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a.
\(-1\le sinx\le1\Rightarrow-7\le y\le-3\)
\(y_{min}=-7\) khi \(sinx=-1\)
\(y_{max}=-3\) khi \(sinx=1\)
b.
\(-1\le cos\left(x+\frac{\pi}{3}\right)\le1\Rightarrow1\le y\le5\)
\(y_{min}=1\) khi \(cos\left(x+\frac{\pi}{3}\right)=-1\)
\(y_{max}=5\) khi \(cos\left(x+\frac{\pi}{3}\right)=1\)
c.
\(0\le1-cosx\le2\Rightarrow-5\le y\le3\sqrt{2}-5\)
\(y_{min}=-5\) khi \(cosx=1\)
\(y_{max}=3\sqrt{2}-5\) khi \(cosx=-1\)
d.
ĐKXĐ: \(0\le sinx\Rightarrow0\le sinx\le1\Rightarrow1\le y\le3\)
\(y_{min}=1\) khi \(sinx=0\)
\(y_{max}=3\) khi \(sinx=1\)
1. Do \(\cos x+2>0\forall x\in R\) \(\Rightarrow\) Hàm số xác định \(\forall x\in R\)
\(y=\dfrac{\sin x+1}{\cos x+2}\)
\(\Leftrightarrow\)\(y\cos x-\sin x=1-2y\)
pt có nghiệm \(\Leftrightarrow y^2+\left(-1\right)^2\ge\left(1-2y\right)^2\)
\(\Leftrightarrow3y^2-4y\le0\)
\(\Leftrightarrow0\le y\le\dfrac{4}{3}\)
2. \(y=\dfrac{\cos x+2\sin x+3}{2\cos x-\sin x+4}\)
\(\Leftrightarrow\left(2y-1\right)\cos x-\left(y+2\right)\sin x=3-4y\)
pt có nghiệm \(\Leftrightarrow\left(2y-1\right)^2+\left(y+2\right)^2\ge\left(3-4y\right)^2\)
\(\Leftrightarrow11y^2-24y+4\le0\)
\(\Leftrightarrow\dfrac{2}{11}\le y\le2\)
kiểm tra giúp mình xem có sai sót gì không...
Khi cho A td KOH thu được ancol đồng đẳng. => Các ancol là no đơn chức mạch hở.
Gọi CT các este: \(C_mH_{2m+1}COOC_{m'}H_{2m'+1};C_nH_{2n-1}COOC_{n'}H_{2n'-1};C_qH_{2q}\left(COOC_{q'}H_{2q'}\right)_2\)
TN2: Đốt hỗn hợp 3 muối.
Đặt \(n_{K_2CO_3}=x;n_{H_2O}=y\left(mol\right)\)
\(BTNT.K\Rightarrow n_{COOK^-}=2n_{K_2CO_3}=2x\left(mol\right)\\ BTNT.O\Rightarrow2n_{COOK^-}+2n_{O_2}=3n_{K_2CO_3}+2n_{CO_2}+n_{H_2O}\\ \Rightarrow x-y=0,3\\ BTKL\Rightarrow m_{M'}+m_{O_2}=m_{K_2CO_3}+m_{CO_2}+m_{H_2O}\\ \Rightarrow138x+18y=99,9\\ \Rightarrow\left\{{}\begin{matrix}x=0,675\\y=0,375\end{matrix}\right.\)
H2 muối gồm: \(C_mH_{2m+1}COOK\text{ }a\text{ }mol;C_nH_{2n-1}COOK\text{ }b\text{ }mol;C_qH_{2q}\left(COOK\right)_2\text{ }c\text{ }mol\)
\(\Rightarrow n_A=a+b+c=0,85\\ BTNT.C\Rightarrow\left(m+1\right)a+\left(n+1\right)b+\left(q+2\right)c=n_{K_2CO_3}+n_{CO_2}=1,75\\ \Rightarrow ma+nb+qc=0,4\\ BTNT.K\Rightarrow a+b+2c=1,35\\ BTNT.H\Rightarrow\left(2m+1\right)a+\left(2n-1\right)b+2qc=2n_{H_2O}=0,75\\ \Rightarrow a-b=-0,05\\ \Rightarrow\left\{{}\begin{matrix}a=0,15\\b=0,2\\c=0,5\end{matrix}\right.\\ \Rightarrow0,15m+0,2n+0,5q=0,4\)
Do \(m;q\ge0\Rightarrow n\le\frac{0,4}{0,2}=2\)
Mà \(n\ge2\Rightarrow n=2\Rightarrow m=q=0\)
\(\text{c) }y=2sin^2x+4\sqrt{3}sinx\cdot cosx+6cos^2x+1\\ =\left(1-cos2x\right)+2\sqrt{3}sin2x+3\left(cos2x+1\right)+1\\ =2cos2x+2\sqrt{3}sin2x+5\)
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 \(Min\text{ }y=1\Leftrightarrow sin2x=-\frac{1}{2}\)
\(Max\text{ }y=9\Leftrightarrow sin2x=\frac{1}{2}\)
e/
\(y=5sinx+6cosx-7\)
\(=\sqrt{61}\left(\frac{5}{\sqrt{61}}sinx+\frac{6}{\sqrt{61}}cosx\right)-7\)
\(=\sqrt{61}\left(sinx.cosa+cosx.sina\right)-7\) (với \(a\in\left(0;\pi\right)\) sao cho \(cosa=\frac{5}{\sqrt{61}}\))
\(=\sqrt{61}.sin\left(x+a\right)-7\)
Do \(-1\le sin\left(x+a\right)\le1\Rightarrow7-\sqrt{61}\le y\le7+\sqrt{61}\)
\(y_{min}=7-\sqrt{61}\) khi \(sin\left(x+a\right)=-1\)
\(y_{max}=7+\sqrt{61}\) khi \(sin\left(x+a\right)=1\)
f/
\(y=2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)+3\)
\(=2sin\left(x+\frac{\pi}{3}\right)+3\)
\(\Rightarrow1\le y\le5\)
\(y_{min}=1\) khi \(sin\left(x+\frac{\pi}{3}\right)=-1\)
\(y_{max}=5\) khi \(x+\frac{\pi}{3}=1\)
c/
\(y=2\left(1-cos2x\right)+sin2x+cos2x\)
\(=sin2x-cos2x+2=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)+2\)
Do \(-1\le sin\left(2x-\frac{\pi}{4}\right)\le1\)
\(\Rightarrow2-\sqrt{2}\le y\le2+\sqrt{2}\)
\(y_{min}=2-\sqrt{2}\) khi \(sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(y_{max}=2+\sqrt{2}\) khi \(sin\left(2x+\frac{\pi}{4}\right)=1\)
d/
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(=1-3sin^2x.cos^2x\)
\(=1-\frac{3}{4}sin^22x\)
Mà \(0\le sin^22x\le1\Rightarrow\frac{1}{4}\le y\le1\)
\(y_{min}=\frac{1}{4}\) khi \(sin^22x=1\)
\(y_{max}=1\) khi \(sin2x=0\)
a) làm tương tự 2 bài mk đã giải nha.
b) \(y=2\cos^2x-2\sqrt{3}\sin x\cos x+1\)
\(=1-\left(\cos2x+\sqrt{3}\sin2x\right)\)
Lại có \(-2\le\cos2x+\sqrt{3}\sin2x\le2\) \(\Rightarrow-1\le y\le3\)
c) Vì \(\left\{{}\begin{matrix}0\le\sqrt[4]{\sin x}\le1\\0\le\sqrt{\cos x}\le1\end{matrix}\right.\)
Do đó \(-1\le y\le1\)