Chọn C

\(y=\sqrt{3}cosx-sinx=2\left(\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx\right)=2cos\left(x+\dfrac{\pi}{6}\right)\)

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`.

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 \(a=-5\)

\(sin\in\left[-1;1\right]\Rightarrow y=3-4sinx\in\left[-1;7\right]\)

\(\Rightarrow y_{min}=-1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=7\Leftrightarrow sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)