Tìm max, min (nếu có) của:
a, y = 1 - \(\sqrt{2cos^2x+1}\)
b, y = \(\dfrac{1}{1+2sin^2x}\)
Tìm min, max của :
1. y = \(\sqrt{4-sin^52x}-8\)
2. y = \(\dfrac{4}{\sqrt{5-2cos^2x.sin^2x}}\)
2.Biểu thức luôn xác định
\(y=\dfrac{4}{\sqrt{5-2cos^2sin^2x}}=\dfrac{4}{\sqrt{5-\dfrac{1}{2}sin^22x}}\)
Có: \(1\ge sin^22x\ge0\)
\(\Leftrightarrow-\dfrac{1}{2}\le-\dfrac{1}{2}sin^22x\le0\)
\(\Leftrightarrow\dfrac{3\sqrt{2}}{2}\le\sqrt{5-\dfrac{1}{2}sin^22x}\le\sqrt{5}\)
\(\Rightarrow\dfrac{4\sqrt{2}}{3}\ge y\ge\dfrac{4\sqrt{5}}{5}\)
miny=\(\dfrac{4\sqrt{5}}{5}\) \(\Leftrightarrow sin2x=0\)\(\Leftrightarrow x=\dfrac{k\pi}{2}\left(k\in Z\right)\)
maxy=\(\dfrac{4\sqrt{2}}{3}\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{-\pi}{4}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
1.Biểu thức luôn xác định
Xét \(sin2x=0\) \(\Leftrightarrow x=\dfrac{k\pi}{2}\left(k\in Z\right)\) khi đó \(y=-6\)
Xét \(sin2x\ne0\)
=> \(1\ge sin^52x\ge-1\)
\(\Leftrightarrow4-1\le4-sin^52x\le4+1\)
\(\Leftrightarrow\sqrt{3}\le\sqrt{4-sin^52x}\le\sqrt{5}\)
\(\Leftrightarrow\sqrt{3}-8\le y\le\sqrt{5}-8\)
\(y=\sqrt{3}-8< -6\) , \(y=\sqrt{5}-8>-6\)
=>min= \(\sqrt{3}-8\) \(\Leftrightarrow sin2x=1\left(tm\right)\) \(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)
maxy=\(\sqrt{5}-8\)\(\Leftrightarrow sin2x=-1\left(tm\right)\) \(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)
(câu này e ko chắc)
Tìm Min và Max(nếu có)
A=2x-\(\sqrt{x}\)
B=x+\(\sqrt{x}\)
C=1+\(\sqrt{2-x}\)
D=\(\sqrt{-x^2+2x+5}\)
E=\(\dfrac{1}{2x-\sqrt{x}+3}\)
F=\(\dfrac{1}{3-\sqrt{1-x^2}}\)
$A=2x-\sqrt{x}=2(x-\frac{1}{2}\sqrt{x}+\frac{1}{4^2})-\frac{1}{8}$
$=2(\sqrt{x}-\frac{1}{4})^2-\frac{1}{8}$
$\geq \frac{-1}{8}$
Vậy $A_{\min}=-\frac{1}{8}$. Giá trị này đạt tại $x=\frac{1}{16}$
$B=x+\sqrt{x}$
Vì $x\geq 0$ nên $B\geq 0+\sqrt{0}=0$
Vậy $B_{\min}=0$. Giá trị này đạt tại $x=0$
Vì $2-x\geq 0$ (theo ĐKXĐ) nên $C=1+\sqrt{2-x}\geq 1$
Vậy $C_{\min}=1$. Giá trị này đạt tại $2-x=0\Leftrightarrow x=2$
tìm max, min
a) y=\(\dfrac{\sqrt{x-1}}{x}\) trên \([1;5]\)
b) y=\(\dfrac{x+3}{\sqrt{x^2+1}}\) trên \([1;3]\)
c) y=\(\sin^2x-\cos x+1\)
d) y=\(\sin^3x-3\sin^2x+2\)
a0
a.
\(y'=\dfrac{2-x}{2x^2\sqrt{x-1}}=0\Rightarrow x=2\)
\(y\left(1\right)=0\) ; \(y\left(2\right)=\dfrac{1}{2}\) ; \(y\left(5\right)=\dfrac{2}{5}\)
\(\Rightarrow y_{min}=y\left(1\right)=0\)
\(y_{max}=y\left(2\right)=\dfrac{1}{2}\)
b.
\(y'=\dfrac{1-3x}{\sqrt{\left(x^2+1\right)^3}}< 0\) ; \(\forall x\in\left[1;3\right]\Rightarrow\) hàm nghịch biến trên [1;3]
\(\Rightarrow y_{max}=y\left(1\right)=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)
\(y_{min}=y\left(3\right)=\dfrac{6}{\sqrt{10}}=\dfrac{3\sqrt{10}}{5}\)
c.
