Chọn C

Đáp án C

\(y=sin^3x+2sin^2x+sinx-2\)

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

 pt \(\Leftrightarrow\)\(y=t^3+2t^2+t-2\)

\(y'=3t^2+4t+1\)

\(y'=0\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=-\dfrac{1}{3}\end{matrix}\right.\)

vậy GTLN của y = 2 với t=1 \(\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

GTNN của y=-58/27  với \(t=-\dfrac{1}{3}\Leftrightarrow sinx=-\dfrac{1}{3}\Leftrightarrow x=sin^{-1}\left(-\dfrac{1}{3}\right)\)

`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.

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

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.

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

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

\(y_{min}=\dfrac{1}{2}\) 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

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

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

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

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