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a/ \(\dfrac{1}{\sqrt{2}}\le\dfrac{2}{\sqrt{5+3cos4x}}\le\sqrt{2}\)
\(\Leftrightarrow2\sqrt{2}\ge\sqrt{5+3cos4x}\ge\sqrt{2}\)
\(\Leftrightarrow8\ge5+3cos4x\ge2\)
\(\Leftrightarrow1\ge cos4x\ge-1\) (đúng)
a/
\(\Leftrightarrow3\left(1-sin^22x\right)+4sin2x-4=0\)
\(\Leftrightarrow-3sin^22x+4sin2x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(\frac{1}{3}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(\frac{1}{3}\right)+k\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)
f/
\(\Leftrightarrow4\left(1-2sin^2\frac{x}{2}\right)-5sin\frac{x}{2}=1\)
\(\Leftrightarrow8sin^2\frac{x}{2}+5sin\frac{x}{2}-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\frac{x}{2}=-1\\sin\frac{x}{2}=\frac{3}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\pi+k4\pi\\x=2arcsin\left(\frac{3}{8}\right)+k4\pi\\x=2\pi-2arcsin\left(\frac{3}{8}\right)+k4\pi\end{matrix}\right.\)
a/ \(0\le cos^2x\le1\Rightarrow2\le y\le\sqrt{7}\)
\(y_{min}=2\) khi \(cos^2x=1\)
\(y_{max}=\sqrt{7}\) khi \(cos^2x=0\)
b/ \(y=\frac{2}{1+tan^2x}=\frac{2}{\frac{1}{cos^2x}}=2cos^2x\le2\)
\(\Rightarrow y_{max}=2\) khi \(cos^2x=1\)
\(y_{min}\) ko tồn tại
c/ \(y=1-cos2x+\sqrt{3}sin2x=2\left(\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x\right)+1\)
\(y=2sin\left(2x-\frac{\pi}{6}\right)+1\)
Do \(-1\le sin\left(2x-\frac{\pi}{6}\right)\le1\Rightarrow-1\le y\le3\)
d/
ĐKXĐ: ...
Biến đôi biểu thức vế trái trước:
\(1+tanx.tan\frac{x}{2}=1+\frac{sinx.sin\frac{x}{2}}{cosx.cos\frac{x}{2}}=\frac{sinx.sin\frac{x}{2}+cosx.cos\frac{x}{2}}{cosx.cos\frac{x}{2}}=\frac{cos\left(x-\frac{x}{2}\right)}{cosx.cos\frac{x}{2}}=\frac{1}{cosx}\)
Do đó pt tương đương:
\(\sqrt{3}\left(1+tan^2x\right)-tanx-2\sqrt{3}=sinx.\frac{1}{cosx}\)
\(\Leftrightarrow\sqrt{3}tan^2x-2tanx-\sqrt{3}=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-\frac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Sử dụng kết quả biến đổi trên làm câu c sẽ lẹ hơn cách cũ
c/
ĐKXĐ: ...
\(\Leftrightarrow2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=2cos^2x-4\)
\(\Leftrightarrow2cos^2x+2cos^2x.tanx.tan\frac{x}{2}=2cos^2x-4\)
\(\Leftrightarrow cos^2x.tanx.tan\frac{x}{2}=-2\)
\(\Leftrightarrow sinx.cosx.tan\frac{x}{2}=-2\)
\(\Leftrightarrow sinx.cosx.\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=-2\)
\(\Leftrightarrow sinx.cosx.\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}.cos\frac{x}{2}}=-1\)
\(\Leftrightarrow cosx\left(\frac{1-cosx}{2}\right)=-1\)
\(\Leftrightarrow cos^2x-cosx-2=0\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pi+k2\pi\)
d/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+\frac{5}{2}=4\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+\frac{5}{2}=4sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow2sin^2\left(x+\frac{\pi}{6}\right)+4sin\left(x+\frac{\pi}{6}\right)-\frac{7}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{-2+\sqrt{11}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{-2-\sqrt{11}}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\\x=\frac{5\pi}{6}-arcsin\left(\frac{-2+\sqrt{11}}{2}\right)+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
1. \(4\cos^2x-6\sin^2x+5\sin2x-4=0\)
\(\Leftrightarrow4\cos^2x-6\sin^2x+10\sin x\cos x-4\left(\cos^2x+\sin^2x\right)=0\)
\(\Leftrightarrow10\sin x\cos x-10\sin^2x=0\)
\(\Leftrightarrow10\sin x\left(\cos x-\sin x\right)=0\)
2. \(\sqrt{3}\cos^2x+2\sin x\cos x-\sqrt{3}\sin^2x-1=0\)
\(\Leftrightarrow\left(\sqrt{3}\cos^2x+\sin x\cos x\right)+\left(\sin x\cos x-\sqrt{3}\sin^2x\right)-1=0\)
\(\Leftrightarrow2\cos x\left(\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)+2\sin x\left(\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x\right)-1=0\)
\(\Leftrightarrow2\cos x.\cos\left(\dfrac{\Pi}{6}-x\right)+2\sin x.\sin\left(\dfrac{\Pi}{6}-x\right)-1=0\)
\(\Leftrightarrow\cos\dfrac{\Pi}{6}+\cos\left(2x-\dfrac{\Pi}{6}\right)+\cos\left(2x-\dfrac{\Pi}{6}\right)-\cos\dfrac{\Pi}{6}-1=0\)
\(\Leftrightarrow\cos\left(2x-\dfrac{\Pi}{6}\right)=\dfrac{1}{2}\)
3. \(2\sin^22x-3\sin2x\cos2x+\cos^22x=2\)
\(\Leftrightarrow2\sin^22x-3\sin2x\cos2x+\cos^22x-2\left(\sin^22x+\cos^22x\right)=0\)
\(\Leftrightarrow3\sin2x\cos2x+\cos^22x=0\)
\(\Leftrightarrow\cos2x\left(3\sin2x+\cos2x\right)=0\)
-TH1: ...
- TH2: \(\cos2x=-3\sin2x\) mà \(\cos^22x+\sin^22x=1\) suy ra ...
4. \(4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}=3\)
\(\Leftrightarrow4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}-3\left(\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\dfrac{1+\cos x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\cos x+\sin x=-1\)
Lời giải:
Ta có:
\(f'(x)=3x^2+2(a-1)x+2\)
Theo định lý về dấu của tam thức bậc 2, để \(f'(x)>0\) với mọi \(x\in\mathbb{R}\) thì \(\Delta'=(a-1)^2-6<0\)
\(\Leftrightarrow -\sqrt{6}< a-1< \sqrt{6}\)
\(\Leftrightarrow 1-\sqrt{6}< a< 1+\sqrt{6}\)
Đáp án B
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 \(sin^2x=1\)