Google
Câu hỏi: Có bao nhiêu số nguyên $y\in \left[ -2022;2022 \right]$ để bất phương trình ${{\left( 3x \right)}^{y+\dfrac{{{\log }_{3}}x}{10}}}\ge {{3}^{\dfrac{11}{10}{{\log }_{3}}x}}$ có nghiệm đúng với mọi số thực $x\in \left( 1;9 \right)$.
A. $4044$.
B. $4026$.
C. $2022$.
D. $2023$.
A. $4044$.
B. $4026$.
C. $2022$.
D. $2023$.
Đặt $t={{\log }_{3}}x\Leftrightarrow x={{3}^{t}}\Rightarrow t\in \left( 0;2 \right)$, khi đó:
${{\left( 3x \right)}^{y+\dfrac{{{\log }_{3}}x}{10}}}\ge {{3}^{\dfrac{11}{10}{{\log }_{3}}x}}\Leftrightarrow {{\left( {{3}^{t+1}} \right)}^{y+\dfrac{t}{10}}}\ge {{3}^{\dfrac{11}{10}t}},\forall t\in \left( 0;2 \right)$
Ta có ${{\left( {{3}^{t+1}} \right)}^{y+\dfrac{t}{10}}}\ge {{3}^{\dfrac{11}{10}t}}\Leftrightarrow \left( t+1 \right)\left( 10y+t \right)\ge 11t$
Do $t+1>0,\forall t\in \left( 0;2 \right)\Rightarrow \left( t+1 \right)\left( 10y+t \right)\ge 11t\Leftrightarrow 10y\ge \dfrac{11t}{t+1}-t,\forall t\in \left( 0;2 \right)$
$\Rightarrow 10y\ge \dfrac{16}{3}\Leftrightarrow y\ge \dfrac{16}{30}\Rightarrow y\in \left\{ 1;2;...;2022 \right\}$.
${{\left( 3x \right)}^{y+\dfrac{{{\log }_{3}}x}{10}}}\ge {{3}^{\dfrac{11}{10}{{\log }_{3}}x}}\Leftrightarrow {{\left( {{3}^{t+1}} \right)}^{y+\dfrac{t}{10}}}\ge {{3}^{\dfrac{11}{10}t}},\forall t\in \left( 0;2 \right)$
Ta có ${{\left( {{3}^{t+1}} \right)}^{y+\dfrac{t}{10}}}\ge {{3}^{\dfrac{11}{10}t}}\Leftrightarrow \left( t+1 \right)\left( 10y+t \right)\ge 11t$
Do $t+1>0,\forall t\in \left( 0;2 \right)\Rightarrow \left( t+1 \right)\left( 10y+t \right)\ge 11t\Leftrightarrow 10y\ge \dfrac{11t}{t+1}-t,\forall t\in \left( 0;2 \right)$
$\Rightarrow 10y\ge \dfrac{16}{3}\Leftrightarrow y\ge \dfrac{16}{30}\Rightarrow y\in \left\{ 1;2;...;2022 \right\}$.
Đáp án C.