Với mọi ta luôn có \(\left\{{}\begin{matrix}\left(y+1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow-1\le\frac{2y}{y^2+1}\le1\)
Đặt \(\frac{2y}{y^2+1}=x\Rightarrow-1\le x\le1\)
\(A=cos2x+cosx+1=2cos^2x+cosx\)
Đặt \(cosx=t\Rightarrow cos1\le t\le1\)
\(A=2t^2+t\)
Xét \(f\left(t\right)=2t^2+t\) trên \(\left[cos1;1\right]\)
\(-\frac{b}{2a}=-\frac{1}{4}\notin\left[cos1;1\right]\)
\(f\left(cos1\right)=2cos^21+cos1\) ; \(f\left(1\right)=3\)
\(\Rightarrow2cos^21+cos1\le A\le3\)
\(A_{max}=3\) khi \(y=0\)
\(A_{min}=2cos^21+cos1\) khi \(y=\pm1\)
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