Question

(理)已知f(x)=ax³+bx²+cx+d(a≠0)是定義在R上的函數(shù),其圖象交x軸于A、B、C三點(diǎn).若點(diǎn)B的坐標(biāo)為(2,0),且f(x)在［-1,0］和［4,5］上有相同的單調(diào)性,在［0,2］和［4,5］上有相反的單調(diào)性.

(1)求c的值.

(2)在函數(shù)f(x)的圖象上是否存在一點(diǎn)M(x₀,y₀),使得f(x)在點(diǎn)M處的切線斜率為3b?若存在,求出點(diǎn)M的坐標(biāo);若不存在,請(qǐng)說明理由.

(3)求|AC|的取值范圍.

(文)已知函數(shù)f(x)=x⁴-4x³+ax²-1在區(qū)間［0,1］單調(diào)遞增,在區(qū)間［1,2)單調(diào)遞減.

(1)求a的值;

(2)若點(diǎn)A(x₀,f(x₀))在函數(shù)f(x)的圖象上,求證點(diǎn)A關(guān)于直線x=1的對(duì)稱點(diǎn)B也在函數(shù)f(x)的圖象上;

(3)是否存在實(shí)數(shù)b,使得函數(shù)g(x)=bx²-1的圖象與函數(shù)f(x)的圖象恰有3個(gè)交點(diǎn),若存在,請(qǐng)求出實(shí)數(shù)b的值;若不存在,試說明理由.

Answer 1

(理)解：(1)f′(x)=3ax²+2bx+c.                                              

依題意f(x)在［-1,0］和［0,2］上有相反的單調(diào)性.

所以x=0是f(x)的一個(gè)極值點(diǎn).

故f′(0)=0,得c=0.                                                           

(2)令f′(x)=0,得3ax²+2bx=0.

解得x₁=0,x₂=.                                                        

因?yàn)閒(x)在［0,2］和［4,5］上有相反的單調(diào)性,

所以f′(x)在［0,2］和［4,5］上有相反的符號(hào).

故2≤≤4-6≤≤-3.                                             

假設(shè)存在點(diǎn)M(x₀,y₀)使得f(x)在點(diǎn)M處的切線斜率為3b,則f′(x₀)=3b,

即3ax₀²+2bx₀-3b=0.

因?yàn)棣?(2b)²-4×3a(-3b)=4b²+36ab=4ab(+9),

且-6≤≤-3,所以3≤+9≤6,且a、b異號(hào).

所以Δ=4ab(+9)＜0.

故不存在點(diǎn)M(x₀,y₀),使得f(x)在點(diǎn)M處的切線斜率為3b.                        

(3)設(shè)A(α,0),C(β,0),依題意可令f(x)=a(x-α)(x-2)(x-β),

即f(x)=a［x³-(2+α+β)x²+(2α+2β+αβ)x-2αβ］

=ax³-a(2+α+β)x²+a(2α+2β+αβ)x-2aαβ.

所以

即.                                                      

所以|AC|=|α-β|=.

因?yàn)?6≤≤-3,所以當(dāng)=-6時(shí),|AC|_max=;

當(dāng)=-3時(shí),|AC|_min=3.

故3≤|AC|≤4.                                                       

(文)解：(1)由函數(shù)f(x)=x⁴-4x³+ax²-1在區(qū)間［0,1］上單調(diào)遞增,在區(qū)間［1,2)上單調(diào)遞減,所以x=1時(shí),取得極大值.

所以f′(1)=0.                                                              

因?yàn)閒′(x)=4x³-12x²+2ax,

所以4-12+2a=0.解得a=4.                                                   

(2)因?yàn)辄c(diǎn)A(x₀,f(x₀))關(guān)于直線x=1的對(duì)稱點(diǎn)B的坐標(biāo)為(2-x₀,f(x₀)),                

且f(2-x₀)=(2-x₀)⁴-4(2-x₀)³+4(2-x₀)²-1

=(2-x₀)²［(2-x₀)-2］²-1

=x₀⁴-4x₀³+4x₀²-1=f(x₀).                                                      

所以點(diǎn)A關(guān)于直線x=1的對(duì)稱點(diǎn)B也在函數(shù)f(x)的圖象上.                      

(3)因?yàn)楹瘮?shù)g(x)=bx²-1的圖象與函數(shù)f(x)的圖象恰有3個(gè)交點(diǎn),

等價(jià)于方程x⁴-4x³+4x²-1=bx²-1恰有3個(gè)不等實(shí)根.                              

由x⁴-4x³+4x²-1=bx²-1,得x⁴-4x³+(4-b)x²=0.

因?yàn)閤=0是其中一個(gè)根,

所以方程x²-4x+(4-b)=0有2個(gè)非零且不等的實(shí)數(shù)根.                            

故由得b＞0且b≠4.