Y'=2sinxcosx+2pcosx=0 y''=2cos(2x)-2psinx y''(π/2)=-2-2p=-1-√30 (对应最小值)
x1=π/2 sinx=-p x2=arcsin(-p)
y(π/2)=1+2p+q = 9
y(arcsin(-p))=p^2-2p^2+q=q-p^2=6 即:
2p+q=8 (1)
-p^2+q=6 (2)
(1)-(2):
p^2+2p-2=0 p = (-1+√3) ≈ 0.732 (1)
q=8-2p q = 10-2√3 ≈ 6.5359 (2)
最后结果如:(1)、(2)所示.