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\(t=\sin x;t\in\left[-1;1\right]\)
Xét hàm f(t) trên [-1;1]
\(f\left(-1\right)=2+3+1=6\)
\(f\left(1\right)=2-3+1=0\)
\(f\left(\frac{3}{4}\right)=2.\frac{9}{16}-3.\frac{3}{4}+1=-\frac{1}{8}\)
\(\Rightarrow\left\{{}\begin{matrix}y_{max}=6;"="\Leftrightarrow\sin x=-1\\y_{min}=-\frac{1}{8};"="\Leftrightarrow\sin x=\frac{3}{4}\end{matrix}\right.\)
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: \(-1\le\sin\left(2x+a\right)\le1\)
=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)
=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)
=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)
=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)
\(y=1-cos2x+2sin2x+6=2sin2x-cos2x+7\)
\(y=\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)+7\)
Đặt \(\dfrac{2}{\sqrt{5}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\)
\(y=\sqrt{5}sin\left(2x-a\right)+7\)
\(\Rightarrow-\sqrt{5}+7\le y\le\sqrt{5}+7\)
\(\text{c) }y=2sin^2x+4\sqrt{3}sinx\cdot cosx+6cos^2x+1\\ =\left(1-cos2x\right)+2\sqrt{3}sin2x+3\left(cos2x+1\right)+1\\ =2cos2x+2\sqrt{3}sin2x+5\)
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 \(Min\text{ }y=1\Leftrightarrow sin2x=-\frac{1}{2}\)
\(Max\text{ }y=9\Leftrightarrow sin2x=\frac{1}{2}\)
\(b\text{) }y=cosx-cos2x+4\\ =cosx-\left(2cos^2x-1\right)+4\\ =-2cos^2x+cosx+5\)
y cosx -1 1/4 1 41/8 2 4
\(\Rightarrow Min\text{ }y=2\Leftrightarrow cosx=-1\)
\(\Rightarrow Max\text{ }y=\frac{41}{8}\Leftrightarrow cosx=\frac{1}{4}\)
a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)
Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).
\(\Rightarrow y=f\left(t\right)=3-t^2\)
\(\Rightarrow y_{min}=minf\left(t\right)=2\)
\(y_{max}=maxf\left(t\right)=3\)
Đặt \(sinx=t\left(t\in\left[-1;1\right]\right)\).
\(\Rightarrow y=f\left(t\right)=-2t^2+3t-1\)
\(\Rightarrow y_{min}=min\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{3}{4}\right)\right\}=f\left(-1\right)=-6\)
\(y_{max}=max\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{3}{4}\right)\right\}=f\left(\dfrac{3}{4}\right)=\dfrac{1}{8}\)