Cho hàm số $f\left( x \right)={{x}^{4}}-2{{x}^{2}}$. Gọi $S$ là...

Đặt \)">t=\cos x+1, t\in \left[ 0;2 \right]y=\left| {{t}^{4}}-2{{t}^{2}}+m \right|t\in \left[ 0;2 \right]f\left( t \right)={{t}^{4}}-2{{t}^{2}}+mt\in \left[ 0;2 \right]{f}'\left( t \right)=4{{t}^{3}}-4t=0\Leftrightarrow \left[ \begin{aligned}
& t=0 \left( \text{nhan} \right) \\
& t=1 \left( \text{nhan} \right) \\
& t=-1\left( \text{loai} \right) \\
\end{aligned} \right.f\left( 0 \right)=mf\left( 1 \right)=-1+mf\left( 2 \right)=8+m\underset{\left[ 0;2 \right]}{\mathop{\max }} f\left( t \right)=8+m\underset{\left[ 0;2 \right]}{\mathop{\min }} f\left( t \right)=m-1\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| f\left( t \right) \right|=\dfrac{\left| m+8+m-1 \right|+\left| m+8-m+1 \right|}{2}=\dfrac{\left| 2m+7 \right|+9}{2}\max y=\underset{\left[ 0;2 \right]}{\mathop{\max }} \left| f\left( t \right) \right|=\dfrac{\left| 2m+7 \right|+9}{2}\ge \dfrac{9}{2}2m+7=0\Leftrightarrow m=-\dfrac{7}{2}$.