( {1 + t} \right)}^2}}} = \frac{{{t^2} + 2t - 10}}{{{{\left( {t + 1} \right)}^2}}} = 0 \Leftrightarrow \left[ \begin{array}{l}t = - 1 - \sqrt {11} \,\,\left( {ktm} \right)\\t = - 1 + \sqrt {11} \,\,\,\left( {tm} \right)\end{array} \right.\)
\(\begin{align} & f\left( 0 \right)=0;\,\,f\left( 10 \right)=0;\,\,f\left( -1+\sqrt{11} \right)=-12+2\sqrt{11}\Rightarrow \underset{\left[ 0;10 \right]}{\mathop{\min }}\,f\left( t \right)=-12+2\sqrt{11} \\ & \Rightarrow -10m\le -12+2\sqrt{11} \\ & \Leftrightarrow m\ge \frac{-12+2\sqrt{11}}{-10}=\frac{6-\sqrt{11}}{5} \\\end{align}\)
Kết hợp điều kiện \(m\in \left[ -2018;2018 \right]\Rightarrow m\in \left[ \frac{6-\sqrt{11}}{5};2018 \right]\) vậy có 2018 giá trị nguyên dương của m thỏa mãn điều kiện bài toán.
Chọn A.
