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Câu hỏi: Cho $\left( {{C}_{1}} \right): y=g\left( x \right)=\dfrac{2}{x}; \left( {{C}_{2}} \right): y=f\left( x \right)=4{{x}^{2}}+bx+c$, $\left( {{C}_{2}} \right)$ tiếp xúc với $Ox$ tại $A\left( \dfrac{3}{2};0 \right)$ và qua $B\left( 2;1 \right)$. Giá trị nhỏ nhất của $\sqrt{2}g\left( \dfrac{4+x+2f\left( x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)} \right)-\sqrt{x}\left| 2x-3 \right|$ trên đoạn $\left( 0;2 \right]$ bằng $\sqrt{\dfrac{a}{b}}$ và $M=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$. Giá trị của M nằm thuộc khoảng nào sau đây:
A. $M\in \left( -2;0 \right)$.
B. $M\in \left( 0;1 \right)$.
C. $M\in \left( 1;2 \right)$.
D. $M\in \left( 2;4 \right)$.
A. $M\in \left( -2;0 \right)$.
B. $M\in \left( 0;1 \right)$.
C. $M\in \left( 1;2 \right)$.
D. $M\in \left( 2;4 \right)$.
Do đồ thị $f\left( x \right)$ tiếp xúc với $Ox$ tại $A\left( \dfrac{3}{2};0 \right)$ và qua $B\left( 2;1 \right)$ nên ta có:
$\left\{ \begin{aligned}
& 9+\dfrac{3}{2}b+c=0 \\
& 16+2b+c=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=-12 \\
& c=9 \\
\end{aligned} \right. $ nên $ f\left( x \right)=4{{x}^{2}}-12x+9={{\left( 2x-3 \right)}^{2}}$.
Ta có:
$\dfrac{4+x+2f\left( x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)}=\dfrac{2\left( 1+f\left( x \right) \right)+\left( 2+x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)}=\dfrac{2}{2+x}+\dfrac{1}{1+f\left( x \right)}=\dfrac{2}{2+x}+\dfrac{1}{1+{{\left( 2x-3 \right)}^{2}}}$
$=\dfrac{1}{1+\dfrac{x}{2}}+\dfrac{1}{1+{{\left( 2x-3 \right)}^{2}}}$
Khi đó: $\sqrt{2}g\left( \dfrac{4+x+2f\left( x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)} \right)-\sqrt{x}\left| 2x-3 \right|$ $=\sqrt{2}\dfrac{2}{\dfrac{1}{1+\dfrac{x}{2}}+\dfrac{1}{1+{{\left( 2x-3 \right)}^{2}}}}-\sqrt{2}.\sqrt{\dfrac{x}{2}}.\left| 2x-3 \right|$.
Đặt $\left\{ \begin{aligned}
& a=\sqrt{\dfrac{x}{2}} \\
& b=\left| 2x-3 \right| \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=2{{a}^{2}} \\
& b=\left| 4{{a}^{2}}-3 \right| \\
\end{aligned} \right. $;$ x\in \left( 0;2 \right]\Rightarrow a\in \left( 0;1 \right]; b\in \left[ 0;3 \right) $
●Nếu $ab>1\Rightarrow a\left| 4{{a}^{2}}-3 \right|>1$
$\Leftrightarrow \left[ \begin{aligned}
& a\left( 4{{a}^{2}}-3 \right)>1 \\
& a\left( 4{{a}^{2}}-3 \right)<-1 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& 4{{a}^{3}}-3a-1>0 \\
& 4{{a}^{3}}-3a+1<0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left[ \begin{aligned}
& \left( a-1 \right){{\left( 2a+1 \right)}^{2}}>0 \\
& \left( a+1 \right){{\left( 2a-1 \right)}^{2}}<0 \\
\end{aligned} \right. $ ( vô lý do $ a\in \left( 0;1 \right]$).
●Nếu $0\le ab\le 1$
$P=\dfrac{2\sqrt{2}}{\dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}}-\sqrt{2}ab$
Áp dụng bổ đề $\dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}\le \dfrac{2}{1+ab}$, ta có:
$P=\dfrac{2\sqrt{2}}{\dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}}-\sqrt{2}ab\ge \dfrac{2\sqrt{2}}{\dfrac{2}{1+ab}}-\sqrt{2}ab=\dfrac{2}{2\sqrt{2}}\left( ab+1 \right)-\sqrt{2}ab=\sqrt{2}$.
Giá trị nhỏ nhất của $\sqrt{2}g\left( \dfrac{4+x+2f\left( x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)} \right)-\sqrt{x}\left| 2x-3 \right|$ trên đoạn $\left( 0;2 \right]$ bằng $\sqrt{2}$ khi $x=2$.
Ta có: $a=2; b=1\Rightarrow M=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\dfrac{3}{5}\in \left( 0;1 \right)$.
Chứng minh bổ đề: $\dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}\le \dfrac{2}{1+ab}$ với $0\le ab\le 1$
$\begin{aligned}
& \dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}\le \dfrac{2}{1+ab}\Leftrightarrow \left( \dfrac{1}{{{a}^{2}}+1}-\dfrac{1}{1+ab} \right)+\left( \dfrac{1}{{{b}^{2}}+1}-\dfrac{1}{1+ab} \right)\le 0 \\
& \Leftrightarrow \dfrac{ab-{{a}^{2}}}{\left( {{a}^{2}}+1 \right)\left( 1+ab \right)}+\dfrac{ab-{{b}^{2}}}{\left( {{b}^{2}}+1 \right)\left( 1+ab \right)}\le 0 \\
& \Leftrightarrow a\left( b-a \right)\left( {{b}^{2}}+1 \right)-b\left( b-a \right)\left( {{a}^{2}}+1 \right)\le 0 \\
& \Leftrightarrow \left( b-a \right)\left( a{{b}^{2}}+a-{{a}^{2}}b-b \right)\le 0 \\
& \Leftrightarrow \left( b-a \right)\left[ ab\left( b-a \right)-\left( b-a \right) \right]\le 0 \\
\end{aligned}$
$\Leftrightarrow {{\left( b-a \right)}^{2}}\left( ab-1 \right)\le 0$ ( luôn đúng khi $0\le ab\le 1$ ).
