a, \(y=2sin^2x-cos2x=1-2cos2x\)

Vì \(cos2x\in\left[-1;1\right]\Rightarrow y=2sin^2x-cos2x\in\left[-1;3\right]\)

\(\Rightarrow\left\{{}\begin{matrix}y_{min}=-1\\y_{max}=3\end{matrix}\right.\)

a.

\(y=sinx.cosx+1=\dfrac{1}{2}sin2x+1\)

\(-1\le sin2x\le1\Rightarrow\dfrac{1}{2}\le y\le\dfrac{3}{2}\)

\(y_{min}=\dfrac{1}{2}\) khi \(sin2x=-1\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)

\(y_{max}=\dfrac{3}{2}\) khi \(sin2x=1\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

b.

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

\(-1\le sin\left(x-\dfrac{\pi}{6}\right)\le1\Rightarrow-4\le y\le0\)

\(y_{min}=-4\) khi \(sin\left(x-\dfrac{\pi}{6}\right)=-1\Rightarrow x=-\dfrac{\pi}{3}+k2\pi\)

\(y_{max}=0\) khi \(sin\left(x-\dfrac{\pi}{6}\right)=1\Rightarrow x=\dfrac{2\pi}{3}+k2\pi\)

Chọn B

1.

\(y=\sqrt{5-2\cos ^2x\sin ^2x}=\sqrt{5-\frac{1}{2}(2\cos x\sin x)^2}=\sqrt{5-\frac{1}{2}\sin ^22x}\)

Dễ thấy:

$\sin ^22x\geq 0\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\leq \sqrt{5}$

Vậy $y_{\max}=\sqrt{5}$

$\sin ^22x\leq 1\Rightarrow y=\sqrt{5-\frac{1}{2}\sin ^22x}\geq \sqrt{5-\frac{1}{2}}=\frac{3\sqrt{2}}{2}$

Vậy $y_{\min}=\frac{3\sqrt{2}}{2}$

2.

$y=1+\frac{1}{2}\sin 2x\cos 2x=1+\frac{1}{4}.2\sin 2x\cos 2x$

$=1+\frac{1}{4}\sin 4x$

Vì $-1\leq \sin 4x\leq 1$

$\Rightarrow \frac{5}{4}\leq 1+\frac{1}{4}\sin 4x\leq \frac{3}{4}$

$\Leftrightarrow \frac{5}{4}\leq y\leq \frac{3}{4}$
Vậy $y_{\max}=\frac{5}{4}; y_{\min}=\frac{3}{4}$

3.

$\sin x\geq -1\Rightarrow \sqrt{1+\sin x}\geq 0$

$\Rightarrow y\geq -3$

Vậy $y_{\min}=-3$

$\sin x\leq 1\Rightarrow \sqrt{1+\sin x}\leq \sqrt{2}$

$\Rightarrow y\leq \sqrt{2}-3$
Vậy $y_{\max}=\sqrt{2}-3$

a) Đặt \(t=sinx+cosx\)

\(\Rightarrow t^2=\left(sinx+cosx\right)^2\overset{bunhiacopxki}{\le}\left(1^2+1^2\right)\left(sinx^2+cosx^2\right)=2\\ \Rightarrow-\sqrt{2}\le t\le\sqrt{2}\\ \Rightarrow-\sqrt{2}+1\le y=t+1\le\sqrt{2}+1\)

Vậy \(Min\text{ }y=-\sqrt{2}+1\Leftrightarrow sinx=cosx=\frac{-1}{\sqrt{2}}\Leftrightarrow x=\pm\frac{3\pi}{4}+k2\pi\)

\(Max\text{ }y=\sqrt{2}+1\Leftrightarrow sinx=cosx=\frac{1}{\sqrt{2}}\Leftrightarrow x=\pm\frac{\pi}{4}+k2\pi\)

\(b\text{) }y=cosx-cos2x+4\\ =cosx-\left(2cos^2x-1\right)+4\\ =-2cos^2x+cosx+5\)

\(\Rightarrow Min\text{ }y=2\Leftrightarrow cosx=-1\)

\(\Rightarrow Max\text{ }y=\frac{41}{8}\Leftrightarrow cosx=\frac{1}{4}\)

\(\text{c) }y=2sin^2x+4\sqrt{3}sinx\cdot cosx+6cos^2x+1\\ =\left(1-cos2x\right)+2\sqrt{3}sin2x+3\left(cos2x+1\right)+1\\ =2cos2x+2\sqrt{3}sin2x+5\)

Đặt \(t=2cos2x+2\sqrt{3}sin2x\)

\(\Rightarrow t^2\le\left[2^2+\left(2\sqrt{3}\right)^2\right]\left(cos^22x+sin^22x\right)=16\\ \Rightarrow-4\le t\le4\\ \Rightarrow1\le y\le9\\ \)

Vậy \(Min\text{ }y=1\Leftrightarrow sin2x=-\frac{1}{2}\)

\(Max\text{ }y=9\Leftrightarrow sin2x=\frac{1}{2}\)

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

Vì \(cos\left(x+\dfrac{\pi}{6}\right)\in\left[-1;1\right]\Rightarrow y=\sqrt{3}cosx-sinx\in\left[-2;2\right]\)

\(\Rightarrow y_{min}=-2\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=-1\Leftrightarrow x+\dfrac{\pi}{6}=\pi+k2\pi\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)

\(y_{max}=2\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=1\Leftrightarrow x+\dfrac{\pi}{6}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+k2\pi\)

Đặt \(sinx=t\left(t\in\left[-1;1\right]\right)\)

\(y=\left|sinx+cos2x\right|=\left|2sin^2x-sinx-1\right|\)

\(\Leftrightarrow y=\left|f\left(t\right)\right|=\left|2t^2-t-1\right|\)

\(f\left(-1\right)=2\Rightarrow y=2\)

\(f\left(1\right)=0\Rightarrow y=0\)

\(f\left(\dfrac{1}{4}\right)=-\dfrac{9}{8}\Rightarrow y=\dfrac{9}{8}\)

\(\Rightarrow y_{min}=0;y_{max}=2\)