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1.
Các hàm \(sinx;sin\frac{x}{2};sin\frac{x}{3};...;sin\frac{x}{10}\) có chu kì lần lượt là \(2\pi;4\pi;6\pi;...;20\pi\)
\(\Rightarrow\) Chu kì của hàm đã cho là \(BCNN\left(2\pi;4\pi;...;20\pi\right)=15120\pi\)
2.
a.
\(y=cos^22x+3cos2x+3\)
\(y=\left(cos2x+1\right)\left(cos2x+2\right)+1\ge1\Rightarrow y_{min}=1\) khi \(cos2x=-1\)
\(y=\left(cos2x-1\right)\left(cos2x+4\right)+7\le7\Rightarrow y_{max}=7\) khi \(cos2x=1\)
b.
Đặt \(a=4sinx-3cosx\Rightarrow a^2\le\left(4^2+\left(-3\right)^2\right)\left(sin^2x+cos^2x\right)=25\)
\(\Rightarrow-5\le a\le5\)
\(y=a^2-4a+1\) với \(a\in\left[-5;5\right]\)
\(y=\left(a-2\right)^2-3\ge-3\Rightarrow y_{min}=-3\) khi \(a=2\)
\(y=\left(a-9\right)\left(a+5\right)+46\le46\Rightarrow y_{max}=46\) khi \(a=-5\)
\(y=\sqrt{3}cosx-sinx=2\left(\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx\right)=2cos\left(x+\dfrac{\pi}{6}\right)\)
Vì \(cos\left(x+\dfrac{\pi}{6}\right)\in\left[-1;1\right]\Rightarrow y=\sqrt{3}cosx-sinx\in\left[-2;2\right]\)
\(\Rightarrow y_{min}=-2\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=-1\Leftrightarrow x+\dfrac{\pi}{6}=\pi+k2\pi\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)
\(y_{max}=2\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=1\Leftrightarrow x+\dfrac{\pi}{6}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+k2\pi\)
a.\(-1\le cosx\le1\Rightarrow-4\le y=3cosx-1\le2\)
b.-1 \(\le sinx\le1\)\(\Rightarrow3\le y=5+2sinx\le7\)
c.\(\sqrt{3-1}\le\sqrt{3+cos2x}\le\sqrt{3+1}\Rightarrow\sqrt{2}\le y\le2\)
d.\(y=\sqrt{5sinx-1}+2\le\sqrt{5.1-1}+2=4\)
\(y=\sqrt{5sinx-1}+2\ge2\) . " = " \(\Leftrightarrow sinx=\dfrac{1}{5}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\dfrac{1}{5}\right)+2k\pi\\x=\pi-arcsin\left(\dfrac{1}{5}\right)+2k\pi\end{matrix}\right.\) ( k thuộc Z )
`y=1/2 sinx +3cosx`
`-\sqrt( (1/2)^2+3^2) <= y <= \sqrt( (1/2)^2+3^2)`
`<=> -\sqrt37/2 <= y <= \sqrt37/2`
`=> y_(min) = -\sqrt37/2`
`y_(max) = \sqrt37/2`.
a/
\(y=\sqrt{5-sin2x}\)
Do \(-1\le sin2x\le1\Rightarrow2\le y\le\sqrt{6}\)
\(y_{min}=2\) khi \(sin2x=1\)
\(y_{max}=\sqrt{6}\) khi \(sin2x=-1\)
b/
\(y=cos^2x-3cosx-4+8=\left(cosx+1\right)\left(cosx-4\right)+8\le8\)
\(y_{max}=8\) khi \(cosx=-1\)
\(y=cos^2x-3cosx+2+2=\left(1-cosx\right)\left(2-cosx\right)+2\ge2\)
\(y_{min}=2\) khi \(cosx=1\)
a.
\(0\le sin^2x\le1\Rightarrow\frac{4}{3}\le y\le4\)
\(y_{max}=4\) khi \(sinx=0\)
\(y_{min}=\frac{4}{3}\) khi \(sin^2x=1\)
b.
Đặt \(4sinx-3cosx=5\left(\frac{4}{5}sinx-\frac{3}{5}cosx\right)=5sin\left(x-a\right)=t\)
\(\Rightarrow-5\le t\le5\)
\(\Rightarrow y=t^2-4t+1=\left(t-2\right)^2-3\ge-3\)
\(y_{min}=-3\) khi \(t=2\)
\(y=t^2-4t-45+46=\left(t-9\right)\left(t+5\right)+46\le46\)
\(y_{max}=46\) khi \(t=-5\)
a) Ta có : -\(\sqrt{a^2+b^2}< =asinx+bcosx< =\sqrt{a^2+b^2}\)
=> \(-\sqrt{12^2+\left(-5\right)^2}< =y< =\sqrt{12^2+\left(-5\right)^2}\)
<=> \(-\sqrt{13}< =y< =\sqrt{13}\)
Vậy min=\(-\sqrt{13}\) ,max=\(\sqrt{13}\)
b) \(-\sqrt{9+16}< =3cosx-4sinx< =\sqrt{9+16}\)
<=> -5 <=3cos x -4sinx <= 5
<=> 0<= y <= 10
Vậy min=0 max=10