设f(x)是定义域为R最,小正周期为 _ 函数若,f(x)5.设f(x)是定义域为R最,小正周期为 _ 函数若,f(x) \$= \left\{ \begin{array} { l } \cos x , - \frac { \pi } { 2 } \leqslant x \leqslant 0 \\ \sin x , 0 x \leqslant \pi \end{array} \right.\$ _ 值等...
$=\dfrac {\sqrt {3}} {2}sin\omega x+\dfrac {1} {2}\left ( {1-cos\omega x} \right )$ $=sin\left ( {\omega x-\dfrac {\pi } {6}} \right )+\dfrac {1} {2}$, $\because $函数f ( x )的最小正周期为2, $\therefore $$T=\dfrac {2\pi } {\omega }=...
设ak≥0,∑∞k=1ak<+∞,f(x)=∑∞k=1akTpk,qk(x). \begin{enumerate} \item[(1)] 求证: $f(x)$是在$\mathbb{R}$上连续的以$2\pi$为周期的周期函数. \item[(2)] 判断并证明: $f(x)$的Fourier级数在$x=0$处的收敛性. \end{enumerate} (2020年电子科大)设0<an<1,an+1=an(1−...
$\therefore $可得$\omega =4$ $\therefore $$f\left ( {x} \right )=3sin\left ( {4x+\dfrac {\pi } {6}} \right )$ $\therefore f ( ( (α ) 4+ (π ) (12)) )=3sin [ (4 ( ( (α ) 4+ (π ) (12)) )+ (π ) 6) ]=3sin ( (α + (π ) 2) )=3...
【答案】C【解析】题意得$f\left(x\right)$的最小正周期$T=\dfrac{\pi }{4}\times 4=\pi $,所以$\omega =2$,$f\left(x\right)=2sin(2x-\dfrac{1}{3}\mathrm{\pi })$,由$-\dfrac{1}{2}\pi +2k\pi \leqslant 2x-\dfrac{1}{3}\pi \leqslant \dfrac{1}{2}\pi +2k...
见解析。(1)解:∵的最大值为2,且A0,∴A=2.∵的最小正周期为8,∴2元 8 0,得0 4.∴TT f(x)=2sin(x+ 44.(2)解法1:∵f(2)=2sin + 2 4 =2cos=√2 4,T f(4)=2sin + =-2sin元=-√2 4 4,∴P(2,√2),Q(4,-2).∴lP=6,|PQ|=2√3,|Q|=3√2.∴o+o-|Q(6...
(Ⅰ$)f\left(x\right)=\sqrt {3}\sin \omega x\cos \omega x-\cos ^{2}\omega x+1=\dfrac{\sqrt {3}}{2}\sin 2\omega x-\dfrac{1+\cos 2\omega x}{2}+1=\sin \left(2\omega x-\dfrac{\pi }{6}\right)+\dfrac{1}{2}\left(\omega \gt 0\right)$, 依题意,$T=\...
(1)f(x)=a⋅ b+((√3))/2=sin ω xcos ω x-√3(sin^2)ωx+((√3))/2=1/2sin2ωx+((√3))/2cos2ωx=sin((2ωx+π/3)),因为函数f(x)的图象上相邻两个对称轴之间的距离是π/2,所以最小正周期T=π ,由2ω =(2π)/T=(2π)/π=2,得ω =1,所以f(x)=si...
已知函数f(x)=2sin (wx+φ )+1,(w > 0,|φ |≤ (π )2), 其图象与直线y=3相邻两个交点的距离为π ,若f(x) > 2 对∀x∈
解:(1)化简可得f(x)=•2sinωxcosωx-(cos2ωx-sin2ωx)+λ =sin2ωx-cos2ωx+λ=2sin(2ωx-)+λ 由函数图象关于直线对称可得2ω•-=kπ+,k∈Z, 解得ω=k+1,结合ω∈(0,2)可得ω=1, ∴f(x)=2sin(2x-)+λ, ∴函数f(x)的最小正周期T==π; (2)∵y=f(x)的图...