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Hàm số xác định trên R khi và chỉ khi:
\(sin^2x+\left(2m-3\right)cosx+3m-2>0;\forall x\in R\)
\(\Leftrightarrow-cos^2x+\left(2m-3\right)cosx+3m-1>0\)
\(\Leftrightarrow t^2-\left(2m-3\right)t-3m+1< 0;\forall t\in\left[-1;1\right]\)
\(\Leftrightarrow t^2+3t+1< m\left(2t+3\right)\)
\(\Leftrightarrow\dfrac{t^2+3t+1}{2t+3}< m\) (do \(2t+3>0;\forall t\in\left[-1;1\right]\))
\(\Leftrightarrow m>\max\limits_{\left[-1;1\right]}\dfrac{t^2+3t+1}{2t+3}\)
Ta có: \(\dfrac{t^2+3t+1}{2t+3}=\dfrac{t^2+t-2+2t+3}{2t+3}=\dfrac{\left(t-1\right)\left(t+2\right)}{2t+3}+1\)
Do \(-1\le t\le1\Rightarrow\dfrac{\left(t-1\right)\left(t+2\right)}{2t+3}\le0\)
\(\Rightarrow\max\limits_{\left[-1;1\right]}\dfrac{t^2+3t+1}{2t+3}=1\)
\(\Rightarrow m>1\)
y = \(\dfrac{sin^2x}{cosx\left(sinx-cosx\right)}+\dfrac{1}{4}\)
y = \(\dfrac{sin^2x}{sinx.cosx-cos^2x}+\dfrac{1}{4}=\dfrac{\dfrac{sin^2x}{cos^2x}}{\dfrac{sinx.cosx}{cos^2x}-1}+\dfrac{1}{4}\)
y = \(\dfrac{tan^2x}{tanx-1}+\dfrac{1}{4}\)
y = \(\dfrac{4tan^2x+tanx-1}{4tanx-4}\). Đặt t = tanx. Do x ∈ \(\left(\dfrac{\pi}{4};\dfrac{\pi}{2}\right)\) nên t ∈ (1 ; +\(\infty\))\
Ta đươc hàm số f(t) = \(\dfrac{4t^2+t-1}{4t-4}\)
⇒ ymin = \(\dfrac{17}{4}\) khi t = 2. hay x = arctan(2) + kπ
1.
ĐK: \(x\ne\dfrac{k\pi}{2}\)
\(cotx-tanx=sinx+cosx\)
\(\Leftrightarrow\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)
\(\Leftrightarrow\dfrac{cos^2x-sin^2x}{sinx.cosx}=sinx+cosx\)
\(\Leftrightarrow\left(\dfrac{cosx-sinx}{sinx.cosx}-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx=sinx.cosx\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow t=\dfrac{1-t^2}{2}\left(t=cosx-sinx,\left|t\right|\le2\right)\)
\(\Leftrightarrow t^2+2t-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow cosx-sinx=-1+\sqrt{2}\)
\(\Leftrightarrow-\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=-1+\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}-1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)
a.
\(\Leftrightarrow m-cosx\ge0\) ; \(\forall x\)
\(\Leftrightarrow m\ge max\left(cosx\right)\)
\(\Leftrightarrow m\ge1\)
b.
\(\Leftrightarrow2sinx-m\ge0\) ; \(\forall x\)
\(\Leftrightarrow m\le2sinx\) ; \(\forall x\)
\(\Leftrightarrow m\le\min\limits_{x\in R}\left(2sinx\right)\)
\(\Leftrightarrow m\le-2\)
c.
\(\Leftrightarrow cosx+m\ne0\) ; \(\forall x\)
\(\Leftrightarrow\left[{}\begin{matrix}m>\max\limits_R\left(cosx\right)\\m< \min\limits_R\left(cosx\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)
Lời giải:
Ta có: \(\frac{1}{\cos x}-\frac{1}{\sin x}=m\)
\(\Leftrightarrow f(x)=\frac{1}{\cos x}-\frac{1}{\sin x}-m=0\)
Ta thấy: Hàm \(f(x)\) liên tục trên đoạn \(\left[\frac{\pi}{6}; \frac{\pi}{3}\right]\), mà:
\(f\left(\frac{\pi}{3}\right)=\frac{6-2\sqrt{3}}{3}-m\)
\(f\left(\frac{\pi}{6}\right)=\frac{-6+2\sqrt{3}}{3}-m\)
\(\Rightarrow f\left(\frac{\pi}{3}\right)f\left(\frac{\pi}{6}\right)=\left(\frac{6-2\sqrt{3}}{3}-m\right)\left(\frac{-6+2\sqrt{3}}{3}-m\right)=-\left(\frac{6-2\sqrt{3}}{3}-m\right)^2\)
\(\Rightarrow f\left(\frac{\pi}{3}\right)f\left(\frac{\pi}{6}\right)\leq 0\)
Do đó tồn tại ít nhất một nghiệm \(c\in \left[\frac{\pi}{6}; \frac{\pi}{3}\right]\)
Ta có đpcm.