2.

\(0\le\left|sinx\right|\le1\Rightarrow1\le y\le3\)

Min và max lần lượt là 3 và 1

3.

\(cos\left(x-\frac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)

8.

\(y=\frac{1}{2}+\frac{1}{2}cos2x+2cos2x=\frac{1}{2}+\frac{5}{2}cos2x\le\frac{1}{2}+\frac{5}{2}.1=3\)

15.

Nó đi qua vô số điểm nên ko có 4 đáp án để chọn thì ko ai có thể trả lời câu này cho bạn cả

18.

\(y=\frac{sinx+2cosx+1}{sinx+cosx+2}\Leftrightarrow y.sinx+y.cosx+2y=sinx+2cosx+1\)

\(\Leftrightarrow\left(y-1\right)sinx+\left(y-2\right)cosx=1-2y\)

\(\left(y-1\right)^2+\left(y-2\right)^2\ge\left(1-2y\right)^2\)

\(\Leftrightarrow2y^2+2y-4\le0\Rightarrow-2\le y\le1\)

\(\Rightarrow y_{max}=1\)

8.

\(y=cos^2x+2\left(2cos^2x-1\right)=5cos^2x-2\)

Do \(0\le cos^2x\le1\Rightarrow-2\le y\le3\)

\(y_{min}=-2;y_{max}=3\)

10.

\(y=2-\left(cosx+1\right)^2\le2\)

\(y_{max}=2\)

14.

Hàm tuần hoàn với chu kì \(T=\pi\)

\(y=1-2\sin^2x-\sin x+3=-2\sin^2x-\sin x+4\)

\(\sin x=t;t\in\left[-1;1\right]\)

Xét hàm f(t) trên [-1;1]

\(f\left(-1\right)=-2+1+4=3\)

\(f\left(1\right)=-2-1+4=1\)

\(f\left(-\frac{1}{4}\right)=-2.\frac{1}{16}+\frac{1}{4}+4=\frac{33}{8}\)

\(\Rightarrow\left\{{}\begin{matrix}y_{max}=\frac{33}{8};"="\Leftrightarrow\sin x=-\frac{1}{4}\Rightarrow x=...\\y_{min}=1;"="\Leftrightarrow\sin x=1\end{matrix}\right.\)

8.

\(y=\left(cosx+1\right)^2-1\ge-1\Rightarrow y_{min}=-1\)

\(y=\left(cosx-1\right)\left(cosx+3\right)+3\le3\Rightarrow y_{max}=3\)

10.

\(y=2-\left(cosx+1\right)^2\le2\Rightarrow y_{max}=2\)

14.

Hàm tuần hoàn với chu kì \(T=\pi\)

15.

Đáp án A đúng

20.

\(-1\le sin\left(\frac{x}{2}+\frac{\pi}{7}\right)\le1\Rightarrow-5\le y\le-1\)

\(y_{max}=-1\) ; \(y_{min}=-5\)

-1<=sin x<=1

=>-1<=-sin x<=1

=>2<=-sin x+3<=4

=>4<=(3-sin x)^2<=16

=>5<=y<=17

y min=5 khi 3-sin x=2

=>sin x=1

=>x=pi/2+k2pi

y max=17 khi 3-sin x=4

=>sin x=-1

=>x=-pi/2+k2pi

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

Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)

2.

$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

Vì: $0\leq \sin ^22x\leq 1$

$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$

Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$

3.

$0\leq |\sin x|\leq 1$

$\Rightarrow 3\geq 3-2|\sin x|\geq 1$

Vậy $y_{\min}=1; y_{\max}=3$