Question

对函数y=f(x)(x大于等于x1小于等于x2)设点A(X1,Y1)B(X2,Y2)

Answer 1

高度为4

f(x)=2cos(2x-π/4)在区间【π/8,9π/8】的

端点是A(π/8,2),B(9π/8,2)

向量ON=λ(π/8,2)+(1−λ)(9π/8,2)=(9π/8-λπ,2)

∴N坐标是(9π/8-λπ,2)

∵点N满足向量ON=λ向量OA+(1-λ)向量OB,λ≥0,∴0≤λ≤1．

∴xM=9π/8-λπ

yM=2cos(9π/4-2λπ)

∴|MN|

=√[0^2+(2cos(9π/4-2λπ)-2)^2]

=|2cos(9π/4-2λπ)-2|

∵-1