Giúp em bài 4 với ạ. Bài này em chưa hiểu nên không biết làm. Mọi người giúp em với ạ. November 24, 2020 by Philomena Giúp em bài 4 với ạ. Bài này em chưa hiểu nên không biết làm. Mọi người giúp em với ạ.
$1) \, y = 2 + \cos^2x – \sin^2x$ $\Leftrightarrow y = 2 + \cos2x$ Ta có: $-1 \leq \cos2x \leq 1$ $\Leftrightarrow 1 \leq 2 + \cos2x \leq 3$ Hay $1 \leq y \leq 3$ Vậy $\min y = 1 \Leftrightarrow \cos2x = -1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$ $\max y = 3 \Leftrightarrow \cos2x = 1 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)$ $2)\, y = \cos^2x – 2\sqrt3\cos x\sin x + 1$ $\Leftrightarrow y = \dfrac{1 + \cos2x}{2} – \sqrt3\sin2x + 1$ $\Leftrightarrow y = \dfrac{1}{2}\cos2x – \sqrt3\sin2x + \dfrac{3}{2}$ $\Leftrightarrow y – \dfrac{3}{2} = \dfrac{1}{2}\cos2x – \sqrt3\sin2x$ Áp dụng bất đẳng thức $Bunyakovsky$ ta được: (Phương trình có nghiệm) $\left(y – \dfrac{3}{2}\right)^2 = \left(\dfrac{1}{2}\cos2x – \sqrt3\sin2x\right)^2 \leq \left(\dfrac{1}{4} + 3\right)(\cos^22x + \sin^22x) = \dfrac{13}{4}$ $\Rightarrow -\dfrac{\sqrt{13}}{2} \leq y – \dfrac{3}{2} \leq \dfrac{\sqrt{13}}{2}$ $\Leftrightarrow \dfrac{3 – \sqrt{13}}{2} \leq y \leq \dfrac{3 + \sqrt{13}}{2}$ Vậy $\min y = \dfrac{3 – \sqrt{13}}{2} \Leftrightarrow \dfrac{1}{2}\cos2x – \sqrt3\sin2x = – \dfrac{\sqrt{13}}{2} \Leftrightarrow x = \dfrac{\pi}{2} – \dfrac{\arccos\dfrac{1}{\sqrt{13}}}{2} + k\pi$ $\max y = \dfrac{3 + \sqrt{13}}{2} \Leftrightarrow \dfrac{1}{2}\cos2x – \sqrt3\sin2x = \dfrac{\sqrt{13}}{2} \Leftrightarrow x = -\dfrac{\arccos\dfrac{1}{\sqrt{13}}}{2} + k\pi \quad (k \in \Bbb Z)$ Reply
Giải thích các bước giải: Ta có: \(\begin{array}{l}1,\\y = 2 + {\cos ^2}x – {\sin ^2}x\\ = 2 + {\cos ^2}x – \left( {1 – {{\cos }^2}x} \right)\,\,\,\,\,\,\,\,\,\,\,\left( {{{\sin }^2}x + {{\cos }^2}x = 1} \right)\\ = 2 + {\cos ^2}x – 1 + {\cos ^2}x\\ = 1 + 2{\cos ^2}x\\ – 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1\\ \Rightarrow 0 \le 2{\cos ^2}x \le 2\\ \Rightarrow 1 \le 1 + 2{\cos ^2}x \le 1 + 2\\ \Leftrightarrow 1 \le y \le 3\\ \Rightarrow \left\{ \begin{array}{l}{y_{\min }} = 1 \Leftrightarrow {\cos ^2}x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\{y_{\max }} = 3 \Leftrightarrow {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 0 \Rightarrow x = k\pi \end{array} \right.\\3,\\cos2x = 2{\cos ^2}x – 1 \Rightarrow {\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}\\y = {\cos ^2}x – 2\sqrt 3 \cos x.\sin x + 1\\ = \dfrac{{\cos 2x + 1}}{2} – \sqrt 3 .\left( {2\cos x.\sin x} \right) + 1\\ = \dfrac{1}{2}\cos 2x – \sqrt 3 \sin 2x + \dfrac{3}{2}\\ – \sqrt {{a^2} + {b^2}} \le a.\sin x + b.\sin y \le \sqrt {{a^2} + {b^2}} \\ \Rightarrow – \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} + \dfrac{3}{2} \le y \le \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} + \dfrac{3}{2}\\ \Leftrightarrow \dfrac{{3 – \sqrt {13} }}{2} \le y \le \dfrac{{3 + \sqrt {13} }}{2}\end{array}\) Reply
