Question

已知函数y=f(x)=-x³+ax²+b(x∈R)的图象是曲线C.

（1）当x=2时，函数f(x)取得极值0，求a、b的值；

（2）在（1）的条件下，求过点P（0，-4）且与曲线C相切的切线方程；

（3）若曲线C上任意两点的连线的斜率都小于1，求实数a的取值范围.

Answer 1

解:（1）∵x=2时，函数f(x)的极值是0，?

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

∴-12+4a=0,-8+4a+b=0.?

∴a=3,b=-4.                                                                                                            ?

经检验a=3,b=-4,函数f(x)在x=2处取得极值是0.                                             ?

（2）设切点P（x₀,y₀）,

f′(x₀)=-3x₀²+6x₀,?

∴切点为P（x₀,y₀）的曲线C的切线方程是y-y₀=(-3x₀²+6x₀)(x-x₀),                           ?

∵点（0，-4）在切线上,

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

解得x₀=0,或x₀=，?

∴切点是（0，-4），或（，-）.                                                                ?

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

（3）解法一：设曲线C上任意两点为A（x₁,y₁）,B(x₂,y₂)(x₁≠x₂).

依题意＜1,?

∴ ＜1.                                                             ?

化简得a(x₁+x₂)-(x₁²+x₁x₂+x₂²)＜1,                                                                            ?

即不等式-x₁²+(a-x₂)x₁-x₂²+ax₂-1＜0对一切x₁∈R恒成立.?

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

即不等式3x₂²-2ax₂-a²+4＞0对一切x₂∈R恒成立.?

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

解得-＜a＜.                                                                                                       ?

解法二：∵曲线C上任意两点的连线的斜率都小于1,

∴f′(x)＜1.                                                                                                              ?

∴-3x²+2ax＜1对一切实数x恒成立,?

即3x²-2ax+1＞0对一切实数x恒成立.                                                                     ?

∴Δ=（-2a）²-4×3×1＜0,?

解得-＜a＜.