19.
\(y'=3x^2+2mx+1-m\)
Hàm số nghịch biến trên khoảng đã cho khi \(\forall x\in\left(-1;0\right)\) ta có:
\(y'\le0\)
\(\Leftrightarrow3x^2+2mx+1-m\le0\)
\(\Leftrightarrow3x^2+1\le m\left(1-2x\right)\)
\(\Leftrightarrow m\ge\dfrac{3x^2+1}{1-2x}\) (do \(1-2x>0;\forall x< 0\))
\(\Leftrightarrow m\ge\max\limits_{\left(-1;0\right)}\dfrac{3x^2+1}{1-2x}\)
Đặt \(f\left(x\right)=\dfrac{3x^2+1}{1-2x}\), xét hàm \(f\left(x\right)\) trên \(\left(-1;0\right)\)
\(f'\left(x\right)=\dfrac{-6x^2+6x+2}{\left(1-2x\right)^2}=0\Rightarrow-3x^2+3x+1=0\Rightarrow x=\dfrac{3-\sqrt{21}}{6}\)
\(f\left(-1\right)=\dfrac{4}{3}\) ; \(f\left(0\right)=1\) ; \(f\left(\dfrac{3-\sqrt{21}}{6}\right)=\dfrac{\sqrt{21}-3}{2}\)
\(\Rightarrow f\left(x\right)< \dfrac{4}{3}\Rightarrow m\ge\dfrac{4}{3}\)
\(\Rightarrow m=\left\{2;3;4;...;2021\right\}\) có \(2021-2+1=2020\) giá trị nguyên
20.
\(y'=3x^2+4mx+1\)
Hàm đồng biến trên khoảng đã cho khi \(\forall x\in\left(0;1\right)\) ta có:
\(y'\ge0\)
\(\Leftrightarrow3x^2+4mx+1\ge0\)
\(\Leftrightarrow4mx\ge-3x^2-1\)
\(\Leftrightarrow m\ge\dfrac{-3x^2-1}{4x}\)
\(\Rightarrow m\ge\max\limits_{\left(0;1\right)}\dfrac{-3x^2-1}{4x}\)
Ta có: \(\dfrac{3x^2+1}{4x}\ge\dfrac{2\sqrt{3x^2}}{4x}=\dfrac{\sqrt{3}}{2}\Rightarrow-\dfrac{3x^2+1}{4x}\le-\dfrac{\sqrt{3}}{2}\)
\(\Rightarrow m\ge-\dfrac{\sqrt{3}}{2}\)
