Question

求函数y=x2+√(1-x2)的值域

Answer 1

1-x^2≥0

-1≤x≤1

设x=sina,-π/2≤a≤π/2

则

y=(sina)^2+cosa=1-(cosa)^2+cosa=-(cosa-1/2)^2+5/4

0≤cosa≤1

1≤y≤5/4

解法2

令√(1-x^2)=t 0≤t≤1

x^2=1-t^2

y=1-t^2+t=-(t-1/2)^2+5/4

1≤y≤5/4