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Câu hỏi: Cho hàm số $f\left( x \right)=m{{x}^{4}}+2\left( m-1 \right){{x}^{2}}$ với $m$ là tham số thực. Nếu $\underset{\left[ 0;2 \right]}{\mathop{\min }} f\left( x \right)=f\left( 1 \right)$ thì $\underset{\left[ 0;2 \right]}{\mathop{\max }} f\left( x \right)$ bằng
A. $2$.
B. $-1$.
C. $4$.
D. $0$.
A. $2$.
B. $-1$.
C. $4$.
D. $0$.
Vì $\underset{\left[ 0;2 \right]}{\mathop{\min }} f\left( x \right)=f\left( 1 \right)$ nên suy ra ${f}'\left( 1 \right)=0$
Ta có ${f}'\left( x \right)=4m{{x}^{3}}+4\left( m-1 \right)x\Rightarrow {f}'\left( 1 \right)=0\Leftrightarrow m=\dfrac{1}{2}$
Với $m=\dfrac{1}{2}$ thì $f\left( x \right)=\dfrac{1}{2}{{x}^{4}}-{{x}^{2}}$
Ta có ${f}'\left( x \right)=2{{x}^{3}}-2x;{f}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=\pm 1 \\
\end{aligned} \right.$
$f\left( 0 \right)=0;f\left( 1 \right)=-\dfrac{1}{2};f\left( 2 \right)=4$.
Vậy $\underset{\left[ 0;2 \right]}{\mathop{\max }} f\left( x \right)=4$.
Ta có ${f}'\left( x \right)=4m{{x}^{3}}+4\left( m-1 \right)x\Rightarrow {f}'\left( 1 \right)=0\Leftrightarrow m=\dfrac{1}{2}$
Với $m=\dfrac{1}{2}$ thì $f\left( x \right)=\dfrac{1}{2}{{x}^{4}}-{{x}^{2}}$
Ta có ${f}'\left( x \right)=2{{x}^{3}}-2x;{f}'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=\pm 1 \\
\end{aligned} \right.$
$f\left( 0 \right)=0;f\left( 1 \right)=-\dfrac{1}{2};f\left( 2 \right)=4$.
Vậy $\underset{\left[ 0;2 \right]}{\mathop{\max }} f\left( x \right)=4$.
Đáp án C.