Google
Câu hỏi: Tìm giá trị nhỏ nhất của hàm số $y=2{{\sin }^{2}}x+2\sin x-1$
A. $-\dfrac{2}{3}$.
B. $-\dfrac{3}{2}$.
C. $\dfrac{2}{3}$.
D. $\dfrac{3}{2}$.
A. $-\dfrac{2}{3}$.
B. $-\dfrac{3}{2}$.
C. $\dfrac{2}{3}$.
D. $\dfrac{3}{2}$.
TXĐ: $D=\mathbb{R}$.
Đặt $\sin x=t$, $\left( -1\le t\le 1 \right)$
Ta có $f\left( x \right)=2{{t}^{2}}+2t-1$ liên tục trên đoạn $\left[ -1;1 \right]$
${f}'\left( x \right)=4t+2=0\Leftrightarrow t=-\dfrac{1}{2}$
$f\left( -1 \right)=-1$ ; $f\left( -\dfrac{1}{2} \right)=-\dfrac{3}{2}$ ; $f\left( 1 \right)=3$.
Suy ra $\underset{\mathbb{R}}{\mathop{\min }} y=\underset{\left[ -1;1 \right]}{\mathop{\min }} f\left( x \right)=-\dfrac{3}{2}\Leftrightarrow t=-\dfrac{1}{2}\Leftrightarrow \sin x=-\dfrac{1}{2}\Leftrightarrow \left[ \begin{aligned}
& x=-\dfrac{\pi }{6}+k2\pi \\
& x=\dfrac{7\pi }{6}+k2\pi \\
\end{aligned} \right. $, $ k\in \mathbb{Z}$.
Đặt $\sin x=t$, $\left( -1\le t\le 1 \right)$
Ta có $f\left( x \right)=2{{t}^{2}}+2t-1$ liên tục trên đoạn $\left[ -1;1 \right]$
${f}'\left( x \right)=4t+2=0\Leftrightarrow t=-\dfrac{1}{2}$
$f\left( -1 \right)=-1$ ; $f\left( -\dfrac{1}{2} \right)=-\dfrac{3}{2}$ ; $f\left( 1 \right)=3$.
Suy ra $\underset{\mathbb{R}}{\mathop{\min }} y=\underset{\left[ -1;1 \right]}{\mathop{\min }} f\left( x \right)=-\dfrac{3}{2}\Leftrightarrow t=-\dfrac{1}{2}\Leftrightarrow \sin x=-\dfrac{1}{2}\Leftrightarrow \left[ \begin{aligned}
& x=-\dfrac{\pi }{6}+k2\pi \\
& x=\dfrac{7\pi }{6}+k2\pi \\
\end{aligned} \right. $, $ k\in \mathbb{Z}$.
Đáp án B.