1. Let x- 1 = cost; Y=sint，x+y >； Change r to f(t)=cost+ 1+sint and find the maximum value. Derivation can be used to find the extreme point or the formula of trigonometric function, and can also be used to find the extreme point;

2.∫0∏/2 f(xcosx)xsinxdx

=-f(xcosx)xdcosx

=-f(xcos x)xd cosx+∫0∏/2 fcos x df(xcos x)x

=-f(xcos x)* xcos x+cosxf(xcos x)* x+∫0∏/2 f(xcos x)xsin xdx = 1

3. It can be judged that this triangle is an isosceles right triangle, and the key is to be familiar with the properties of hyperbola. Let the intersection of PQ and X axis be m, then MF = MPMF = C-A 2/C, and MP=-ab/c is obtained from asymptote equation Y =-BX/A and X = A 2/C;

Get: c 2 = a 2-ab. Then: c 2 = a 2+b 2, and a=b is obtained, so e= root 2.