Đề thi THPT QG - 2021 - mã 101

Trên đoạn \(\left[ {0;3} \right]\), hàm số \(y =  - {x^3} + 3x\) đạt giá trị lớn nhất tại điểm

$ - \dfrac{1}{2}; - \dfrac{{\sqrt 3 }}{2}$

$ - \dfrac{{\sqrt 3 }}{2}; - 1$

$ - \dfrac{{\sqrt 3 }}{2}; - 2$

$ - \dfrac{{\sqrt 2 }}{2}; - \dfrac{{\sqrt 3 }}{2}$

$f\left( x \right) \geqslant  - 2,\forall x \in \left[ {1;3} \right]$ 

$f\left( 1 \right) = f\left( 3 \right) =  - 2$

$f\left( x \right) <  - 2,\forall x \in \left[ {1;3} \right]$                     

$f\left( x \right) \leqslant  - 2,\forall x \in \left[ {1;3} \right]$

$f\left( 0 \right) < 5$

$f\left( 2 \right) \geqslant 5$ 

$f\left( 1 \right) = 5$      

$f\left( 0 \right) = 5$

\(\mathop {\min }\limits_{\left[ { - 3;\,\,1} \right]} g\left( x \right) = g\left( { - 1} \right)\).

\(\mathop {\min }\limits_{\left[ { - 3;\,\,1} \right]} g\left( x \right) = g\left( 1 \right)\)

\(\mathop {\min }\limits_{\left[ { - 3;\,\,1} \right]} g\left( x \right) = g\left( { - 3} \right)\)

$\mathop {\min }\limits_{\left[ { - 3;\,\,1} \right]} g\left( x \right) = \dfrac{{g\left( { - 3} \right) + g\left( 1 \right)}}{2}$

$ - 1$

$1$

$\pi $

$0$

\( - 16\).

\(4\).

\(\dfrac{{ - 1}}{{16}}\).

\(\dfrac{1}{4}\).

$m = 2$

$m = 5$

$m = 3$

$m = 4$