Chọn B

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

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

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

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

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

21.
a) `2sin(x-30^@)-1=0`
`<=>sin(x-30^@)=1/2`
`<=> sin(x-30^@)=sin30^@`
`<=>[(x-30^@=30^@+k360^@),(x-30^@=180^@-30^@+k360^@):}`
`<=> [(x=60^@+k360^@),(x=180^@+k360^@):}`
b) `5sin^2x+3cosx+3=0`
`<=>5(1-cos^2x)+3cosx+3=0`
`<=>-5cos^2x+3cosx+8=0`
`<=>(cosx+1)(cosx=8/5)=0`
`<=>[(cosx=-1),(cosx=8/5\ (VN)):}`
`<=>x=180^@+k360^@`
22.
`-1<=sin2x<=1`
`<=>2<=3+sin2x<=4`
`=> y_(min)=2 ; y_(max)=4`

Xét tính chẵn lẻ:

a) TXĐ: D = R \ {π/2 + kπ| k nguyên}

Với mọi x thuộc D ta có (-x) thuộc D và

\(f\left(-x\right)=\frac{3\tan^3\left(-x\right)-5\sin\left(-x\right)}{2+\cos\left(-x\right)}=-\frac{3\tan^3x-5\sin x}{2+\cos x}=-f\left(x\right)\)

Vậy hàm đã cho là hàm lẻ

b) TXĐ: D = R \ \(\left\{\pm\sqrt{2};\pm1\right\}\)

Với mọi x thuộc D ta có (-x) thuộc D và

\(f\left(-x\right)=\frac{\sin\left(-x\right)}{\left(-x\right)^4-3\left(-x\right)^2+2}=-\frac{\sin x}{x^4-3x^2+2}=-f\left(x\right)\)

Vậy hàm đã cho là hàm lẻ

Tìm GTLN, GTNN:

TXĐ: D = R

a)  Ta có (\(\left(\sin x+\cos x\right)^2=1+\sin2x\)

Với mọi x thuộc D ta có\(-1\le\sin2x\le1\Leftrightarrow0\le1+\sin2x\le2\Leftrightarrow0\le\left(\sin x+\cos x\right)^2\le2\)

\(\Leftrightarrow0\le\left|\sin x+\cos x\right|\le\sqrt{2}\Leftrightarrow-\sqrt{2}\le\sin x+\cos x\le\sqrt{2}\)

Vậy  \(Min_{f\left(x\right)}=-\sqrt{2}\) khi \(\sin2x=-1\Leftrightarrow2x=-\frac{\pi}{2}+k2\pi\Leftrightarrow x=-\frac{\pi}{4}+k\pi\)

\(Max_{f\left(x\right)}=\sqrt{2}\) khi\(\sin2x=1\Leftrightarrow x=\frac{\pi}{4}+k\pi\)

b) Với mọi x thuộc D ta có: 

\(-1\le\cos x\le1\Leftrightarrow-2\le2\cos x\le2\Leftrightarrow1\le2\cos x+3\le5\)

\(\Leftrightarrow1\le\sqrt{2\cos x+3}\le\sqrt{5}\Leftrightarrow5\le\sqrt{2\cos x+3}+4\le\sqrt{5}+4\)

Vậy\(Min_{f\left(x\right)}=5\)  khi \(\cos x=-1\Leftrightarrow x=\pi+k2\pi\)

\(Max_{f\left(x\right)}=\sqrt{5}+4\)  khi \(\cos x=1\Leftrightarrow x=k2\pi\)

c) \(y=\sin^4x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x\cos^2x\)\(=1-\frac{1}{2}\left(2\sin x\cos x\right)^2=1-\frac{1}{2}\sin^22x\)

Với mọi x thuộc D ta có: \(0\le\sin^22x\le1\Leftrightarrow-\frac{1}{2}\le-\frac{1}{2}\sin^22x\le0\Leftrightarrow\frac{1}{2}\le1-\frac{1}{2}\sin^22x\le1\)

Đến đây bạn tự xét dấu '=' xảy ra khi nào nha :p

1.

\(3cos2x-7=2m\)

\(\Leftrightarrow cos2x=\dfrac{2m-7}{3}\)

Phương trình đã cho có nghiệm khi:

\(-1\le\dfrac{2m-7}{3}\le1\)

\(\Leftrightarrow2\le m\le5\)

2.

\(2cos^2x-\sqrt{3}cosx=0\)

\(\Leftrightarrow cosx\left(2cosx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pm\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\) Có 4 nghiệm \(\dfrac{\pi}{2};\dfrac{3\pi}{2};\dfrac{\pi}{6};\dfrac{11\pi}{6}\) thuộc đoạn \(\left[0;2\pi\right]\)

\(y\le\sqrt{2\left(6-2x+3+2x\right)}=3\sqrt{2}\)

\(y_{max}=3\sqrt{2}\) khi \(x=\dfrac{3}{4}\)

\(y\ge\sqrt{6-2x+3+2x}=3\)

\(y_{min}=3\) khi \(\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Đáp án đúng : D

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

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

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

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

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

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

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

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