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F x Ax sinx Cosx

由f(x)=ax+sinx+cosx,得f′(x)=a+cosx-sinx,设A(x1,y1),B(x2,y2),则f′(x1)=a+cosx1-sinx1,f′(x2)=a+cosx2-sinx2.由f′(x1)f′(x2)=-1,得a2+[(cosx1-sinx1)+(cosx2-sinx2)]a+(cosx1-sinx1)(cosx2-sinx2)+1=0.令m=cosx...

证: f'(x)=cosx+sinx- 2/π =√2sin(x+π/4)- 2/π 令f'(x)≥0 √2sin(x+π/4)≥2/π sin(x+π/4)≥√2/π x∈(0,π) 函数f(x)的单调递增区间为(0,3π/4 -arcsin(√2/π)] 函数f(x)的单调递减区间为[3π/4 -arcsin(√2/π),π) 只需考察两边界 f(0)=sin0-cos0-(2/π...

(Ⅰ)∵函数f(x)=sinx-cosx+1.设函数F(x)=sinx-cosx+1-ax,∴F′(x)=cosx+sinx-a∵f(x)≥ax在[0,π]上恒成立,∴函数F(x)=sinx-cosx+1-ax≥F(0)=0,∴只需F′(x)=cosx+sinx-a≥0恒成立,即:a≤(sinx+cosx)min,∵sinx+cosx=2sin(x+π4),...

f'(x)=cosx+sinx-a=0 √2sin(x+π/4)=a ∴-√2

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C ∵f′(x)=[(ax+b)sinx]′+[(cx+d)cosx]′=(ax+b)′sinx+(ax+b)(sinx)′+(cx+d)′cosx+(cx+d)(cosx)′=asinx+(ax+b)cosx+ccosx-(cx+d)sinx=(a-d-cx)sinx+(ax+b+c)cosx.为使f′(x)=xcosx,应满足 解方程组,得 从而可知,f(x)...

解析:∵f(x)≤1+sinx 设h(x)=ax+cosx-sinx-1=√2π/4时 ∴h(x)在x1处取极大值,在x2处取极小值 要满足f(x)≤1+sinx,只须满足h(x1)

(Ⅰ)当a=π2时,f(x)=(x-π2)sinx+cosx,x∈(0,π).f′(x)=(x-π2)cosx,由f′(x)=0得x=π2,f(x),f′(x)的情况如下:x(0,π2)π2(π2,π)x-π2-0+cosx+0-f′(x)-0-f(x)↓↓因为f(0)=1,f(π)=-1,所以函数f(x)的值域为(-1,...

(1)由f(x)=sinx-cosx+x+1=2sin(x?π4)+x+1,0<x<2π,知f′(x)=1+2sin(x+π4).令f′(x)=0,从而可得sin(x+π4)=-22,解得x=π,或x=3π2,当x变化时,f′(x),f(x)变化情况如下表: x (0,π) π (π,3π2) 3π2 f′(x) + 0 - 0 + ...

(1)f(x)=sinx-cosx-ax (0

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