来自唐俊的问题
已知x∈[0,1],则函数y=x+2−−−−−√−1−x−−−−−√的值域是()。A.[2√−1,3√−1]B.[1,3√]C.[2√−1,3√]D.[0,2√−1]
已知x∈[0,1],则函数y=x+2−−−−−√−1−x−−−−−√的值域是( )。A.[2√−1,3√−1]B.[1,3√]C.[2√−1,3√]D.[0,2√−1]


已知x∈[0,1],则函数y=x+2−−−−−√−1−x−−−−−√的值域是()。A.[2√−1,3√−1]B.[1,3√]C.[2√−1,3√]D.[0,2√−1]
已知x∈[0,1],则函数y=x+2−−−−−√−1−x−−−−−√的值域是( )。A.[2√−1,3√−1]B.[1,3√]C.[2√−1,3√]D.[0,2√−1]
本题主要考查函数的值域。由于函数表达式均为根式,故对其进行平方处理,得y2=3−2−x2−x+2−−−−−−−−−−√,当x∈[0,1]时,−x2−x+2∈[0,2],所以y2∈[3−22√,3],由于x>0,x+2−−−−−√>1−x−−