$\left\{ \begin{aligned}
& 9+\dfrac{3}{2}b+c=0 \\
& 16+2b+c=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& b=-12 \\
& c=9 \\
\end{aligned} \right. $ nên $ f\left( x \right)=4{{x}^{2}}-12x+9={{\left( 2x-3 \right)}^{2}}$.
Ta có:
$\dfrac{4+x+2f\left( x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)}=\dfrac{2\left( 1+f\left( x \right) \right)+\left( 2+x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)}=\dfrac{2}{2+x}+\dfrac{1}{1+f\left( x \right)}=\dfrac{2}{2+x}+\dfrac{1}{1+{{\left( 2x-3 \right)}^{2}}}$
$=\dfrac{1}{1+\dfrac{x}{2}}+\dfrac{1}{1+{{\left( 2x-3 \right)}^{2}}}$
Khi đó: $\sqrt{2}g\left( \dfrac{4+x+2f\left( x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)} \right)-\sqrt{x}\left| 2x-3 \right|$ $=\sqrt{2}\dfrac{2}{\dfrac{1}{1+\dfrac{x}{2}}+\dfrac{1}{1+{{\left( 2x-3 \right)}^{2}}}}-\sqrt{2}.\sqrt{\dfrac{x}{2}}.\left| 2x-3 \right|$.
Đặt $\left\{ \begin{aligned}
& a=\sqrt{\dfrac{x}{2}} \\
& b=\left| 2x-3 \right| \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& x=2{{a}^{2}} \\
& b=\left| 4{{a}^{2}}-3 \right| \\
\end{aligned} \right. $;$ x\in \left( 0;2 \right]\Rightarrow a\in \left( 0;1 \right]; b\in \left[ 0;3 \right) $
●Nếu $ab>1\Rightarrow a\left| 4{{a}^{2}}-3 \right|>1$
$\Leftrightarrow \left[ \begin{aligned}
& a\left( 4{{a}^{2}}-3 \right)>1 \\
& a\left( 4{{a}^{2}}-3 \right)<-1 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& 4{{a}^{3}}-3a-1>0 \\
& 4{{a}^{3}}-3a+1<0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left[ \begin{aligned}
& \left( a-1 \right){{\left( 2a+1 \right)}^{2}}>0 \\
& \left( a+1 \right){{\left( 2a-1 \right)}^{2}}<0 \\
\end{aligned} \right. $ ( vô lý do $ a\in \left( 0;1 \right]$).
●Nếu $0\le ab\le 1$
$P=\dfrac{2\sqrt{2}}{\dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}}-\sqrt{2}ab$
Áp dụng bổ đề $\dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}\le \dfrac{2}{1+ab}$, ta có:
$P=\dfrac{2\sqrt{2}}{\dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}}-\sqrt{2}ab\ge \dfrac{2\sqrt{2}}{\dfrac{2}{1+ab}}-\sqrt{2}ab=\dfrac{2}{2\sqrt{2}}\left( ab+1 \right)-\sqrt{2}ab=\sqrt{2}$.
Giá trị nhỏ nhất của $\sqrt{2}g\left( \dfrac{4+x+2f\left( x \right)}{\left( 1+f\left( x \right) \right)\left( 2+x \right)} \right)-\sqrt{x}\left| 2x-3 \right|$ trên đoạn $\left( 0;2 \right]$ bằng $\sqrt{2}$ khi $x=2$.
Ta có: $a=2; b=1\Rightarrow M=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\dfrac{3}{5}\in \left( 0;1 \right)$.
Chứng minh bổ đề: $\dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}\le \dfrac{2}{1+ab}$ với $0\le ab\le 1$
$\begin{aligned}
& \dfrac{1}{{{a}^{2}}+1}+\dfrac{1}{{{b}^{2}}+1}\le \dfrac{2}{1+ab}\Leftrightarrow \left( \dfrac{1}{{{a}^{2}}+1}-\dfrac{1}{1+ab} \right)+\left( \dfrac{1}{{{b}^{2}}+1}-\dfrac{1}{1+ab} \right)\le 0 \\
& \Leftrightarrow \dfrac{ab-{{a}^{2}}}{\left( {{a}^{2}}+1 \right)\left( 1+ab \right)}+\dfrac{ab-{{b}^{2}}}{\left( {{b}^{2}}+1 \right)\left( 1+ab \right)}\le 0 \\
& \Leftrightarrow a\left( b-a \right)\left( {{b}^{2}}+1 \right)-b\left( b-a \right)\left( {{a}^{2}}+1 \right)\le 0 \\
& \Leftrightarrow \left( b-a \right)\left( a{{b}^{2}}+a-{{a}^{2}}b-b \right)\le 0 \\
& \Leftrightarrow \left( b-a \right)\left[ ab\left( b-a \right)-\left( b-a \right) \right]\le 0 \\
\end{aligned}$
$\Leftrightarrow {{\left( b-a \right)}^{2}}\left( ab-1 \right)\le 0$ ( luôn đúng khi $0\le ab\le 1$ ).
Đáp án B.