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Câu hỏi: Gọi $M,m$ lần lượt là giá trị lớn nhất, giá trị nhỏ nhất của hàm số $f\left( x \right)={{x}^{2}}-\ln \left( 1-2x \right)$ trên đoạn $\left[ -2;0 \right].$ Biết $M+m=a+b\ln 2+c\ln 5\left( a,b,c\in Q \right).$ Khi đó tổng $a+b+c$ bằng
A. $\dfrac{9}{4}$
B. $\dfrac{17}{4}$
C. $-\dfrac{3}{4}$
D. $\dfrac{15}{4}$
A. $\dfrac{9}{4}$
B. $\dfrac{17}{4}$
C. $-\dfrac{3}{4}$
D. $\dfrac{15}{4}$
TXĐ: $D=\left( -\infty ;\dfrac{1}{2} \right].$
Ta có: $f'\left( x \right)=2x+\dfrac{2}{1-2x}=\dfrac{-4{{x}^{2}}+2x+2}{1-2x}.$
$f'\left( x \right)=0\Leftrightarrow -4{{x}^{2}}+2x+2=0\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=-\dfrac{1}{2} \\
\end{aligned} \right.. $Tính: $ f\left( -2 \right)=4-\ln 5;f\left( 0 \right)=0;f\left( -\dfrac{1}{2} \right)=\dfrac{1}{4}-\ln 2.$
$\Rightarrow m=\underset{\left[ -2;0 \right]}{\mathop{\min }} f\left( x \right)=f\left( -\dfrac{1}{2} \right)=\dfrac{1}{4}-\ln 2$ khi $x=-\dfrac{1}{2}.$
$M=\underset{\left[ -2;0 \right]}{\mathop{\max }} f\left( x \right)=f\left( -2 \right)=4-\ln 5\Rightarrow M+m=4-\ln 5+\dfrac{1}{4}-\ln 2=\dfrac{17}{4}-\ln 2-\ln 5.$
$\Rightarrow a=\dfrac{17}{4},b=-1,c=-1\Rightarrow a+b+c=\dfrac{9}{4}.$
Ta có: $f'\left( x \right)=2x+\dfrac{2}{1-2x}=\dfrac{-4{{x}^{2}}+2x+2}{1-2x}.$
$f'\left( x \right)=0\Leftrightarrow -4{{x}^{2}}+2x+2=0\Leftrightarrow \left[ \begin{aligned}
& x=1 \\
& x=-\dfrac{1}{2} \\
\end{aligned} \right.. $Tính: $ f\left( -2 \right)=4-\ln 5;f\left( 0 \right)=0;f\left( -\dfrac{1}{2} \right)=\dfrac{1}{4}-\ln 2.$
$\Rightarrow m=\underset{\left[ -2;0 \right]}{\mathop{\min }} f\left( x \right)=f\left( -\dfrac{1}{2} \right)=\dfrac{1}{4}-\ln 2$ khi $x=-\dfrac{1}{2}.$
$M=\underset{\left[ -2;0 \right]}{\mathop{\max }} f\left( x \right)=f\left( -2 \right)=4-\ln 5\Rightarrow M+m=4-\ln 5+\dfrac{1}{4}-\ln 2=\dfrac{17}{4}-\ln 2-\ln 5.$
$\Rightarrow a=\dfrac{17}{4},b=-1,c=-1\Rightarrow a+b+c=\dfrac{9}{4}.$
Đáp án A.