Google
Câu hỏi: Cho số phức z thỏa mãn $\left| z-1-i \right|=1$. Khi $3\left| z \right|+2\left| z-4-4i \right|$ đạt giá trị lớn nhất, giá trị $\left| z \right|$ bằng
A. 3.
B. 2.
C. $\sqrt{2}+1$.
D. $\sqrt{3}$.
A. 3.
B. 2.
C. $\sqrt{2}+1$.
D. $\sqrt{3}$.
Đặt $z=a+bi\Rightarrow \left| z-1-i \right|=1\Leftrightarrow {{\left( a-1 \right)}^{2}}+{{\left( b-1 \right)}^{2}}=1\Leftrightarrow {{a}^{2}}+{{b}^{2}}=2a+2b-1$.
Khi đó $3\left| z \right|+2\left| z-4-4i \right|=3\sqrt{{{a}^{2}}+{{b}^{2}}}+2\sqrt{{{\left( a-4 \right)}^{2}}+{{\left( b-4 \right)}^{2}}}=3\sqrt{2a+2b-1}+2\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)-8a-8b+32}$
$=3\sqrt{2a+2b-1}+2\sqrt{\left( 2a+2b-1 \right)-8a-8b+32}=3\sqrt{2a+2b-1}+2\sqrt{-6a-6b+31}$
$=\sqrt{3}\sqrt{6a+6b-3}+2\sqrt{-6a-6b+31}\le \sqrt{\left( {{\sqrt{3}}^{2}}+{{2}^{2}} \right)\left( 6a+6b-3-6a-6b+31 \right)}=14$.
Dấu "=" xảy ra khi $\left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=2a+2b-1 \\
& \dfrac{\sqrt{6a+6b-3}}{\sqrt{3}}=\dfrac{\sqrt{-6a-6b+31}}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a+b=\dfrac{5}{2} \\
& {{a}^{2}}+{{b}^{2}}=4 \\
\end{aligned} \right.\Rightarrow \left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=2$.
Khi đó $3\left| z \right|+2\left| z-4-4i \right|=3\sqrt{{{a}^{2}}+{{b}^{2}}}+2\sqrt{{{\left( a-4 \right)}^{2}}+{{\left( b-4 \right)}^{2}}}=3\sqrt{2a+2b-1}+2\sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)-8a-8b+32}$
$=3\sqrt{2a+2b-1}+2\sqrt{\left( 2a+2b-1 \right)-8a-8b+32}=3\sqrt{2a+2b-1}+2\sqrt{-6a-6b+31}$
$=\sqrt{3}\sqrt{6a+6b-3}+2\sqrt{-6a-6b+31}\le \sqrt{\left( {{\sqrt{3}}^{2}}+{{2}^{2}} \right)\left( 6a+6b-3-6a-6b+31 \right)}=14$.
Dấu "=" xảy ra khi $\left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=2a+2b-1 \\
& \dfrac{\sqrt{6a+6b-3}}{\sqrt{3}}=\dfrac{\sqrt{-6a-6b+31}}{2} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a+b=\dfrac{5}{2} \\
& {{a}^{2}}+{{b}^{2}}=4 \\
\end{aligned} \right.\Rightarrow \left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=2$.
Đáp án B.