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\(sin\left(x+\frac{\pi}{6}\right)=1\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\Rightarrow x=\frac{\pi}{3}+k2\pi\)
\(\Delta y=4\sqrt{2\left(x+\Delta x\right)-6}-4\sqrt{2x-6}=\frac{8\Delta x}{\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}}\)
\(f'\left(x\right)=\lim\limits_{\Delta\rightarrow0}\frac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\rightarrow0}\frac{8\Delta x}{\Delta x\left(\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}\right)}\)
\(=\lim\limits_{\Delta x\rightarrow0}\frac{8}{\sqrt{2x+2\Delta x-6}+\sqrt{2x-6}}=\frac{8}{2\sqrt{2x-6}}=\frac{4}{\sqrt{2x-6}}\)
b/ \(f'\left(5\right)=\frac{4}{\sqrt{2.5-6}}=2\) ; \(f\left(5\right)=4\sqrt{2.5-6}=8\)
Pt tiếp tuyến: \(y=2\left(x-5\right)+8=2x-2\)
c/ \(f'\left(x\right)>4\Leftrightarrow\frac{4}{\sqrt{2x-6}}>4\Leftrightarrow\frac{1}{\sqrt{2x-6}}>1\)
\(\Leftrightarrow\sqrt{2x-6}< 1\Leftrightarrow2x-6< 1\Rightarrow x< \frac{7}{2}\)
\(\Rightarrow3< x< \frac{7}{2}\)
a/
\(y=\frac{1}{sinx}+\frac{1}{cosx}\ge\frac{4}{sinx+cosx}=\frac{4}{\sqrt{2}sin\left(x+\frac{\pi}{4}\right)}\ge\frac{4}{\sqrt{2}}=2\sqrt{2}\)
\(y_{min}=2\sqrt{2}\) khi \(\left\{{}\begin{matrix}sinx=cosx\\sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{4}\)
\(y_{max}\) không tồn tại (y dần tới dương vô cùng khi x gần tới 0 hoặc \(\frac{\pi}{2}\))
b/
\(y=\frac{1}{1-cosx}+\frac{1}{1+cosx}=\frac{1+cosx+1-cosx}{1-cos^2x}=\frac{2}{sin^2x}\)
Hàm số ko tồn tại cả min lẫn max ( \(0< y< \infty\))
c/
Do \(tan^2x\) ko tồn tại max (tiến tới vô cực) trên khoảng đã cho nên hàm ko tồn tại max
\(y=2+\frac{sin^4x+cos^4x}{\left(sinx.cosx\right)^2}+\frac{1}{sin^4x+cos^4x}\ge2+2\sqrt{\frac{sin^4x+cos^4x}{\frac{1}{4}sin^22x.\left(sin^4x+cos^4x\right)}}\)
\(y\ge2+\frac{4}{sin2x}\ge2+\frac{4}{1}=6\)
\(y_{min}=6\) khi \(\left\{{}\begin{matrix}sin2x=1\\sin^4x+cos^4x=sinx.cosx\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{4}\)
Lời giải:
\(C=\lim\limits_{x\to +\infty}\left[x\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x})...(1+\frac{a_n}{x})}-x\right]\)
\(=\lim\limits_{x\to +\infty}x\left[\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}).....(1+\frac{a_n}{x})}-1\right]\)
\(=\lim\limits _{x\to +\infty}\frac{\sqrt[n]{(1+\frac{a_1}{x})(1+\frac{a_2}{x}).....(1+\frac{a_n}{x})}-1}{(1+\frac{a_1}{x})(1+\frac{a_2}{x})..(1+\frac{a_n}{x})-1}.\frac{(1+\frac{a_1}{x})(1+\frac{a_2}{x})...(1+\frac{a_n}{x})-1}{\frac{1}{x}}\)
\(=\lim\limits _{x\to +\infty}(A.B)=\lim\limits_{x\to +\infty}A.\lim\limits_{x\to +\infty}B\)
Với $A$. Đặt \(\sqrt[n]{\prod_{i=1}^n (1+\frac{a_i}{x})}=u\). \(x\to +\infty\Rightarrow \frac{a_i}{x}\to 0\Rightarrow 1+\frac{a_i}{x}\to 1\Rightarrow u\to 1\)
