核心提示:已知函数f(x)=|x3|+|xa|,g(x)=x3+1,若函数y=f(g(x))的图象为轴对称图形,则实数a的值可能是___.考点:函数的图象分析:由题意,化简y=f(g(x))=|x3+1-3|+...
已知函数f(x)=|x−3|+|x−a|,g(x)=x3+1,若函数y=f(g(x))的图象为轴对称图形,则实数a的值可能是___.
考点:
函数的图象
分析:
由题意,化简y=f(g(x))=|x3+1-3|+|x3+1-a|=|x3-2|+|x3+1-a|,讨论a的取值以去绝对值号,从而确定对称轴的可能取值,再取点检验,最后再判断即可.
解答:
y=f(g(x))=|x3+1−3|+|x3+1−a|
=|x3−2|+|x3+1−a|,
当1−a=−2,即a=3时,
y=2|x3−2|不是轴对称图形,
当1−a<−2;即a>3时,
由y=⎧⎩⎨⎪⎪1+a−2x3,x<2√3a−3,2√3⩽x⩽a−1−−−−−√32x3−1−a,x>a−1−−−−−√3;
若存在对称轴,则对称轴应为x=2√3+a−1−−−−−√32;
而令y=1+a解得,x=0或x=a+1−−−−−√3;
则a+1−−−−−√3=2√3+a−1−−−−−√3;
即a+1−−−−−√3−a−1−−−−−√3=2√3;
∵a>3;
∴a+1−−−−−√3−a−1−−−−−√3<2√3;
故不成立;
当1−a>−2;即a<3时,
由y=⎧⎩⎨⎪⎪1+a−2x3,x2√3;
若存在对称轴,则对称轴应为x=2√3+a−1−−−−−√32;
而令y=1+a解得,x=0或x=a+1−−−−−√3;
则a+1−−−−−√3=2√3+a−1−−−−−√3;
即a+1−−−−−√3−a−1−−−−−√3=2√3;
∵a<3;
∴当a=1或a=−1时,等号成立;
经检验,当a=1时,y=f(g(x))=|x3−2|+|x3|不对称,
当a=−1时,y=f(g(x))=|x3−2|+|x3+2|对称;
故答案为:−1.