Tìm GTLN GTNN của hàm: y= \(\sin^4x-2\cos^2x+1\)
Tìm GTLN và GTNN của hàm số : 1. y = sinx + 2cosx +1 / 2sinx + cosx + 3
2.y= 2sin^2sinx - 3 sinx cosx + cos^2 x
Giải phương trình : 1. 2sin^2 * 2x + sin7x -1 = sinx
2.cos 4x + 12 sin^2 x -1 = 0
Tìm GTNN và GTLN của hàm số sau:
1.\(y=cosx+cos\left(x-\dfrac{\pi}{3}\right)\)
2.\(y=sin^4x+cos^4x\)
3.\(y=3-2\left|sinx\right|\)
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 $y_{\min}=1; y_{\max}=3$
1.
\(y=\cos x+\cos (x-\frac{\pi}{3})=\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)
\(=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)
\(y^2=(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)^2\leq (\cos ^2x+\sin ^2x)(\frac{9}{4}+\frac{3}{4})\)
\(\Leftrightarrow y^2\leq 3\Rightarrow -\sqrt{3}\leq y\leq \sqrt{3}\)
Vậy $y_{\min}=-\sqrt{3}; y_{max}=\sqrt{3}$
tìm GTLN và GTNN
1. y=\(2\sin^3x+\sin x\)
2. y=\(\cos^2x-2\sin x\)
3. y=\(\sin^2x+\cos^4x\)
4. y=\(\sin^4x+\cos^4x+\sin x\times\cos x\)
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 \(sinx=-1\)
Tìm GTLN/GTNN của hàm số: \(y=sin^25x+cos^25x+3sin2x-2\)
`y=sin^2 5x+cos^2 5x+3sin 2x-2` `TXĐ: R`
`y=1+3sin 2x-2`
`y=3sin 2x-1`
Ta có: `-1 <= sin 2x <= 1`
`<=>-3 <= 3sin 2x <= 3`
`<=>-4 <= y <= 2`
`=> y_[mi n]=4<=>sin 2x =-1<=>x=-\pi/4 + k\pi` `(k in ZZ)`
`y_[max] = 2<=>sin 2x=1<=>x=\pi/4+k\pi` `(k in ZZ)`
Tìm GTLN, GTNN:
a, \(y=4\sin^2x-4\sin x+3\).
b, \(y=\cos^2x+2\sin x+2\).
c, \(y=\sin^4x-2\cos^2x+1\).
a.
Tìm min:
$y=(4\sin ^2x-4\sin x+1)+2=(2\sin x-1)^2+2$
Vì $(2\sin x-1)^2\geq 0$ với mọi $x$ nên $y=(2\sin x-1)^2+2\geq 0+2=2$
Vậy $y_{\min}=2$
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Mặt khác:
$y=4\sin x(\sin x+1)-8(\sin x+1)+11$
$=(\sin x+1)(4\sin x-8)+11$
$=4(\sin x+1)(\sin x-2)+11$
Vì $\sin x\in [-1;1]\Rightarrow \sin x+1\geq 0; \sin x-2<0$
$\Rightarrow 4(\sin x+1)(\sin x-2)\leq 0$
$\Rightarrow y=4(\sin x+1)(\sin x-2)+11\leq 11$
Vậy $y_{\max}=11$
b.
$y=\cos ^2x+2\sin x+2=1-\sin ^2x+2\sin x+2$
$=3-\sin ^2x+2\sin x$
$=4-(\sin ^2x-2\sin x+1)=4-(\sin x-1)^2\leq 4-0=4$
Vậy $y_{\max}=4$.
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Mặt khác:
$y=3-\sin ^2x+2\sin x = (1-\sin ^2x)+(2+2\sin x)$
$=(1-\sin x)(1+\sin x)+2(1+\sin x)=(1+\sin x)(1-\sin x+2)$
$=(1+\sin x)(3-\sin x)$
Vì $\sin x\in [-1;1]$ nên $1+\sin x\geq 0; 3-\sin x>0$
$\Rightarrow y=(1+\sin x)(3-\sin x)\geq 0$
Vậy $y_{\min}=0$
c.
$y=\sin ^4x-2\cos ^2x+1=\sin ^4x-2(1-\sin ^2x)+1$
$=\sin ^4x+2\sin ^2x-1$
$=(\sin ^4x-1)+(2\sin ^2x-2)+2$
$=(\sin ^2x-1)(\sin ^2x+1)+2(\sin ^2x-1)+2$
$=(\sin ^2x-1)(\sin ^2x+3)+2$
Vì $\sin x\in [-1;1]$ nên $\sin ^2x\leq 1$
$\Rightarrow (\sin ^2x-1)(\sin ^2x+3)\leq 0$
$\Rightarrow y=(\sin ^2x-1)(\sin ^2x+3)+2\leq 2$
Vậy $y_{\max}=2$
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$y=\sin ^4x+2\sin ^2x-1=\sin ^2x(\sin ^2x+2)-1$
Vì $\sin ^2x\geq 0$ nên $\sin ^2x(\sin ^2x+2)\geq 0$
$\Rightarrow y=\sin ^2x(\sin ^2x+2)-1\geq 0-1=-1$
Vậy $y_{\min}=-1$
Tìm GTLN, GTNN của các hàm số :
a) \(y=sin\left(1-x^2\right)\)
b) \(y=cos\sqrt{2-x^2}\)
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 \(x=0\)
tìm GTLN và GTNN
1.y=\(3\sin^2x-2\)
2.y=\(2\sin^3x+\sin x\)
3.y=\(\cos^2x-2\sin x\)
4.y=\(\sin^2x+\cos^4x\)
5.y=\(\sin^4x+\cos^4x+\sin x\times\cos x\)
Tìm GTLN và GTNN của hàm số y = 2 sin x + cos x + 3 2 cos x - sin x + 4 là:
A. m i n y = - 3 2 - 1 , m a x y = 3 2 + 1
B. m i n y = - 3 2 - 1 , m a x y = 3 2 - 1
C. m i n y = - 3 2 , m a x y = 3 2 - 1
D. m i n y = - 3 2 - 2 , m a x y = 3 2 - 1