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\(lim\left(\sqrt[3]{n^3+4}-\sqrt[3]{n^3-1}\right)\)
\(=lim\left(\sqrt[3]{1+\dfrac{4}{n^3}}-\sqrt[3]{1-\dfrac{1}{n^3}}\right)=\sqrt[3]{1}-\sqrt[3]{1}=0\)
20: \(\lim\limits_{x\rightarrow+\infty}x^3+2x-1=\lim\limits_{x\rightarrow+\infty}\left[x^3\left(1+\dfrac{2}{x^2}-\dfrac{1}{x^3}\right)\right]\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow+\infty}x^3=+\infty\\\lim\limits_{x\rightarrow+\infty}1+\dfrac{2}{x^2}-\dfrac{1}{x^3}=1\end{matrix}\right.\)
12.
\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\le\sqrt[]{2}\)
\(\Rightarrow M=\sqrt{2}\)
13.
Pt có nghiệm khi:
\(5^2+m^2\ge\left(m+1\right)^2\)
\(\Leftrightarrow2m\le24\)
\(\Rightarrow m\le12\)
14.
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=k2\pi\)
15.
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(3\right)+k\pi\end{matrix}\right.\)
Đáp án A
16.
\(\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
\(\left[{}\begin{matrix}2\pi\le\dfrac{\pi}{3}+k2\pi\le2018\pi\\2\pi\le\pi+k2\pi\le2018\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1\le k\le1008\\1\le k\le1008\end{matrix}\right.\)
Có \(1008+1008=2016\) nghiệm
7.
Hàm có đúng 1 điểm gián đoạn khi và chỉ khi \(x^2-2\left(m+2\right)x+4=0\) có đúng 1 nghiệm
\(\Rightarrow\Delta'=\left(m+2\right)^2-4=0\)
\(\Leftrightarrow m^2+4m=0\Rightarrow\left[{}\begin{matrix}m=-4\\m=0\end{matrix}\right.\)
\(-4+0=-4\)
8.
Hàm gián đoạn khi \(x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Nên hàm đồng biến trên các khoảng \(\left(-\infty;-3\right);\left(-3;1\right);\left(1;+\infty\right)\) và các tập con của chúng
A đúng
1.
\(sin^2x-4sinx.cosx+3cos^2x=0\)
\(\Rightarrow\dfrac{sin^2x}{cos^2x}-\dfrac{4sinx}{cosx}+\dfrac{3cos^2x}{cos^2x}=0\)
\(\Rightarrow tan^2x-4tanx+3=0\)
2.
\(\Leftrightarrow\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
3.
\(\Leftrightarrow2^2+m^2\ge1\)
\(\Leftrightarrow m^2\ge-3\) (luôn đúng)
Pt có nghiệm với mọi m (đề bài sai)
4.
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=1\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)
6.
ĐKXĐ: \(cosx\ne0\)
Nhân 2 vế với \(cos^2x\)
\(sin^2x-4cosx+5cos^2x=0\)
\(\Leftrightarrow1-cos^2x-4cosx+5cos^2x=0\)
\(\Leftrightarrow\left(2cosx-1\right)^2=0\)
\(\Leftrightarrow cosx=\dfrac{1}{2}\Rightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)
6.
\(cos^2x+\sqrt{3}sinx.cosx-1=0\)
\(\Leftrightarrow-sin^2x+\sqrt{3}sinx.cosx=0\)
\(\Leftrightarrow sinx\left(sinx-\sqrt{3}cosx\right)=0\)
\(\Leftrightarrow sinx\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)=0\)
\(\Leftrightarrow sinx.sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sin\left(x-\dfrac{\pi}{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
3.
\(y=\dfrac{1-sin^24x}{5}=\dfrac{cos^24x}{5}\)
\(cos4x\in\left[-1;1\right]\Rightarrow cos^24x\in\left[0;1\right]\Rightarrow y\in\left[0;\dfrac{1}{5}\right]\Rightarrow\left\{{}\begin{matrix}y_{min}=0\\y_{max}=\dfrac{1}{5}\end{matrix}\right.\)
6.
\(y=sinx+cosx+2=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+2\)
\(sin\left(x+\dfrac{\pi}{4}\right)\in\left[-1;1\right]\Rightarrow y=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+2\in\left[-\sqrt{2}+2;\sqrt{2}+2\right]\)
\(\Rightarrow y_{min}=-\sqrt{2}+2\)
\(y_{max}=\sqrt{2}+2\)
b.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=-cosx\)
\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(x+\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\pi+k2\pi\\2x+\dfrac{\pi}{6}=-x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
c.
\(\Leftrightarrow2cos4x.sin3x=2sin4x.cos4x\)
\(\Leftrightarrow cos4x\left(sin4x-sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin4x=sin3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\4x=3x+k2\pi\\4x=\pi-3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=k2\pi\\x=\dfrac{\pi}{7}+\dfrac{k2\pi}{7}\end{matrix}\right.\)
2.
\(f\left(x\right)=\dfrac{1}{2}-\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x-5\)
\(=-\dfrac{9}{2}-\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)\)
\(=-\dfrac{9}{2}-cos\left(2x-\dfrac{\pi}{3}\right)\)
Do \(-1\le-cos\left(2x-\dfrac{\pi}{3}\right)\le1\Rightarrow-\dfrac{11}{2}\le y\le-\dfrac{7}{2}\)
\(y_{min}=-\dfrac{11}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=1\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)
\(y_{max}=-\dfrac{7}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=-1\Rightarrow x=\dfrac{2\pi}{3}+k\pi\)