f'(x)=cosx+sinx+1=√2sin(x+π/4)+1
π/4≤x ≤ π/2 √2/2≤sin(x+π/4)≤1 所以f'(x)在此区间大于0,是单调递增函数
f(0)=-1+a0 所以0点在此区间只有一个
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