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a: \(y=\sqrt{2}sin\left(x+\dfrac{pi}{4}\right)\)
\(-1< =sin\left(x+\dfrac{pi}{4}\right)< =1\)
=>\(-\sqrt{2}< =y< =\sqrt{2}\)
\(y_{min}=-\sqrt{2}\) khi sin(x+pi/4)=-1
=>x+pi/4=-pi/2+k2pi
=>x=-3/4pi+k2pi
\(y_{max}=\sqrt{2}\) khi sin(x+pi/4)=1
=>x+pi/4=pi/2+k2pi
=>x=pi/4+k2pi
b: \(y=sinx\cdot cos\left(\dfrac{pi}{3}\right)+cosx\cdot sin\left(\dfrac{pi}{3}\right)+3\)
\(=sin\left(x+\dfrac{pi}{3}\right)+3\)
-1<=sin(x+pi/3)<=1
=>-1+3<=sin(x+pi/3)+3<=4
=>2<=y<=4
y min=2 khi sin(x+pi/3)=-1
=>x+pi/3=-pi/2+k2pi
=>x=-5/6pi+k2pi
y max=4 khi sin(x+pi/3)=1
=>x+pi/3=pi/2+k2pi
=>x=pi/6+k2pi
c: \(y=2\cdot\left(sin2x\cdot\dfrac{\sqrt{3}}{2}-cos2x\cdot\dfrac{1}{2}\right)\)
\(=2sin\left(2x-\dfrac{pi}{6}\right)\)
-1<=sin(2x-pi/6)<=1
=>-2<=y<=2
y min=-2 khi sin(2x-pi/6)=-1
=>2x-pi/6=-pi/2+k2pi
=>2x=-1/3pi+k2pi
=>x=-1/6pi+kpi
y max=2 khi sin(2x-pi/6)=1
=>2x-pi/6=pi/2+k2pi
=>2x=2/3pi+k2pi
=>x=1/3pi+kpi
2. ĐKXĐ:
a. \(\left\{{}\begin{matrix}cosx\ne0\\2-cosx+tan^2x\ge0\left(luôn-đúng\right)\end{matrix}\right.\)
\(\Rightarrow x\ne\frac{\pi}{2}+k\pi\)
(BPT dưới luôn đúng do \(\left\{{}\begin{matrix}tan^2x\ge0\\2-cosx>0\end{matrix}\right.\) với mọi x)
b. \(sin2x-sinx+3\ge0\)
\(\Leftrightarrow\left(sin2x+2\right)+\left(1-sinx\right)\ge0\)
Do \(\left\{{}\begin{matrix}sin2x\ge-1\\sinx\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}sin2x+2>0\\1-sinx\ge0\end{matrix}\right.\)
\(\Rightarrow\) BPT luôn thỏa mãn hay hàm số xác định trên R
1.
\(\Leftrightarrow f\left(x\right)=sin^4x+cos^4x-2m.sinx.cosx\ge0\) ;\(\forall x\in R\)
\(f\left(x\right)=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-2m.sinx.cosx\)
\(=-\frac{1}{2}sin^22x-m.sin2x+1\)
Đặt \(sin2x=t\Rightarrow\left|t\right|\le1\)
\(f\left(t\right)=-\frac{1}{2}t^2-mt+1\ge0\) ; \(\forall t\in\left[-1;1\right]\)
\(\Leftrightarrow\min\limits_{\left[-1;1\right]}f\left(t\right)\ge0\)
\(a=-\frac{1}{2}< 0\Rightarrow\min\limits f\left(t\right)\) xảy ra tại 1 trong 2 đầu mút
\(f\left(-1\right)=m+\frac{1}{2}\) ; \(f\left(1\right)=\frac{1}{2}-m\)
TH1: \(\left\{{}\begin{matrix}m+\frac{1}{2}\ge\frac{1}{2}-m\\\frac{1}{2}-m\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ge0\\m\le\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow0\le m\le\frac{1}{2}\)
TH2: \(\left\{{}\begin{matrix}\frac{1}{2}-m\ge m+\frac{1}{2}\\m+\frac{1}{2}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\le0\\m\ge-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow-\frac{1}{2}\le m\le\frac{1}{2}\)
D
sao ra z