TÌM GTLN GTNN:
a. y=cos x - \(\sqrt{3}\)sin x
b. y= sin2x-cos2x+1
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20 tháng 8 2023
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
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 7 2021
24.
\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)
\(y_{max}=4\)
26.
\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)
Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)
\(y_{max}=\sqrt{2}\)
b.
\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2020
a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)
\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)
Do \(-1\le sin\left(2x+\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le1\)
b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)
\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)
\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)
c/ \(y=1-sin2x+2cos2x+\frac{3}{2}sin2x=\frac{1}{2}sin2x+2cos2x+1\)
\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)
\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)
LT
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2021
\(y=2\left(\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x\right)=2sin\left(2x+\dfrac{\pi}{3}\right)\)
\(-1\le sin\left(2x+\dfrac{\pi}{3}\right)\le1\Rightarrow-2\le y\le2\)
\(y_{min}=-2\) khi \(sin\left(2x+\dfrac{\pi}{3}\right)=-1\Rightarrow x=-\dfrac{5\pi}{12}+k\pi\)
\(y_{max}=2\) khi \(sin\left(2x+\dfrac{\pi}{3}\right)=1\Rightarrow x=\dfrac{\pi}{12}+k\pi\)
HA
1
20 tháng 8 2023
a: -1<=cos2x<=1
=>3>=-3cos2x>=-3
=>7>=-3cos2x+4>=1
=>7>=y>=1
\(y_{min}=1\) khi \(cos2x=1\)
=>2x=k2pi
=>x=kpi
\(y_{max}=-1\) khi cos2x=-1
=>2x=pi+k2pi
=>x=pi/2+kpi
b: \(0< =sin^2x< =1\)
=>\(3< =sin^2x+3< =4\)
=>3<=y<=4
y min=3 khi sin^2x=0
=>sinx=0
=>x=kpi
y max=4 khi sin^2x=1
=>cos^2x=0
=>x=pi/2+kpi
c: \(y=sin2x+3\)
-1<=sin2x<=1
=>-1+3<=sin2x+3<=1+3
=>2<=y<=4
\(y_{min}=2\) khi sin 2x=-1
=>2x=-pi/2+k2pi
=>x=-pi/4+kpi
y max=4 khi sin2x=1
=>2x=pi/2+k2pi
=>x=pi/4+kpi
A
1
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2021
a.
\(-1\le sin\left(1-x^2\right)\le1\)
\(\Rightarrow y_{min}=-1\) khi \(1-x^2=-\dfrac{\pi}{2}+k2\pi\) \(\Rightarrow x^2=\dfrac{\pi}{2}+1+k2\pi\) (\(k\ge0\))
\(y_{max}=1\) khi \(1-x^2=\dfrac{\pi}{2}+k2\pi\Rightarrow x^2=1-\dfrac{\pi}{2}+k2\pi\) (\(k\ge1\))
b.
Đặt \(\sqrt{2-x^2}=t\Rightarrow t\in\left[0;\sqrt{2}\right]\subset\left[0;\pi\right]\)
\(y=cost\) nghịch biến trên \(\left[0;\pi\right]\Rightarrow\) nghịch biến trên \(\left[0;\sqrt{2}\right]\)
\(\Rightarrow y_{max}=y\left(0\right)=cos0=1\) khi \(x^2=2\Rightarrow x=\pm\sqrt{2}\)
\(y_{min}=y\left(\sqrt{2}\right)=cos\sqrt{2}\) khi \(x=0\)
a) Ta có:
\(y=2\left(\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)=2sin\left(\dfrac{\pi}{6}-x\right)\)
\(\Rightarrow-2\le y\le2\) (Do \(-1\le sin\alpha\le1\))
Vậy min y = -2 , max y = 2