若a≥0，f(x)=(a+cosx)(a+sinx)的最大值为252，则a为________

f(x)=a^2+a(sinx+cosx)+sinxcosx.
令t=sinx+cosx. .. 则sinx*cosx=(t^2-1)/2
t取值范围为[-sqrt2,sqrt2]
f(x)=f(t)=a^2+at+t^2/2-1/2=1/2（t+a)^2+a^2/2-1
a>0. 所以maxf(x)=f(t=sqrt2)=a^2+(sqrt2)a+1/2.

f(x)=(a+cosx)(a+sinx)=a^2+a(sinx+cosx)+sinxcosx
=a^2+a/2^(1/2)*sin(x+pi/4)+1/2*sin2x
当x=pi/4时，f(x)取得最大a^2+a/2^(1/2)+1/2
所以当a=-1/4*2^(1/2)+1/4*4026^(1/2)时，最大值为252

f(x)=(a+cosx)(a+sinx)
=a^2+a(sinx+cosx)+sinxcosx
=a^2+(a√2)*sin(x+45°)+0.5sin2x
x=45°,最大值f(x)=a^2+(a√2)+0.5
a^2+(a√2)+0.5=252
a^2+(√2)a-251.5=0
a>0
a=(√1008-√2)/2