2 câu trả lời
Đáp án:
\(\begin{array}{l}
Min = - 1 \Leftrightarrow \sin x = - 1 \Leftrightarrow x = - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
Max = 3 \Leftrightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
y = 2{\sin ^2}x + 2\sin x - 1\\
= {\left( {\sqrt 2 \sin x} \right)^2} + 2\sqrt 2 \sin x.\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} - \dfrac{3}{2}\\
= {\left( {\sqrt 2 \sin x + \dfrac{1}{{\sqrt 2 }}} \right)^2} - \dfrac{3}{2}\\
Do: - 1 \le \sin x \le 1\\
\to - \sqrt 2 \le \sqrt 2 \sin x \le \sqrt 2 \\
\to - \sqrt 2 + \dfrac{1}{{\sqrt 2 }} \le \sqrt 2 \sin x + \dfrac{1}{{\sqrt 2 }} \le \sqrt 2 + \dfrac{1}{{\sqrt 2 }}\\
\to \dfrac{1}{2} \le {\left( {\sqrt 2 \sin x + \dfrac{1}{{\sqrt 2 }}} \right)^2} \le \dfrac{9}{2}\\
\to - 1 \le {\left( {\sqrt 2 \sin x + \dfrac{1}{{\sqrt 2 }}} \right)^2} - \dfrac{3}{2} \le 3\\
\to Min = - 1 \Leftrightarrow \sin x = - 1 \Leftrightarrow x = - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
Max = 3 \Leftrightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)
\end{array}\)
Đặt t=Sinx (Với -1≤t≤1)
\(\begin{array}{l}
y = 2.{{\mathop{\rm Sin}\nolimits} ^2}x + 2{\mathop{\rm S}\nolimits} {\rm{inx}} - 1\\
\Leftrightarrow y = 2{t^2} + 2t - 1(*)\\
y' = 4t + 2\\
y' = 0 \Leftrightarrow 4t + 2 = 0\\
\to t = \frac{{ - 1}}{2}\\
f(1) = 3\\
f( - 1) = - 1\\
f(\frac{{ - 1}}{2}) = \frac{{ - 3}}{2}
\end{array}\)
Thay tất cả các t vào (*)
\( \to Min = \frac{{ - 3}}{2},\max = 3\)