函数y=sin2x+sinx-1的值域为( ) A.[-1,1] B.[-,-1] C.[-,1] D.[-1,] 答案:C [解析] 本题考查了换元法,一元二次函数闭区间上的最值问题,通过sinx=t换元转化为t的二次函数的最值问题,体现了换元思想和转化的思想,令t=sinx∈[-1,1],y=t2+t-1,(-1≤t≤1),显然-≤y≤1,选C.14、As a Senior 3 student,if you
want to get a good result,it is important to
a good state of mind for the coming entrance examination.
A.keep
up
B.carry
out
C.keep
off
D.keep out