Lời giải của Tự Học 365

Ta có \(g\left( x \right) = 1 + 2\int\limits_0^x {f\left( t \right)dt} \)  suy ra \(\left\{ \begin{array}{l}g\left( x \right) - 1 = 2\int\limits_0^x {f\left( t \right)dt} \\g\left( 0 \right) = 1 + \int\limits_0^0 {f\left( t \right)dt} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}g'\left( x \right) = 2f\left( x \right) \Rightarrow f\left( x \right) = \dfrac{{g'\left( x \right)}}{2}\\g\left( 0 \right) = 1\end{array} \right.\)

Mà \(g\left( x \right) \ge {\left[ {f\left( x \right)} \right]^3} \Leftrightarrow g\left( x \right) \ge {\left[ {\dfrac{{g'\left( x \right)}}{2}} \right]^3} \Leftrightarrow \sqrt[3]{{g\left( x \right)}} \ge \dfrac{{g'\left( x \right)}}{2} \Leftrightarrow \dfrac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}} \le 2\)

Với \(t \in \left[ {0;1} \right]\), Lấy tích phân hai vế ta được

\(\begin{array}{l}\int\limits_0^t {\dfrac{{g'\left( x \right)}}{{\sqrt[3]{{g\left( x \right)}}}}} dx \le \int\limits_0^t {2dx}  \Leftrightarrow \int\limits_0^t {{{\left[ {g\left( x \right)} \right]}^{\dfrac{{ - 1}}{3}}}} d\left( {g\left( x \right)} \right) \le 2t\\ \Leftrightarrow 2t \ge \dfrac{3}{2}\left. {{{\left[ {g\left( x \right)} \right]}^{\dfrac{2}{3}}}} \right|_0^t \Leftrightarrow \dfrac{4}{3}t \ge \sqrt[3]{{{g^2}\left( t \right)}} - \sqrt[3]{{{g^2}\left( 0 \right)}}\end{array}\)

mà \(g\left( 0 \right) = 1\) nên \(\sqrt[3]{{{g^2}\left( t \right)}} \le \dfrac{4}{3}t + 1 \Rightarrow \sqrt[3]{{{g^2}\left( x \right)}} \le \dfrac{4}{3}x + 1\)

Từ đó ta có \(\int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \int\limits_0^1 {\left( {\dfrac{4}{3}x + 1} \right)dx}  \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \left. {\left( {\dfrac{2}{3}{x^2} + x} \right)} \right|_0^1 \Leftrightarrow \int\limits_0^1 {\sqrt[3]{{{g^2}\left( x \right)}}} \,dx \le \dfrac{5}{3}\)

Hay giá trị lớn nhất cần tìm là \(\dfrac{5}{3}.\)

Đáp án cần chọn là: b