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F x Ax sinx Cosx

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

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...

f(x)=ax+sinx+cosx, f'(x)=a+cosx-sinx, f(x)的图象上存在不同的两点A(x1,*),B(x2,*),使得曲线y=f(x)在点A,B处的切线互相垂直, (a+cosx1-sinx1)(a+cosx2-sinx2)=-1, a^2+a(cosx1-sinx1+cosx2-sinx2)+(cosx1-sinx1)(cosx2-sinx2)+1=0, 设u=cosx1...

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

(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),...

设函数f(x)=ax+sinx+cosx.若函数f(x)的图象上存在不同的两点A,B,使得曲线y= f(x)在点A,B处的切线互相垂直,则实数a的取值范围为 f(x)=ax+sinx+cosx, f'(x)=a+cosx-sinx, f(x)的图象上存在不同的两点A(x1,*),B(x2,*),使得曲线y=f(x)在点...

解: f'(x) =[(a-sinx)/cosx]' =[(a-sinx)'*cosx-(a-sinx)(cosx)']/cos²x =[-cos²x+(a-sinx)sinx]/cos²x =(asinx-1)/cos²x 依题意,f'(x)在(π/6,π/3)上恒大于0 ∴ asinx-1在(π/6,π/3)上恒大于0 ∴a*sin(π/6)>0 ∴a>2 即,a的...

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:利用零点分区去绝对值 ①由sinx-cosx>0得 √2sin(x-π/4)>0 解得 2kπ

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...

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