Question

两道数学题目

Answer 1

1.f(x)=ax^3-3x^2,f'(x)=3ax^2-6x
(1)f'(2)=12a-12=0,a=1,检验得,x=2处是极值点,∴a=1
(2)g(x)=ax^3-3x^2+3ax^2-6x=ax^3+3(a-1)x^2-6x,g'(x)=3ax^2+6(a-1)x-6
a=0时,g(x)=-3x^2-6x,g'(x)=-6x-6<0,g(x)在[0,2]单调减,在x=0处取最大值
a<0时,g'(x)的对称轴-6(a-1)/6a<0,g'(x)在[0,2]上单调减,g'(0)=-6<0,
∴g'(x)<0
g(x)在[0,2]上单调减,在x=0处取最大值
a>0时,g'(x)的对称轴-6(a-1)/6a=(1-a)/a=1/a-1,g'(0)=-6<0,
g'(x)=0的解一正一负,
∴g'(x)在[0,2]上至多有一个根,
g(x)在[0,2]上单调减或先单调减再单调增,只要保证在x=2处取值不大于x=0处即可
g(0)=0,g(2)=20a-24<=0,0<a<=6/5
综上,a的范围为a<=6/5
2.a=2,b=1,椭圆方程x^2/4+y^2=1,AB:x+2y-2=0,EF:y=kx,D(2/(2k+1),2k/(2k+1))
(1)向量ED=6向量DF,设E((2-6t)/(2k+1),k(2-6t)/(2k+1)),F((2+t)/(2k+1),k(2+t)/(2k+1))
EF关于原点对称,(2-6t)/(2k+1)+(2+t)/(2k+1)=0,t=4/5
E(-14/5(2k+1),-14k/5(2k+1)),F(14/5(2k+1),14k/5(2k+1))
将E或F代入椭圆方程49/25(2k+1)^2+196/25(2k+1)^2=1,
49(1+k^2)=25(4k^2+4k+1)24k^2-25k+6=(3k-2)(8k-3)=0,k=2/3或3/8
(2)y=kx,E(-2/√(1+4k^2),-2k/√(1+4k^2)),F(2/√(1+4k^2),2k/(1+4k^2))
EF=√(1+k^2)*4/√(1+4k^2),A到EF距离2k/√(1+k^2),B到EF距离1/√(1+k^2)
AEBF面积=√(1+k^2)*2/√(1+4k^2)*(2k+1)/√(1+k^2)/2
=2(2k+1)/√(1+4k^2)=2√[(4k^2+4k+1)/(1+4k^2)]=2√[1+4k/(4k^2+1)]
=2√[1+4/(4k+1/k)]
4k+1/k>=2√(4k*1/k)=4(取等4k=1/k,k=1/2)
AEBF面积<=2√[1+4/4]=2√2所以AEBF面积的最大值为2√2