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36.
\(sin^2x-cos^2x\ne0\Leftrightarrow cos2x\ne0\)
\(\Leftrightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
37.
\(cos3x\ne cosx\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{k\pi}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\)
38.
\(\left\{{}\begin{matrix}x\ge0\\sin\pi x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\pi x\ne k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne k\end{matrix}\right.\)
39.
\(\left\{{}\begin{matrix}cos\left(x-\frac{\pi}{3}\right)\ne0\\tan\left(x-\frac{\pi}{3}\right)\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\\x-\frac{\pi}{3}\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{5\pi}{6}+k\pi\\x\ne-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
33.
\(\left\{{}\begin{matrix}cosx\ne0\\cos\frac{x}{2}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)
34.
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\cotx\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\frac{\pi}{4}+k\pi\end{matrix}\right.\)
35.
\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\)
\(\Leftrightarrow x\ne k\pi\)
23.
\(tan^2x\ge0\Rightarrow y\le2\)
\(y_{max}=2\) khi \(tanx=0\)
\(y_{min}\) không tồn tại
24.
\(-1\le cosx\le1\Rightarrow0< 1+cosx\le2\)
\(\Rightarrow y\ge\frac{1}{2}\)
\(y_{min}=\frac{1}{2}\) khi \(cosx=1\)
\(y_{max}\) ko tồn tại
19.
\(y=\sqrt{5-\frac{1}{2}\left(2sinxcosx\right)^2}=\sqrt{5-\frac{1}{2}sin^22x}\)
\(0\le sin^22x\le1\Rightarrow\frac{3\sqrt{2}}{2}\le y\le\sqrt{5}\)
\(y_{min}=\frac{3\sqrt{2}}{2}\) khi \(sin^22x=1\)
\(y_{max}=\sqrt{5}\) khi \(sin^22x=0\)
21.
\(y=2sin^2x-\left(1-2sin^2x\right)=4sin^2x-1\)
\(0\le sin^2x\le1\Rightarrow-1\le y\le3\)
\(y_{min}=-1\) khi \(sin^2x=0\)
\(y_{max}=3\) khi \(sin^2x=1\)
1/ \(y'=\frac{\sqrt{9-x^2}-x\left(\sqrt{9-x^2}\right)'}{9-x^2}=\frac{\sqrt{9-x^2}+\frac{x^2}{\sqrt{9-x^2}}}{9-x^2}=\frac{9}{\left(9-x^2\right)\sqrt{9-x^2}}\)
2/ \(y'=\frac{\left(\sqrt{x^2+x+3}\right)'.\left(2x+1\right)-2\sqrt{x^2+x+3}}{\left(2x+1\right)^2}=\frac{\frac{\left(2x+1\right)}{2\sqrt{x^2+x+3}}.\left(2x+1\right)-2\sqrt{x^2+x+3}}{\left(2x+1\right)^2}\)
\(=\frac{\left(2x+1\right)^2-4\left(x^2+x+3\right)}{2\left(2x+1\right)^2\sqrt{x^2+x+3}}=\frac{-11}{2\left(2x+1\right)^2\sqrt{x^2+x+3}}\)
3/ \(y'=3\left(1+tan^23x\right)=3+3tan^23x\)
4/ \(y'=\frac{\left(cosx-sinx\right)\left(sinx-cosx\right)-\left(cosx+sinx\right)\left(sinx+cosx\right)}{\left(sinx-cosx\right)^2}\)
\(=-\frac{\left(sinx-cosx\right)^2+\left(sinx+cosx\right)^2}{\left(sinx-cosx\right)^2}=-\frac{sin^2x+cos^2x-2sinxcosx+sin^2x+cos^2x+2sinxcosx}{sin^2x+cos^2x-2sinxcosx}\)
\(=\frac{-2}{1-sin2x}\)
5/ \(y'=4x+\frac{1}{2\sqrt{x}}-\frac{\pi}{2}cos\left(\frac{\pi x}{2}\right)\)
6/ \(y'=3sin^2\left(1-3x\right).\left(sin\left(1-3x\right)\right)'=3sin^2\left(1-3x\right).cos\left(1-3x\right).\left(1-3x\right)'\)
\(=-9sin^2\left(1-3x\right).cos\left(1-3x\right)\)
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 \(sin2x=0\)
1.
Hàm tuần hoàn với chu kì \(2\pi\) nên ta chỉ cần xét trên đoạn \(\left[0;2\pi\right]\)
\(y'=\frac{-4}{\left(cosx-2\right)^2}.sinx=0\Leftrightarrow x=k\pi\)
\(\Rightarrow x=\left\{0;\pi;2\pi\right\}\)
\(y\left(0\right)=-3\) ; \(y\left(\pi\right)=\frac{1}{3}\) ; \(y\left(2\pi\right)=-3\)
\(\Rightarrow\left\{{}\begin{matrix}M=\frac{1}{3}\\m=-3\end{matrix}\right.\)
\(\Rightarrow9M+m=0\)
2.
\(\Leftrightarrow y.cosx+y.sinx+2y=2k.cosx+k+1\)
\(\Leftrightarrow y.sinx+\left(y-2k\right)cosx=k+1-2y\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\Rightarrow y^2+\left(y-2k\right)^2\ge\left(k+1-2y\right)^2\)
\(\Leftrightarrow2y^2-4k.y+4k^2\ge4y^2-4\left(k+1\right)y+\left(k+1\right)^2\)
\(\Leftrightarrow2y^2-4y-3k^2+2k+1\le0\)
\(\Leftrightarrow2\left(y-1\right)^2\le3k^2-2k+1\)
\(\Leftrightarrow y\le\sqrt{\frac{3k^2-2k+1}{2}}+1\)
\(y_{max}=f\left(k\right)=\frac{1}{\sqrt{2}}\sqrt{3k^2-2k+1}+1\)
\(f\left(k\right)=\frac{1}{\sqrt{2}}\sqrt{3\left(k-\frac{1}{3}\right)^2+\frac{2}{3}}+1\ge\frac{1}{\sqrt{3}}+1\)
Dấu "=" xảy ra khi và chỉ khi \(k=\frac{1}{3}\)
Đáp án A
1.
ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow sinx.cosx\ne0\)
\(\Leftrightarrow sin2x\ne0\Leftrightarrow x\ne\frac{k\pi}{2}\)
2.
ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\\sin3x\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{\pi}{6}+\frac{k2\pi}{3}\end{matrix}\right.\)
3.
Do \(sin6x< 2\) với mọi x nên hàm số xác định trên R
4.
Hàm số xác định khi và chỉ khi \(cosx\ge1\Leftrightarrow cosx=1\)
\(\Leftrightarrow x=k2\pi\)
Đáp án A
Do \(-1\le cosx\le1\Rightarrow\left\{{}\begin{matrix}2+cosx>0\\2-cosx>0\end{matrix}\right.\)
\(\Rightarrow\frac{2+cosx}{2-cosx}>0\) \(\forall x\in R\)