题目

f(x)=1+tanX/1+(tanx)^2,x属于[派/12,派/2],求f(x)取值范围

提问时间：2021-01-29

答案

f(x)=(1+tanX)/(1+tan²x),x属于[π/12,π/2],求f(x)取值范围
f(x)=[1+(sinx/cosx)]/sec²x=(cosx+sinx)/secx=(cosx+sinx)cosx
=(√2)sin(x+π/4)cosx=(√2)×(1/2)[sin(π/4)+sin(2x+π/4)]
=(√2/2)[(√2/2)+sin(2x+π/4)]=(1/2)+(√2/2)sin(2x+π/4)
故在区间[π/12,π/2]内,maxf(x)=f(π/8)=(1/2)+(√2/2)sin(π/2)=(1/2)(1+√2)；
minf(x)=f(π/2)=(1/2)+(√2/2)sin(π+π/4)=(1/2)-(√2/2)sin(π/4)=1/2-1/2=0
即值域为[0,(1/2)(1+√2)]