\(y=1-cos^2x-cosx+1=-cos^2x-cosx+2\)
Đặt \(cosx=t\Rightarrow t\in\left[-1;1\right]\)
\(y=f\left(t\right)=-t^2-t+2\)
\(f'\left(t\right)=-2t-1=0\Rightarrow t=-\dfrac{1}{2}\)
\(f\left(-1\right)=2\) ; \(f\left(1\right)=0\) ; \(f\left(-\dfrac{1}{2}\right)=\dfrac{9}{4}\)
\(\Rightarrow y_{min}=0\) ; \(y_{max}=\dfrac{9}{4}\)
d.
Đặt \(sinx=t\Rightarrow t\in\left[-1;1\right]\)
\(y=f\left(t\right)=t^3-3t^2+2\Rightarrow f'\left(t\right)=3t^2-6t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\notin\left[-1;1\right]\end{matrix}\right.\)
\(f\left(-1\right)=-2\) ; \(f\left(1\right)=0\) ; \(f\left(0\right)=2\)
\(\Rightarrow y_{min}=-2\) ; \(y_{max}=2\)
Bài 1: a) Cho cotx=3. Tìm:
\(B=\dfrac{2sin^2x+3sinx.cosx}{1-2cos^2x}\)
b) Cho tanx=-3; \(\dfrac{3\pi}{2}< x< 2\pi\)
Tìm: A=\(\sqrt{10}cosx-2sinx+3\)
b:
3/2pi<x<2pi
=>cosx>0; sin x<0
\(1+tan^2x=\dfrac{1}{cos^2x}\)
=>\(\dfrac{1}{cos^2x}=1+\left(-3\right)^2=10\)
=>cosx=1/căn 10
=>sin x=-3/căn 10
\(A=\sqrt{10}\cdot\dfrac{1}{\sqrt{10}}-2\cdot\dfrac{-3}{\sqrt{10}}+3=4+\dfrac{6}{\sqrt{10}}=\dfrac{4\sqrt{10}+6}{\sqrt{10}}\)
a: cot x=3 nên cosx/sinx=3
=>cosx=3*sinx
\(B=\dfrac{2sin^2x+3sinx\cdot3\cdot sinx}{1-2\cdot\left(3\cdot sinx\right)^2}=\dfrac{11sin^2x}{sin^2x+cos^2x-18sin^2x}\)
\(=\dfrac{11sin^2x}{-17sin^2x+9sin^2x}=\dfrac{-11}{8}\)
Tìm Min,Max của các hàm số
a/ y= 2sin bình x+ cos bình 2x+2
b/ y=4sin2x + 5cos2x -2
c/ y= 3sin(2x-(pi/3))-2cos(2x-(pi/3))+1
d/ y=(2sin bình 3x+4sin3x.cos3x+1)/(sin6x+4cos6x+10)
Giusp mình với ạ!!! Thực sự mình rất cầN!!! Mình cảm ơn!
Tìm giá trị lớn nhất và giá trị nhỏ nhất của các hàm số sau:
1,\(y=5-3cosx\)
2,\(y=3cos^2x-2cosx+2\)
3,\(y=cos^2x+2cos2x\)
4,\(y=\sqrt{5-2sin^2x.cos^2x}\)
5,\(y=cos2x-cos\left(2x-\dfrac{\pi}{3}\right)\)
6,\(y=\sqrt{3}sinx-cosx-2\)
7,\(y=2cos^2x-sin2x+5\)
8,\(y=2sin^2x-sin2x+10\)
9,\(y=sin^6x+cos^6x\)
Tìm min, max
a, y= \(4sin^2x-5sinx.cosx+cos^2x+10\)
b, y= \(\dfrac{sin^2x-2sin2x+1}{3+sin^2x+2cos^2x}\)
c, y= \(2sinx+3cosx+4\)
a.
\(y=2\left(1-cos2x\right)-\dfrac{5}{2}sin2x+\dfrac{1}{2}+\dfrac{1}{2}cos2x+10\)
\(=-\dfrac{1}{2}\left(5sin2x+3cos2x\right)+\dfrac{25}{2}\)
\(=-\dfrac{\sqrt{34}}{2}\left(\dfrac{5}{\sqrt{34}}sin2x+\dfrac{3}{\sqrt{34}}cos2x\right)+\dfrac{25}{2}\)
Đặt \(\dfrac{5}{\sqrt{34}}=cosa\)
\(\Rightarrow y=-\dfrac{\sqrt{34}}{2}\left(sin2x.cosa+cos2x.sina\right)+\dfrac{25}{2}\)
\(=-\dfrac{\sqrt{34}}{2}sin\left(2x+a\right)+\dfrac{25}{2}\)
Do \(-1\le sin\left(2x+a\right)\le1\)
\(\Rightarrow\dfrac{25-\sqrt{34}}{2}\le y\le\dfrac{25+\sqrt{34}}{2}\)
b.