$1) \, y = 2 + \cos^2x – \sin^2x$
$\Leftrightarrow y = 2 + \cos2x$
Ta có:
$-1 \leq \cos2x \leq 1$
$\Leftrightarrow 1 \leq 2 + \cos2x \leq 3$
Hay $1 \leq y \leq 3$
Vậy $\min y = 1 \Leftrightarrow \cos2x = -1 \Leftrightarrow x = \dfrac{\pi}{2} + k\pi$
$\max y = 3 \Leftrightarrow \cos2x = 1 \Leftrightarrow x = k\pi \quad (k \in \Bbb Z)$
$2)\, y = \cos^2x – 2\sqrt3\cos x\sin x + 1$
$\Leftrightarrow y = \dfrac{1 + \cos2x}{2} – \sqrt3\sin2x + 1$
$\Leftrightarrow y = \dfrac{1}{2}\cos2x – \sqrt3\sin2x + \dfrac{3}{2}$
$\Leftrightarrow y – \dfrac{3}{2} = \dfrac{1}{2}\cos2x – \sqrt3\sin2x$
Áp dụng bất đẳng thức $Bunyakovsky$ ta được:
(Phương trình có nghiệm)
$\left(y – \dfrac{3}{2}\right)^2 = \left(\dfrac{1}{2}\cos2x – \sqrt3\sin2x\right)^2 \leq \left(\dfrac{1}{4} + 3\right)(\cos^22x + \sin^22x) = \dfrac{13}{4}$
$\Rightarrow -\dfrac{\sqrt{13}}{2} \leq y – \dfrac{3}{2} \leq \dfrac{\sqrt{13}}{2}$
$\Leftrightarrow \dfrac{3 – \sqrt{13}}{2} \leq y \leq \dfrac{3 + \sqrt{13}}{2}$
Vậy $\min y = \dfrac{3 – \sqrt{13}}{2} \Leftrightarrow \dfrac{1}{2}\cos2x – \sqrt3\sin2x = – \dfrac{\sqrt{13}}{2} \Leftrightarrow x = \dfrac{\pi}{2} – \dfrac{\arccos\dfrac{1}{\sqrt{13}}}{2} + k\pi$
$\max y = \dfrac{3 + \sqrt{13}}{2} \Leftrightarrow \dfrac{1}{2}\cos2x – \sqrt3\sin2x = \dfrac{\sqrt{13}}{2} \Leftrightarrow x = -\dfrac{\arccos\dfrac{1}{\sqrt{13}}}{2} + k\pi \quad (k \in \Bbb Z)$
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
y = 2 + {\cos ^2}x – {\sin ^2}x\\
= 2 + {\cos ^2}x – \left( {1 – {{\cos }^2}x} \right)\,\,\,\,\,\,\,\,\,\,\,\left( {{{\sin }^2}x + {{\cos }^2}x = 1} \right)\\
= 2 + {\cos ^2}x – 1 + {\cos ^2}x\\
= 1 + 2{\cos ^2}x\\
– 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1\\
\Rightarrow 0 \le 2{\cos ^2}x \le 2\\
\Rightarrow 1 \le 1 + 2{\cos ^2}x \le 1 + 2\\
\Leftrightarrow 1 \le y \le 3\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 1 \Leftrightarrow {\cos ^2}x = 0 \Leftrightarrow x = \dfrac{\pi }{2} + k\pi \\
{y_{\max }} = 3 \Leftrightarrow {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 0 \Rightarrow x = k\pi
\end{array} \right.\\
3,\\
cos2x = 2{\cos ^2}x – 1 \Rightarrow {\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}\\
y = {\cos ^2}x – 2\sqrt 3 \cos x.\sin x + 1\\
= \dfrac{{\cos 2x + 1}}{2} – \sqrt 3 .\left( {2\cos x.\sin x} \right) + 1\\
= \dfrac{1}{2}\cos 2x – \sqrt 3 \sin 2x + \dfrac{3}{2}\\
– \sqrt {{a^2} + {b^2}} \le a.\sin x + b.\sin y \le \sqrt {{a^2} + {b^2}} \\
\Rightarrow – \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} + \dfrac{3}{2} \le y \le \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} + \dfrac{3}{2}\\
\Leftrightarrow \dfrac{{3 – \sqrt {13} }}{2} \le y \le \dfrac{{3 + \sqrt {13} }}{2}
\end{array}\)