Đề 003-TN THPT QG

Đề 003-TN THPT QG

Câu 48.    Biết giá trị lớn nhất của hàm số $y=f\left( x \right)=\left| \frac{2}{5}{{x}^{3}}-3x+3m-1 \right|+\frac{9}{5}x$ trên $\left[ 0\,;\,3 \right]$ bằng $12.$Tính tổng tất cả các giá trị của tham số thực $m.$

A. $-\frac{4}{5}.$            

B. $\frac{6}{5}.$          

C. $\frac{2}{5}.$          

D. $0.$

Hướng dẫn và Lời giải

Chọn C

* TH1: $\frac{2}{5}{{x}^{3}}-3x+3m-1\ge 0$

$\Leftrightarrow 3m\ge -\frac{2}{5}{{x}^{3}}+3x+1$, $\forall x\in \left[ 0\,;\,3 \right]\Leftrightarrow m\ge \frac{1+\sqrt{10}}{3}\,\,\,\left( 1 \right)$

Khi đó: $f\left( x \right)=\frac{2}{5}{{x}^{3}}-\frac{6x}{5}+3m-1$

${f}’\left( x \right)=\frac{6}{5}{{x}^{2}}-\frac{6}{5}$

$f’\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = – 1 \notin \left[ {0{\mkern 1mu} ;{\mkern 1mu} 3} \right]\\ x = 1 \in \left[ {0{\mkern 1mu} ;{\mkern 1mu} 3} \right] \end{array} \right.$

* $f\left( 0 \right)=3m-1$; $f\left( 1 \right)=3m-\frac{9}{5}$; $f\left( 3 \right)=3m+\frac{31}{5}$

Theo giả thiết, ta có: $\underset{\left[ 0\,;\,3 \right]}{\mathop{max}}\,f\left( x \right)=3m+\frac{31}{5}=12\Leftrightarrow m=\frac{29}{15}$ (thỏa $\left( 1 \right)$)

 * TH2: $\frac{2}{5}{{x}^{3}}-3x+3m-1<0$

$\Leftrightarrow 3m<-\frac{2}{5}{{x}^{3}}+3x+1$, $\forall x\in \left[ 0\,;\,3 \right]\Leftrightarrow m<-\frac{4}{5}$

Khi đó: $f\left( x \right)=-\frac{2}{5}{{x}^{3}}+\frac{24}{5}x-3m+1$

${f}’\left( x \right)=-\frac{6}{5}{{x}^{2}}+3+\frac{9}{5}=-\frac{6}{5}{{x}^{2}}+\frac{24}{4}$

$f’\left( 0 \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = – 2 \notin \left[ {0{\mkern 1mu} ;{\mkern 1mu} 3} \right]\\ x = 2 \in \left[ {0{\mkern 1mu} ;{\mkern 1mu} 3} \right] \end{array} \right.$

* $f\left( 0 \right)=-3m+1$; $f\left( 3 \right)=\frac{23}{5}-3m$; $f\left( 2 \right)=\frac{37}{5}-3m$

Theo giả thiết, ta có: $\underset{\left[ 0\,;\,3 \right]}{\mathop{max}}\,f\left( x \right)=\frac{37}{5}-3m=12\Leftrightarrow m=-\frac{23}{15}$ (thỏa $\left( 1 \right)$)

* Tổng hai giá trị của tham số $m$là: $\frac{29}{15}-\frac{23}{15}=\frac{2}{5}$

Cách 2:

Ta có: $\left| \frac{2}{5}{{x}^{3}}-3x+3m-1 \right|+\frac{9}{5}x\le 12$, $\forall x\in \left[ 0\,;\,3 \right]$$\Leftrightarrow \left| \frac{2}{5}{{x}^{3}}-3x+3m-1 \right|\le 12-\frac{9}{5}x$, $\forall x\in \left[ 0\,;\,3 \right]$

$ \Leftrightarrow \left\{ \begin{array}{l} \frac{2}{5}{x^3} – 3x + 3m – 1 \ge \frac{9}{5}x – 12\\ \frac{2}{5}{x^3} – 3x + 3m – 1 \le 12 – \frac{9}{5}x \end{array} \right.$

* Đặt $g\left( x \right)=-\frac{2}{5}{{x}^{3}}+\frac{24}{5}x-11$

${g}’\left( x \right)=-\frac{6}{5}{{x}^{2}}+\frac{24}{5}$

$ \Leftrightarrow \left\{ \begin{array}{l} 3m \ge – \frac{2}{5}{x^3} + \frac{{24}}{5}x – 11\\ 3m \le – \frac{2}{5}{x^3} + \frac{6}{5}x + 13 \end{array} \right.$

$g\left( 0 \right)=-11$; $g\left( 3 \right)=-\frac{37}{5}$

$\Rightarrow \underset{\left[ 0\,;\,3 \right]}{\mathop{max}}\,g\left( x \right)=g\left( 2 \right)=-\frac{23}{5}$

* Đặt $h\left( x \right)=-\frac{2}{5}{{x}^{3}}+\frac{6}{5}x+13$

${h}’\left( x \right)=-\frac{6}{5}{{x}^{2}}+\frac{6}{5}$

$g’\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 2 \in \left[ {0{\mkern 1mu} ;{\mkern 1mu} 3} \right] \Rightarrow g\left( 2 \right) = – \frac{{23}}{5}\\ x = – 2 \notin \left[ {0{\mkern 1mu} ;{\mkern 1mu} 3} \right] \end{array} \right.$

$h\left( 0 \right)=13$; $h\left( 3 \right)=\frac{29}{5}$

$\Rightarrow \underset{\left[ 0\,;\,3 \right]}{\mathop{\min }}\,h\left( x \right)=\frac{29}{5}$

$h’\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 1 \in \left[ {0{\mkern 1mu} ;{\mkern 1mu} 3} \right] \Rightarrow h\left( 1 \right) = \frac{{69}}{5}\\ x = – 1 \notin \left[ {0{\mkern 1mu} ;{\mkern 1mu} 3} \right] \end{array} \right.$

$h\left( 0 \right)=13$; $h\left( 3 \right)=\frac{29}{5}$

$\Rightarrow \underset{\left[ 0\,;\,3 \right]}{\mathop{\min }}\,h\left( x \right)=\frac{29}{5}$

$\left( * \right) \Rightarrow \left\{ \begin{array}{l} 3m \ge \mathop {max}\limits_{\left[ {0;3} \right]} {\mkern 1mu} g\left( x \right) = – \frac{{23}}{5}\\ 3m \le \mathop {\min }\limits_{\left[ {0;3} \right]5} {\mkern 1mu} h\left( x \right) = \frac{{29}}{5} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \ge – \frac{{23}}{{15}}\\ m \le \frac{{29}}{{15}} \end{array} \right.$

Vậy

$\mathop {max}\limits_{\left[ {0;3} \right]} {\mkern 1mu} f\left( x \right) = 12 \Leftrightarrow \left[ \begin{array}{l} m = – \frac{{23}}{{15}}\\ m = \frac{{29}}{{15}} \end{array} \right.$

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