tim gia tri lo nhat nho nhat
a, y=2cos2x+2sin2x tren r
b,y=can3cos3x-sin3x
c, y=sinx+cosx-sinx*cosx tren r
Moi nguoi giup minh voi:
Tim m thuoc R de h/s :
Y=sinx+cosx+ms dong bien tren R.
Lời giải:
Để hàm \(y=\sin x+\cos x+mx\) đồng biến trên R thì \(y'=\cos x-\sin x+m>0\) với mọi $x$
\(\Leftrightarrow m>\sin x-\cos x\) với mọi \(x\in\mathbb{R}\)
\(\Leftrightarrow m>\max (\sin x-\cos x)\)
Xét hàm \(f(x)=\sin x-\cos x\Rightarrow f'(x)=\cos x+\sin x=0\)
\(\Leftrightarrow \sin x=-\cos x\). Kết hợp với \(\sin ^2x+\cos^2x=1\)
\(\Rightarrow \)\(\left[{}\begin{matrix}\sin x=\dfrac{1}{\sqrt{2}},\cos x=\dfrac{-1}{\sqrt{2}}\Rightarrow f\left(x\right)=\sqrt{2}\\\sin x=\dfrac{-1}{2},\cos x=\dfrac{1}{\sqrt{2}}\Rightarrow f\left(x\right)=-\sqrt{2}\end{matrix}\right.\)
Do đó \(\max (\sin x-\cos x)=\sqrt{2}\)
Vậy \(m>\sqrt{2}\)
Rút gọn
A= \(\frac{cosx-cos2x-cos3x+cos4x}{sinx-sin2x-sin3x+sin4x}\)
B= sinx(1+2cos2x+2cos4x+2cos6x)
\(A=\frac{cosx-cos3x+cos4x-cos2x}{sinx-sin3x+sin4x-sin2x}=\frac{2sin2x.sinx-2sin3x.sinx}{-2cos2x.sinx+2cos3x.sinx}\)
\(=\frac{sin2x-sin3x}{cos3x-cos2x}=\frac{-2cos\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}{-2sin\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}=cot\left(\frac{5x}{2}\right)\)
\(B=sinx+2cos2x.sinx+2cos4x.sinx+2cos6x.sinx\)
\(=sinx+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)
\(=sin7x\)
1) Gia tri lon nhat cua ham so: y = \(\dfrac{cosx+2sinx+3}{2cosx-sinx+4}\)
A. 0 B. 3-2\(\sqrt{3}\) C. \(2-2\sqrt{2}\) D. -1
Chứng minh các hàm số sau tuần hoàn, tìm chu kì T:
\(a,y=\left|sinx\right|\)
\(b,y=cosx+sinx\)
\(c,sin3x\)
\(d,y=\left|cosx\right|\)
giải pt : \(\dfrac{2cos2x+1}{\sqrt{3}sinx+cosx}\)=2cosx-1
tìm txđ hàm số D: y=\(\dfrac{2+3sinx}{2sin2x+\sqrt{2}}\)
Cho hàm số: y = ex cosx Khi đó: y'/ex?
A. cosx - sinx
B. sinx - cosx
C. sinx + cosx
D. cosx
\(y'=\left(e^x\right)'.cosx+e^x.\left(cosx\right)'=e^x\left(cosx-sinx\right)\)
=> Chọn A
giải các pt
a) \(sinx+\sqrt{3}cosx=2sin\left(x+\frac{\pi}{6}\right)\)
b) \(\sqrt{3}sinx+cosx=2sin\frac{\pi}{12}\)
c) \(cosx-\sqrt{3}sinx=2cos3x\)
d) \(sin3x-\sqrt{3}cos3x=2sin2x\)
a/
\(\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{6}\right)\)
\(\Rightarrow x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
b/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=sin\frac{\pi}{12}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=sin\frac{\pi}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{12}+k2\pi\\x+\frac{\pi}{6}=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=cos3x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos3x\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=3x+k2\pi\\x+\frac{\pi}{3}=-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)
d/
\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=sin2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=2x+k2\pi\\3x-\frac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{15}+\frac{k2\pi}{5}\end{matrix}\right.\)
Tìm chu kì:
a, y= Sin3x × Cos3x
b, y= Sinx (2x - n/6)
c, y= căn2/2×Sinx + căn2/2×Cosx
d, y= Cosbình4x
tim gtln ginn
a=-4cos^2 x+2sinx +3 tren r
b, 4cos^2 x+cosx -1 tren r
c,y=tan^2 x-2tanx tren[-pi/4;pi/3)