Question

已知函數y=f(x)=-x³+ax²+b(x∈R)的圖象是曲線C.

（1）當x=2時，函數f(x)取得極值0，求a、b的值；

（2）在（1）的條件下，求過點P（0，-4）且與曲線C相切的切線方程；

（3）若曲線C上任意兩點的連線的斜率都小于1，求實數a的取值范圍.

Answer 1

解:（1）∵x=2時，函數f(x)的極值是0，?

∴f′(2)=0,f(2)=0.                                                                                              ?

∴-12+4a=0,-8+4a+b=0.?

∴a=3,b=-4.                                                                                                            ?

經檢驗a=3,b=-4,函數f(x)在x=2處取得極值是0.                                             ?

（2）設切點P（x₀,y₀）,

f′(x₀)=-3x₀²+6x₀,?

∴切點為P（x₀,y₀）的曲線C的切線方程是y-y₀=(-3x₀²+6x₀)(x-x₀),                           ?

∵點（0，-4）在切線上,

∴-4-（-x₀³+3x₀²-4）=(-3x₀²+6x₀)(-x₀).                                                                      ?

解得x₀=0,或x₀=，?

∴切點是（0，-4），或（，-）.                                                                ?

∴所求切線方程是y+4=0,或9x-4y-16=0.                                                           ?

（3）解法一：設曲線C上任意兩點為A（x₁,y₁）,B(x₂,y₂)(x₁≠x₂).

依題意＜1,?

∴ ＜1.                                                             ?

化簡得a(x₁+x₂)-(x₁²+x₁x₂+x₂²)＜1,                                                                            ?

即不等式-x₁²+(a-x₂)x₁-x₂²+ax₂-1＜0對一切x₁∈R恒成立.?

∴Δ₁=(a-x₂)²+4(-x₂²+ax₂-1)＜0,                                                                                ?

即不等式3x₂²-2ax₂-a²+4＞0對一切x₂∈R恒成立.?

∴Δ₂=(-2a)²-12(-a²+4)＜0.                                                                                       ?

解得-＜a＜.                                                                                                       ?

解法二：∵曲線C上任意兩點的連線的斜率都小于1,

∴f′(x)＜1.                                                                                                              ?

∴-3x²+2ax＜1對一切實數x恒成立,?

即3x²-2ax+1＞0對一切實數x恒成立.                                                                     ?

∴Δ=（-2a）²-4×3×1＜0,?

解得-＜a＜.