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Câu hỏi: Có bao nhiêu giá trị nguyên $m\in \left[ -10;10 \right]$ để phương trình
${{2018}^{\sin \left( 2x-\dfrac{\pi }{3} \right)-\dfrac{m}{2}}}.{{\log }_{2019}}\left( \sin 2x-m+12 \right)={{\log }_{2019}}\left( \sqrt{3}\cos 2x+12 \right)$ có 4 nghiệm thuộc $\left[ \dfrac{\pi }{6};\dfrac{5\pi }{3} \right]?$
A. 3.
B. 1.
C. 9.
D. 2.
Ta có:
$\sin \left( 2x-\dfrac{\pi }{3} \right)-\dfrac{m}{2}=\dfrac{1}{2}\sin 2x-\dfrac{\sqrt{3}}{2}\cos 2x-\dfrac{m}{2}=\dfrac{\sin 2x-\sqrt{3}\cos 2x-m}{2}.$
Đặt $\left\{ \begin{aligned}
& u=\sin 2x-m+12 \\
& v=\sqrt{3}\cos 2x+12 \\
\end{aligned} \right.,$ khi đó phương trình có dạng:
${{2018}^{\dfrac{u-v}{2}}}.{{\log }_{2019}}u={{\log }_{2019}}v\Leftrightarrow \dfrac{{{\left( \sqrt{2018} \right)}^{u}}}{{{\left( \sqrt{2018} \right)}^{v}}}.{{\log }_{2019}}u={{\log }_{2019}}v$
$\Leftrightarrow {{\left( \sqrt{2018} \right)}^{u}}.{{\log }_{019}}u={{\left( \sqrt{2018} \right)}^{v}}{{\log }_{2019}}v\Leftrightarrow f\left( u \right)=f\left( v \right)$ trong đó $f\left( t \right)={{\left( \sqrt{2018} \right)}^{t}}.{{\log }_{2019}}t$
$\Leftrightarrow u=v\Leftrightarrow \sin 2x-m+12=\sqrt{3}\cos 2x+12\Leftrightarrow \sin \left( 2x-\dfrac{\pi }{3} \right)=\dfrac{m}{2}$ (*)
Do $x\in \left[ \dfrac{\pi }{6};\dfrac{5\pi }{3} \right]\Rightarrow \left( 2x-\dfrac{\pi }{3} \right)\in \left[ 0;3\pi \right]$ nên để (*) có 4 nghiệm thì: $0\le \dfrac{m}{2}<1$
$\Leftrightarrow 0\le m<2\xrightarrow{m\in \mathbb{Z}}m\in \left\{ 0;1 \right\}:$ có giá trị m thỏa mãn.
${{2018}^{\sin \left( 2x-\dfrac{\pi }{3} \right)-\dfrac{m}{2}}}.{{\log }_{2019}}\left( \sin 2x-m+12 \right)={{\log }_{2019}}\left( \sqrt{3}\cos 2x+12 \right)$ có 4 nghiệm thuộc $\left[ \dfrac{\pi }{6};\dfrac{5\pi }{3} \right]?$
A. 3.
B. 1.
C. 9.
D. 2.
Ta có:
$\sin \left( 2x-\dfrac{\pi }{3} \right)-\dfrac{m}{2}=\dfrac{1}{2}\sin 2x-\dfrac{\sqrt{3}}{2}\cos 2x-\dfrac{m}{2}=\dfrac{\sin 2x-\sqrt{3}\cos 2x-m}{2}.$
Đặt $\left\{ \begin{aligned}
& u=\sin 2x-m+12 \\
& v=\sqrt{3}\cos 2x+12 \\
\end{aligned} \right.,$ khi đó phương trình có dạng:
${{2018}^{\dfrac{u-v}{2}}}.{{\log }_{2019}}u={{\log }_{2019}}v\Leftrightarrow \dfrac{{{\left( \sqrt{2018} \right)}^{u}}}{{{\left( \sqrt{2018} \right)}^{v}}}.{{\log }_{2019}}u={{\log }_{2019}}v$
$\Leftrightarrow {{\left( \sqrt{2018} \right)}^{u}}.{{\log }_{019}}u={{\left( \sqrt{2018} \right)}^{v}}{{\log }_{2019}}v\Leftrightarrow f\left( u \right)=f\left( v \right)$ trong đó $f\left( t \right)={{\left( \sqrt{2018} \right)}^{t}}.{{\log }_{2019}}t$
$\Leftrightarrow u=v\Leftrightarrow \sin 2x-m+12=\sqrt{3}\cos 2x+12\Leftrightarrow \sin \left( 2x-\dfrac{\pi }{3} \right)=\dfrac{m}{2}$ (*)
Do $x\in \left[ \dfrac{\pi }{6};\dfrac{5\pi }{3} \right]\Rightarrow \left( 2x-\dfrac{\pi }{3} \right)\in \left[ 0;3\pi \right]$ nên để (*) có 4 nghiệm thì: $0\le \dfrac{m}{2}<1$
$\Leftrightarrow 0\le m<2\xrightarrow{m\in \mathbb{Z}}m\in \left\{ 0;1 \right\}:$ có giá trị m thỏa mãn.
Đáp án D.