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Question

TÌM GIÁ TRỊ LỚN NHẤT,GTNN CỦA CÁC HÀM SỐ SAU:

a. y=3-2sin2x

b. y=4 cos(x-$\frac{\pi}{3}$)+1

c. y=$\sqrt{cosx+1}+2$

d. 3cos2 x+5

e. y=1-3$\left|cos2x\right|$

g. y=$\sqrt{sin^22x+3}-2$

MỌ...

Nguyễn Việt Lâm · Accepted Answer

a/ $-1\le sin2x\le1\Rightarrow1\le y\le5$

$y_{min}=1$ khi $sin2x=1$

$y_{max}=5$ khi $sin2x=-1$

b/ $-1\le cos\left(x-\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le5$

$y_{min}=-3$ khi $cos\left(x-\frac{\pi}{3}\right)=-1$

$y_{max}=5$ khi $cos\left(x-\frac{\pi}{3}\right)=1$

c/ $-1\le cosx\le1\Rightarrow0\le\sqrt{cosx+1}\le\sqrt{2}$

$\Rightarrow2\le y\le2+\sqrt{2}$

$y_{min}=2$ khi $cosx=-1$

$y_{max}=2+\sqrt{2}$ khi $cosx=1$Câu trả lời: a/ $-1\le sin2x\le1\Rightarrow1\le y\le5$

$y_{min}=1$ khi $sin2x=1$

$y_{max}=5$ khi $sin2x=-1$

b/ $-1\le cos\left(x-\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le5$

$y_{min}=-3$ khi $cos\left(x-\frac{\pi}{3}\right)=-1$

$y_{max}=5$ khi $cos\left(x-\frac{\pi}{3}\right)=1$

c/ $-1\le cosx\le1\Rightarrow0\le\sqrt{cosx+1}\le\sqrt{2}$

$\Rightarrow2\le y\le2+\sqrt{2}$

$y_{min}=2$ khi $cosx=-1$

$y_{max}=2+\sqrt{2}$ khi $cosx=1$

Nguyễn Việt Lâm · Answer

d/

$0\le cos^2x\le1\Rightarrow5\le y\le8$

$y_{min}=5$ khi $cosx=0$

$y_{max}=8$ khi $cosx=\pm1$

e/

$0\le\left|cos2x\right|\le1$

$\Rightarrow-2\le y\le1$

$y_{min}=-2$ khi $cos2x=\pm1$

$y_{max}=1$ khi $cos2x=0$

f/

$0\le sin^23x\le1\Rightarrow\sqrt{3}-2\le y\le0$

$y_{min}=\sqrt{3}-2$ khi $sin2x=0$

$y_{max}=0$ khi $sin2x=\pm1$Câu trả lời: d/