Question

已知函数f(x)=ae^x-x^2-(2a+1)x,若函数f(x)在区间(0,ln2)上有最值,则实数a的取值

Answer 1

f(x)=ae^x-x^2-(2a+1)x
f'(x)=ae^x-2x-(2a+1)
函数f(x)在区间(0,ln2)上最值→区间(0,ln2)上存在驻点，即f'(x)=0 在(0,ln2)有解
f'(ln2)=-2ln2-1<0
当f'(0)=-a-1>0→a<-1时f'(ln2)·f'(0)<0,f'(x)在(0,ln2)上存在零点
f''(x)=ae^x-2
-1≤a≤0时 f''(x)<0,f'(x)单调递减，f'(0)<0，f'(x)在(0,ln2)上无零点;
0<a≤1，f'(x)驻点x=ln(2/a)≥ln(2),区间内f'(x)<0,f'(x)在(0,ln2)上无零点;
1<a<2时，区间内包含f'(x)的驻点<0,f''(x)>0为极小值，f'(x)在(0,ln2)上无零点
a≥2时，f'(x)驻点x=ln(2/a)<0,区间内f'(x)>0,f'(x)<f'(ln2)<0,f'(x)在(0,ln2)上无零点

∴a<1