Cho hàm số f(x) = x^3 -3x+m+2. Có bao nhiêu số

* Hướng dẫn giải

Đáp án C

Đặt $g\left(x\right)=x^3-3x+2\Rightarrow \underset{\left[-1;3\right]}{\max }g\left(x\right)=20;\underset{\left[-1;3\right]}{\min }g\left(x\right)=0$. Khi đó $f\left(x\right)=m+g\left(x\right)$.

Ta có:

$\begin{array}{l}f\left(a\right)+f\left(b\right)>f\left(c\right),\forall a,b,c\in \left[-1;3\right]\\ \Rightarrow m>g\left(c\right)-\left(g\left(a\right)+g\left(b\right)\right),\forall a,b,c\in \left[-1;3\right]\\ \Rightarrow m>\underset{\left[-1;3\right]}{\max }g\left(x\right)-2\underset{\left[-1;3\right]}{\min }g\left(x\right)\Rightarrow m>20.\end{array}$

$\begin{array}{l}{f}^2\left(a\right)+f^2\left(b\right)>f^2\left(c\right),\forall a,b,c\in \left[-1;3\right]\\ \Rightarrow {\left(m+g\left(a\right)\right)}^2+{\left(m+g\left(b\right)\right)}^2>{\left(m+g\left(c\right)\right)}^2,\forall a,b,c\in \left[-1;3\right]\\ \Rightarrow m^2+2\left(g\left(a\right)+g\left(b\right)-g\left(c\right)\right)m+g^2\left(a\right)+g^2\left(b\right)-g^2\left(c\right)>0,\forall a,b,c\in \left[-1;3\right]\\ \Rightarrow {\left(m+g\left(a\right)+g\left(b\right)-g\left(c\right)\right)}^2-2g\left(a\right)g\left(b\right)+2g\left(a\right)g\left(c\right)+2g\left(b\right)g\left(c\right)-2g^2\left(c\right)>0,\forall a,b,c\in \left[-1;3\right]\\ \Rightarrow {\left(m+g\left(a\right)+g\left(b\right)-g\left(c\right)\right)}^2>2\left(g\left(a\right)-g\left(c\right)\right)\left(g\left(b\right)-g\left(c\right)\right),\forall a,b,c\in \left[-1;3\right]\\ \Rightarrow {\left(m+g\left(a\right)+g\left(b\right)-g\left(c\right)\right)}^2>2{\left(\underset{\left[-1;3\right]}{\max }g\left(x\right)-\underset{\left[-1;3\right]}{\min }g\left(x\right)\right)}^2,\forall a,b,c\in \left[-1;3\right]\\ m>\underset{\left[-1;3\right]}{\max }g\left(x\right)+20\sqrt{2}-2\underset{\left[-1;3\right]}{\min }g\left(x\right)\Rightarrow m\ge 49.\end{array}$