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2.
$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$
$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$
Vì: $0\leq \sin ^22x\leq 1$
$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$
Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$
3.
$0\leq |\sin x|\leq 1$
$\Rightarrow 3\geq 3-2|\sin x|\geq 1$
Vậy $y_{\min}=1; y_{\max}=3$
a.
\(-1\le sin\left(1-x^2\right)\le1\)
\(\Rightarrow y_{min}=-1\) khi \(1-x^2=-\dfrac{\pi}{2}+k2\pi\) \(\Rightarrow x^2=\dfrac{\pi}{2}+1+k2\pi\) (\(k\ge0\))
\(y_{max}=1\) khi \(1-x^2=\dfrac{\pi}{2}+k2\pi\Rightarrow x^2=1-\dfrac{\pi}{2}+k2\pi\) (\(k\ge1\))
b.
Đặt \(\sqrt{2-x^2}=t\Rightarrow t\in\left[0;\sqrt{2}\right]\subset\left[0;\pi\right]\)
\(y=cost\) nghịch biến trên \(\left[0;\pi\right]\Rightarrow\) nghịch biến trên \(\left[0;\sqrt{2}\right]\)
\(\Rightarrow y_{max}=y\left(0\right)=cos0=1\) khi \(x^2=2\Rightarrow x=\pm\sqrt{2}\)
\(y_{min}=y\left(\sqrt{2}\right)=cos\sqrt{2}\) khi \(x=0\)
24.
\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)
\(y_{max}=4\)
26.
\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)
Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)
\(y_{max}=\sqrt{2}\)
b.
\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
a.
\(y=\dfrac{3}{2}sin2x-2\left(cos^2x-sin^2x\right)+5=\dfrac{3}{2}sin2x-2cos2x+5\)
\(=\dfrac{5}{2}\left(\dfrac{3}{5}sin2x-\dfrac{4}{5}cos2x\right)+5=\dfrac{5}{2}sin\left(2x-a\right)+5\) (với \(cosa=\dfrac{3}{5}\))
\(\Rightarrow-\dfrac{5}{2}+5\le y\le\dfrac{5}{2}+5\)
b.
\(\Leftrightarrow y.sinx-2y.cosx+4y=3sinx-cosx+1\)
\(\Leftrightarrow\left(y-3\right)sinx+\left(1-2y\right)cosx=1-4y\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(y-3\right)^2+\left(1-2y\right)^2\ge\left(1-4y\right)^2\)
\(\Leftrightarrow11y^2+2y-9\le0\)
\(\Leftrightarrow-1\le y\le\dfrac{9}{11}\)
c.
Do \(x^2+y^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}x=sina\\y=cosa\end{matrix}\right.\)
\(\Rightarrow y=\dfrac{2\left(sin^2a+6sina.cosa\right)}{1+2sina.cosa+cos^2a}=\dfrac{1-cos2a+6sin2a}{1+sin2a+\dfrac{1+cos2a}{2}}=\dfrac{2-2cos2a+12sin2a}{3+2sin2a+cos2a}\)
\(\Leftrightarrow3y+2y.sin2a+y.cos2a=2-2cos2a+12sin2a\)
\(\Leftrightarrow\left(2y-12\right)sin2a+\left(y+2\right)cos2a=2-3y\)
Theo điều kiện có nghiệm của pt bậc nhất theo sin2a, cos2a:
\(\left(2y-12\right)^2+\left(y+2\right)^2\ge\left(2-3y\right)^2\)
\(\Leftrightarrow y^2+8y-36\le0\)
\(\Rightarrow-4-2\sqrt{13}\le y\le-4+2\sqrt{13}\)
1. Ta có: \(-1\le sinx\le1\)
\(\Rightarrow-3\le y\le3\) (hàm đã cho đồng biến trên \(\left[-\frac{\pi}{2};\frac{\pi}{2}\right]\)
\(y_{min}=-3\) khi \(sinx=-1\)
\(y_{max}=3\) khi \(sinx=1\)
2.