\(y=\dfrac{sin^2x-2sin2x+1}{3+sin^2x+2cos^2x}=\dfrac{2sin^2x-4sin2x+2}{6+2\left(sin^2x+cos^2x\right)+2cos^2x}\)
\(=\dfrac{1-cos2x-4sin2x+2}{8+1+cos2x}=\dfrac{3-4sin2x-cos2x}{9+cos2x}\)
\(\Rightarrow9y+y.cos2x=3-4sin2x-cos2x\)
\(\Rightarrow4sin2x+\left(y+1\right)cos2x=3-9y\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(4^2+\left(y+1\right)^2\ge\left(3-9y\right)^2\)
\(\Leftrightarrow80y^2-56y-8\le0\)
\(\Rightarrow\dfrac{7-\sqrt{89}}{20}\le y\le\dfrac{7+\sqrt{89}}{20}\)
c.
\(y=2sinx+3cosx+4\)
\(=\sqrt{13}\left(\dfrac{2}{\sqrt{13}}sinx+\dfrac{3}{\sqrt{13}}cosx\right)+4\)
Đặt \(\dfrac{2}{\sqrt{13}}=cosa\)
\(\Rightarrow y=\sqrt{13}\left(sinx.cosa+cosx.sina\right)+4\)
\(=\sqrt{13}sin\left(x+a\right)+4\)
Do \(-1\le sin\left(x+a\right)\le1\)
\(\Rightarrow-\sqrt{13}+4\le y\le\sqrt{13}+4\)
Tìm min, max
a) \(y=\sqrt{7-3cos^2x}\)
b) \(y=\frac{2}{1+tan^2x}\)
c) \(y=2sin^2x+\sqrt{3}sin2x\)
a/ \(0\le cos^2x\le1\Rightarrow2\le y\le\sqrt{7}\)
\(y_{min}=2\) khi \(cos^2x=1\)
\(y_{max}=\sqrt{7}\) khi \(cos^2x=0\)
b/ \(y=\frac{2}{1+tan^2x}=\frac{2}{\frac{1}{cos^2x}}=2cos^2x\le2\)
\(\Rightarrow y_{max}=2\) khi \(cos^2x=1\)
\(y_{min}\) ko tồn tại
c/ \(y=1-cos2x+\sqrt{3}sin2x=2\left(\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x\right)+1\)
\(y=2sin\left(2x-\frac{\pi}{6}\right)+1\)
Do \(-1\le sin\left(2x-\frac{\pi}{6}\right)\le1\Rightarrow-1\le y\le3\)
1. Tìm Min, Max của :
B = cos 2x + \(\sqrt{1+2sin^2x}\)
Lời giải:
Ta có:
\(B=\cos 2x+\sqrt{1+2\sin ^2x}=\cos ^2x-\sin ^2x+\sqrt{1+2\sin ^2x}\)
\(=1-2\sin ^2x+\sqrt{1+2\sin ^2x}\)
Đặt \(\sin ^2x=t(t\in [0;1])\). Khi đó:
\(B=1-2t+\sqrt{1+2t}\)
\(B'=\frac{1}{\sqrt{1+2t}}-2=0\Leftrightarrow t=-\frac{3}{8}\) (loại)
Lập bảng biến thiên suy ra:
\(B_{\max}=B(0)=2\)
\(B_{\min}=B(1)=\sqrt{3}-1\)
Tìm Min, Max:
\(y=2Sin^2x+3SinxCosx+Cos^2x\)
\(y=2sin^2x+3sinx.cosx+cos^2x\)
\(=-\left(1-2sin^2x\right)+\dfrac{3}{2}sin2x+\dfrac{1}{2}\left(2cos^2x-1\right)+\dfrac{1}{2}\)
\(=-cos2x+\dfrac{3}{2}sin2x+\dfrac{1}{2}cos2x+\dfrac{1}{2}\)
\(=\dfrac{3}{2}sin2x-\dfrac{1}{2}cos2x+\dfrac{1}{2}\)
\(=\dfrac{\sqrt{10}}{2}\left(\dfrac{3}{\sqrt{10}}sin2x-\dfrac{1}{\sqrt{10}}cos2x\right)+\dfrac{1}{2}\)
\(=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\)
Vì \(sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)\in\left[-1;1\right]\)
\(\Rightarrow y=\dfrac{\sqrt{10}}{2}sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)+\dfrac{1}{2}\in\left[-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2};\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\right]\)
\(\Rightarrow y_{min}=-\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=-1\Leftrightarrow...\)
\(y_{max}=\dfrac{\sqrt{10}}{2}+\dfrac{1}{2}\Leftrightarrow sin\left(2x-arccos\dfrac{3}{\sqrt{10}}\right)=1\Leftrightarrow...\)