\(\lim\limits_{x\to +\infty}A=\lim\limits_{u\to 1}\frac{u-1}{u^n-1}=\lim\limits_{u\to 1}\frac{1}{u^{n-1}+...+1}=\frac{1}{n}\)
Với $B$
\(\lim\limits _{x\to +\infty}B=\lim\limits _{x\to +\infty}\frac{1+\frac{a_1+a_2+..+a_n}{x}+\frac{a_1a_2+a_2a_3+...+a_{n-1}a_n}{x^2}+....-1}{\frac{1}{x}}\)
\(=\lim\limits _{x\to +\infty}\left(a_1+a_2+...+a_n+\frac{a_1a_2+...+a_{n-1}a_n}{x}+...\right)=a_1+a_2+..+a_n\)
Do đó: $C=\frac{a_1+a_2+...+a_n}{n}$
Đáp án C
Bài 1:
\(\lim\limits _{x\to 1}\frac{4x^6-5x^5+x}{(1-x)^2}=\lim\limits _{x\to 1}\frac{x(x-1)^2(4x^3+3x^2+2x+1)}{(1-x)^2}\)
\(=\lim\limits _{x\to 1}x(4x^3+3x^2+2x+1)=1(4.1^3+3.1^2+2.1+1)=10\)
Bài 3:
\(\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-(ax+b)]=0\)
\(\Rightarrow \lim\limits _{x\to +\infty}\frac{\sqrt{9x^2-4x+3}-(ax+b)}{x}=0\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\left(\sqrt{9-\frac{4}{x}+\frac{3}{x^2}}-a+\frac{b}{x}\right)=0\)
\(\Leftrightarrow a=3\)
Thay $a=3$ vào đk ban đầu:
\(\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-3x-b]=0\)
\(\Leftrightarrow \lim\limits _{x\to +\infty} (\sqrt{9x^2-4x+3}-3x)=b\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4x+3}{\sqrt{9x^2-4x+3}+3x}=b\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4+\frac{3}{x}}{\sqrt{9-\frac{4}{x}+\frac{3}{x}}+3}=b\)
\(\Leftrightarrow \frac{-4}{6}=b\Leftrightarrow b=-\frac{2}{3}\)
3.
\(x-2y+1=0\Leftrightarrow y=\frac{1}{2}x+\frac{1}{2}\)
\(y'=\frac{2}{\left(x+1\right)^2}\Rightarrow\frac{2}{\left(x+1\right)^2}=\frac{1}{2}\)
\(\Rightarrow\left(x+1\right)^2=4\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-3\Rightarrow y=3\end{matrix}\right.\)
Có 2 tiếp tuyến: \(\left[{}\begin{matrix}y=\frac{1}{2}\left(x-1\right)+1\\y=\frac{1}{2}\left(x+3\right)+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{2}x+\frac{1}{2}\left(l\right)\\y=\frac{1}{2}x+\frac{9}{2}\end{matrix}\right.\)
4.
\(\lim\limits\frac{\sqrt{2n^2+1}-3n}{n+2}=\lim\limits\frac{\sqrt{2+\frac{1}{n^2}}-3}{1+\frac{2}{n}}=\sqrt{2}-3\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
5.
\(\lim\limits_{x\rightarrow a}\frac{2\left(x^2-a^2\right)+a\left(a+1\right)-\left(a+1\right)x}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+2a\right)-\left(a+1\right)\left(x-a\right)}{\left(x-a\right)\left(x+a\right)}\)
\(=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+a-1\right)}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{2x+a-1}{x+a}=\frac{3a-1}{2a}\)
1.
\(f'\left(x\right)=-3x^2+6mx-12=3\left(-x^2+2mx-4\right)=3g\left(x\right)\)
Để \(f'\left(x\right)\le0\) \(\forall x\in R\) \(\Leftrightarrow g\left(x\right)\le0;\forall x\in R\)
\(\Leftrightarrow\Delta'=m^2-4\le0\Rightarrow-2\le m\le2\)
\(\Rightarrow m=\left\{-1;0;1;2\right\}\)
2.
\(f'\left(x\right)=\frac{m^2-20}{\left(2x+m\right)^2}\)
Để \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)