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Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{\left( m+1 \right)\sqrt{-2x+3}-1}{-\sqrt{-2x+3}+\dfrac{2}{m}}$ ( $m\ne 0$ và là tham số thực). Tập hợp $m$ để hàm số đã cho nghịch biến trên khoảng $\left( -\dfrac{1}{2};1 \right)$ có dạng $S=\left( -\infty ;a \right)\cup \left( b;c \right]\cup \left[ d;+\infty \right)$ với $a,b,c,d$ là các số thực. Tính $P=a-b+c-d$.
A. $-3$.
B. $-1$.
C. $0$.
D. $2$.
A. $-3$.
B. $-1$.
C. $0$.
D. $2$.
Đặt $t=\sqrt{-2x+3}$, $x\in \left( -\dfrac{1}{2};1 \right)$.
Ta có: ${{{t}'}_{x}}=\dfrac{-1}{\sqrt{-2x+3}}<0,\forall x\in \left( -\dfrac{1}{2};1 \right)$ nên với $x\in \left( -\dfrac{1}{2};1 \right)$ thì $t\in \left( 1;2 \right)$
Ta có ${{{y}'}_{x}}={{{y}'}_{t}}.{{{t}'}_{x}}=\dfrac{\left( m+1 \right)\dfrac{2}{m}-1}{{{\left( -t+\dfrac{2}{m} \right)}^{2}}}.\left( \dfrac{-1}{t} \right)$.
Hàm số đã cho nghịch biến trên khoảng $\left( -\dfrac{1}{2};1 \right)$ thì $\left\{ \begin{aligned}
& \dfrac{\left( m+1 \right)\dfrac{2}{m}-1}{{{\left( -t+\dfrac{2}{m} \right)}^{2}}}.\left( \dfrac{-1}{t} \right)<0 \\
& \left[ \begin{aligned}
& \dfrac{2}{m}<1 \\
& \dfrac{2}{m}>2 \\
\end{aligned} \right. \\
\end{aligned} \right., \forall t\in \left( 1;2 \right)$
$\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{\left( m+1 \right)\dfrac{2}{m}-1}{{{\left( -t+\dfrac{2}{m} \right)}^{2}}}.\left( \dfrac{-1}{t} \right)<0 \\
& \left[ \begin{aligned}
& \dfrac{2}{m}\le 1 \\
& \dfrac{2}{m}\ge 2 \\
\end{aligned} \right. \\
\end{aligned} \right., \forall t\in \left( 1;2 \right) \Leftrightarrow \left\{ \begin{aligned}
& 1-\left( m+1 \right)\dfrac{2}{m}<0 \\
& \left[ \begin{aligned}
& m\in \left( -\infty ;0 \right)\cup \left[ 2;+\infty \right) \\
& m\in \left( 0;1 \right] \\
\end{aligned} \right. \\
\end{aligned} \right., \forall t\in \left( 1;2 \right) $ $ \Leftrightarrow \dfrac{\left( m+1 \right)\dfrac{2}{m}-1}{{{\left( -t+\dfrac{2}{m} \right)}^{2}}}.\left( \dfrac{-1}{t} \right)<0, \forall t\in \left( 1;2 \right) \Leftrightarrow \left\{ \begin{aligned}
& 1-\left( m+1 \right)\dfrac{2}{m}<0 \\
& -t+\dfrac{2}{m}\ne 0 \\
\end{aligned} \right. ,\forall t\in \left( 1;2 \right) $ $ \Leftrightarrow \left\{ \begin{aligned}
& \dfrac{-m-2}{m}<0 \\
& m\in \left( -\infty ;0 \right)\cup \left( 0;1 \right]\cup \left[ 2;+\infty \right) \\
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& m\in \left( -\infty ;-2 \right)\cup \left( 0;+\infty \right) \\
& m\in \left( -\infty ;0 \right)\cup \left( 0;1 \right]\cup \left[ 2;+\infty \right) \\
\end{aligned} \right.$
$\Leftrightarrow m\in \left( -\infty ;-2 \right)\cup \left( 0;1 \right]\cup \left[ 2;+\infty \right)$.
Suy ra $a=-2, b=0, c=1, d=2.$
Vậy $P=a-b+c-d=-3.$
Ta có: ${{{t}'}_{x}}=\dfrac{-1}{\sqrt{-2x+3}}<0,\forall x\in \left( -\dfrac{1}{2};1 \right)$ nên với $x\in \left( -\dfrac{1}{2};1 \right)$ thì $t\in \left( 1;2 \right)$
Ta có ${{{y}'}_{x}}={{{y}'}_{t}}.{{{t}'}_{x}}=\dfrac{\left( m+1 \right)\dfrac{2}{m}-1}{{{\left( -t+\dfrac{2}{m} \right)}^{2}}}.\left( \dfrac{-1}{t} \right)$.
Hàm số đã cho nghịch biến trên khoảng $\left( -\dfrac{1}{2};1 \right)$ thì $\left\{ \begin{aligned}
& \dfrac{\left( m+1 \right)\dfrac{2}{m}-1}{{{\left( -t+\dfrac{2}{m} \right)}^{2}}}.\left( \dfrac{-1}{t} \right)<0 \\
& \left[ \begin{aligned}
& \dfrac{2}{m}<1 \\
& \dfrac{2}{m}>2 \\
\end{aligned} \right. \\
\end{aligned} \right., \forall t\in \left( 1;2 \right)$
$\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{\left( m+1 \right)\dfrac{2}{m}-1}{{{\left( -t+\dfrac{2}{m} \right)}^{2}}}.\left( \dfrac{-1}{t} \right)<0 \\
& \left[ \begin{aligned}
& \dfrac{2}{m}\le 1 \\
& \dfrac{2}{m}\ge 2 \\
\end{aligned} \right. \\
\end{aligned} \right., \forall t\in \left( 1;2 \right) \Leftrightarrow \left\{ \begin{aligned}
& 1-\left( m+1 \right)\dfrac{2}{m}<0 \\
& \left[ \begin{aligned}
& m\in \left( -\infty ;0 \right)\cup \left[ 2;+\infty \right) \\
& m\in \left( 0;1 \right] \\
\end{aligned} \right. \\
\end{aligned} \right., \forall t\in \left( 1;2 \right) $ $ \Leftrightarrow \dfrac{\left( m+1 \right)\dfrac{2}{m}-1}{{{\left( -t+\dfrac{2}{m} \right)}^{2}}}.\left( \dfrac{-1}{t} \right)<0, \forall t\in \left( 1;2 \right) \Leftrightarrow \left\{ \begin{aligned}
& 1-\left( m+1 \right)\dfrac{2}{m}<0 \\
& -t+\dfrac{2}{m}\ne 0 \\
\end{aligned} \right. ,\forall t\in \left( 1;2 \right) $ $ \Leftrightarrow \left\{ \begin{aligned}
& \dfrac{-m-2}{m}<0 \\
& m\in \left( -\infty ;0 \right)\cup \left( 0;1 \right]\cup \left[ 2;+\infty \right) \\
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& m\in \left( -\infty ;-2 \right)\cup \left( 0;+\infty \right) \\
& m\in \left( -\infty ;0 \right)\cup \left( 0;1 \right]\cup \left[ 2;+\infty \right) \\
\end{aligned} \right.$
$\Leftrightarrow m\in \left( -\infty ;-2 \right)\cup \left( 0;1 \right]\cup \left[ 2;+\infty \right)$.
Suy ra $a=-2, b=0, c=1, d=2.$
Vậy $P=a-b+c-d=-3.$
Đáp án A.