\(y=1-sin^2x-2sinx=2-\left(sinx+1\right)^2\)
Do \(-1\le sinx\le1\Rightarrow0\le sinx+1\le2\)
\(\Rightarrow-2\le y\le2\)
\(y_{min}=-2\) khi \(sinx=1\)
\(y_{max}=2\) khi \(sinx=-1\)
3.
\(y=1-cos^2x+cos^4x=\left(cos^2x-\frac{1}{2}\right)^2+\frac{3}{4}\)
\(\Rightarrow y\ge\frac{3}{4}\Rightarrow y_{min}=\frac{3}{4}\) khi \(cos^2x=\frac{1}{2}\)
\(y=1+cos^2x\left(cos^2x-1\right)\le1\) do \(cos^2x-1\le0\)
\(\Rightarrow y_{max}=1\) khi \(\left[{}\begin{matrix}cos^2x=1\\cos^2x=0\end{matrix}\right.\)
4.
\(y=\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2+sinx.cosx\)
\(y=1-\frac{1}{2}sin^22x+\frac{1}{2}sin2x\)
\(y=\frac{9}{8}-\frac{1}{2}\left(sinx-\frac{1}{2}\right)^2\le\frac{9}{8}\)
\(y_{max}=\frac{9}{8}\) khi \(sinx=\frac{1}{2}\)
\(y=\frac{1}{2}\left(sinx+1\right)\left(2-sinx\right)\ge0;\forall x\)
\(\Rightarrow y_{min}=0\) khi \(sinx=-1\)
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\Rightarrow\left\{{}\begin{matrix}-\sqrt{2}\le t\le\sqrt{2}\\2sinx.cosx=t^2-1\end{matrix}\right.\)
\(\Rightarrow y=t+t^2-1-1=t^2+t-2\)
Xét hàm \(f\left(t\right)=t^2+t-2\) trên \(\left[-\sqrt{2};\sqrt{2}\right]\)
\(-\frac{b}{2a}=-\frac{1}{2}\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(f\left(-\frac{1}{2}\right)=-\frac{9}{4}\) ; \(f\left(-\sqrt{2}\right)=-\sqrt{2}\) ; \(f\left(\sqrt{2}\right)=\sqrt{2}\)
\(\Rightarrow y_{max}=\sqrt{2}\) khi \(t=\sqrt{2}\)
\(y_{min}=-\frac{9}{4}\) khi \(t=-\frac{1}{2}\)
Câu 1:
\(y=S\left(\frac{3-S^2}{2}\right)=\frac{3}{2}S-\frac{1}{2}S^3\)
Khi \(S\rightarrow+\infty\) thì \(y\rightarrow-\infty\)
Khi \(S\rightarrow-\infty\) thì \(y\rightarrow+\infty\)
Hàm số không có GTLN và GTNN
Câu 2:
\(y=sin^4x+cos^4x+2sin^2x.cos^2x-2sin^2x.cos^2x\)
\(y=\left(sin^2x+cos^2x\right)^2-\frac{1}{2}\left(2sinx.cosx\right)^2\)
\(y=1-\frac{1}{2}sin^22x\)
Do \(0\le sin^22x\le1\)
\(\Rightarrow y_{max}=1\) khi \(sin2x=0\)
\(y_{min}=\frac{1}{2}\) khi \(sin2x=\pm1\)
Câu 3:
\(y=sin^6x+cos^6x+3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\)
\(y=1-\frac{3}{4}sin^22x\)
Do \(0\le sin^22x\le1\)
\(\Rightarrow y_{max}=1\) khi \(sin2x=0\)
\(y_{min}=\frac{1}{4}\) khi \(sin2x=\pm1\)
Câu 4:
\(y=\frac{cosx+2sinx+3}{2cosx-sinx+4}\)
\(\Leftrightarrow2y.cosx-y.sinx+4y=cosx+2sinx+3\)
\(\Leftrightarrow\left(y+2\right)sinx+\left(1-2y\right)cosx=4y-3\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(y+2\right)^2+\left(1-2y\right)^2\ge\left(4y-3\right)^2\)
\(\Leftrightarrow11y^2-24y+4\le0\)
\(\Leftrightarrow\frac{2}{11}\le y\le2\)
Đáp án C
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