Câu 15: \(y=cosx+sinx=\sqrt{2}\cdot sin\left(x+\dfrac{\Omega}{4}\right)\)
\(-1< =sin\left(x+\dfrac{\Omega}{4}\right)< =1\)
=>\(-\sqrt{2}< =\sqrt{2}\cdot sin\left(x+\dfrac{\Omega}{4}\right)< =\sqrt{2}\)
=>\(-\sqrt{2}< =y< =\sqrt{2}\)
\(y_{min}=-\sqrt{2}\) khi \(sin\left(x+\dfrac{\Omega}{4}\right)=-1\)
=>\(x+\dfrac{\Omega}{4}=-\dfrac{\Omega}{2}+k2\Omega\)
=>\(x=-\dfrac{3}{4}\Omega+k2\Omega\)
\(y_{max}=\sqrt{2}\) khi \(sin\left(x+\dfrac{\Omega}{4}\right)=1\)
=>\(x+\dfrac{\Omega}{4}=\dfrac{\Omega}{2}+k2\Omega\)
=>\(x=\dfrac{\Omega}{4}+k2\Omega\)
Câu 17:
Phương trình hoành độ giao điểm là:
\(sin\left(2x-\dfrac{\Omega}{3}\right)=sin\left(\dfrac{\Omega}{4}-2x\right)\)
=>\(\left[{}\begin{matrix}2x-\dfrac{\Omega}{3}=\dfrac{\Omega}{4}-2x+k2\Omega\\2x-\dfrac{\Omega}{3}=\Omega-\dfrac{\Omega}{4}+2x+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x+2x=\dfrac{\Omega}{4}+\dfrac{\Omega}{3}+k2\Omega\\2x-2x=\dfrac{3}{4}\Omega+\dfrac{\Omega}{3}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{7}{12}\Omega+k2\Omega\\0x=\dfrac{13}{12}\Omega+k2\Omega\left(loại\right)\end{matrix}\right.\)
=>\(x=\dfrac{7}{48}\Omega+\dfrac{k\Omega}{2}\)
Thay \(x=\dfrac{7}{48}\Omega+\dfrac{k\Omega}{2}\) vào \(y=sin\left(2x-\dfrac{\Omega}{3}\right)\), ta được:
\(y=sin\left[2\left(\dfrac{7}{48}\Omega+\dfrac{k\Omega}{2}\right)-\dfrac{\Omega}{3}\right]=sin\left(\dfrac{7}{24}\Omega+k\Omega-\dfrac{\Omega}{3}\right)\)
\(=sin\left(-\dfrac{1}{24}\Omega+k\Omega\right)\)
13.
ĐKXĐ: \(1-x^2\ge0\Rightarrow-1\le x\le1\)
15.
Do \(\left\{{}\begin{matrix}0\le sin^2x\le1\\0\le cos^2x\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(sin^2x\right)^2\le sin^2x\\\left(cos^2x\right)^2\le cos^2x\end{matrix}\right.\)
\(\Rightarrow\left(sin^2x\right)^2+\left(cos^2x\right)^2\le sin^2x+cos^2x\)
\(\Rightarrow sin^4x+cos^4x\le1\)
Vậy \(y_{max}=1\)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}sin^2x=0\\sin^2x=1\end{matrix}\right.\)
\(y=sin^4x+cos^4x\ge\dfrac{1}{2}\left(sin^2x+cos^2x\right)^2=\dfrac{1}{2}\)
\(y_{min}=\dfrac{1}{2}\) khi \(sin^2x=cos^2x\)
